$$ (\Psi_{\varepsilon}(\mathbf{x}))_{\varepsilon} = (\Psi_{\varepsilon}(\mathbf{x}, \mathbf{x}_0))_{\varepsilon} = $$

$$ = \left( \frac{\mu(\mathbf{x}_0)}{\sqrt[4]{2\pi\eta_{\varepsilon}}} \Psi_{\varepsilon}(\mathbf{x}) \exp\left[ \frac{\langle \mathbf{x} - \mathbf{x}_0, \mathbf{x} - \mathbf{x}_0 \rangle}{2\eta_{\varepsilon}} \right] \right)_{\varepsilon} = $$

$$ = \left( \frac{1}{(\sqrt[4]{2\pi\eta_{\varepsilon}})^d} \exp\left[ \frac{\langle \mathbf{x} - \mathbf{x}_0, \mathbf{x} - \mathbf{x}_0 \rangle}{2\eta_{\varepsilon}} \right] \exp\left[ \frac{i}{\hbar_{\varepsilon}} S(\mathbf{x}) \right] \right)_{\varepsilon}, \quad (4.3.2) $$

where

$$ (\eta_{\varepsilon})_{\varepsilon} \approx 0 \text{ and } \left(\frac{\hbar_{\varepsilon}}{\eta_{\varepsilon}}\right)_{\varepsilon} \approx 0, $$

**Definition 4.3.1.** Quantum trajectory $\{(x_{\varepsilon,i}(t, \mathbf{x}_0; \hbar))\}_{\varepsilon}^{d}\}_{i=1}^{d}$ with initial data $\{(x_{\varepsilon,i}(0, \mathbf{x}_0; \hbar))\}_{\varepsilon}^{d}\}_{i=1}^{d} = \mathbf{x}_0$ is defined by following formula

$$ (x_{\varepsilon,i}(t, \mathbf{x}_0; \hbar))_{\varepsilon} = \left( \int_{\mathbb{R}^d} x_i |\Psi_{\varepsilon}(\mathbf{x}, t, \mathbf{x}_0; \hbar)|^2 d^d x \right)_{\varepsilon}. \quad (4.3.3) $$

## V. Schrödinger's cat paradox resolution using GRW collapse model.

In this subsection an possible solution of the Schrödinger's cat paradox is considered. We pointed out that: the collapsed state of the cat always shows definite and predictable measurement outcomes even if Schrödinger's cat consists of a superposition: $|\text{cat}\rangle = c_1|\text{live cat}\rangle + c_2|\text{death cat}\rangle$.

Contrary to van Kampen's [61] and some others' opinions, "looking" at the outcome changes nothing, beyond informing the observer of what has already happened.

We remain: there are widespread claims that Schrödinger's cat is not in a definite alive or dead state but is, instead, in a superposition of the two. van Kampen, for
example, writes “The whole system is in a superposition of two states: one in which no decay has occurred and. . .one in which it has occurred. Hence, the state of the cat also consists of a superposition: $|cat\rangle = c_1|live cat\rangle + c_2|death cat\rangle$. The state remains a superposition until an observer looks at the cat” [61]. The canonical formulation [62]:

$|cat\rangle = c_1|live cat\rangle|undecayed nucleus\rangle + c_2|death cat\rangle|decayed nucleus\rangle$ completely obscures the unitary Schrödinger evolution which by using GRW collapse model, predicts specific nonlocal entanglement. The cat state must be

written as:

$$|cat(t)\rangle = c_1|live cat(t)\rangle|undecayed nucleus (t)\rangle + c_2|death cat(t)\rangle|decayed nucleus(t)\rangle.$$

This entangled state actually is the collapsed state of both the cat and the nucleus, showing definite outcomes at each instant $t \ge T_{col}$.

## V.I.Schrödinger's cat paradox

As Weinberg recently reminded us [63], the measurement problem remains a fundamental conundrum. During measurement the state vector of the microscopic system collapses in a probabilistic way to one of a number of classical states, in a way that is unexplained, and cannot be described by the time-dependent Schrödinger equation. To review the essentials, it is sufficient to consider two-state systems. Suppose a nucleus **n**, whose Hilbert space is spanned by orthonormal states $|s_i(t)\rangle$, $i = 1,2$, where $|s_1(t)\rangle = |undecayed nucleus at instant t\rangle$ and $|s_2(t)\rangle = |decayed nucleus at instant t\rangle$ is in the superposition state,

$$|\Psi_t\rangle_n = c_1|s_1(t)\rangle + c_2|s_2(t)\rangle, |c_1|^2 + |c_2|^2 = 1. \qquad (5.1.1)$$

An measurement apparatus **A**, which may be microscopic or macroscopic, is designed to distinguish between states $|s_i(t)\rangle$ by transitioning at each instant $t$ into state $|a_i(t)\rangle$ if it finds **n** is in $|s_i(t)\rangle$, $i = 1,2$. Assume the detector is reliable, implying the $|a_1(t)\rangle$ and $|a_2(t)\rangle$ are orthonormal at each instant $t$-i.e., $\langle a_1(t)|a_2(t)\rangle = 0$ and that the measurement interaction does not disturb states $|s_i\rangle$-i.e., the measurement is “ideal”. When **A** measures $|\Psi_t\rangle_n$, the Schrödinger equation’s unitary time evolution then leads to the “measurement state” (MS) $|\Psi_t\rangle_{nA}$:

$$|\Psi_t\rangle_{nA} = c_1|s_1(t)\rangle|a_1(t)\rangle + c_2|s_2(t)\rangle|a_2(t)\rangle, |c_1|^2 + |c_2|^2 = 1. \qquad (5.1.2)$$

of the composite system **nA** following the measurement.

Standard formalism of continuous quantum measurements [62]-[68] leads to a definite but unpredictable measurement outcome, either $|a_1(t)\rangle$ or $|a_2(t)\rangle$ and that $|\Psi_t\rangle_n$ suddenly “collapses” at instant $t'$ into the corresponding state $|s_i(t')\rangle$. But
example, writes “The whole system is in a superposition of two states: one in which no decay has occurred and. . .one in which it has occurred. Hence, the state of the cat also consists of a superposition: $|cat\rangle = c_1|live cat\rangle + c_2|death cat\rangle$. The state remains a superposition until an observer looks at the cat” [61]. The canonical formulation [62]:

$|cat\rangle = c_1|live cat\rangle|undecayed nucleus\rangle + c_2|death cat\rangle|decayed nucleus\rangle$ completely obscures the unitary Schrödinger evolution which by using GRW collapse model, predicts specific nonlocal entanglement. The cat state must be

written as:

$$|cat(t)\rangle = c_1|live cat(t)\rangle|undecayed nucleus (t)\rangle + c_2|death cat(t)\rangle|decayed nucleus(t)\rangle.$$

This entangled state actually is the collapsed state of both the cat and the nucleus, showing definite outcomes at each instant $t \ge T_{col}$.

## V.I.Schrödinger's cat paradox

As Weinberg recently reminded us [63], the measurement problem remains a fundamental conundrum. During measurement the state vector of the microscopic system collapses in a probabilistic way to one of a number of classical states, in a way that is unexplained, and cannot be described by the time-dependent Schrödinger equation. To review the essentials, it is sufficient to consider two-state systems. Suppose a nucleus $n$, whose Hilbert space is spanned by orthonormal states $|s_i(t)\rangle$, $i = 1,2$, where $|s_1(t)\rangle = |undecayed nucleus at instant t\rangle$ and $|s_2(t)\rangle = |decayed nucleus at instant t\rangle$ is in the superposition state,

$$|\Psi_t\rangle_n = c_1|s_1(t)\rangle + c_2|s_2(t)\rangle, |c_1|^2 + |c_2|^2 = 1. \qquad (5.1.1)$$

An measurement apparatus $A$, which may be microscopic or macroscopic, is designed to distinguish between states $|s_i(t)\rangle$ by transitioning at each instant $t$ into state $|a_i(t)\rangle$ if it finds $n$ is in $|s_i(t)\rangle$, $i = 1,2$. Assume the detector is reliable, implying the $|a_1(t)\rangle$ and $|a_2(t)\rangle$ are orthonormal at each instant $t$-i.e., $\langle a_1(t)|a_2(t)\rangle = 0$ and that the measurement interaction does not disturb states $|s_i\rangle$-i.e., the measurement is “ideal”. When $A$ measures $|\Psi_t\rangle_n$, the Schrödinger equation’s unitary time evolution then leads to the “measurement state” (MS) $|\Psi_t\rangle_{nA}$:

$$|\Psi_t\rangle_{nA} = c_1|s_1(t)\rangle|a_1(t)\rangle + c_2|s_2(t)\rangle|a_2(t)\rangle, |c_1|^2 + |c_2|^2 = 1. \qquad (5.1.2)$$

of the composite system $nA$ following the measurement.

Standard formalism of continuous quantum measurements [62]-[68] leads to a definite but unpredictable measurement outcome, either $|a_1(t)\rangle$ or $|a_2(t)\rangle$ and that $|\Psi_t\rangle_n$ suddenly “collapses” at instant $t'$ into the corresponding state $|s_i(t')\rangle$. But
unfortunately equation (1.2) does not appear to resemble such a collapsed state at
instant $t'$?

The measurement problem is as follows [62]-[68]:

(I) How do we reconcile canonical collapse models postulate's

(II) How do we reconcile the measurement postulate's definite outcomes with the
"measurement state" $|\Psi_t\rangle_{n,A}$ at each instant $t$ and

(III) how does the outcome become irreversibly recorded in light of the Schrödinger
equation's unitary and, hence, reversible evolution?

This paper deals with only the special case of the measurement problem, known as
Schrödinger's cat paradox. For a good and complete explanation of this paradox
see

Hobson [62] and Leggett [68].

<figure><img src="image_57.png" /><figcaption>Pic.5.1.1.Schrödinger's cat.</figcaption></figure>

**Schrödinger's cat:** a cat, a flask of poison, and a radioactive source are placed in a sealed box. If an internal monitor detects radioactivity (i.e. a single atom decaying), the flask is shattered, releasing the poison that kills the cat. The Copenhagen interpretation of quantum mechanics implies that after a while, the cat is simultaneously alive and dead. Yet, when one looks in the box, one sees the cat either alive or dead, not both alive and dead. This poses the question of when exactly quantum superposition ends and reality collapses into one possibility or the other. This subsection presents an theoretical approach of the MS that resolves the problem of definite outcomes for Schrödinger's "cat". It shows that the MS actually is the collapsed state of both Schrödinger's "cat" and nucleus, even though it evolved purely unitarily.

**The canonical collapse models.**

In order to appreciate how canonical collapse models work, and what they are able
to achieve, we briefly review the GRW model. Let us consider a system of *n*
particles which, only for the sake of simplicity, we take to be scalar and spinless;
the GRW model is defined by the following postulates: (1) The state of the system
is represented by a wave function $\psi_t(\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_n)$ belonging to the Hilbert space
unfortunately equation (1.2) does not appear to resemble such a collapsed state at
instant $t'$?

The measurement problem is as follows [62]-[68]:

(I) How do we reconcile canonical collapse models postulate's

(II) How do we reconcile the measurement postulate's definite outcomes with the
"measurement state" $|\Psi_t\rangle_{n,A}$ at each instant $t$ and

(III) how does the outcome become irreversibly recorded in light of the Schrödinger
equation's unitary and, hence, reversible evolution?

This paper deals with only the special case of the measurement problem, known as
Schrödinger's cat paradox. For a good and complete explanation of this paradox
see

Hobson [62] and Leggett [68].

<figure><img src="image_58.png" /><figcaption>Pic.5.1.1.Schrödinger's cat.</figcaption></figure>

**Schrödinger's cat:** a cat, a flask of poison, and a radioactive source are placed in a sealed box. If an internal monitor detects radioactivity (i.e. a single atom decaying), the flask is shattered, releasing the poison that kills the cat. The Copenhagen interpretation of quantum mechanics implies that after a while, the cat is simultaneously alive and dead. Yet, when one looks in the box, one sees the cat either alive or dead, not both alive and dead. This poses the question of when exactly quantum superposition ends and reality collapses into one possibility or the other. This subsection presents an theoretical approach of the MS that resolves the problem of definite outcomes for Schrödinger's "cat". It shows that the MS actually is the collapsed state of both Schrödinger's "cat" and nucleus, even though it evolved purely unitarily.

**The canonical collapse models.**

In order to appreciate how canonical collapse models work, and what they are able
to achieve, we briefly review the GRW model. Let us consider a system of *n*
particles which, only for the sake of simplicity, we take to be scalar and spinless;
the GRW model is defined by the following postulates: (1) The state of the system
is represented by a wave function $\psi_t(\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_n)$ belonging to the Hilbert space
$\mathcal{L}_2(\mathbb{R}^{3n})$. (2) At random times, the wave function experiences a sudden jump of the form: $\psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n) \to \psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n; \tilde{\mathbf{x}}_m)$, where

$$ \psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n; \tilde{\mathbf{x}}_m) = \frac{\mathfrak{R}_m(\tilde{\mathbf{x}}_m)\psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n)}{\|\mathfrak{R}_m(\tilde{\mathbf{x}}_m)\psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n)\|_2}, \quad (5.1.3) $$

where $\psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n)$ is the state vector of the whole system at time $t$, immediately prior to the jump process and $\mathfrak{R}_n(\tilde{\mathbf{x}}_m)$ is a linear operator which is conventionally chosen equal to:

$$ \mathfrak{R}_m(\tilde{\mathbf{x}}_m) = (\pi r_c^2)^{-3/4} \exp\left[-\frac{(\hat{\mathbf{x}}_m - \tilde{\mathbf{x}}_m)^2}{2r_c^2}\right], \quad (5.1.4) $$

where $r_c$ is a new parameter of the model which sets the width of the localization process, and $\hat{\mathbf{x}}_m$ is the position operator associated to the $m$-th particle of the system and the random variable $\tilde{\mathbf{x}}_m$ corresponds to the place where the jump occurs. (3) It is assumed that the jumps are distributed in time like a Poissonian process with frequency $\lambda = \lambda_{GRW}$ this is the second new parameter of the model.

(4) Between two consecutive jumps, the state vector evolves according to the standard Schrödinger equation.

The 1-particle master equation of the GRW model takes the form

$$ \frac{d}{dt} \rho(t) = -\frac{i}{\hbar} \left[ \hat{\mathbf{H}}, \rho(t) \right] - T[\rho(t)]. \quad (5.1.5) $$

Here $\hat{\mathbf{H}}$ is the standard quantum Hamiltonian of the particle, and $T[\cdot]$ represents the effect of the spontaneous collapses on the particle's wave function. In the position representation, this operator becomes:

$$ \langle \mathbf{x} | T[\rho(t)] | \mathbf{y} \rangle = \lambda \left\{ 1 - \exp \left[ - \frac{(\mathbf{x} - \mathbf{y})^2}{4r_c^2} \right] \right\} \langle \mathbf{x} | \rho(t) | \mathbf{y} \rangle. \quad (5.1.6) $$

**Remark 1.1.** We note that GRW collapse model follows from the more general

S. Weinberg formalism [1].

Another modern approach to stochastic reduction is to describe it using a stochastic nonlinear Schrödinger equation, an elegant simple example of which is the following one particle case known as Quantum Mechanics with Universal Position Localization [QMUPL]:

$$ d|\psi_t(x)\rangle = \left[-\frac{i}{\hbar}\hat{\mathbf{H}} - k(\hat{\mathbf{x}} - \langle x_t \rangle)^2 dt\right] |\psi_t(x)\rangle dt + \sqrt{2k}(\hat{q} - \langle q_t \rangle) dW_t |\psi_t(x)\rangle. \quad (5.1.7) $$
$\mathcal{L}_2(\mathbb{R}^{3n})$. (2) At random times, the wave function experiences a sudden jump of the form: $\psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n) \to \psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n; \tilde{\mathbf{x}}_m)$, where

$$ \psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n; \tilde{\mathbf{x}}_m) = \frac{\Re_m(\tilde{\mathbf{x}}_m)\psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n)}{\|\Re_m(\tilde{\mathbf{x}}_m)\psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n)\|_2}, \quad (5.1.3) $$

where $\psi_t(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n)$ is the state vector of the whole system at time $t$, immediately prior to the jump process and $\Re_n(\tilde{\mathbf{x}}_m)$ is a linear operator which is conventionally chosen equal to:

$$ \Re_m(\tilde{\mathbf{x}}_m) = (\pi r_c^2)^{-3/4} \exp\left[-\frac{(\hat{\mathbf{x}}_m - \tilde{\mathbf{x}}_m)^2}{2r_c^2}\right], \quad (5.1.4) $$

where $r_c$ is a new parameter of the model which sets the width of the localization process, and $\hat{\mathbf{x}}_m$ is the position operator associated to the $m$-th particle of the system and the random variable $\tilde{\mathbf{x}}_m$ corresponds to the place where the jump occurs. (3) It is assumed that the jumps are distributed in time like a Poissonian process with frequency $\lambda = \lambda_{GRW}$ this is the second new parameter of the model.

(4) Between two consecutive jumps, the state vector evolves according to the standard Schrödinger equation.

The 1-particle master equation of the GRW model takes the form

$$ \frac{d}{dt} \rho(t) = -\frac{i}{\hbar} \left[ \hat{\mathbf{H}}, \rho(t) \right] - T[\rho(t)]. \quad (5.1.5) $$

Here $\hat{\mathbf{H}}$ is the standard quantum Hamiltonian of the particle, and $T[\cdot]$ represents the effect of the spontaneous collapses on the particle's wave function. In the position representation, this operator becomes:

$$ \langle \mathbf{x} | T[\rho(t)] | \mathbf{y} \rangle = \lambda \left\{ 1 - \exp \left[ - \frac{(\mathbf{x} - \mathbf{y})^2}{4r_c^2} \right] \right\} \langle \mathbf{x} | \rho(t) | \mathbf{y} \rangle. \quad (5.1.6) $$

**Remark 1.1.** We note that GRW collapse model follows from the more general

S. Weinberg formalism [1].

Another modern approach to stochastic reduction is to describe it using a stochastic nonlinear Schrödinger equation, an elegant simple example of which is the following one particle case known as Quantum Mechanics with Universal Position Localization [QMUPL]:

$$ d|\psi_t(x)\rangle = \left[-\frac{i}{\hbar}\hat{\mathbf{H}} - k(\hat{\mathbf{x}} - \langle x_t \rangle)^2 dt\right] |\psi_t(x)\rangle dt + \sqrt{2k}(\hat{q} - \langle q_t \rangle)dW_t|\psi_t(x)\rangle. \quad (5.1.7) $$
Here $\hat{x}$ is the position operator, $\langle x_t \rangle = \langle \psi_t | \hat{x} | \psi_t \rangle$ it is its expectation value, and $k$ is a constant, characteristic of the model, which sets the strength of the collapse mechanics, and it is chosen proportional to the mass $m$ of the particle according to the formula: $k = (m/m_0)\lambda_0$, where $m_0$ is the nucleon's mass and $\lambda_0$ measures the collapse strength. It is easy to see that Eq.(5.1.5) contains both non-linear and stochastic terms, which are necessary to induce the collapse of the wave function.

For an example let us consider a free particle ($\hat{H} = p^2/2m$), and a Gaussian state:

$$ \psi_t(x) = \exp\{-a_t(x - \bar{x}_t)^2 + i\bar{k}_t x\}. \quad (5.1.8) $$

It is easy to see that $\psi_t(x)$ given by Eq.(5.1.6) is solution of Eq.(5.1.5), where

$$ \frac{da_t}{dt} = k - \frac{2i\hbar}{m}a_t^2, \quad \frac{d\bar{x}_t}{dt} = \frac{\hbar}{m}\bar{k}_t + \frac{\sqrt{k}}{2\text{Re}(a_t)}\dot{W}_t, \quad \frac{d\bar{k}_t}{dt} = -\sqrt{k}\frac{\text{Im}(a_t)}{\text{Re}(a_t)}\dot{W}_t. \quad (5.1.9) $$

The CSL model is defined by the following stochastic differential equation in the Fock space:

$$ d|\psi_t(\mathbf{x})\rangle = \left[ -\frac{i}{\hbar} \hat{H} - k(\hat{M}(\mathbf{x}) - \langle M_t(\mathbf{x}) \rangle)^2 dt \right] |\psi_t(\mathbf{x})\rangle dt + \sqrt{2k} (\hat{M}(\mathbf{x}) - \langle M_t(\mathbf{x}) \rangle) dW_t |\psi_t(\mathbf{x})\rangle. \quad (5.1.10) $$

## V.II. Generalized Gamow theory of the alpha decay via tunneling using GRW collapse model.

By 1928, George Gamow had solved the theory of the alpha decay via tunneling [69]. The alpha particle is trapped in a potential well by the nucleus. Classically, it is forbidden to escape, but according to the (then) newly discovered principles of quantum mechanics, it has a tiny (but non-zero) probability of "tunneling" through the barrier and appearing on the other side to escape the nucleus. Gamow solved a model potential for the nucleus and derived, from first principles, a relationship between the half-life of the decay, and the energy of the emission. The $\alpha$-particle has total energy $E$ and is incident on the barrier from the right to left.

<figure><img src="image_58.png" /><figcaption>Pic.5.2.1. The particle has total energy $E$ and is incident on the barrier $V(x)$ from right to left.</figcaption></figure>
Here $\hat{x}$ is the position operator, $\langle x_t \rangle = \langle \psi_t | \hat{x} | \psi_t \rangle$ it is its expectation value, and $k$ is a constant, characteristic of the model, which sets the strength of the collapse mechanics, and it is chosen proportional to the mass $m$ of the particle according to the formula: $k = (m/m_0)\lambda_0$, where $m_0$ is the nucleon's mass and $\lambda_0$ measures the collapse strength. It is easy to see that Eq.(5.1.5) contains both non-linear and stochastic terms, which are necessary to induce the collapse of the wave function.

For an example let us consider a free particle ($\hat{H} = p^2/2m$), and a Gaussian state:

$$ \psi_t(x) = \exp\{-a_t(x - \bar{x}_t)^2 + i\bar{k}_t x\}. \quad (5.1.8) $$

It is easy to see that $\psi_t(x)$ given by Eq.(5.1.6) is solution of Eq.(5.1.5), where

$$ \frac{da_t}{dt} = k - \frac{2i\hbar}{m}a_t^2, \quad \frac{d\bar{x}_t}{dt} = \frac{\hbar}{m}\bar{k}_t + \frac{\sqrt{k}}{2\text{Re}(a_t)}\dot{W}_t, \quad \frac{d\bar{k}_t}{dt} = -\sqrt{k}\frac{\text{Im}(a_t)}{\text{Re}(a_t)}\dot{W}_t. \quad (5.1.9) $$

The CSL model is defined by the following stochastic differential equation in the Fock space:

$$ d|\psi_t(\mathbf{x})\rangle = \left[ -\frac{i}{\hbar} \hat{H} - k \left( \hat{M}(\mathbf{x}) - \langle M_t(\mathbf{x}) \rangle \right)^2 dt \right] |\psi_t(\mathbf{x})\rangle dt + \sqrt{2k} \left( \hat{M}(\mathbf{x}) - \langle M_t(\mathbf{x}) \rangle \right) dW_t |\psi_t(\mathbf{x})\rangle. \quad (5.1.10) $$

## V.II. Generalized Gamow theory of the alpha decay via tunneling using GRW collapse model.

By 1928, George Gamow had solved the theory of the alpha decay via tunneling [69]. The alpha particle is trapped in a potential well by the nucleus. Classically, it is forbidden to escape, but according to the (then) newly discovered principles of quantum mechanics, it has a tiny (but non-zero) probability of "tunneling" through the barrier and appearing on the other side to escape the nucleus. Gamow solved a model potential for the nucleus and derived, from first principles, a relationship between the half-life of the decay, and the energy of the emission. The $\alpha$-particle has total energy $E$ and is incident on the barrier from the right to left.

<figure><img src="image_59.png" /><figcaption>Pic.5.2.1. The particle has total energy $E$ and is incident on the barrier $V(x)$ from right to left.</figcaption></figure>
The Schrödinger equation in each of regions **I** = {$x|x < 0$}, **II** = {$x|0 ≤ x ≤ l$} and **III** = {$x|x > l$} takes the following form

$$ \frac{\partial^2 \Psi(x)}{\partial x^2} + \frac{2m}{\hbar^2} [E - U(x)] \Psi(x) = 0, \qquad (5.2.1) $$

where

$$ U(x) = \begin{cases} 0 & \text{for } x < 0 \\ U_0 & \text{for } 0 \le x \le l \\ 0 & \text{for } x > l \end{cases} \qquad (5.2.2) $$

The solutions reads [8]:

$$ \begin{aligned} \Psi_{\text{III}}(x) &= C_+ \exp(ikx) + C_- \exp(-ikx), \\ \Psi_{\text{II}}(x) &= B_+ \exp(k'x) + B_- \exp(-k'x), \end{aligned} \qquad (5.2.3) $$

$$ \Psi_{\text{I}}(x) = A \cos(kx) = \frac{A}{2} [\exp(ikx) + \exp(-ikx)], $$

where

$$ k = \frac{2\pi}{\hbar} \sqrt{2mE}, \quad k' = \frac{2\pi}{\hbar} \sqrt{2m(U_0 - E)}. \qquad (5.2.4) $$

At the boundary $x = 0$ we have the following boundary conditions:

$$ \Psi_{\text{I}}(0) = \Psi_{\text{II}}(0), \quad \left. \frac{\partial \Psi_{\text{I}}(x)}{\partial x} \right|_{x=0} = \left. \frac{\partial \Psi_{\text{II}}(x)}{\partial x} \right|_{x=0}. \qquad (5.2.5) $$

At the boundary $x = l$ we have the following boundary conditions

$$ \Psi_{\text{II}}(l) = \Psi_{\text{III}}(l), \quad \left. \frac{\partial \Psi_{\text{II}}(x)}{\partial x} \right|_{x=l} = \left. \frac{\partial \Psi_{\text{III}}(x)}{\partial x} \right|_{x=l}. \qquad (5.2.6) $$

From the boundary conditions (5.2.5)-(5.2.6) one obtains [69]:

$$ \begin{aligned} B_+ &= \frac{A}{2} \left(1 + i \frac{k}{k'}\right), & B_- &= \frac{A}{2} \left(1 - i \frac{k}{k'}\right), \\ C_+ &= A[\mathrm{ch}(k'l) + i\mathrm{Dsh}(k'l)], & C_- &= i(\mathrm{ASsh}(k'l) \exp(ikl)), \\ D &= \frac{1}{2} \left(\frac{k}{k'} - \frac{k'}{k}\right), & S &= \frac{1}{2} \left(\frac{k}{k'} + \frac{k'}{k}\right). \end{aligned} \qquad (5.2.7) $$

From (5.2.7) one obtain the conservation law
The Schrödinger equation in each of regions **I** = {$x|x < 0$}, **II** = {$x|0 ≤ x ≤ l$} and **III** = {$x|x > l$} takes the following form

$$ \frac{\partial^2 \Psi(x)}{\partial x^2} + \frac{2m}{\hbar^2} [E - U(x)] \Psi(x) = 0, \qquad (5.2.1) $$

where

$$ U(x) = \begin{cases} 0 & \text{for } x < 0 \\ U_0 & \text{for } 0 \le x \le l \\ 0 & \text{for } x > l \end{cases} \qquad (5.2.2) $$

The solutions reads [8]:

$$ \Psi_{\text{III}}(x) = C_+ \exp(ikx) + C_- \exp(-ikx), $$

$$ \Psi_{\text{II}}(x) = B_+ \exp(k'x) + B_- \exp(-k'x), \qquad (5.2.3) $$

$$ \Psi_{\text{I}}(x) = A \cos(kx) = \frac{A}{2} [\exp(ikx) + \exp(-ikx)], $$

where

$$ k = \frac{2\pi}{\hbar} \sqrt{2mE}, \quad k' = \frac{2\pi}{\hbar} \sqrt{2m(U_0 - E)}. \qquad (5.2.4) $$

At the boundary $x = 0$ we have the following boundary conditions:

$$ \Psi_{\text{I}}(0) = \Psi_{\text{II}}(0), \quad \left. \frac{\partial \Psi_{\text{I}}(x)}{\partial x} \right|_{x=0} = \left. \frac{\partial \Psi_{\text{II}}(x)}{\partial x} \right|_{x=0}. \qquad (5.2.5) $$

At the boundary $x = l$ we have the following boundary conditions

$$ \Psi_{\text{II}}(l) = \Psi_{\text{III}}(l), \quad \left. \frac{\partial \Psi_{\text{II}}(x)}{\partial x} \right|_{x=l} = \left. \frac{\partial \Psi_{\text{III}}(x)}{\partial x} \right|_{x=l}. \qquad (5.2.6) $$

From the boundary conditions (5.2.5)-(5.2.6) one obtains [69]:

$$ B_+ = \frac{A}{2} \left(1 + i \frac{k}{k'}\right), \quad B_- = \frac{A}{2} \left(1 - i \frac{k}{k'}\right), \qquad (5.2.7) $$

$$ C_+ = A[\mathrm{ch}(k'l) + i\mathrm{Dsh}(k'l)], \quad C_- = i(AS\mathrm{sh}(k'l)\exp(ikl)), $$

$$ D = \frac{1}{2} \left( \frac{k}{k'} - \frac{k'}{k} \right), \quad S = \frac{1}{2} \left( \frac{k}{k'} + \frac{k'}{k} \right). $$

From (5.2.7) one obtain the conservation law
$$|A|^2 = |C_+|^2 - |C_-|^2.$$

Let us introduce now a function $E_{II}(x, l) = \theta_2(x, l)E_2(x, l)$ where

$$E_2(x, l) = \begin{cases} (\pi r_c^2)^{-1/4} \exp\left(-\frac{x^2}{2r_c^2}\right) & \text{for } -\infty < x < \frac{l}{2} \\ (\pi r_c^2)^{-1/4} \exp\left(-\frac{(x-l)^2}{2r_c^2}\right) & \text{for } \frac{l}{2} \le x < \infty \end{cases} \qquad (5.2.8) \\ \\ \theta_2(x, l) = \begin{cases} 1 & \text{for } x \in [0, l] \\ 0 & \text{for } x \notin [0, l] \end{cases}$$

**Assumption 5.2.1.** We assume now that:

(i) at instant $t = 0$ the wave function $\Psi_I(x)$ experiences a sudden jump of the form

$$\Psi_I(x) \to \Psi_I^{\#}(x) = \frac{\Re_I(\hat{x})\Psi_I(x)}{\|\Re_I(\hat{x})\Psi_I(x)\|_2}, \qquad (5.2.9)$$

where $\Re_I(\hat{x})$ is a linear operator which is chosen equal to:

$$\Re_I(\hat{x}) = (\pi r_c^2)^{-1/4} \theta_1(\hat{x}, l) \exp\left[-\frac{\hat{x}^2}{2r_c^2}\right]; \qquad (5.2.10)$$

where

$$\theta_1(x, l) = \begin{cases} 1 & \text{for } x \in [-l, 0], \\ 0 & \text{for } x \notin [-l, 0]. \end{cases}$$

**Remark 5.2.1.** Note that: $\text{supp}(\Psi_I^{\#}(x)) \subseteq [-l, 0]$

(ii) at instant $t = 0$ the wave function $\Psi_{II}(x)$ experiences a sudden jump of the form

$$\Psi_{II}(x) \to \Psi_{II}^{\#}(x) = \frac{\Re_{II}(\hat{x})\Psi_{II}(x)}{\|\Re_{II}(\hat{x})\Psi_{II}(x)\|_2}, \qquad (5.2.11)$$

where $\Re_{II}(\hat{x})$ is a linear operator which is chosen equal to:

$$\Re_{II}(\hat{x}) = E_{II}(\hat{x}, l); \qquad (5.2.12)$$

**Remark 5.2.2.** Note that: $\text{supp}(\Psi_{II}^{\#}(x)) \subseteq [0, l]$.

(iii) at instant $t = 0$ the wave function $\Psi_{III}(x)$ experiences a sudden jump of the form
$$|A|^2 = |C_+|^2 - |C_-|^2.$$

Let us introduce now a function $E_{II}(x, l) = \theta_2(x, l)E_2(x, l)$ where

$$E_2(x, l) = \begin{cases} (\pi r_c^2)^{-1/4} \exp\left(-\frac{x^2}{2r_c^2}\right) & \text{for } -\infty < x < \frac{l}{2} \\ (\pi r_c^2)^{-1/4} \exp\left(-\frac{(x-l)^2}{2r_c^2}\right) & \text{for } \frac{l}{2} \le x < \infty \end{cases} \qquad (5.2.8) \\ \theta_2(x, l) = \begin{cases} 1 & \text{for } x \in [0, l] \\ 0 & \text{for } x \notin [0, l] \end{cases}$$

**Assumption 5.2.1.** We assume now that:

(i) at instant $t = 0$ the wave function $\Psi_I(x)$ experiences a sudden jump of the form

$$\Psi_I(x) \to \Psi_I^{\#}(x) = \frac{\Re_I(\hat{x})\Psi_I(x)}{\|\Re_I(\hat{x})\Psi_I(x)\|_2}, \qquad (5.2.9)$$

where $\Re_I(\hat{x})$ is a linear operator which is chosen equal to:

$$\Re_I(\hat{x}) = (\pi r_c^2)^{-1/4} \theta_1(\hat{x}, l) \exp\left[-\frac{\hat{x}^2}{2r_c^2}\right]; \qquad (5.2.10)$$

where

$$\theta_1(x, l) = \begin{cases} 1 & \text{for } x \in [-l, 0], \\ 0 & \text{for } x \notin [-l, 0]. \end{cases}$$

**Remark 5.2.1.** Note that: $\text{supp}(\Psi_I^{\#}(x)) \subseteq [-l, 0]$

(ii) at instant $t = 0$ the wave function $\Psi_{II}(x)$ experiences a sudden jump of the form

$$\Psi_{II}(x) \to \Psi_{II}^{\#}(x) = \frac{\Re_{II}(\hat{x})\Psi_{II}(x)}{\|\Re_{II}(\hat{x})\Psi_{II}(x)\|_2}, \qquad (5.2.11)$$

where $\Re_{II}(\hat{x})$ is a linear operator which is chosen equal to:

$$\Re_{II}(\hat{x}) = E_{II}(\hat{x}, l); \qquad (5.2.12)$$

**Remark 5.2.2.** Note that: $\text{supp}(\Psi_{II}^{\#}(x)) \subseteq [0, l]$.

(iii) at instant $t = 0$ the wave function $\Psi_{III}(x)$ experiences a sudden jump of the form
$$ \Psi_{\text{III}}(x) \rightarrow \Psi_{\text{III}}^{\#}(x) = \frac{\Re_{\text{III}}(\hat{x})\Psi_{\text{III}}(x)}{\|\Re_{\text{III}}(\hat{x})\Psi_{\text{III}}(x)\|_2}, \qquad (5.2.13) $$

where $\Re_{\text{III}}(\hat{x})$ is a linear operator which is chosen equal to:

$$ \Re_{\text{III}}(\hat{x}) = (\pi r_c^2)^{-1/4} \exp\left[-\frac{(\hat{x} - l)^2}{2r_c^2}\right]. \qquad (5.2.14) $$

**Remark 5.2.3.** Note that. We have choose operators (5.2.10),(5.2.12) and (5.2.14) such that the boundary conditions (5.2.5),(5.2.6) is satisfied.

**Definition 5.2.1.** Let $\Psi(x)$ be an solution of the Schrödinger equation (5.2.1). The stationary Schrödinger equation (5.2.1) is a weakly well preserved in region $\Gamma \subseteq \mathbb{R}$ by collapsed wave function $\Psi^{\#}(x)$ if there exist an wave function $\Psi(x)$ such that the estimate

$$ \int_{\Gamma} \left\{ \frac{\partial^2 \Psi^{\#}(x)}{\partial x^2} + \frac{2m}{\hbar^2} [E - U(x)] \Psi^{\#}(x) \right\} dx = O(\hbar^{2+\alpha}), \qquad (5.2.15) $$

where the inequality $\alpha \ge 1$, is satisfied.

**Proposition 5.2.1.** The Schrödinger equation in each of regions I, II, III is a weakly

well preserved by collapsed wave function $\Psi_I^{\#}(x)$, $\Psi_{II}^{\#}(x)$ and $\Psi_{III}^{\#}(x)$ correspondingly.

**Proof.** See Appendix CAT-B.

**Definition 5.2.2.** Let us consider the time-dependent Schrödinger equation:

$$ i\hbar \frac{\partial \Psi(\mathbf{x}, t)}{\partial t} = \hat{\mathbf{H}} \Psi(\mathbf{x}, t), \qquad (5.2.16) $$
$$ t \in [0, T], \mathbf{x} \in \mathbb{R}^{3n}. $$

The time-dependent Schrödinger equation (5.2.16) is a weakly well preserved by corresponding to $\Psi(\mathbf{x}, t)$ collapsed wave function $\Psi^{\#}(\mathbf{x}, t)$
$$ \Psi_{\text{III}}(x) \to \Psi_{\text{III}}^{\#}(x) = \frac{\Re_{\text{III}}(\hat{x})\Psi_{\text{III}}(x)}{\|\Re_{\text{III}}(\hat{x})\Psi_{\text{III}}(x)\|_2}, \qquad (5.2.13) $$

where $\Re_{\text{III}}(\hat{x})$ is a linear operator which is chosen equal to:

$$ \Re_{\text{III}}(\hat{x}) = (\pi r_c^2)^{-1/4} \exp\left[-\frac{(\hat{x} - l)^2}{2r_c^2}\right]. \qquad (5.2.14) $$

**Remark 5.2.3.** Note that. We have choose operators (5.2.10),(5.2.12) and (5.2.14) such that the boundary conditions (5.2.5),(5.2.6) is satisfied.

**Definition 5.2.1.** Let $\Psi(x)$ be an solution of the Schrödinger equation (5.2.1). The stationary Schrödinger equation (5.2.1) is a weakly well preserved in region $\Gamma \subseteq \mathbb{R}$ by collapsed wave function $\Psi^{\#}(x)$ if there exist an wave function $\Psi(x)$ such that the estimate

$$ \int_{\Gamma} \left\{ \frac{\partial^2 \Psi^{\#}(x)}{\partial x^2} + \frac{2m}{\hbar^2} [E - U(x)] \Psi^{\#}(x) \right\} dx = O(\hbar^{2+\alpha}), \qquad (5.2.15) $$

where the inequality $\alpha \ge 1$, is satisfied.

**Proposition 5.2.1.** The Schrödinger equation in each of regions I, II, III is a weakly

well preserved by collapsed wave function $\Psi_I^{\#}(x)$, $\Psi_{II}^{\#}(x)$ and $\Psi_{III}^{\#}(x)$ correspondingly.

**Proof.** See Appendix CAT-B.

**Definition 5.2.2.** Let us consider the time-dependent Schrödinger equation:

$$ i\hbar \frac{\partial \Psi(\mathbf{x}, t)}{\partial t} = \hat{\mathbf{H}} \Psi(\mathbf{x}, t), \qquad (5.2.16) $$
$$ t \in [0, T], \mathbf{x} \in \mathbb{R}^{3n}. $$

The time-dependent Schrödinger equation (5.2.16) is a weakly well preserved by corresponding to $\Psi(\mathbf{x}, t)$ collapsed wave function $\Psi^{\#}(\mathbf{x}, t)$
$$ \Psi^{\#}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n, t) = $$

$$ \Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n, t; \tilde{\mathbf{x}}_{m_1}, \dots, \tilde{\mathbf{x}}_{m_k}) = $$

$$ = \frac{\Re_{m_1, \dots, m_k}(\tilde{\mathbf{x}}_{m_1}, \dots, \tilde{\mathbf{x}}_{m_k}) \Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n, t)}{\| \Re_{m_1, \dots, m_k}(\tilde{\mathbf{x}}_{m_1}, \dots, \tilde{\mathbf{x}}_{m_k}) \Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n, t) \|_2}, $$

$$ \Re_{m_1, \dots, m_k}(\tilde{\mathbf{x}}_{m_1}, \dots, \tilde{\mathbf{x}}_{m_k}) = \prod_{i=1}^{k} \Re_{m_i}(\tilde{\mathbf{x}}_{m_i}) $$

in region $\Gamma \subseteq \mathbb{R}^{3d}$ if there exist an wave function $\Psi(\mathbf{x}, t)$ such that the estimate

$$ \int_{\Gamma} \left\{ i\hbar \frac{\partial \Psi^{\#}(\mathbf{x}, t)}{\partial t} - \hat{H} \Psi^{\#}(\mathbf{x}, t) \right\} d^{3d}x = O(\hbar^{\alpha}), \quad (5.2.17) $$

$$ t \in [0, T], \mathbf{x} \in \mathbb{R}^{3d}, $$

where the inequality $\alpha \ge 1$, is satisfied.

**Definition 5.2.3.** Let $\Psi^{\#}(\mathbf{x}, t) = \Psi^{\#}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t)$ be a function

$\Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t; \tilde{\mathbf{x}}_1, \dots, \tilde{\mathbf{x}}_d)$. Let us consider the Probability Current Law

$$ \frac{\partial}{\partial t} P(\Gamma, t) + \int_{\partial \Gamma} \mathbf{J}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t) \cdot \mathbf{n} d^{2d}x = O(\hbar^{\alpha}), \quad (5.2.18) $$

$$ \mathbf{J}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t) = \Psi(\mathbf{x}, t) \overline{\nabla \Psi(\mathbf{x}, t)} - \overline{\Psi(\mathbf{x}, t)} \nabla \Psi(\mathbf{x}, t), $$
$$ t \in [0, T], \mathbf{x} \in \mathbb{R}^{3d}, $$

corresponding to Schrödinger equation (5.2.16). Probability Current Law (5.2.18) is

a weakly well preserved by corresponding to $\Psi(\mathbf{x}, t)$ collapsed wave function $\Psi^{\#}(\mathbf{x}, t)$ in region $\Gamma \subseteq \mathbb{R}^{3d}$ if there exist an wave function $\Psi(\mathbf{x}, t)$ such that the estimate

$$ \frac{\partial}{\partial t} P(\Gamma, t) + \int_{\partial \Gamma} \mathbf{J}^{\#}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t) \cdot \mathbf{n} d^{2d}x = O(\hbar^{\alpha}), \quad (5.2.19) $$

$$ \mathbf{J}^{\#}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t) = \Psi^{\#}(\mathbf{x}, t) \overline{\nabla \Psi^{\#}(\mathbf{x}, t)} - \overline{\Psi^{\#}(\mathbf{x}, t)} \nabla \Psi^{\#}(\mathbf{x}, t) $$
$$ = O(\hbar^{\alpha}), $$
$$ t \in [0, T], \mathbf{x} \in \mathbb{R}^{3d}, $$

where the inequality $\alpha \ge 1$, is satisfied.

**Proposition 5.2.2.** Assume that there exist an wave function $\Psi(\mathbf{x}, t)$ such that the estimate (5.2.17) is satisfied. Then Probability Current Law (2.18) is a weakly well
$$ \Psi^{\#}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n, t) = $$

$$ \Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n, t; \tilde{\mathbf{x}}_{m_1}, \dots, \tilde{\mathbf{x}}_{m_k}) = $$

$$ = \frac{\Re_{m_1, \dots, m_k}(\tilde{\mathbf{x}}_{m_1}, \dots, \tilde{\mathbf{x}}_{m_k}) \Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n, t)}{\| \Re_{m_1, \dots, m_k}(\tilde{\mathbf{x}}_{m_1}, \dots, \tilde{\mathbf{x}}_{m_k}) \Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n, t) \|_2}, $$

$$ \Re_{m_1, \dots, m_k}(\tilde{\mathbf{x}}_{m_1}, \dots, \tilde{\mathbf{x}}_{m_k}) = \prod_{i=1}^{k} \Re_{m_i}(\tilde{\mathbf{x}}_{m_i}) $$

in region $\Gamma \subseteq \mathbb{R}^{3d}$ if there exist an wave function $\Psi(\mathbf{x}, t)$ such that the estimate

$$ \int_{\Gamma} \left\{ i\hbar \frac{\partial \Psi^{\#}(\mathbf{x}, t)}{\partial t} - \hat{H} \Psi^{\#}(\mathbf{x}, t) \right\} d^{3d}x = O(\hbar^{\alpha}), \quad (5.2.17) $$

$$ t \in [0, T], \mathbf{x} \in \mathbb{R}^{3d}, $$

where the inequality $\alpha \ge 1$, is satisfied.

**Definition 5.2.3.** Let $\Psi^{\#}(\mathbf{x}, t) = \Psi^{\#}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t)$ be a function

$\Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t; \tilde{\mathbf{x}}_1, \dots, \tilde{\mathbf{x}}_d)$. Let us consider the Probability Current Law

$$ \frac{\partial}{\partial t} P(\Gamma, t) + \int_{\partial \Gamma} \mathbf{J}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t) \cdot \mathbf{n} d^{2d}x = O(\hbar^{\alpha}), \quad (5.2.18) $$

$$ \mathbf{J}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t) = \Psi(\mathbf{x}, t) \overline{\nabla \Psi(\mathbf{x}, t)} - \overline{\Psi(\mathbf{x}, t)} \nabla \Psi(\mathbf{x}, t), \\ t \in [0, T], \mathbf{x} \in \mathbb{R}^{3d}, $$

corresponding to Schrödinger equation (5.2.16). Probability Current Law (5.2.18) is

a weakly well preserved by corresponding to $\Psi(\mathbf{x}, t)$ collapsed wave function $\Psi^{\#}(\mathbf{x}, t)$ in region $\Gamma \subseteq \mathbb{R}^{3d}$ if there exist an wave function $\Psi(\mathbf{x}, t)$ such that the estimate

$$ \frac{\partial}{\partial t} P(\Gamma, t) + \int_{\partial \Gamma} \mathbf{J}^{\#}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t) \cdot \mathbf{n} d^{2d}x = O(\hbar^{\alpha}), \quad (5.2.19) $$

$$ \mathbf{J}^{\#}(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_d, t) = \Psi^{\#}(\mathbf{x}, t) \overline{\nabla \Psi^{\#}(\mathbf{x}, t)} - \overline{\Psi^{\#}(\mathbf{x}, t)} \nabla \Psi^{\#}(\mathbf{x}, t) \\ = O(\hbar^{\alpha}), \\ t \in [0, T], \mathbf{x} \in \mathbb{R}^{3d}, $$

where the inequality $\alpha \ge 1$, is satisfied.

**Proposition 5.2.2.** Assume that there exist an wave function $\Psi(\mathbf{x}, t)$ such that the estimate (5.2.17) is satisfied. Then Probability Current Law (2.18) is a weakly well
preserved by corresponding to $\Psi(\mathbf{x},t)$ collapsed wave function $\Psi^{\#}(\mathbf{x},t)$ in region
$\Gamma \subseteq \mathbb{R}^{3d}$, i.e. the estimate (5.2.19) is satisfied on the wave function $\Psi^{\#}(\mathbf{x},t)$.

V.III. Schrödinger's Cat paradox resolution

In this section we shall consider the problem of the collapse of the cat state vector
on the basis of two different hypotheses:

(A) The canonical postulate of QM is correct in all cases.

(B) The canonical interpretation of the wave function $\psi = c_1\psi_1 + c_2\psi_2$ is correct only when the supports the wave functions $\psi_1$ and $\psi_2$ essentially overlap. When the wave functions $\psi_1$ and $\psi_2$ have separated supports (as in the case of the experiment that we are considering in section II) we claim that canonical interpretation of the wave function $\psi = c_1\psi_1 + c_2\psi_2$ is no longer valid for a such cat state, for details see Appendix CAT-C.

V.III.1. Considerationtion of the Schrödinger's cat
paradox using canonical von Neumann postulate

Let $|s_1(t)\rangle$ and $|s_2(t)\rangle$ be

$$
|s_1(t)\rangle = |\text{undecayed nucleus at instant } t\rangle, \qquad (5.3.1)
$$

$$
|s_2(t)\rangle = |\text{decayed nucleus at instant } t\rangle.
$$

In a good approximation we assume now that

$$
|s_1(0)\rangle = \int_{-\infty}^{+\infty} \Psi_{II}^{\#}(x) |x\rangle dx \qquad (5.3.2)
$$

and

$$
|s_2(0)\rangle = \int_{-\infty}^{+\infty} \Psi_I^{\#}(x) |x\rangle dx. \tag{5.3.3}
$$

**Remark 5.3.1.** Note that: (i) $|s_2(0)\rangle = |\text{decayed nucleus at instant } 0\rangle =$
$= |\text{free } \alpha\text{-particle at instant } 0\rangle$. (ii) Feynman propagator of a free $\alpha$-particle are [70]:

$$
K_2(x,t,x_0) = \left( \frac{m}{2\pi i \hbar t} \right)^{1/2} \exp \left\{ \frac{i}{\hbar} \left[ \frac{m(x-x_0)^2}{2t} \right] \right\}. \quad (5.3.4)
$$

Therefore from Eq.(5.3.3),Eq.(5.2.9) and Eq.(5.3.4) we obtain
preserved by corresponding to $\Psi(x, t)$ collapsed wave function $\Psi^#(x, t)$ in region
$\Gamma \subseteq \mathbb{R}^{3d}$, i.e. the estimate (5.2.19) is satisfied on the wave function $\Psi^#(x, t)$.

V.III. Schrödinger’s Cat paradox resolution

In this section we shall consider the problem of the collapse of the cat state vector
on the basis of two different hypotheses:

(A) The canonical postulate of QM is correct in all cases.

(B) The canonical interpretation of the wave function $\psi = c_1\psi_1 + c_2\psi_2$ is correct only when the supports the wave functions $\psi_1$ and $\psi_2$ essentially overlap. When the wave functions $\psi_1$ and $\psi_2$ have separated supports (as in the case of the experiment that we are considering in section II) we claim that canonical interpretation of the wave function $\psi = c_1\psi_1 + c_2\psi_2$ is no longer valid for a such cat state, for details see Appendix CAT-C.

V.III.1. Considerationtion of the Schrödinger’s cat
paradox using canonical von Neumann postulate

Let $|s_1(t)\rangle$ and $|s_2(t)\rangle$ be

$$
|s_1(t)\rangle = |\text{undecayed nucleus at instant } t\rangle, \qquad (5.3.1)
$$

$$
|s_2(t)\rangle = |\text{decayed nucleus at instant } t\rangle.
$$

In a good approximation we assume now that

$$
|s_1(0)\rangle = \int_{-\infty}^{+\infty} \Psi_{II}^{\#}(x)|x\rangle dx \qquad (5.3.2)
$$

and

$$
|s_2(0)\rangle = \int_{-\infty}^{+\infty} \Psi_I^\#(x) |x\rangle dx. \qquad (5.3.3)
$$

**Remark 5.3.1.** Note that: (i) $|s_2(0)\rangle = |\text{decayed nucleus at instant } 0\rangle =$
$= |\text{free } \alpha\text{-particle at instant } 0\rangle$. (ii) Feynman propagator of a free $\alpha$-particle are [70]:

$$
K_2(x,t,x_0) = \left(\frac{m}{2\pi i \hbar t}\right)^{1/2} \exp\left\{\frac{i}{\hbar}\left[\frac{m(x-x_0)^2}{2t}\right]\right\}. \quad (5.3.4)
$$

Therefore from Eq.(5.3.3),Eq.(5.2.9) and Eq.(5.3.4) we obtain
$$
\begin{align*}
& |s_2(t)\rangle = \int_{-\infty}^{+\infty} \Psi_I^{\#}(x,t) |x\rangle dx, \\
& \Psi_I^{\#}(x,t) = \int_{-\infty}^{0} \Psi_I^{\#}(x_0) K_2(x,t,x_0) dx_0 = \\
& (\pi r_c^2)^{-1/4} \times \left(\frac{m}{2\pi i \hbar t}\right)^{1/2} \times \int_{-\infty}^{0} \theta_1(x_0,l) \exp\left(-\frac{x_0^2}{2r_c^2}\right) \exp\left(-i \frac{2\pi}{\hbar} \sqrt{2mE} x_0\right) \times \\
& \qquad \times \exp\left\{\frac{i}{\hbar} \left[\frac{m(x-x_0)^2}{2t}\right]\right\} dx_0 = \tag{5.3.5} \\
& (\pi r_c^2)^{-1/4} \times \left(\frac{m}{2\pi i \hbar \epsilon t}\right)^{1/2} \times \int_{-l}^{0} \theta_1(x_0,l) \exp\left(-\frac{x_0^2}{2r_c^2}\right) \times \\
& \qquad \times \exp\left\{\frac{i}{\hbar} \left[\frac{m(x-x_0)^2}{2t} - \pi\sqrt{4mE} x_0\right]\right\} dx_0 = \\
& (\pi r_c^2)^{-1/4} \times \left(\frac{m}{2\pi i \hbar t}\right)^{1/2} \times \int_{-l}^{0} \theta_1(x_0,l) \exp\left(-\frac{x_0^2}{2r_c^2}\right) \times \exp\left\{\frac{i}{\hbar}[S(t,x,x_0)]\right\} dx_0,
\end{align*}
$$

where

$$
S(t,x,x_0) = \frac{m(x-x_0)^2}{2t} - \pi\sqrt{8mE}x_0. \tag{5.3.6}
$$

We assume now that

$$
h \ll 2r_c^2 \ll l^2 < 1. \tag{5.3.7}
$$

Oscillatory integral in RHS of Eq.(5.3.5) is calculated now directly using stationary phase approximation. The phase term $S(x,x_0)$ given by Eq.(5.3.6) is stationary when

$$
\frac{\partial S(t, x, x_0)}{\partial x_0} = - \frac{m(x - x_0)}{t} - \pi \sqrt{8mE} = 0. \quad (5.3.8)
$$

Therefore

$$
\begin{align}
& -\frac{m(x-x_0)}{t} - \pi\sqrt{8mE} = 0, \tag{5.3.9} \\
& -(x-x_0) = \pi t \sqrt{8E/m}, \nonumber
\end{align}
$$

and thus stationary point $x_0(t,x)$ are
$$
\begin{align*}
& |s_2(t)\rangle = \int_{-\infty}^{+\infty} \Psi_I^{\#}(x,t) |x\rangle dx, \\
& \Psi_I^{\#}(x,t) = \int_{-\infty}^{0} \Psi_I^{\#}(x_0) K_2(x,t,x_0) dx_0 = \\
& (\pi r_c^2)^{-1/4} \times \left(\frac{m}{2\pi i \hbar t}\right)^{1/2} \times \int_{-\infty}^{0} \theta_1(x_0,l) \exp\left(-\frac{x_0^2}{2r_c^2}\right) \exp\left(-i \frac{2\pi}{\hbar} \sqrt{2mE} x_0\right) \times \\
& \qquad \times \exp\left\{\frac{i}{\hbar} \left[ \frac{m(x-x_0)^2}{2t} \right]\right\} dx_0 = \tag{5.3.5} \\
& (\pi r_c^2)^{-1/4} \times \left(\frac{m}{2\pi i \hbar \epsilon t}\right)^{1/2} \times \int_{-l}^{0} \theta_1(x_0,l) \exp\left(-\frac{x_0^2}{2r_c^2}\right) \times \\
& \qquad \times \exp\left\{\frac{i}{\hbar} \left[ \frac{m(x-x_0)^2}{2t} - \pi \sqrt{4mE} x_0 \right] \right\} dx_0 = \\
& (\pi r_c^2)^{-1/4} \times \left(\frac{m}{2\pi i \hbar t}\right)^{1/2} \times \int_{-l}^{0} \theta_1(x_0,l) \exp\left(-\frac{x_0^2}{2r_c^2}\right) \times \exp\left\{\frac{i}{\hbar} [S(t,x,x_0)]\right\} dx_0,
\end{align*}
$$

where

$$
S(t,x,x_0) = \frac{m(x-x_0)^2}{2t} - \pi \sqrt{8mE} x_0. \tag{5.3.6}
$$

We assume now that

$$
\hbar \ll 2r_c^2 \ll l^2 < 1. \tag{5.3.7}
$$

Oscillatory integral in RHS of Eq.(5.3.5) is calculated now directly using stationary phase approximation. The phase term $S(x,x_0)$ given by Eq.(5.3.6) is stationary when

$$
\frac{\partial S(t, x, x_0)}{\partial x_0} = - \frac{m(x - x_0)}{t} - \pi \sqrt{8mE} = 0. \quad (5.3.8)
$$

Therefore

$$
\begin{align}
& -\frac{m(x-x_0)}{t} - \pi\sqrt{8mE} = 0, \tag{5.3.9} \\
& -(x-x_0) = \pi t \sqrt{8E/m}, \nonumber
\end{align}
$$

and thus stationary point $x_0(t,x)$ are
$$x_0(t,x) = \pi t \sqrt{8E/m} + x. \qquad (5.3.10)$$

Thus from Eq.(5.3.5) and Eq.(5.3.10) using stationary phase approximation we obtain

$$|s_2(t)\rangle = |s_2(t)\rangle = \int_{-\infty}^{+\infty} \Psi_I^{\#}(x,t)|x\rangle dx, \qquad (5.3.11) \\ \Psi_I^{\#}(x,t) = \\ (\pi r_c^2)^{-1/4} \times \theta_1(x_0(t,x),l) \exp\left[-\frac{x_0^2(t,x)}{2r_c^2}\right] \times \exp\left\{\frac{i}{\hbar}[S(t,x,x_0(t,x))]\right\} + O(\hbar),$$

where

$$S(x,x_0(t,x)) = \frac{m(x - x_0(t,x))^2}{2t} - \pi \sqrt{8mE} x_0(t,x). \qquad (5.3.12)$$

From Eq.(5.3.11) we obtain

$$\langle s_2(t) | s_2(t) \rangle \approx (\pi r_c^2)^{-1/2} \times \theta_1(x + \pi t \sqrt{8E/m}, l) \exp\left[-\frac{(x + \pi t \sqrt{8E/m})^2}{r_c^2}\right]. \qquad (5.3.13)$$

**Remark 5.3.2.** From the inequality (5.3.7) and Eq.(5.3.13) follows that $\alpha$-particle at each instant $t \ge 0$ moves quasiclassically from right to left by the law

$$x(t) = -\pi t \sqrt{8E/m}, \qquad (5.3.14)$$

i.e. i.e.,estimating the position $x(t,x_0,t_0;\hbar)$ at each instant $t \ge 0$ with final error $r_c$ gives $|\langle x \rangle(t) - x(t)| \le r_c, i = 1,...,d$ with a probability

$$P\{|\langle x \rangle(t, 0, 0; \hbar) - x(t)| \le r_c\} = 1.$$

**Remark 5.3.3.** We assume now that a distance between radioactive source and internal monitor which detects a single atom decaying (see Pic.1) is equal to $L$.

**Proposition 5.3.1.** After $\alpha$-decay at instant $t = 0$ the collapse: $|\text{live cat}\rangle \rightarrow |\text{death cat}\rangle$ arises at instant

$$T = \frac{L}{\pi \sqrt{8E/m}} \qquad (5.3.15)$$

with a probability $P_T(|\text{death cat}\rangle)$ to observe a state $|\text{death cat}\rangle$ at instant $T$ is $P_T(|\text{death cat}\rangle) = 1$.

**Proof.** Note that. In this case Schrödinger's cat in fact permorm the single measurement of $\alpha$-particle position with accuracy of $\delta x = l$ at instant $t = T$ (given by Eq.(5.3.15)) by internal monitor (see Pic.5.1.1). The probability of getting a result $L$

with accuracy of $\delta x = l$ given by
$$x_0(t,x) = \pi t \sqrt{8E/m} + x. \qquad (5.3.10)$$

Thus from Eq.(5.3.5) and Eq.(5.3.10) using stationary phase approximation we obtain

$$|s_2(t)\rangle = |s_2(t)\rangle = \int_{-\infty}^{+\infty} \Psi_I^{\#}(x,t)|x\rangle dx, \qquad (5.3.11) \\ \Psi_I^{\#}(x,t) = \\ (\pi r_c^2)^{-1/4} \times \theta_1(x_0(t,x),l) \exp\left[-\frac{x_0^2(t,x)}{2r_c^2}\right] \times \exp\left\{\frac{i}{\hbar}[S(t,x,x_0(t,x))]\right\} + O(\hbar),$$

where

$$S(x,x_0(t,x)) = \frac{m(x - x_0(t,x))^2}{2t} - \pi \sqrt{8mE} x_0(t,x). \qquad (5.3.12)$$

From Eq.(5.3.11) we obtain

$$\langle s_2(t) | s_2(t) \rangle \approx (\pi r_c^2)^{-1/2} \times \theta_1(x + \pi t \sqrt{8E/m}, l) \exp\left[-\frac{(x + \pi t \sqrt{8E/m})^2}{r_c^2}\right]. \qquad (5.3.13)$$

**Remark 5.3.2.** From the inequality (5.3.7) and Eq.(5.3.13) follows that $\alpha$-particle at each instant $t \ge 0$ moves quasiclassically from right to left by the law

$$x(t) = -\pi t \sqrt{8E/m}, \qquad (5.3.14)$$

i.e. i.e.,estimating the position $x(t,x_0,t_0;\hbar)$ at each instant $t \ge 0$ with final error $r_c$ gives $|\langle x \rangle(t) - x(t)| \le r_c, i = 1,...,d$ with a probability

$$\mathbf{P}\{|\langle x \rangle(t, 0, 0; \hbar) - x(t)| \le r_c\} = 1.$$

**Remark 5.3.3.** We assume now that a distance between radioactive source and internal monitor which detects a single atom decaying (see Pic.1) is equal to $L$.

**Proposition 5.3.1.** After $\alpha$-decay at instant $t = 0$ the collapse: $|\text{live cat}\rangle \rightarrow |\text{death cat}\rangle$ arises at instant

$$T = \frac{L}{\pi \sqrt{8E/m}} \qquad (5.3.15)$$

with a probability $P_T(|\text{death cat}\rangle)$ to observe a state $|\text{death cat}\rangle$ at instant $T$ is $P_T(|\text{death cat}\rangle) = 1$.

**Proof.** Note that. In this case Schrödinger's cat in fact permorm the single measurement of $\alpha$-particle position with accuracy of $\delta x = l$ at instant $t = T$ (given by Eq.(5.3.15)) by internal monitor (see Pic.5.1.1). The probability of getting a result $L$

with accuracy of $\delta x = l$ given by
$$ \int_{|L-x| \le l/2} |\langle x || c_2 ||^2 s_2(T) \rangle|^2 dx = 1. \qquad (5.3.16) $$

Therefore at instant $T$ the $\alpha$-particle kills Schrödinger's cat with a probability $P_T(|\text{death cat}\rangle) = 1$.

**Remark 5.3.4.** Note that. When Schrödinger's cat has permormed this measurement the immediate post measurement state of $\alpha$-particle (by von Neumann postulate C.4, see Appendix CAT-C) will end up in the state

$$ |\Psi_T\rangle = \frac{\int_{|L-x| \le l/2} |x\rangle\langle x||s_2(T)\rangle dx}{\sqrt{\int_{|L-x| \le l/2} |\langle x||s_2(T)\rangle|^2 dx}} \in S_\Theta, \Theta = \{x||L-x| \le l/2\} \qquad (5.3.17) $$

From Eq.(5.3.17) one obtains

$$ \langle x' || \Psi_T \rangle = \int_{|L-x| \le l/2} \langle x' || x \rangle \langle x || s_2(T) \rangle dx = \int_{|L-x| \le l/2} \delta(x' - x) \langle x || s_2(T) \rangle dx = \Psi_I^{\#}(x', t). \qquad (5.3.18) $$

Therefore the state $|\Psi_T\rangle$ again kills Schrödinger's cat with a probability $P_T(|\text{death cat}\rangle) = 1$.

Suppose now that a nucleus **n**, whose Hilbert space is spanned by orthonormal states $|s_i(t)\rangle$, $i = 1,2$, where $|s_1(t)\rangle = |\text{undecayed nucleus at instant } t\rangle$ and $|s_2(t)\rangle = |\text{decayed nucleus at instant } t\rangle$ is in the superposition state

$$ |\Psi_t\rangle_n = c_1|s_1(t)\rangle + c_2|s_2(t)\rangle, |c_1|^2 + |c_2|^2 = 1. \qquad (5.3.19) $$

**Remark 5.3.5.** Note that: (i) $|s_1(0)\rangle = |\text{undecayed nucleus at instant } t=0\rangle = |\alpha\text{-particle iside region } (0,l] \text{ at instant } t=0\rangle$. (ii) Feynman propagator of $\alpha$-particle inside region $(0,l]$ are [70]:

$$ K_2(x, t, x_0) = \left(\frac{m}{2\pi i\hbar t}\right)^{1/2} \exp\left\{\frac{i}{\hbar}[S(t, x, x_0)]\right\}, \qquad (5.3.20) $$

where

$$ S(t, x, x_0) = \frac{m(x - x_0)^2}{2t} + mt(U_0 - E). \qquad (5.3.21) $$

Therefore from Eq.(5.2.11)-Eq.(5.2.12) and Eq.(5.3.20)-Eq.(5.3.21) we obtain

$$ \begin{gathered} |s_1(t)\rangle = \int_{-\infty}^{+\infty} \Psi_{II}^{\#}(x, t)|x\rangle dx, \\ \Psi_{II}^{\#}(x, t) = \int_{0}^{l} \Psi_{II}^{\#}(x_0) K_2(x, t, x_0) dx_0 = \qquad (5.3.22) \\ \left(\frac{m}{2\pi i\hbar t}\right)^{1/2} \int_{0}^{l} E(x_0, l) \Psi_{II}(x_0) \theta_l(x_0) \exp\left\{\frac{i}{\hbar}[S(t, x, x_0)]\right\} dx_0, \end{gathered} $$

where
$$ \int_{|L-x| \le l/2} |\langle x || c_2 ||^2 s_2(T) \rangle|^2 dx = 1. \qquad (5.3.16) $$

Therefore at instant $T$ the $\alpha$-particle kills Schrödinger's cat with a probability $P_T(|\text{death cat}\rangle) = 1$.

**Remark 5.3.4.** Note that. When Schrödinger's cat has permormed this measurement the immediate post measurement state of $\alpha$-particle (by von Neumann postulate C.4, see Appendix CAT-C) will end up in the state

$$ |\Psi_T\rangle = \frac{\int_{|L-x| \le l/2} |x\rangle\langle x||s_2(T)\rangle dx}{\sqrt{\int_{|L-x| \le l/2} |\langle x||s_2(T)\rangle|^2 dx}} \in S_\Theta, \Theta = \{x||L-x| \le l/2\} \qquad (5.3.17) $$

From Eq.(5.3.17) one obtains

$$ \langle x' || \Psi_T \rangle = \int_{|L-x| \le l/2} \langle x' || x \rangle \langle x || s_2(T) \rangle dx = \int_{|L-x| \le l/2} \delta(x' - x) \langle x || s_2(T) \rangle dx = \Psi_I^{\#}(x', t). \qquad (5.3.18) $$

Therefore the state $|\Psi_T\rangle$ again kills Schrödinger's cat with a probability $P_T(|\text{death cat}\rangle) = 1$.

Suppose now that a nucleus **n**, whose Hilbert space is spanned by orthonormal states $|s_i(t)\rangle$, $i = 1,2$, where $|s_1(t)\rangle = |\text{undecayed nucleus at instant } t\rangle$ and $|s_2(t)\rangle = |\text{decayed nucleus at instant } t\rangle$ is in the superposition state

$$ |\Psi_t\rangle_n = c_1|s_1(t)\rangle + c_2|s_2(t)\rangle, |c_1|^2 + |c_2|^2 = 1. \qquad (5.3.19) $$

**Remark 5.3.5.** Note that: (i) $|s_1(0)\rangle = |\text{undecayed nucleus at instant } t=0\rangle = |\alpha\text{-particle iside region } (0,l] \text{ at instant } t=0\rangle$. (ii) Feynman propagator of $\alpha$-particle inside region $(0,l]$ are [70]:

$$ K_2(x, t, x_0) = \left(\frac{m}{2\pi i\hbar t}\right)^{1/2} \exp\left\{\frac{i}{\hbar}[S(t, x, x_0)]\right\}, \qquad (5.3.20) $$

where

$$ S(t, x, x_0) = \frac{m(x - x_0)^2}{2t} + mt(U_0 - E). \qquad (5.3.21) $$

Therefore from Eq.(5.2.11)-Eq.(5.2.12) and Eq.(5.3.20)-Eq.(5.3.21) we obtain

$$ \begin{gathered} |s_1(t)\rangle = \int_{-\infty}^{+\infty} \Psi_{II}^{\#}(x, t)|x\rangle dx, \\ \Psi_{II}^{\#}(x, t) = \int_{0}^{l} \Psi_{II}^{\#}(x_0) K_2(x, t, x_0) dx_0 = \qquad (5.3.22) \\ \left(\frac{m}{2\pi i\hbar t}\right)^{1/2} \int_{0}^{l} E(x_0, l) \Psi_{II}(x_0) \theta_l(x_0) \exp\left\{\frac{i}{\hbar}[S(t, x, x_0)]\right\} dx_0, \end{gathered} $$

where
$$ \theta_l(x) = \begin{cases} 1 & \text{for } x \in [0, l] \\ 0 & \text{for } x \notin [0, l] \end{cases} $$

**Remark 5.3.6.** We assume for simplification now that

$$ k' \le 1. \qquad (5.3.23) $$

Therefore oscillatory integral in RHS of Eq.(5.3.22) is calculated now directly using stationary phase approximation. The phase term $S(x, x_0)$ given by Eq.(5.3.21) is stationary when

$$ \frac{\partial S(t, x, x_0)}{\partial x_0} = - \frac{m(x - x_0)}{t} = 0. \qquad (5.3.24) $$

and thus stationary point $x_0(t, x)$ are

$$ \begin{aligned} -x + x_0 &= 0 \\ x_0(t, x) &= x. \end{aligned} \qquad (5.3.25) $$

Thus from Eq.(5.3.22) and Eq.(5.3.25) using stationary phase approximation we obtain

$$ \begin{aligned} \Psi_{\Pi}^{\#}(x,t) &= E(x_0(t,x),l)\Psi_{\Pi}(x_0(t,x))\theta_l(x_0(t,x))\exp\left\{\frac{i}{\hbar}[S(t,x,x_0(t,x))]\right\} + O(\hbar) = \\ &= E(x,l)\Psi_{\Pi}(x)\theta_l(x)\exp\left\{\frac{i}{\hbar}[mt(U_0 - E)]\right\} + O(\hbar) = \\ &E(x,l)\theta_l(x)O(1)\exp\left\{\frac{i}{\hbar}[mt(U_0 - E)]\right\} + O(\hbar). \end{aligned} \qquad (5.3.26) $$

Therefore from Eq.(3.22) and Eq.(3.26) we obtain

$$ \langle s_1(t) | s_1(t) \rangle = |\Psi_{\Pi}^{\#}(x,t)|^2 = E^2(x,l)\theta_l(x)O(1) + O(\hbar). \qquad (3.27) $$

**Remark 5.3.7.** Note that for each instant $t > 0$ : $\text{supp}(\Psi_{\Pi}^{\#}(x,t)) \cap \text{supp}(\Psi_{I}^{\#}(x,t)) = \emptyset$.

**Remark 5.3.8.** Note that. From Eq.(5.3.11),Eq.(5.3.13), Eq.(5.3.19), Eq.(5.3.22)-Eq.(5.3.27) and Eq.(A.13),see Appendix CAT-A, by Remark 5.3.7 we obtain
$$ \theta_l(x) = \begin{cases} 1 & \text{for } x \in [0, l] \\ 0 & \text{for } x \notin [0, l] \end{cases} $$

**Remark 5.3.6.** We assume for simplification now that

$$ k' \le 1. \qquad (5.3.23) $$

Therefore oscillatory integral in RHS of Eq.(5.3.22) is calculated now directly using stationary phase approximation. The phase term $S(x, x_0)$ given by Eq.(5.3.21) is stationary when

$$ \frac{\partial S(t, x, x_0)}{\partial x_0} = - \frac{m(x - x_0)}{t} = 0. \qquad (5.3.24) $$

and thus stationary point $x_0(t, x)$ are

$$ \begin{aligned} -x + x_0 &= 0 \\ x_0(t, x) &= x. \end{aligned} \qquad (5.3.25) $$

Thus from Eq.(5.3.22) and Eq.(5.3.25) using stationary phase approximation we obtain

$$ \begin{aligned} \Psi_{\Pi}^{\#}(x,t) &= E(x_0(t,x),l)\Psi_{\Pi}(x_0(t,x))\theta_l(x_0(t,x))\exp\left\{\frac{i}{\hbar}[S(t,x,x_0(t,x))]\right\} + O(\hbar) = \\ &= E(x,l)\Psi_{\Pi}(x)\theta_l(x)\exp\left\{\frac{i}{\hbar}[mt(U_0 - E)]\right\} + O(\hbar) = \\ &E(x,l)\theta_l(x)O(1)\exp\left\{\frac{i}{\hbar}[mt(U_0 - E)]\right\} + O(\hbar). \end{aligned} \qquad (5.3.26) $$

Therefore from Eq.(3.22) and Eq.(3.26) we obtain

$$ \langle s_1(t)|s_1(t) \rangle = |\Psi_{\Pi}^{\#}(x,t)|^2 = E^2(x,l)\theta_l(x)O(1) + O(\hbar). \qquad (3.27) $$

**Remark 5.3.7.** Note that for each instant $t > 0$ : $\text{supp}(\Psi_{\Pi}^{\#}(x,t)) \cap \text{supp}(\Psi_{I}^{\#}(x,t)) = \emptyset$.

**Remark 5.3.8.** Note that. From Eq.(5.3.11),Eq.(5.3.13), Eq.(5.3.19), Eq.(5.3.22)-Eq.(5.3.27) and Eq.(A.13),see Appendix CAT-A, by Remark 5.3.7 we obtain
$$
\begin{align}
\mathbf{n} \langle \Psi_t | \hat{x} | \Psi_t \rangle_{\mathbf{n}} &= |c_1|^2 \langle s_1(t) | \hat{x} | s_1(t) \rangle + |c_2|^2 \langle s_2(t) | \hat{x} | s_2(t) \rangle + \nonumber \\
&\qquad c_1 c_2^* \langle s_2(t) | \hat{x} | s_1(t) \rangle + c_1^* c_2 \langle s_2(t) | \hat{x} | s_1(t) \rangle^* = \tag{5.3.28}
\end{align}
$$

$$
|c_1|^2 \langle s_1(t) | \hat{x} | s_1(t) \rangle + |c_2|^2 \langle s_2(t) | \hat{x} | s_2(t) \rangle = |c_1|^2 l + |c_2|^2 T \pi \sqrt{8E/m}.
$$

**Proposition 5.3.2.** (i) Suppose that a nucleus **n** is in the superposition state |Ψ<sub>t</sub>⟩<sub>n</sub> (|Ψ<sub>t</sub>⟩<sub>n</sub>-particle) given by Eq.(5.3.19). Then the collaps: |live cat⟩ → |death cat⟩ arises at instant

$$
T_{\text{col}} \approx \frac{L \pm l}{|c_2|^2 \sqrt{8\pi^2 E/m}}. \qquad (5.3.29)
$$

with a probability $\mathbf{P}_{T_{\text{col}}}(|\text{death cat}\rangle)$ to observe a state $|\text{death cat}\rangle$ at instant $T_{\text{col}}$ is
$\mathbf{P}_{T_{\text{col}}}(|\text{death cat}\rangle) = |c_2|^2$.

(ii) Assume now a Schrödinger's cat has performed the single measurement of $|\Psi_t\rangle_n$-particle position with accuracy of $\delta x = l$ at instant $T = T_{\text{col}}$ (given by Eq.(5.3.29)) by internal monitor (see Pic.5.1.1) and the result $x \approx L \pm l$ is not observed by Schrödinger's cat. Then the collaps: $|\text{live cat}\rangle \to |\text{death cat}\rangle$ never arises at any instant $T > T_{\text{col}}$ and a probability $\mathbf{P}_{T>T_{\text{col}}}(|\text{death cat}\rangle)$ to observe a state $|\text{death cat}\rangle$ at instant $T > T_{\text{col}}$ is $\mathbf{P}_{T>T_{\text{col}}}(|\text{death cat}\rangle) = 0$.

**Proof.** (i) Note that for $t > 0$ the marginal density matrix $\rho(t)$ is diagonal

$$
\rho(t) = \begin{pmatrix} |c_1|^2 \int |\Psi_{\mathbf{II}}^{\#}(x,t)|^2 dx & 0 \\ 0 & |c_2|^2 \int |\Psi_{\mathbf{I}}^{\#}(x,t)|^2 dx \end{pmatrix}
$$

In this case a Schrödinger's cat in fact perform the single measurement of
$|\Psi_t\rangle_n$-particle position with accuracy of $\delta x = l$ at instant $t = T_{\text{col}}$ (given by
Eq.(5.3.29)) by internal monitor (see Pic.5.1.1). The probability of getting a result $L$

at instant $T \approx T_{\text{col}}$ with accuracy of $\delta x = l$ given by

$$
\begin{align*}
\int_{|L-x| \le l/2} |\langle x || \Psi_T \rangle_n|^2 dx &= \int_{|L-x| \le l/2} |\langle x | c_1 | s_1(T) \rangle + c_2 |s_2(T) \rangle|^2 dx = \\
&\qquad \int_{|L-x| \le l/2} |c_1 \langle x || s_1(T) \rangle + c_2 \langle x || s_2(T) \rangle|^2 dx = \tag{5.3.30} \\
&\qquad \int_{|L-x| \le l/2} |c_1^2 \Psi_{II}^{\#2}(x,T) + c_2^2 \Psi_{I}^{\#2}(x,T) + 2c_1 c_2 \Psi_{I}^{\#}(x,T) \Psi_{II}^{\#}(x,T)| dx.
\end{align*}
$$

From Eq.(5.3.30) by Remark 5.3.7 and Eq.(5.3.13) one obtains

$$
\int_{|L-x| \le l/2} |\langle x || \Psi_T \rangle_n|^2 dx = \int_{|L-x| \le l/2} |c_2^2 \Psi_I^{\#2}(x,T)| dx = |c_2|^2 \int_{|L-x| \le l/2} |\Psi_I^{\#}(x,T)|^2 dx = |c_2|^2. \quad (5.3.31)
$$

Note that. When Schrödinger's cat has permormed this measurement and the
result $x \approx L \pm l$ is observed, then the immediate post measurement state of
$\alpha$-particle (by
$$
\begin{align}
\mathbf{n} \langle \Psi_t | \hat{x} | \Psi_t \rangle_{\mathbf{n}} &= |c_1|^2 \langle s_1(t) | \hat{x} | s_1(t) \rangle + |c_2|^2 \langle s_2(t) | \hat{x} | s_2(t) \rangle + \nonumber \\
&\qquad c_1 c_2^* \langle s_2(t) | \hat{x} | s_1(t) \rangle + c_1^* c_2 \langle s_2(t) | \hat{x} | s_1(t) \rangle^* = \tag{5.3.28}
\end{align}
$$

$$
|c_1|^2 \langle s_1(t) | \hat{x} | s_1(t) \rangle + |c_2|^2 \langle s_2(t) | \hat{x} | s_2(t) \rangle = |c_1|^2 l + |c_2|^2 T \pi \sqrt{8E/m}.
$$

**Proposition 5.3.2.** (i) Suppose that a nucleus **n** is in the superposition state |Ψ<sub>*t*</sub>⟩<sub>**n**</sub> (|Ψ<sub>*t*</sub>⟩<sub>**n**</sub>-particle) given by Eq.(5.3.19). Then the collaps: |live cat⟩ → |death cat⟩ arises at instant

$$
T_{\text{col}} \approx \frac{L \pm l}{|c_2|^2 \sqrt{8\pi^2 E/m}}. \qquad (5.3.29)
$$

with a probability $\mathbf{P}_{T_{\text{col}}}(|\text{death cat}\rangle)$ to observe a state $|\text{death cat}\rangle$ at instant $T_{\text{col}}$ is
$\mathbf{P}_{T_{\text{col}}}(|\text{death cat}\rangle) = |c_2|^2$.

(ii) Assume now a Schrödinger's cat has performed the single measurement of $|\Psi_t\rangle_n$-particle position with accuracy of $\delta x = l$ at instant $T = T_{\text{col}}$ (given by Eq.(5.3.29)) by internal monitor (see Pic.5.1.1) and the result $x \approx L \pm l$ is not observed by Schrödinger's cat. Then the collaps: $|\text{live cat}\rangle \to |\text{death cat}\rangle$ never arises at any instant $T > T_{\text{col}}$ and a probability $\mathbf{P}_{T>T_{\text{col}}}(|\text{death cat}\rangle)$ to observe a state $|\text{death cat}\rangle$ at instant $T > T_{\text{col}}$ is $\mathbf{P}_{T>T_{\text{col}}}(|\text{death cat}\rangle) = 0$.

**Proof.** (i) Note that for $t > 0$ the marginal density matrix $\rho(t)$ is diagonal

$$
\rho(t) = \begin{pmatrix} |c_1|^2 \int |\Psi_{\mathbf{II}}^{\#}(x,t)|^2 dx & 0 \\ 0 & |c_2|^2 \int |\Psi_{\mathbf{I}}^{\#}(x,t)|^2 dx \end{pmatrix}
$$

In this case a Schrödinger's cat in fact perform the single measurement of
$|\Psi_t\rangle_n$-particle position with accuracy of $\delta x = l$ at instant $t = T_{\text{col}}$ (given by
Eq.(5.3.29)) by internal monitor (see Pic.5.1.1). The probability of getting a result $L$

at instant $T \approx T_{\text{col}}$ with accuracy of $\delta x = l$ given by

$$
\begin{align}
\int_{|L-x|\le l/2} |\langle x || \Psi_T \rangle_n|^2 dx &= \int_{|L-x|\le l/2} |\langle x | c_1 | s_1(T) \rangle + c_2 | s_2(T) \rangle|^2 dx = \nonumber \\
&\quad \int_{|L-x|\le l/2} |c_1 \langle x || s_1(T) \rangle + c_2 \langle x || s_2(T) \rangle|^2 dx = \tag{5.3.30} \\
&\quad \int_{|L-x|\le l/2} |c_1^2 \Psi_{II}^{\#2}(x,T) + c_2^2 \Psi_{I}^{\#2}(x,T) + 2c_1 c_2 \Psi_{I}^{\#}(x,T) \Psi_{II}^{\#}(x,T)| dx. \nonumber
\end{align}
$$

From Eq.(5.3.30) by Remark 5.3.7 and Eq.(5.3.13) one obtains

$$
\int_{|L-x| \le l/2} |\langle x || \Psi_T \rangle_n|^2 dx = \int_{|L-x| \le l/2} |c_2^2 \Psi_I^{\#2}(x,T)| dx = |c_2|^2 \int_{|L-x| \le l/2} |\Psi_I^{\#}(x,T)|^2 dx = |c_2|^2. \quad (5.3.31)
$$

Note that. When Schrödinger's cat has permormed this measurement and the
result $x \approx L \pm l$ is observed, then the immediate post measurement state of
$\alpha$-particle (by
von Neumann postulate C.4, see Appendix CAT-C) will end up in the state

$$
\begin{align}
|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}} &= \frac{\int_{|L-x|\le l/2} |x\rangle\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}} dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} = \nonumber \\
&\qquad \frac{\int_{|L-x|\le l/2} |x\rangle\langle x|(c_1|s_1(T_{\text{col}})\rangle + c_2|s_2(T_{\text{col}})\rangle)dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} = \tag{5.3.32} \\
&\qquad \frac{c_1 \int_{|L-x|\le l/2} |x\rangle\langle x|s_1(T_{\text{col}})\rangle + c_2 \int_{|L-x|\le l/2} |x\rangle\langle x|s_2(T_{\text{col}})\rangle dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} \in \mathbf{S}_{\Theta}, \Theta = \{x||L-x| \le l/2\}. \nonumber
\end{align}
$$

From Eq. (5.3.32) by Eq. (5.3.31) and by Remark 5.3.7 one obtains

$$
\begin{align}
|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}} &= \frac{\int_{|L-x|\le l/2} |x\rangle\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}} dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} = \nonumber \\
&\quad \frac{\int_{|L-x|\le l/2} |x\rangle\langle x|(c_1|s_1(T_{\text{col}})\rangle + c_2|s_2(T_{\text{col}})\rangle)dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} = \tag{5.3.32} \\
&= \frac{c_2}{|c_2|} \int_{|L-x|\le l/2} |x\rangle\langle x|s_2(T_{\text{col}})\rangle dx. \nonumber
\end{align}
$$

Obviously by Remark 5.3.4 the state $|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}$ kills Schrödinger's cat with a
probability $P_{T_{\text{col}}}(|\text{death cat}\rangle) = 1$.

**Proof.** (ii) The probability of getting a result *L* at any instant *T* > *T*<sub>col</sub> with accuracy of δx = *l* by Eq.(5.3.31) and Eq.(5.3.13) given by

$$
\begin{align*}
\int_{|L-x|\le l/2} |\langle x|\Psi_T\rangle_n|^2 dx &= \int_{|L-x|\le l/2} |c_2^2 \Psi_I^{\#2}(x,T)| dx = |c_2|^2 \int_{|L-x|\le l/2} |\Psi_I^{\#}(x,T)|^2 dx = \\
&\approx (\pi r_c^2)^{-1/2} \times \theta_1(x + \pi T \sqrt{8E/m}, l) \exp\left[-\frac{(x + \pi T \sqrt{8E/m})^2}{r_c^2}\right] = 0.
\end{align*}
$$

Thus standard formalism of continuous quantum measurements [65],[67] leads to a
definite but unpredictable measurement outcomes, either $|s_1(t)\rangle$ or $|s_2(t)\rangle$ and thus
$|\Psi_t\rangle_n$ suddenly “collapses” at unpredictable instant $t'$ into the corresponding state
$|s_i(t')\rangle$, $i = 1,2$.

V.III.2.Resolution of the Schrödinger’s cat paradox
using generalized von Neumann postulate.
von Neumann postulate C.4, see Appendix CAT-C) will end up in the state

$$
\begin{align}
|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}} &= \frac{\int_{|L-x|\le l/2} |x\rangle\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}} dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} = \nonumber \\
&\qquad \frac{\int_{|L-x|\le l/2} |x\rangle\langle x|(c_1|s_1(T_{\text{col}})\rangle + c_2|s_2(T_{\text{col}})\rangle)dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} = \tag{5.3.32} \\
&\qquad \frac{c_1 \int_{|L-x|\le l/2} |x\rangle\langle x|s_1(T_{\text{col}})\rangle + c_2 \int_{|L-x|\le l/2} |x\rangle\langle x|s_2(T_{\text{col}})\rangle dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} \in \mathbf{S}_{\Theta}, \Theta = \{x||L-x| \le l/2\}. \nonumber
\end{align}
$$

From Eq. (5.3.32) by Eq. (5.3.31) and by Remark 5.3.7 one obtains

$$
\begin{align}
|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}} &= \frac{\int_{|L-x|\le l/2} |x\rangle\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}} dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} = \nonumber \\
&\quad \frac{\int_{|L-x|\le l/2} |x\rangle\langle x|(c_1|s_1(T_{\text{col}})\rangle + c_2|s_2(T_{\text{col}})\rangle)dx}{\sqrt{\int_{|L-x|\le l/2} |\langle x|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}|^2 dx}} = \tag{5.3.32} \\
&= \frac{c_2}{|c_2|} \int_{|L-x|\le l/2} |x\rangle\langle x|s_2(T_{\text{col}})\rangle dx. \nonumber
\end{align}
$$

Obviously by Remark 5.3.4 the state $|\Psi_{T_{\text{col}}}\rangle_{\mathbf{n}}$ kills Schrödinger's cat with a
probability $P_{T_{\text{col}}}(|\text{death cat}\rangle) = 1$.

**Proof.** (ii) The probability of getting a result *L* at any instant *T* > *T*<sub>col</sub> with accuracy of δx = *l* by Eq.(5.3.31) and Eq.(5.3.13) given by

$$
\begin{align*}
\int_{|L-x|\le l/2} |\langle x|\Psi_T\rangle_n|^2 dx &= \int_{|L-x|\le l/2} |c_2^2 \Psi_I^{\#2}(x,T)| dx = |c_2|^2 \int_{|L-x|\le l/2} |\Psi_I^{\#}(x,T)|^2 dx = \\
&\approx (\pi r_c^2)^{-1/2} \times \theta_1(x + \pi T \sqrt{8E/m}, l) \exp\left[-\frac{(x + \pi T \sqrt{8E/m})^2}{r_c^2}\right] = 0.
\end{align*}
$$

Thus standard formalism of continuous quantum measurements [65],[67] leads to a
definite but unpredictable measurement outcomes, either $|s_1(t)\rangle$ or $|s_2(t)\rangle$ and thus
$|\Psi_t\rangle_n$ suddenly “collapses” at unpredictable instant $t'$ into the corresponding state
$|s_i(t')\rangle$, $i = 1,2$.

V.III.2.Resolution of the Schrödinger’s cat paradox
using generalized von Neumann postulate.
**Proposition 5.3.3.** Suppose that a nucleus **n** is in the superposition state given by Eq.(5.3.19). the collaps: |live cat〉 → |death cat〉 arises at instant

$$T_{\text{col}} = \frac{L}{|c_2|^2 \sqrt{8\pi^2 E/m}}. \qquad (5.3.33)$$

with a probability $P_{T_{\text{col}}}(|\text{death cat}\rangle)$ to observe a state $|\text{death cat}\rangle$ at instant $T_{\text{col}}$ is $P_{T_{\text{col}}}(|\text{death cat}\rangle) = 1$.

**Proof.** Let us consider now a state $|\Psi_t\rangle_n$ given by Eq.(5.3.19). This state consists of a summ of two wave packets $c_1\Psi_{\text{II}}^{\#}(x,t)$ and $c_2\Psi_{\text{I}}^{\#}(x,t)$. Wave packet $c_1\Psi_{\text{II}}^{\#}(x,t)$ present an $\alpha_{\text{II}}$-particle which lives in region **II** with a probability $|c_1|^2$ (see Pic.5.2.1). Wave packet $c_2\Psi_{\text{I}}^{\#}(x,t)$ present an $\alpha_{\text{I}}$-particle which lives in region **I** with a probability $|c_2|^2$ (see Pic.5.2.1) and moves from the right to the left. Note that $\text{I} \cap \text{II} = \emptyset$. From Eq.(5.3.28) follows that $\alpha_{\text{I}}$-particle at each instant $t \ge 0$ moves quasiclassically from right to left by the law

$$x(t) = -|c_2|^2 \pi t \sqrt{8E/m}, \qquad (5.3.34)$$

From Eq.(5.3.34) one obtains

$$T = T_{\text{col}} \approx \frac{L}{|c_2| \sqrt{8\pi^2 E/m}}. \qquad (5.3.35)$$

Note that. In this case Schrödinger's cat in fact permorm a single measurement of $|\Psi_t\rangle_n$-particle position with accuracy of $\delta x = l$ at instant $t = T = T_{\text{col}}$ (given by Eq.(5.3.35)) by internal monitor (see Pic.5.1.1). The probability of getting the result

$L$ at instant $t = T_{\text{col}}$ with accuracy of $\delta x = l$ by Remark 5.3.7 and by postulate C.V.2 (see Appendix CAT-C) and by postulate C.IV.3 (see appendix CAT-C) given by

$$\begin{align*}
\int_{|L-x| \le l/2} [ |\langle x | c_1 | s_1(T_{\text{col}}) \rangle|^2 * |\langle x | c_2 | s_2(T_{\text{col}}) \rangle|^2 ] dx = \\
\int_{|L-x| \le l/2} |c_2|^{-2} |c_1|^{-2} [ |\langle x | c_1 |^{-2} | s_1(T_{\text{col}}) \rangle|^2 * |\langle x | c_2 |^{-2} | s_2(T_{\text{col}}) \rangle|^2 ] dx = \tag{5.3.36} \\
\int_{|L-x| \le l/2} |c_2|^{-2} |c_1|^{-2} [ |\Psi_{\text{I}}^{\#}(x|c_2|^{-2}, T_{\text{col}})|^2 * |\Psi_{\text{II}}^{\#}(x|c_1|^{-2}, T_{\text{col}})|^2 ] dx = 1.
\end{align*}$$

Note that. When Schrödinger's cat has permormed this measurement and the

result $x \approx L \pm l$ is observed, then the immediate post measurement state of $\alpha$-particle (by generalized von Neumann postulate C.V.3,see Appendix CAT-C) will end up in the state

$$\begin{align}
|\Psi_{T_{\text{col}}}\rangle_n &= \frac{\int_{|L-x| \le l/2} |x\rangle\langle x| \Psi_{T_{\text{col}}}\rangle_n dx}{\sqrt{\int_{|L-x| \le l/2} [|\langle x|s_1(T_{\text{col}})\rangle|^2 + |\langle x|s_2(T_{\text{col}})\rangle|^2] dx}} = \nonumber \\
&\qquad \frac{\int_{|L-x| \le l/2} |x\rangle\langle x| s_2^{c_2}(T_{\text{col}}) dx}{\sqrt{\int_{|L-x| \le l/2} [|\langle x|s_2(T_{\text{col}})\rangle|^2] dx}} \in \mathbf{H}_{\Theta}, \Theta = \{x||L-x| \le l/2\}. \tag{5.3.37}
\end{align}$$

The staite $|\Psi_{T_{\text{col}}}\rangle_n$ again kills Schrödinger's cat with a probability $P_{T_{\text{col}}}(|\text{death cat}\rangle) = 1$.
**Proposition 5.3.3.** Suppose that a nucleus **n** is in the superposition state given by Eq.(5.3.19). the collaps: |live cat〉 → |death cat〉 arises at instant

$$T_{\text{col}} = \frac{L}{|c_2|^2 \sqrt{8\pi^2 E/m}}. \qquad (5.3.33)$$

with a probability $P_{T_{\text{col}}}(|\text{death cat}\rangle)$ to observe a state $|\text{death cat}\rangle$ at instant $T_{\text{col}}$ is $P_{T_{\text{col}}}(|\text{death cat}\rangle) = 1$.

**Proof.** Let us consider now a state $|\Psi_t\rangle_n$ given by Eq.(5.3.19). This state consists of a summ of two wave packets $c_1\Psi_{\text{II}}^{\#}(x,t)$ and $c_2\Psi_{\text{I}}^{\#}(x,t)$. Wave packet $c_1\Psi_{\text{II}}^{\#}(x,t)$ present an $\alpha_{\text{II}}$-particle which lives in region **II** with a probability $|c_1|^2$ (see Pic.5.2.1). Wave packet $c_2\Psi_{\text{I}}^{\#}(x,t)$ present an $\alpha_{\text{I}}$-particle which lives in region **I** with a probability $|c_2|^2$ (see Pic.5.2.1) and moves from the right to the left. Note that $\text{I} \cap \text{II} = \emptyset$. From Eq.(5.3.28) follows that $\alpha_{\text{I}}$-particle at each instant $t \ge 0$ moves quasiclassically from right to left by the law

$$x(t) = -|c_2|^2 \pi t \sqrt{8E/m}, \qquad (5.3.34)$$

From Eq.(5.3.34) one obtains

$$T = T_{\text{col}} \approx \frac{L}{|c_2| \sqrt{8\pi^2 E/m}}. \qquad (5.3.35)$$

Note that. In this case Schrödinger's cat in fact permorm a single measurement of $|\Psi_t\rangle_n$-particle position with accuracy of $\delta x = l$ at instant $t = T = T_{\text{col}}$ (given by Eq.(5.3.35)) by internal monitor (see Pic.5.1.1). The probability of getting the result

$L$ at instant $t = T_{\text{col}}$ with accuracy of $\delta x = l$ by Remark 5.3.7 and by postulate C.V.2 (see Appendix CAT-C) and by postulate C.IV.3 (see appendix CAT-C) given by

$$\begin{align*}
& \int_{|L-x| \le l/2} [ |\langle x | c_1 | s_1(T_{\text{col}}) \rangle|^2 * |\langle x | c_2 | s_2(T_{\text{col}}) \rangle|^2 ] dx = \\
& \int_{|L-x| \le l/2} |c_2|^{-2} |c_1|^{-2} [ |\langle x | c_1 |^{-2} | s_1(T_{\text{col}}) \rangle|^2 * |\langle x | c_2 |^{-2} | s_2(T_{\text{col}}) \rangle|^2 ] dx = \tag{5.3.36} \\
& \int_{|L-x| \le l/2} |c_2|^{-2} |c_1|^{-2} [ |\Psi_{\text{I}}^{\#}(x|c_2|^{-2}, T_{\text{col}})|^2 * |\Psi_{\text{II}}^{\#}(x|c_1|^{-2}, T_{\text{col}})|^2 ] dx = 1.
\end{align*}$$

Note that. When Schrödinger's cat has permormed this measurement and the

result $x \approx L \pm l$ is observed, then the immediate post measurement state of $\alpha$-particle (by generalized von Neumann postulate C.V.3,see Appendix CAT-C) will end up in the state

$$\begin{align}
|\Psi_{T_{\text{col}}}\rangle_{\text{n}} &= \frac{\int_{|L-x| \le l/2} |x\rangle\langle x| \Psi_{T_{\text{col}}}\rangle_{\text{n}} dx}{\sqrt{\int_{|L-x| \le l/2} [ |\langle x| s_1(T_{\text{col}})\rangle|^2 + |\langle x| s_2(T_{\text{col}})\rangle|^2 ] dx}} = \nonumber \\
&\qquad \frac{\int_{|L-x| \le l/2} |x\rangle\langle x| s_2^{c_2}(T_{\text{col}})\rangle dx}{\sqrt{\int_{|L-x| \le l/2} [ |\langle x| s_2(T_{\text{col}})\rangle|^2 ] dx}} \in \mathbf{H}_{\Theta}, \Theta = \{x||L-x| \le l/2\}. \tag{5.3.37}
\end{align}$$

The staite $|\Psi_{T_{\text{col}}}\rangle_n$ again kills Schrödinger's cat with a probability $P_{T_{\text{col}}}(|\text{death cat}\rangle) = 1$.
Thus is the collapsed state of the cat always shows definite and predictable outcomes even if cat also consists of a superposition:

$$|\text{cat}\rangle = c_1|\text{live cat}\rangle + c_2|\text{death cat}\rangle.$$

Contrary to van Kampen's [61] and some others' opinions, "looking" at the outcome changes nothing, beyond informing the observer of what has already happened. We remain: there are widespread claims that Schrödinger's cat is not in a definite alive or dead state but is, instead, in a superposition of the two. van Kampen, for example, writes "The whole system is in a superposition of two states: one in which no decay has occurred and one in which it has occurred. Hence, the state of the cat also consists of a superposition: $|\text{cat}\rangle = c_1|\text{live cat}\rangle + c_2|\text{death cat}\rangle$. The state remains a superposition until an observer looks at the cat" [61].

# Appendix. CAT-A.

The time-dependent Schrodinger equation governs the time evolution of a quantum mechanical system:

$$i\hbar \frac{\partial \Psi(\mathbf{x}, t)}{\partial t} = \hat{\mathbf{H}} \Psi(\mathbf{x}, t). \qquad (A.1)$$

The average, or expectation, value $\langle x_i \rangle$ of an observable $x_i$ corresponding to a quantum mechanical operator $\hat{x}_i$ is given by:

$$ \langle x_i \rangle (t, \mathbf{x}_0, t_0; \hbar) = \frac{\int_{\mathbb{R}^d} x_i |\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 d^d\mathbf{x}}{\int_{\mathbb{R}^d} |\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 d^d\mathbf{x}} \qquad (A.2) \\ i = 1, \dots, d. $$

**Remark A.1.** We assume now that: the solution $\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)$ of the time-dependent Schrödinger equation (A.1) has a good approximation by a delta function such that

$$ |\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 \approx \prod_{i=1}^{d} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)), \qquad (A.3) \\ x_i(t, \mathbf{x}_0, t_0) = x_{i,0}, \\ i = 1, \dots, d. $$

**Remark A.2.** Note that under conditions given by Eq. (A.3) QM-system which governed by Schrödinger equation Eq. (A.1) completely evolve quasiclassically i.e. estimating the position $\{x_i(t, \mathbf{x}_0, t_0; \hbar)\}_{i=1}^d$ at each instant $t$ with final error $\delta$ gives $|\langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) - x_i(t, \mathbf{x}_0, t_0)| \le \delta, i = 1, \dots, d$ with a probability
Thus is the collapsed state of the cat always shows definite and predictable outcomes even if cat also consists of a superposition:

$$|\text{cat}\rangle = c_1|\text{live cat}\rangle + c_2|\text{death cat}\rangle.$$

Contrary to van Kampen's [61] and some others' opinions, "looking" at the outcome changes nothing, beyond informing the observer of what has already happened. We remain: there are widespread claims that Schrödinger's cat is not in a definite alive or dead state but is, instead, in a superposition of the two. van Kampen, for example, writes "The whole system is in a superposition of two states: one in which no decay has occurred and one in which it has occurred. Hence, the state of the cat also consists of a superposition: $|\text{cat}\rangle = c_1|\text{live cat}\rangle + c_2|\text{death cat}\rangle$. The state remains a superposition until an observer looks at the cat" [61].

## Appendix. CAT-A.

The time-dependent Schrodinger equation governs the time evolution of a quantum mechanical system:

$$i\hbar \frac{\partial \Psi(\mathbf{x}, t)}{\partial t} = \hat{\mathbf{H}} \Psi(\mathbf{x}, t). \qquad (A.1)$$

The average, or expectation, value $\langle x_i \rangle$ of an observable $x_i$ corresponding to a quantum mechanical operator $\hat{x}_i$ is given by:

$$ \langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) = \frac{\int_{\mathbb{R}^d} x_i |\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 d^d\mathbf{x}}{\int_{\mathbb{R}^d} |\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 d^d\mathbf{x}} \qquad (A.2) \\ i = 1, \dots, d. $$

**Remark A.1.** We assume now that: the solution $\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)$ of the time-dependent Schrödinger equation (A.1) has a good approximation by a delta function such that

$$ \begin{gathered} |\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 \approx \prod_{i=1}^{d} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)), \qquad (A.3) \\ x_i(t, \mathbf{x}_0, t_0) = x_{i,0}, \\ i = 1, \dots, d. \end{gathered} $$

**Remark A.2.** Note that under conditions given by Eq. (A.3) QM-system which governed by Schrödinger equation Eq. (A.1) completely evolve quasiclassically i.e. estimating the position $\{x_i(t, \mathbf{x}_0, t_0; \hbar)\}_{i=1}^d$ at each instant $t$ with final error $\delta$ gives $|\langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) - x_i(t, \mathbf{x}_0, t_0)| \le \delta, i = 1, \dots, d$ with a probability
$$ \mathbf{P}\{|\langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) - x_i(t, \mathbf{x}_0, t_0)| \le \delta\} \approx 1. $$

Thus from Eq.(A.2) and Eq.(A.3) we obtain

$$ \langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) \approx $$

$$ \begin{aligned} & \approx \frac{\int_{\mathbb{R}^d} x_i \prod_{i=1}^{d=1} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)) d^d x}{\int_{\mathbb{R}^d} \prod_{i=1}^{d=1} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)) d^d x} = x_i(t, \mathbf{x}_0, t_0). \\ & \qquad i = 1, \dots, d. \end{aligned} \quad (A.4) $$

Thus under condition given by Eq.(A.3) one obtain

$$ \begin{aligned} \langle x_{i,t} \rangle(t, \mathbf{x}_0, t_0; \hbar) &\approx x_i(t, \mathbf{x}_0, t_0), \\ i &= 1, \dots, d. \end{aligned} \quad (A.5) $$

**Remark A.3.** Let $\Psi_i(\mathbf{x}, t, \mathbf{x}_0, t_0), i = 1, 2$ be the solutions of the time-dependent Schrödinger equation (A.1). We assume now that $\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ is a linear superposition such that

$$ \begin{aligned} \Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= c_1 \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2 \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0). \\ |c_1|^2 + |c_2|^2 &= 1. \end{aligned} \quad (A.6) $$

Then we obtain

$$ \begin{aligned} |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 &= (\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0) \Phi^*(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)) = \\ &= ([c_1 \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2 \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)]) \times \\ & \quad \times ([c_1^* \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2^* \Psi_2^*(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)]) = \\ &= |c_1|^2 (|\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2) + c_1 c_2 (\Psi_1^*(\mathbf{x}, t, \mathbf{x}_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)) + \\ & \quad |c_2|^2 (|\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2) + c_1 c_2^* (\Psi_1(\mathbf{x}, t, \mathbf{x}_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0)). \end{aligned} \quad (A.7) $$

**Definition A.1.** Let $\langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ be a vector-function
$$ \mathbf{P}\{|\langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) - x_i(t, \mathbf{x}_0, t_0)| \le \delta\} \approx 1. $$

Thus from Eq.(A.2) and Eq.(A.3) we obtain

$$ \langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) \approx $$

$$ \begin{aligned} & \approx \frac{\int_{\mathbb{R}^d} x_i \prod_{i=1}^{d=1} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)) d^d x}{\int_{\mathbb{R}^d} \prod_{i=1}^{d=1} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)) d^d x} = x_i(t, \mathbf{x}_0, t_0). \\ & \qquad i = 1, \dots, d. \end{aligned} \quad (A.4) $$

Thus under condition given by Eq.(A.3) one obtain

$$ \begin{aligned} \langle x_{i,t} \rangle(t, \mathbf{x}_0, t_0; \hbar) &\approx x_i(t, \mathbf{x}_0, t_0), \\ i &= 1, \dots, d. \end{aligned} \quad (A.5) $$

**Remark A.3.** Let $\Psi_i(\mathbf{x}, t, \mathbf{x}_0, t_0), i = 1, 2$ be the solutions of the time-dependent Schrödinger equation (A.1). We assume now that $\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ is a linear superposition such that

$$ \begin{aligned} \Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= c_1 \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2 \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0). \\ |c_1|^2 + |c_2|^2 &= 1. \end{aligned} \quad (A.6) $$

Then we obtain

$$ \begin{aligned} |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 &= (\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0) \Phi^*(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)) = \\ &= ([c_1 \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2 \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)]) \times \\ &\quad \times ([c_1^* \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2^* \Psi_2^*(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)]) = \\ &= |c_1|^2 (|\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2) + c_1 c_2 (\Psi_1^*(\mathbf{x}, t, \mathbf{x}_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)) + \\ &\quad |c_2|^2 (|\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2) + c_1 c_2^* (\Psi_1(\mathbf{x}, t, \mathbf{x}_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0)). \end{aligned} \quad (A.7) $$

**Definition A.1.** Let $\langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ be a vector-function
$$ \langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) : [0, T] \times \mathbb{R}^d \times \mathbb{R}^d \times [0, T] \rightarrow \mathbb{R}^d $$

$$ \langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \{ \langle x_1 \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0), \dots, \langle x_d \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \}, \quad (A.8) $$

where

$$ \begin{align} \langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d x = \nonumber \\ &= |c_1|^2 \int_{\mathbb{R}^d} x_i |\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2 d^d x + \nonumber \\ &\quad + |c_2|^2 \int_{\mathbb{R}^d} x_i |\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2 d^d x + \tag{A.9} \\ &\quad + c_1^* c_2 \int_{\mathbb{R}^d} x_i \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d x + \nonumber \\ &\quad + c_1 c_2^* \int_{\mathbb{R}^d} x_i \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d x. \nonumber \end{align} $$

**Definition A.2.** Let $\Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ be a vector-function

$$ \Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0) : [0, T] \times \mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{R}^d $$

$$ (\Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0)) = \{ \delta_1(t, \mathbf{x}_0, \mathbf{y}_0, t_0), \dots, \delta_d(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \}, \quad (A.10) $$

where
$$ \langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) : [0, T] \times \mathbb{R}^d \times \mathbb{R}^d \times [0, T] \rightarrow \mathbb{R}^d $$

$$ \langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \{ \langle x_1 \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0), \dots, \langle x_d \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \}, \quad (A.8) $$

where

$$ \begin{align} \langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d x = \nonumber \\ &= |c_1|^2 \int_{\mathbb{R}^d} x_i |\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2 d^d x + \nonumber \\ &\quad + |c_2|^2 \int_{\mathbb{R}^d} x_i |\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2 d^d x + \tag{A.9} \\ &\quad + c_1^* c_2 \int_{\mathbb{R}^d} x_i \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d x + \nonumber \\ &\quad + c_1 c_2^* \int_{\mathbb{R}^d} x_i \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d x. \nonumber \end{align} $$

**Definition A.2.** Let $\Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ be a vector-function

$$ \Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0) : [0, T] \times \mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{R}^d $$

$$ (\Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0)) = \{ \delta_1(t, \mathbf{x}_0, \mathbf{y}_0, t_0), \dots, \delta_d(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \}, \quad (A.10) $$

where
$$
\begin{align}
\delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \delta[x_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0)] = \nonumber \\
&= c_1^* c_2 \int_{\mathbb{R}^d} x_i \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d\mathbf{x} + \tag{A.11} \\
&\quad + c_1 c_2^* \int_{\mathbb{R}^d} x_i \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d\mathbf{x}. \nonumber
\end{align}
$$

Substituting Eqs.(A.11) into Eqs.(A.9) gives

$$
\begin{align}
\langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d\mathbf{x} = \nonumber \\
&= |c_1|^2 \int_{\mathbb{R}^d} x_i |\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2 d^d\mathbf{x} + \tag{A.12} \\
&\quad + |c_2|^2 \int_{\mathbb{R}^d} x_i |\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2 d^d\mathbf{x} + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \nonumber \\
&= |c_1|^2 \langle x_i \rangle(t, \mathbf{x}_0, t_0) + |c_2|^2 \langle x_i \rangle(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0). \nonumber
\end{align}
$$

Substitution equations (A.5) into equations (A.12) gives

$$
\begin{align}
\langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d\mathbf{x} = \nonumber \\
&= |c_1|^2 \langle x_i \rangle(t, \mathbf{x}_0, t_0) + |c_2|^2 \langle x_i \rangle(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \tag{A.13} \\
&\approx |c_1|^2 x_i(t, \mathbf{x}_0, t_0) + |c_2|^2 x_i(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0). \nonumber
\end{align}
$$
$$
\begin{align}
\delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \delta[x_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0)] = \nonumber \\
&= c_1^* c_2 \int_{\mathbb{R}^d} x_i \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d\mathbf{x} + \tag{A.11} \\
&\quad + c_1 c_2^* \int_{\mathbb{R}^d} x_i \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d\mathbf{x}. \nonumber
\end{align}
$$

Substituting Eqs.(A.11) into Eqs.(A.9) gives

$$
\begin{align}
\langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d\mathbf{x} = \nonumber \\
&= |c_1|^2 \int_{\mathbb{R}^d} x_i |\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2 d^d\mathbf{x} + \tag{A.12} \\
&\quad + |c_2|^2 \int_{\mathbb{R}^d} x_i |\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2 d^d\mathbf{x} + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \nonumber \\
&= |c_1|^2 \langle x_i \rangle(t, \mathbf{x}_0, t_0) + |c_2|^2 \langle x_i \rangle(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0). \nonumber
\end{align}
$$

Substitution equations (A.5) into equations (A.12) gives

$$
\begin{align}
\langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d\mathbf{x} = \nonumber \\
&= |c_1|^2 \langle x_i \rangle(t, \mathbf{x}_0, t_0) + |c_2|^2 \langle x_i \rangle(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \tag{A.13} \\
&\approx |c_1|^2 x_i(t, \mathbf{x}_0, t_0) + |c_2|^2 x_i(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0). \nonumber
\end{align}
$$
# Appendix. CAT-B.

The Schrödinger equation (2.1) in region I = {x|x < 0} has the following form

$$h^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2mE\Psi_I(x) = 0. \qquad (B.1)$$

From Schrödinger equation (B.1) follows

$$h^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I(x) dx = 0. \qquad (B.2)$$

Let $\Psi_I^{\#}(x)$ be a function

$$\Psi_I^{\#}(x) = \phi(x)\Psi_I(x), \qquad (B.3)$$

where

$$\phi(x) = (\pi r_c^2)^{-1/4} \exp\left(\frac{x^2}{2r_c^2}\right) \qquad (B.4)$$

see Eq.(2.9). Note that

$$\frac{\partial^2[\phi(x)\Psi_I(x)]}{\partial x^2} = \frac{\partial}{\partial x}\left[\Psi_I(x)\frac{\partial\phi(x)}{\partial x} + \phi(x)\frac{\partial\Psi_I(x)}{\partial x}\right] = 2\frac{\partial\Psi_I(x)}{\partial x}\frac{\partial\phi(x)}{\partial x} + \Psi_I(x)\frac{\partial^2\phi(x)}{\partial x^2} + \phi(x)\frac{\partial^2\Psi_I(x)}{\partial x^2}. \qquad (B.5)$$

Therefore substitution (B.2) into LHS of the Schrödinger equation (B.1) gives
# Appendix. CAT-B.

The Schrödinger equation (2.1) in region I = {x|x < 0} has the following form

$$h^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2mE\Psi_I(x) = 0. \qquad (B.1)$$

From Schrödinger equation (B.1) follows

$$h^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I(x) dx = 0. \qquad (B.2)$$

Let $\Psi_I^{\#}(x)$ be a function

$$\Psi_I^{\#}(x) = \phi(x)\Psi_I(x), \qquad (B.3)$$

where

$$\phi(x) = (\pi r_c^2)^{-1/4} \exp\left(\frac{x^2}{2r_c^2}\right) \qquad (B.4)$$

see Eq.(2.9). Note that

$$\frac{\partial^2[\phi(x)\Psi_I(x)]}{\partial x^2} = \frac{\partial}{\partial x}\left[\Psi_I(x)\frac{\partial\phi(x)}{\partial x} + \phi(x)\frac{\partial\Psi_I(x)}{\partial x}\right] = 2\frac{\partial\Psi_I(x)}{\partial x}\frac{\partial\phi(x)}{\partial x} + \Psi_I(x)\frac{\partial^2\phi(x)}{\partial x^2} + \phi(x)\frac{\partial^2\Psi_I(x)}{\partial x^2}. \qquad (B.5)$$

Therefore substitution (B.2) into LHS of the Schrödinger equation (B.1) gives
$$
\begin{align*}
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I^{\#}(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I^{\#}(x) dx = \\
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \phi(x) \Psi_I(x)}{\partial x^2} dx + 2Em \int_{-\infty}^{0} \phi(x) \Psi_I(x) dx = \\
& 2h^2 \int_{-\infty}^{0} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx + h^2 \int_{-\infty}^{0} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx + \\
& + \int_{-\infty}^{0} \phi(x) \left\{ h^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2Em \int_{-\infty}^{0} \Psi_I(x) \right\} dx.
\end{align*}
\tag{B.6}
$$

Note that

$$
\int_{-\infty}^{0} \phi(x) \left\{ h^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2Em \int_{-\infty}^{0} \Psi_I(x) \right\} dx = 0. \qquad (B.7)
$$

Therefore from Eq.(B.6) and Eq.(2.3)-Eq.(2.4) one obtain

$$
\begin{align*}
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I^{\#}(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I^{\#}(x) dx = \\
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \phi(x) \Psi_I(x)}{\partial x^2} dx + 2Em \int_{-\infty}^{0} \phi(x) \Psi_I(x) dx = \tag{B.8} \\
& = 2h^2 \int_{-\infty}^{\infty} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx + h^2 \int_{-\infty}^{\infty} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx.
\end{align*}
$$

From Eq.(B.6) one obtain
$$
\begin{align*}
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I^{\#}(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I^{\#}(x) dx = \\
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \phi(x) \Psi_I(x)}{\partial x^2} dx + 2Em \int_{-\infty}^{0} \phi(x) \Psi_I(x) dx = \tag{B.6} \\
& 2h^2 \int_{-\infty}^{0} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx + h^2 \int_{-\infty}^{0} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx + \\
& \qquad + \int_{-\infty}^{0} \phi(x) \left\{ h^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2Em \int_{-\infty}^{0} \Psi_I(x) \right\} dx.
\end{align*}
$$

Note that

$$
\int_{-\infty}^{0} \phi(x) \left\{ h^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2Em \int_{-\infty}^{0} \Psi_I(x) \right\} dx = 0. \quad (B.7)
$$

Therefore from Eq.(B.6) and Eq.(2.3)-Eq.(2.4) one obtain

$$
\begin{align*}
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I^{\#}(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I^{\#}(x) dx = \\
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \phi(x) \Psi_I(x)}{\partial x^2} dx + 2Em \int_{-\infty}^{0} \phi(x) \Psi_I(x) dx = \tag{B.8} \\
& = 2h^2 \int_{-\infty}^{\infty} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx + h^2 \int_{-\infty}^{\infty} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx.
\end{align*}
$$

From Eq.(B.6) one obtain
$$
\begin{align*}
\frac{\partial \phi(x)}{\partial x} = (\pi r_c^2)^{-1/4} \frac{\partial}{\partial x} \exp\left[-\frac{x^2}{2r_c^2}\right] &= -(\pi r_c^2)^{-1/4} r_c^{-2} x \exp\left[-\frac{x^2}{2r_c^2}\right], \\
\frac{\partial^2 \phi(x)}{\partial x^2} &= -(\pi r_c^2)^{-1/4} r_c^{-2} \exp\left[-\frac{x^2}{2r_c^2}\right] + \tag{B.9} \\
&\quad + (\pi r_c^2)^{-1/4} r_c^{-4} x^2 \exp\left[-\frac{x^2}{2r_c^2}\right].
\end{align*}
$$

From Eq.(B.9) and Eq.(2.3)-Eq.(2.4) one obtain

$$
\begin{align*}
& h^2 \int_{-\infty}^{0} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx = \\
& - \frac{h^2}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} \frac{\partial \exp(ikx)}{\partial x} x \exp\left[-\frac{x^2}{2r_c^2}\right] dx = \tag{B.10} \\
& - \frac{2\pi \sqrt{2mE} h}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} x \exp\left(i \frac{2\pi \sqrt{2mE}}{h} x\right) \exp\left[-\frac{x^2}{2r_c^2}\right] dx, \\
& k = \frac{2\pi}{h} \sqrt{2mE}.
\end{align*}
$$

and

$$
\begin{align}
& h^2 \int_{-\infty}^{0} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx = - \frac{h^2}{(\pi r_c^2)^{3/4} r_c^2} \int_{-\infty}^{0} \exp(ikx) \exp\left[-\frac{x^2}{2r_c^2}\right] dx + \tag{B.11} \\
& \qquad + \frac{h^2}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} x^2 \exp(ikx) \exp\left[-\frac{x^2}{2r_c^2}\right] dx. \nonumber
\end{align}
$$
$$
\begin{align*}
\frac{\partial \phi(x)}{\partial x} = (\pi r_c^2)^{-1/4} \frac{\partial}{\partial x} \exp\left[-\frac{x^2}{2r_c^2}\right] &= -(\pi r_c^2)^{-1/4} r_c^{-2} x \exp\left[-\frac{x^2}{2r_c^2}\right], \\
\frac{\partial^2 \phi(x)}{\partial x^2} &= -(\pi r_c^2)^{-1/4} r_c^{-2} \exp\left[-\frac{x^2}{2r_c^2}\right] + \tag{B.9} \\
&\quad + (\pi r_c^2)^{-1/4} r_c^{-4} x^2 \exp\left[-\frac{x^2}{2r_c^2}\right].
\end{align*}
$$

From Eq.(B.9) and Eq.(2.3)-Eq.(2.4) one obtain

$$
\begin{align*}
& h^2 \int_{-\infty}^{0} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx = \\
& - \frac{h^2}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} \frac{\partial \exp(ikx)}{\partial x} x \exp\left[-\frac{x^2}{2r_c^2}\right] dx = \tag{B.10} \\
& - \frac{2\pi \sqrt{2mE} h}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} x \exp\left(i \frac{2\pi \sqrt{2mE}}{h} x\right) \exp\left[-\frac{x^2}{2r_c^2}\right] dx, \\
& k = \frac{2\pi}{h} \sqrt{2mE}.
\end{align*}
$$

and

$$
\begin{equation}
\begin{split}
& h^2 \int_{-\infty}^{0} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx = - \frac{h^2}{(\pi r_c^2)^{3/4} r_c^2} \int_{-\infty}^{0} \exp(ikx) \exp\left[-\frac{x^2}{2r_c^2}\right] dx + \\
& \qquad + \frac{h^2}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} x^2 \exp(ikx) \exp\left[-\frac{x^2}{2r_c^2}\right] dx.
\end{split}
\tag{B.11}
\end{equation}
$$
# Appendix. CAT-C. Generalized Postulates for Continuous Valued Observables.

Suppose we have an $n$-dimensional quantum system.

**I. Then we claim the following:**

**C.I.** Any given $n$-dimensional quantum system is identified by a set $\Re$
$\Re \triangleq \langle \mathbf{H}, \mathfrak{I}, \mathfrak{R}, \mathcal{L}_{2,1}, \mathbf{G}, |\psi_t\rangle \rangle$ where:

(i) $\mathbf{H}$ that is some infinite-dimensional complex Hilbert space, (ii) $\mathfrak{I} = (\Omega, \mathcal{F}, \mathbf{P})$ that is complete probability space, (iii) $\Re = (\mathbb{R}^n, \Sigma)$ that is measurable space, (iv) $\mathcal{L}_{2,1}(\Omega)$ that is complete space of random variables $X: \Omega \to \mathbb{R}^n$ such that

$\int_{\Omega} \|X(\omega)\| d\mathbf{P} < \infty, \int_{\Omega} \|X(\omega)\|^2 d\mathbf{P} < \infty$ and (v) $\mathbf{G}: \mathbf{H} \to \mathcal{L}_{2,1}(\Omega)$ that is one to one correspondence such that

$$ \left| \langle \bar{\psi} | \hat{Q} | \psi \rangle \right| = \int_{\Omega} \left( \mathbf{G} \left[ \hat{Q} | \psi \rangle \right] (\omega) \right) d\mathbf{P} = \mathbf{E}_{\Omega} \left( \mathbf{G} \left[ \hat{Q} | \psi \rangle \right] (\omega) \right) \quad (C.1) $$

for any $|\psi\rangle \in \mathbf{H}$ and for any Hermitian operator $\hat{Q}: \mathbf{H} \to \mathbf{H}$,

(vi) $|\psi_t\rangle$ is a continuous vector function $|\psi_t\rangle: \mathbb{R}_+ \to \mathbf{H}$ which represented the evolution of the quantum system $\Re$.

**C.I.2.** For any $|\psi_1\rangle, |\psi_2\rangle \in \mathbf{H}$ and for any Hermitian operator $\hat{Q}: \mathbf{H} \to \mathbf{H}$ such that
$$ \langle \bar{\psi}_1 | \hat{Q} | \psi_2 \rangle = \langle \psi_2 | \hat{Q} | \psi_1 \rangle = 0: $$

$$ \mathbf{G} \left[ \hat{Q} ( |\psi_1\rangle + |\psi_2\rangle ) \right] (\omega) = \mathbf{G} \left[ \hat{Q} |\psi_1\rangle \right] (\omega) + \mathbf{G} \left[ \hat{Q} |\psi_2\rangle \right] (\omega). \quad (C.2) $$

**C.I.3.** Suppose that the evolution of the quantum system is represented by continuous vector function $|\psi_t\rangle: \mathbb{R}_+ \to \mathbf{H}$. Then any process of continuous measurements on measuring observable $\hat{Q}$ for the system in state $|\psi_t\rangle$ one can to describe by an continuous $\mathbb{R}^n$-valued stochastic processes $X_t(\omega) = X_t(\omega; |\psi_t\rangle)$ given on probability space $(\Omega, \mathcal{F}, \mathbf{P})$ and a measurable space $(\mathbb{R}^n, \Sigma)$.

**Remark C.1.** We assume now for short but without loss of generality that $n = 1$.

**Remark C.2.** Let $X(\omega)$ be random variable $X(\omega) \in \mathcal{L}_{2,1}(\Omega)$ such that $X(\omega) = \mathbf{G}[|\psi\rangle](\omega)$, then we denote such random variable by $X_{|\psi\rangle}(\omega)$. The probability density of random variable $X_{|\psi\rangle}(\omega)$ we denote by $p_{|\psi\rangle}(q), q \in \mathbb{R}$.

**Definition C.1.** The classical pure states correspond to vectors $v \in \mathbf{H}$ of norm $\|v\|=1$. Thus the set of all classical pure states corresponds to the unit sphere $S^\infty \subset \mathbf{H}$ in a Hilbert space $\mathbf{H}$.

**Definition C.2.** The projective Hilbert space $P(\mathbf{H})$ of a complex Hilbert space $\mathbf{H}$ is the set of equivalence classes $[v]$ of vectors $v$ in $\mathbf{H}$, with $v \neq 0$, for the equivalence relation given by $v \sim_P w \Leftrightarrow v = \lambda w$ for some non-zero complex number $\lambda \in \mathbb{C}$. The equivalence classes for the relation $\sim_P$ are also called rays or projective rays.

**Remark C.3.** The physical significance of the projective Hilbert space $P(\mathbf{H})$ is that in canonical quantum theory, the states $|\psi\rangle$ and $\lambda|\psi\rangle$ represent the same physical state of the quantum system, for any $\lambda \neq 0$. It is conventional to choose a state $|\psi\rangle$ from the ray $[|\psi\rangle]$ so that it has unit norm $\sqrt{\langle\psi|\psi\rangle} = 1$.

**Remark C.4.** In contrast with canonical quantum theory we have used instead
# Appendix. CAT-C. Generalized Postulates for Continuous Valued Observables.

Suppose we have an $n$-dimensional quantum system.

**I. Then we claim the following:**

**C.I.** Any given $n$-dimensional quantum system is identified by a set $\Re$
$\Re \triangleq \langle \mathbf{H}, \mathfrak{I}, \mathfrak{R}, \mathcal{L}_{2,1}, \mathbf{G}, |\psi_t\rangle \rangle$ where:

(i) $\mathbf{H}$ that is some infinite-dimensional complex Hilbert space, (ii) $\mathfrak{I} = (\Omega, \mathcal{F}, \mathbf{P})$ that is complete probability space, (iii) $\Re = (\mathbb{R}^n, \Sigma)$ that is measurable space, (iv) $\mathcal{L}_{2,1}(\Omega)$ that is complete space of random variables $X: \Omega \to \mathbb{R}^n$ such that

$\int_{\Omega} \|X(\omega)\| d\mathbf{P} < \infty, \int_{\Omega} \|X(\omega)\|^2 d\mathbf{P} < \infty$ and (v) $\mathbf{G}: \mathbf{H} \to \mathcal{L}_{2,1}(\Omega)$ that is one to one correspondence such that

$$ \left| \langle \bar{\psi} | \hat{Q} | \psi \rangle \right| = \int_{\Omega} \left( \mathbf{G} \left[ \hat{Q} | \psi \rangle \right] (\omega) \right) d\mathbf{P} = \mathbf{E}_{\Omega} \left( \mathbf{G} \left[ \hat{Q} | \psi \rangle \right] (\omega) \right) \quad (C.1) $$

for any $|\psi\rangle \in \mathbf{H}$ and for any Hermitian operator $\hat{Q}: \mathbf{H} \to \mathbf{H}$,

(vi) $|\psi_t\rangle$ is a continuous vector function $|\psi_t\rangle: \mathbb{R}_+ \to \mathbf{H}$ which represented the evolution of the quantum system $\Re$.

**C.I.2.** For any $|\psi_1\rangle, |\psi_2\rangle \in \mathbf{H}$ and for any Hermitian operator $\hat{Q}: \mathbf{H} \to \mathbf{H}$ such that
$$ \langle \bar{\psi}_1 | \hat{Q} | \psi_2 \rangle = \langle \psi_2 | \hat{Q} | \psi_1 \rangle = 0 : $$

$$ \mathbf{G} \left[ \hat{Q} ( |\psi_1\rangle + |\psi_2\rangle ) \right] (\omega) = \mathbf{G} \left[ \hat{Q} |\psi_1\rangle \right] (\omega) + \mathbf{G} \left[ \hat{Q} |\psi_2\rangle \right] (\omega). \quad (C.2) $$

**C.I.3.** Suppose that the evolution of the quantum system is represented by continuous vector function $|\psi_t\rangle: \mathbb{R}_+ \to \mathbf{H}$. Then any process of continuous measurements on measuring observable $\hat{Q}$ for the system in state $|\psi_t\rangle$ one can to describe by an continuous $\mathbb{R}^n$-valued stochastic processes $X_t(\omega) = X_t(\omega; |\psi_t\rangle)$ given on probability space $(\Omega, \mathcal{F}, \mathbf{P})$ and a measurable space $(\mathbb{R}^n, \Sigma)$.

**Remark C.1.** We assume now for short but without loss of generality that $n = 1$.

**Remark C.2.** Let $X(\omega)$ be random variable $X(\omega) \in \mathcal{L}_{2,1}(\Omega)$ such that $X(\omega) = \mathbf{G}[|\psi\rangle](\omega)$, then we denote such random variable by $X_{|\psi\rangle}(\omega)$. The probability density of random variable $X_{|\psi\rangle}(\omega)$ we denote by $p_{|\psi\rangle}(q), q \in \mathbb{R}$.

**Definition C.1.** The classical pure states correspond to vectors $v \in \mathbf{H}$ of norm $\|v\|=1$. Thus the set of all classical pure states corresponds to the unit sphere $S^\infty \subset \mathbf{H}$ in a Hilbert space $\mathbf{H}$.

**Definition C.2.** The projective Hilbert space $P(\mathbf{H})$ of a complex Hilbert space $\mathbf{H}$ is the set of equivalence classes $[v]$ of vectors $v$ in $\mathbf{H}$, with $v \neq 0$, for the equivalence relation given by $v \sim_P w \Leftrightarrow v = \lambda w$ for some non-zero complex number $\lambda \in \mathbb{C}$. The equivalence classes for the relation $\sim_P$ are also called rays or projective rays.

**Remark C.3.** The physical significance of the projective Hilbert space $P(\mathbf{H})$ is that in canonical quantum theory, the states $|\psi\rangle$ and $\lambda|\psi\rangle$ represent the same physical state of the quantum system, for any $\lambda \neq 0$. It is conventional to choose a state $|\psi\rangle$ from the ray $[|\psi\rangle]$ so that it has unit norm $\sqrt{\langle\psi|\psi\rangle} = 1$.

**Remark C.4.** In contrast with canonical quantum theory we have used instead
contrary to ~p equivalence relation ~q, a Hilbert space H, see Definition C.7.

**Definition C.3.** The non-classical pure states correspond to the vectors $v \in H$ of a norm $||v|| \neq 1$. Thus the set of all non-classical pure states corresponds to the set $H \setminus S^\infty \subset H$ in the Hilbert space $H$.

Suppose we have an observable $Q$ of a quantum system that is found through an exhaustive series of measurements, to have a set $\mathfrak{T}$ of values $q \in \mathfrak{T}$ such that $\mathfrak{T} = \bigcup_{i=1}^m (\theta_1^i, \theta_2^i)$, $m \ge 2$, $(\theta_1^i, \theta_2^i) \cap (\theta_1^j, \theta_2^j) = \emptyset, i \neq j$. Note that in practice any observable $Q$ is measured to an accuracy $\delta q$ determined by the measuring device. We represent now by $|q\rangle$ the idealized state of the system in the limit $\delta q \to 0$, for which the observable definitely has the value $q$.

## II. Then we claim the following:

**C.II.1.** The states $\{|q\rangle : q \in \mathfrak{T}\}$ form a complete set of $\delta$-function normalized basis states for the state space $H_{\mathfrak{T}}$ of the system.

That the states $\{|q\rangle : q \in \mathfrak{T}\}$ form a complete set of basis states means that any state $|\psi[\mathfrak{T}]\rangle \in H_{\mathfrak{T}}$ of the system can be expressed as: $|\psi[\mathfrak{T}]\rangle = \int_{\mathfrak{T}} c_{\psi[\mathfrak{T}]}(q)dq$, where $\text{supp}(c_{\psi[\mathfrak{T}]}(q)) \subseteq \mathfrak{T}$ and while $\delta$-function normalized means that $\langle q|q'\rangle = \delta(q-q')$ from which follows $c_{\psi[\mathfrak{T}]}(q) = \langle q|\psi[\mathfrak{T}]\rangle$ so that $|\psi[\mathfrak{T}]\rangle = \int_{\mathfrak{T}}|q\rangle\langle q|\psi[\mathfrak{T}]\rangle dq$.

The completeness condition can then be written as $\int_{\mathfrak{T}}|q\rangle\langle q|dq = \hat{1}_{H_{\mathfrak{T}}}$.

**C.II.2.** For the system in state $|\psi[\mathfrak{T}]\rangle$ the probability $P(q, q+dq; |\psi[\mathfrak{T}]\rangle)$ of obtaining the result $q \in \mathfrak{T}$ lying in the range $(q, q+dq) \subset \mathfrak{T}$ on measuring observable $Q$ is given by

$$P(q, q+dq; |\psi[\mathfrak{T}]\rangle) = p_{|\psi[\mathfrak{T}]\rangle}(q)dq \quad (C.3)$$

for any $|\psi[\mathfrak{T}]\rangle \in H_{\mathfrak{T}}$.

**Remark C.5.** Note that in general case $p_{|\psi[\mathfrak{T}]\rangle}(q) \neq |c_{\psi[\mathfrak{T}]}(q)|^2$.

**C.II.3.** The observable $Q_{\mathfrak{T}}$ is represented by a Hermitian operator $\hat{Q}_{\mathfrak{T}} : H_{\mathfrak{T}} \to H_{\mathfrak{T}}$ whose eigenvalues are the possible results $\{q : q \in \mathfrak{T}\}$, of a measurement of $Q_{\mathfrak{T}}$, and the associated eigenstates are the states $\{|q\rangle : q \in \mathfrak{T}\}$, i.e. $\hat{Q}_{\mathfrak{T}}|q\rangle = q|q\rangle$, $q \in \mathfrak{T}$.

**Remark C.6.** Note that the spectral decomposition of the operator $\hat{Q}_{\mathfrak{T}}$ is then

$$\hat{Q}_{\mathfrak{T}} = \int_{\mathfrak{T}} q|q\rangle\langle q|dq. \quad (C.3)$$

**Definition C.4.** A connected set in $\mathbb{R}$ is a set $X \subset \mathbb{R}$ that cannot be partitioned into two nonempty subsets which are open in the relative topology induced on the set. Equivalently, it is a set which cannot be partitioned into two nonempty subsets such that each subset has no points in common with the set closure of the other.

**Definition C.5.** The well localized pure states $|\psi[\Theta]\rangle$ with a support $\Theta = (\theta_1, \theta_2)$ correspond to vectors of norm 1 and such that: $\text{supp}(c_{\psi[\Theta]}(q)) = \Theta$ is a connected set in $\mathbb{R}$ Thus the set of all well localized pure states corresponds to the unit sphere $S^\infty \subset H$ in the Hilbert space $H_\Theta \subset H$.

Suppose we have an observable $Q_\Theta$ of a system that is found through an exhaustive series of measurements, to have a continuous range of values $q : \theta_1 < q < \theta_2$.

## III. Then we claim the following:

**C.III.1.** For the system in well localized pure state $|\psi[\Theta]\rangle$ such that:

(i) $|\psi[\Theta]\rangle \in S_\Theta^\infty$ and
contrary to ~p equivalence relation ~q, a Hilbert space H, see Definition C.7.

**Definition C.3.** The non-classical pure states correspond to the vectors $v \in H$ of a norm $||v|| \neq 1$. Thus the set of all non-classical pure states corresponds to the set $H \setminus S^\infty \subset H$ in the Hilbert space $H$.

Suppose we have an observable $Q$ of a quantum system that is found through an exhaustive series of measurements, to have a set $\mathfrak{T}$ of values $q \in \mathfrak{T}$ such that $\mathfrak{T} = \bigcup_{i=1}^m (\theta_1^i, \theta_2^i)$, $m \ge 2$, $(\theta_1^i, \theta_2^i) \cap (\theta_1^j, \theta_2^j) = \emptyset, i \neq j$. Note that in practice any observable $Q$ is measured to an accuracy $\delta q$ determined by the measuring device. We represent now by $|q\rangle$ the idealized state of the system in the limit $\delta q \to 0$, for which the observable definitely has the value $q$.

## II. Then we claim the following:

**C.II.1.** The states $\{|q\rangle : q \in \mathfrak{T}\}$ form a complete set of $\delta$-function normalized basis states for the state space $H_{\mathfrak{T}}$ of the system.

That the states $\{|q\rangle : q \in \mathfrak{T}\}$ form a complete set of basis states means that any state $|\psi[\mathfrak{T}]\rangle \in H_{\mathfrak{T}}$ of the system can be expressed as: $|\psi[\mathfrak{T}]\rangle = \int_{\mathfrak{T}} c_{\psi[\mathfrak{T}]}(q)dq$, where $\text{supp}(c_{\psi[\mathfrak{T}]}(q)) \subseteq \mathfrak{T}$ and while $\delta$-function normalized means that $\langle q|q'\rangle = \delta(q-q')$ from which follows $c_{\psi[\mathfrak{T}]}(q) = \langle q|\psi[\mathfrak{T}]\rangle$ so that $|\psi[\mathfrak{T}]\rangle = \int_{\mathfrak{T}}|q\rangle\langle q|\psi[\mathfrak{T}]\rangle dq$.

The completeness condition can then be written as $\int_{\mathfrak{T}}|q\rangle\langle q|dq = \hat{1}_{H_{\mathfrak{T}}}$.

**C.II.2.** For the system in state $|\psi[\mathfrak{T}]\rangle$ the probability $P(q, q+dq; |\psi[\mathfrak{T}]\rangle)$ of obtaining the result $q \in \mathfrak{T}$ lying in the range $(q, q+dq) \subset \mathfrak{T}$ on measuring observable $Q$ is given by

$$P(q, q+dq; |\psi[\mathfrak{T}]\rangle) = p_{|\psi[\mathfrak{T}]\rangle}(q)dq \quad (C.3)$$

for any $|\psi[\mathfrak{T}]\rangle \in H_{\mathfrak{T}}$.

**Remark C.5.** Note that in general case $p_{|\psi[\mathfrak{T}]\rangle}(q) \neq |c_{\psi[\mathfrak{T}]}(q)|^2$.

**C.II.3.** The observable $Q_{\mathfrak{T}}$ is represented by a Hermitian operator $\hat{Q}_{\mathfrak{T}} : H_{\mathfrak{T}} \to H_{\mathfrak{T}}$ whose eigenvalues are the possible results $\{q : q \in \mathfrak{T}\}$, of a measurement of $Q_{\mathfrak{T}}$, and the associated eigenstates are the states $\{|q\rangle : q \in \mathfrak{T}\}$, i.e. $\hat{Q}_{\mathfrak{T}}|q\rangle = q|q\rangle$, $q \in \mathfrak{T}$.

**Remark C.6.** Note that the spectral decomposition of the operator $\hat{Q}_{\mathfrak{T}}$ is then

$$\hat{Q}_{\mathfrak{T}} = \int_{\mathfrak{T}} q|q\rangle\langle q|dq. \quad (C.3)$$

**Definition C.4.** A connected set in $\mathbb{R}$ is a set $X \subset \mathbb{R}$ that cannot be partitioned into two nonempty subsets which are open in the relative topology induced on the set. Equivalently, it is a set which cannot be partitioned into two nonempty subsets such that each subset has no points in common with the set closure of the other.

**Definition C.5.** The well localized pure states $|\psi[\Theta]\rangle$ with a support $\Theta = (\theta_1, \theta_2)$ correspond to vectors of norm 1 and such that: $\text{supp}(c_{\psi[\Theta]}(q)) = \Theta$ is a connected set in $\mathbb{R}$ Thus the set of all well localized pure states corresponds to the unit sphere $S^\infty \subset H$ in the Hilbert space $H_\Theta \subset H$.

Suppose we have an observable $Q_\Theta$ of a system that is found through an exhaustive series of measurements, to have a continuous range of values $q$ : $\theta_1 < q < \theta_2$.

## III. Then we claim the following:

**C.III.1.** For the system in well localized pure state $|\psi[\Theta]\rangle$ such that:

(i) $|\psi[\Theta]\rangle \in S^\infty$ and
(ii) $\text{supp}(c_{\psi[\Theta]}(q)) \triangleq \{q|c_{\psi[\Theta]}(q) \neq 0\}$ is a connected set in $\mathbb{R}$, then the probability $P(q, q + dq; |\psi[\Theta]\rangle\rangle)$ of obtaining the result $q$ lying in the range $(q, q + dq)$ on measuring observable $Q_\Theta$ is given by

$$P(q, q + dq; |\psi[\Theta]\rangle\rangle) = |\langle q|\psi[\Theta]\rangle|^2 dq = |c_{\psi[\Theta]}(q)|^2 dq. \qquad (C.4)$$

**C.III.2.** $p_{|\psi[\Theta]\rangle\rangle}(q)dq = |\langle q|\psi[\Theta]\rangle|^2 dq = |c_{\psi[\Theta]}(q)|^2 dq$.

**C.III.3.** Let $|\psi[\Theta_1]\rangle$ and $|\psi[\Theta_2]\rangle$ be well localized pure states with $\Theta_1 = (\theta_1^1, \theta_2^1)$ and $\Theta_2 = (\theta_1^2, \theta_2^2)$ correspondingly. Let $X_1(\omega) = X_{|\psi[\Theta_1]\rangle\rangle}(\omega)$ and $X_2(\omega) = X_{|\psi[\Theta_2]\rangle\rangle}(\omega)$ correspondingly. Assume that $\bar{\Theta}_1 \cap \bar{\Theta}_2 = \emptyset$ (here the closure of $\Theta_i, i = 1,2$ is denoted by $\bar{\Theta}_{i, i = 1,2}$) then random variables $X_1(\omega)$ and $X_2(\omega)$ are independent.

**C.III.4.** If the system is in well localized pure state $|\psi[\Theta]\rangle$ the state $|\psi[\Theta]\rangle$ described by a wave function $\psi(q, \Theta) = \langle q|\psi[\Theta]\rangle$ and the value of observable $Q_\Theta$ is measured once each

on many identically prepared system, the average value of all the measurements will be

$$\langle Q_{\Theta} \rangle = \frac{\int_{\Theta} q |\psi(q, \Theta)|^2 dq}{\int_{\Theta} |\psi(q, \Theta)|^2 dq}. \qquad (C.5)$$

The completeness condition can then be written as $\int_{\Theta} |q\rangle\langle q| dq = \hat{1}_{H_\Theta}$.

Completeness means that for any state $|\psi[\Theta]\rangle\rangle \in S_\Theta^\infty$ it must be the case that $\int_{\Theta} |\langle q|\psi[\Theta]\rangle|^2 dq \neq 0$, i.e. there must be a non-zero probability to get some result on measuring observable $Q_\Theta$.

**C.III.5.** (von Neumann measurement postulate) Assume that

(i) $|\psi\rangle\rangle \in S_\Theta^\infty$ and (ii) $\text{supp}(c_\psi(q)) = \Theta$ is a connected set in $\mathbb{R}$. Then if on performing a measurement of $Q_\Theta$ with an accuracy $\delta q$, the result is obtained in the range $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q)$, then the system will end up in the state

$$\frac{\hat{P}(q, \delta q) |\psi[\Theta]\rangle\rangle}{\sqrt{\langle \psi | \hat{P}(q, \delta q) | \psi[\Theta] \rangle\rangle}} = \frac{\int_{|q-q'| \le \delta q/2} |q'\rangle\langle q'| |\psi[\Theta]\rangle\rangle dq'}{\sqrt{\int_{|q-q'| \le \delta q/2} |\langle q'| \psi[\Theta] \rangle|^2 dq'}}. \qquad (C.6)$$

## IV. We claim the following:

**C.IV.1** For the system in state $|\psi^a[\Theta]\rangle = a|\psi[\Theta]\rangle\rangle \in H_\Theta$, where: (i) $|\psi[\Theta]\rangle\rangle \in S_\Theta^\infty, |a| \neq 1$,

(ii) $\text{supp}(c_{\psi[\Theta]}(q))$ is a connected set in $\mathbb{R}$ and (iii) $|\psi[\Theta]\rangle\rangle = \int_{\theta_1}^{\theta_2} c_{\psi[\Theta]}(q)|q\rangle\rangle dq$

$$G[\hat{Q}_{\Theta}|\psi^a[\Theta]\rangle\rangle] = |a|^2 G[\hat{Q}_{\Theta}|\psi[\Theta]\rangle\rangle]. \qquad (C.6)$$
(ii) $\text{supp}(c_{\psi[\Theta]}(q)) \triangleq \{q|c_{\psi[\Theta]}(q) \neq 0\}$ is a connected set in $\mathbb{R}$, then the probability $P(q, q + dq; |\psi[\Theta]\rangle\rangle)$ of obtaining the result $q$ lying in the range $(q, q + dq)$ on measuring observable $Q_\Theta$ is given by

$$P(q, q + dq; |\psi[\Theta]\rangle\rangle) = |\langle q|\psi[\Theta]\rangle|^2 dq = |c_{\psi[\Theta]}(q)|^2 dq. \qquad (C.4)$$

**C.III.2.** $p_{|\psi[\Theta]\rangle}(q)dq = |\langle q|\psi[\Theta]\rangle|^2 dq = |c_{\psi[\Theta]}(q)|^2 dq$.

**C.III.3.** Let $|\psi[\Theta_1]\rangle$ and $|\psi[\Theta_2]\rangle$ be well localized pure states with $\Theta_1 = (\theta_1^1, \theta_2^1)$ and $\Theta_2 = (\theta_1^2, \theta_2^2)$ correspondingly. Let $X_1(\omega) = X_{|\psi[\Theta_1]\rangle}(\omega)$ and $X_2(\omega) = X_{|\psi[\Theta_2]\rangle}(\omega)$ correspondingly. Assume that $\bar{\Theta}_1 \cap \bar{\Theta}_2 = \emptyset$ (here the closure of $\Theta_i, i = 1,2$ is denoted by $\bar{\Theta}_{i, i = 1,2}$) then random variables $X_1(\omega)$ and $X_2(\omega)$ are independent.

**C.III.4.** If the system is in well localized pure state $|\psi[\Theta]\rangle$ the state $|\psi[\Theta]\rangle$ described by a wave function $\psi(q, \Theta) = \langle q|\psi[\Theta]\rangle$ and the value of observable $Q_\Theta$ is measured once each

on many identically prepared system, the average value of all the measurements will be

$$\langle Q_{\Theta} \rangle = \frac{\int_{\Theta} q |\psi(q, \Theta)|^2 dq}{\int_{\Theta} |\psi(q, \Theta)|^2 dq}. \qquad (C.5)$$

The completeness condition can then be written as $\int_{\Theta} |q\rangle\langle q| dq = \hat{1}_{H_\Theta}$.

Completeness means that for any state $|\psi[\Theta]\rangle \in S_\Theta^\infty$ it must be the case that $\int_{\Theta} |\langle q|\psi[\Theta]\rangle|^2 dq \neq 0$, i.e. there must be a non-zero probability to get some result on measuring observable $Q_\Theta$.

**C.III.5.** (von Neumann measurement postulate) Assume that

(i) $|\psi\rangle \in S_\Theta^\infty$ and (ii) $\text{supp}(c_\psi(q)) = \Theta$ is a connected set in $\mathbb{R}$. Then if on performing a measurement of $Q_\Theta$ with an accuracy $\delta q$, the result is obtained in the range $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q)$, then the system will end up in the state

$$\frac{\hat{P}(q, \delta q) |\psi[\Theta]\rangle}{\sqrt{\langle \psi | \hat{P}(q, \delta q) | \psi[\Theta] \rangle}} = \frac{\int_{|q-q'| \le \delta q/2} |q'\rangle\langle q'| |\psi[\Theta]\rangle dq'}{\sqrt{\int_{|q-q'| \le \delta q/2} |\langle q'| \psi[\Theta] \rangle|^2 dq'}}. \qquad (C.6)$$

## IV. We claim the following:

**C.IV.1** For the system in state $|\psi^a[\Theta]\rangle = a|\psi[\Theta]\rangle \in H_\Theta$, where: (i) $|\psi[\Theta]\rangle \in S_\Theta^\infty, |a| \neq 1$,

(ii) $\text{supp}(c_{\psi[\Theta]}(q))$ is a connected set in $\mathbb{R}$ and (iii) $|\psi[\Theta]\rangle = \int_{\theta_1}^{\theta_2} c_{\psi[\Theta]}(q) |q\rangle dq$

$$G[\hat{Q}_{\Theta}|\psi^a[\Theta]\rangle] = |a|^2 G[\hat{Q}_{\Theta}|\psi[\Theta]\rangle]. \qquad (C.6)$$
**C.IV.2.** Assume that the system in state $|\psi^a[\Theta]\rangle = a|\psi[\Theta]\rangle \in \mathbf{H}_\Theta$, where (i) $|\psi[\Theta]\rangle \in \mathbf{S}_\Theta^\infty$, $|a| \neq 1$, (ii) $\text{supp}(c_{\psi[\Theta]}(q))$ is a connected set in $\mathbb{R}$ and (iii) $|\psi[\Theta]\rangle = \int_{\Theta_1}^{\Theta_2} c_{\psi[\Theta]}(q)|q\rangle dq$.

Then if the system is in state $|\psi^a[\Theta]\rangle$ described by a wave function $\psi^a(q; \Theta) = \langle q | \psi^a[\Theta] \rangle$ and the value of observable $Q_\Theta$ is measured once each on many identically prepared system, the average value of all the measurements will be

$$ \langle Q_\Theta \rangle = \int_\Theta q |\psi^a(q; \Theta)|^2 dq. \qquad (C.7) $$

**C.IV.3.** The probability $P(q, q + dq; |\psi^a[\Theta]\rangle) dq$ of obtaining the result $q$ lying in the range $(q, q + dq)$ on measuring $Q_\Theta$ is

$$ P(q, q + dq; |\psi^a[\Theta]\rangle) dq = |a|^{-2} |c_{\psi[\Theta]}(q|a|^{-2})|^2 dq. \qquad (C.8) $$

**Remark C.7.** Note that C.IV.3 immediately follows from C.IV.1 and C.III.2.

**C.IV.4. (Generalized von Neumann measurement postulate)** If on performing a measurement of observable $Q_\Theta$ with an accuracy $\delta q$, the result is obtained in the range $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q)$, then the system immediately after measurement will end up in the state

$$ \frac{\hat{P}(q, \delta q) |\psi^a[\Theta]\rangle}{\sqrt{\langle \psi | \hat{P}(q, \delta q) |\psi[\Theta]\rangle}} = \frac{\int_{|q-q'| \le \delta q/2} |q'\rangle \langle q'| \|\psi^a[\Theta]\rangle dq'}{\sqrt{\int_{|q-q'| \le \delta q/2} |\langle q' | \psi^a[\Theta] \rangle|^2 dq'}} = \\ \frac{a \int_{|q-q'| \le \delta q/2} |q'\rangle \langle q'| \|\psi[\Theta]\rangle dq'}{\sqrt{\int_{|q-q'| \le \delta q/2} |\langle q' | \psi[\Theta] \rangle|^2 dq'}} \in \mathbf{H}_\Theta. \qquad (C.9) $$

**C.V.1.** Let $|\Psi^{a_1, a_2}[\Theta_1, \Theta_2]\rangle = |\psi_1^{a_1}[\Theta_1]\rangle + |\psi_2^{a_2}[\Theta_2]\rangle \in \mathbf{H}_{1,2} \triangleq \mathbf{H}_{\Theta_1} \oplus \mathbf{H}_{\Theta_2} \subseteq \mathbf{H}$, where

(i) $|\psi_i^{a_i}[\Theta_i]\rangle = a_i |\psi_i[\Theta_i]\rangle \in \mathbf{H}_{\Theta_i}, |\psi_i\rangle = |\psi_i[\Theta_i]\rangle \in \mathbf{S}_{\Theta_i}^\infty, |a_i| \neq 1, i = 1,2$;

(ii) $\text{supp}(c_{\psi_i[\Theta_i]}(q)), i = 1,2$ is a connected sets in $\mathbb{R}$;

(iii) $(\text{supp}(c_{\psi_1[\Theta_1]}(q))) \cap (\text{supp}(c_{\psi_2[\Theta_2]}(q))) = \emptyset$ and

(iv) $|\psi_i[\Theta_i]\rangle = \int_{\Theta_1}^{\Theta_2} c_{\psi_i[\Theta_i]}(q)|q\rangle dq, i = 1,2$.

Then if the system is in a state $|\Psi^{a_1, a_2}[\Theta_1, \Theta_2]\rangle$ described by a wave function $\Psi^{a_1, a_2}(q; \Theta_1, \Theta_2) = \langle q | \Psi^{a_1, a_2}[\Theta_1, \Theta_2] \rangle, q \in \Theta_1 \cup \Theta_2$ and the value of observable $Q_{\Theta_1, \Theta_2}$ is measured once each on many identically prepared system, the average value of all the measurements will be
**C.IV.2.** Assume that the system in state $|\psi^a[\Theta]\rangle = a|\psi[\Theta]\rangle \in \mathbf{H}_\Theta$, where (i) $|\psi[\Theta]\rangle \in \mathbf{S}_\Theta^\infty$, $|a| \neq 1$, (ii) $\text{supp}(c_{\psi[\Theta]}(q))$ is a connected set in $\mathbb{R}$ and (iii) $|\psi[\Theta]\rangle = \int_{\Theta_1}^{\Theta_2} c_{\psi[\Theta]}(q)|q\rangle dq$.

Then if the system is in state $|\psi^a[\Theta]\rangle$ described by a wave function $\psi^a(q; \Theta) = \langle q || \psi^a[\Theta] \rangle$ and the value of observable $Q_\Theta$ is measured once each on many identically prepared system, the average value of all the measurements will be

$$ \langle Q_\Theta \rangle = \int_\Theta q |\psi^a(q; \Theta)|^2 dq. \qquad (C.7) $$

**C.IV.3.** The probability $P(q, q + dq; |\psi^a[\Theta]\rangle) dq$ of obtaining the result $q$ lying in the range $(q, q + dq)$ on measuring $Q_\Theta$ is

$$ P(q, q + dq; |\psi^a[\Theta]\rangle) dq = |a|^{-2} |c_{\psi[\Theta]}(q|a|^{-2})|^2 dq. \qquad (C.8) $$

**Remark C.7.** Note that C.IV.3 immediately follows from C.IV.1 and C.III.2.

**C.IV.4. (Generalized von Neumann measurement postulate)** If on performing a measurement of observable $Q_\Theta$ with an accuracy $\delta q$, the result is obtained in the range $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q)$, then the system immediately after measurement will end up in the state

$$ \frac{\hat{P}(q, \delta q) |\psi^a[\Theta]\rangle}{\sqrt{\langle \psi | \hat{P}(q, \delta q) |\psi[\Theta]\rangle}} = \frac{\int_{|q-q'| \le \delta q/2} |q'\rangle \langle q'| \|\psi^a[\Theta]\rangle dq'}{\sqrt{\int_{|q-q'| \le \delta q/2} |\langle q' | \psi^a[\Theta] \rangle|^2 dq'}} = \\ \frac{a \int_{|q-q'| \le \delta q/2} |q'\rangle \langle q'| \|\psi[\Theta]\rangle dq'}{\sqrt{\int_{|q-q'| \le \delta q/2} |\langle q' | \psi[\Theta] \rangle|^2 dq'}} \in \mathbf{H}_\Theta. \qquad (C.9) $$

**C.V.1.** Let $|\Psi^{a_1, a_2}[\Theta_1, \Theta_2]\rangle = |\psi_1^{a_1}[\Theta_1]\rangle + |\psi_2^{a_2}[\Theta_2]\rangle \in \mathbf{H}_{1,2} \triangleq \mathbf{H}_{\Theta_1} \oplus \mathbf{H}_{\Theta_2} \subseteq \mathbf{H}$, where

(i) $|\psi_i^{a_i}[\Theta_i]\rangle = a_i|\psi_i[\Theta_i]\rangle \in \mathbf{H}_{\Theta_i}, |\psi_i\rangle = |\psi_i[\Theta_i]\rangle \in \mathbf{S}_{\Theta_i}^\infty, |a_i| \neq 1, i = 1,2$;

(ii) $\text{supp}(c_{\psi_i[\Theta_i]}(q)), i = 1,2$ is a connected sets in $\mathbb{R}$;

(iii) $(\text{supp}(c_{\psi_1[\Theta_1]}(q))) \cap (\text{supp}(c_{\psi_2[\Theta_2]}(q))) = \emptyset$ and

(iv) $|\psi_i[\Theta_i]\rangle = \int_{\Theta_1}^{\Theta_2} c_{\psi_i[\Theta_i]}(q)|q\rangle dq, i = 1,2$.

Then if the system is in a state $|\Psi^{a_1, a_2}[\Theta_1, \Theta_2]\rangle$ described by a wave function $\Psi^{a_1, a_2}(q; \Theta_1, \Theta_2) = \langle q || \Psi^{a_1, a_2}[\Theta_1, \Theta_2] \rangle, q \in \Theta_1 \cup \Theta_2$ and the value of observable $Q_{\Theta_1, \Theta_2}$ is measured once each on many identically prepared system, the average value of all the measurements will be
$$ \langle Q_{\Theta_1, \Theta_2} \rangle = \int_{\Theta_1 \cup \Theta_2} q |\Psi^{a_1, a_2}(q; \Theta_1, \Theta_2)|^2 dq. \qquad (C.10) $$

**C.V. 2.** The probability of getting a result $q$ with an accuracy $\delta q$ such that $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q) \in \text{supp}(c_{\psi_1}(q))$ or $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q) \in \text{supp}(c_{\psi_2}(q))$ given by

$$ \int_{|q-q'| \le \delta q/2} \left[ (|\langle q' ||\Psi_1^{a_1}[\Theta_1] \rangle|^2) * (|\langle q' ||\Psi_2^{a_2}[\Theta_2] \rangle|^2) \right] dq'. \qquad (C.11) $$

**Remark C.8.** Note that C.IV.3 immediately follows from C.III.3.

**C.V. 3.** Assume that the system is initially in the state $|\Psi^{a_1, a_2}[\Theta_1, \Theta_2]\rangle$. If on performing a measurement of $Q_{\Theta_1, \Theta_2}$ with an accuracy $\delta q$, the result is obtained in the range $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q)$, then the state of the system immediately after measurement given by

$$ \frac{\hat{P}(q_i, \delta q) |\Psi^{a_1, a_2}[\Theta_1, \Theta_2] \rangle}{\sqrt{\langle \psi | \hat{P}(q_i, \delta q) | \psi \rangle}} = \frac{\int_{|q_i-q'| \le \delta q/2} (|q'\rangle\langle q' ||\psi_1^{a_1}[\Theta_1] \rangle + |q'\rangle\langle q' ||\psi_2^{a_2}[\Theta_2] \rangle) dq'}{\sqrt{\int_{|q_i-q'| \le \delta q/2} [|\langle q' ||\psi_1[\Theta_1] \rangle|^2 + |\langle q' ||\psi_2[\Theta_2] \rangle|^2] dq'}} = \qquad (C.12) $$

$$ \frac{\int_{|q_i-q'| \le \delta q/2} (a_1|q'\rangle\langle q' ||\psi_1[\Theta_1] \rangle + a_2|q'\rangle\langle q' ||\psi_2[\Theta_2] \rangle) dq'}{\sqrt{\int_{|q_i-q'| \le \delta q/2} [|\langle q' ||\psi_1[\Theta_1] \rangle|^2 + |\langle q' ||\psi_2[\Theta_2] \rangle|^2] dq'}} \in \mathbf{H}_{\Theta_i}, \\ q_i \in \Theta_i, i = 1, 2. $$

**Definition C.6.** Let $H_{1,2}$ be $H_{1,2} \triangleq H_{\Theta_1} \oplus H_{\Theta_2}$.

**Definition C.7.** Let $|\psi^a\rangle$ be a state $|\psi^a\rangle = a|\psi\rangle$, where $|\psi\rangle \in S^\infty, |a| \neq 1$ and $|\psi\rangle = \int_{\Theta_1}^{\Theta_2} c_\psi(q)|q\rangle dq$. Let $|\psi_a\rangle$ be an state such that $|\psi_a\rangle \in S^\infty$. States $|\psi^a\rangle$ and $|\psi_a\rangle$ is a $Q$-equivalent: $|\psi^a\rangle \sim_Q |\psi_a\rangle$ iff

$$ P(q, q + dq; |\psi^a\rangle) = |a|^{-2} |c_\psi(q|a|^{-2})|^2 dq = P(qq + dq; |\psi_a\rangle) dq \qquad (C.13) $$

**C.V.** For any state $|\psi^a\rangle = a|\psi\rangle$, where $|\psi\rangle \in S^\infty, |a| \neq 1$ and $|\psi\rangle = \int_{\Theta_1}^{\Theta_2} c_\psi(q)|q\rangle dq$ there exist an state $|\psi_a\rangle \in S^\infty$ such that: $|\psi^a\rangle \sim_Q |\psi_a\rangle$.

**Definition C.8.** Let $|\psi^a\rangle$ be a state $|\psi^a\rangle = a|\psi\rangle$, where $|\psi\rangle \in S^\infty, |a| \neq 1$ and $|\psi\rangle = \int_{\Theta_1}^{\Theta_2} c_\psi(q)|q\rangle dq$. Let $|\psi_a\rangle$ be an state such that $|\psi_a\rangle \in S^\infty$. States $|\psi^a\rangle$ and $|\psi_a\rangle$ is a $\hat{Q}$-equivalent ($|\psi^a\rangle \sim_{\hat{Q}} |\psi_a\rangle$) iff: $\langle \psi^a | \hat{Q} | \psi^a \rangle = \langle \psi_a | \hat{Q} | \psi_a \rangle$.

**C.VI.** For any state $|\psi^a\rangle = a|\psi\rangle$, where $|\psi\rangle \in S^\infty, |a| \neq 1$ and $|\psi\rangle = \int_{\Theta_1}^{\Theta_2} c_\psi(q)|q\rangle dq$ there exist an state $|\psi_a\rangle \in S^\infty$ such that: $|\psi^a\rangle \sim_{\hat{Q}} |\psi_a\rangle$.
$$ \langle Q_{\Theta_1, \Theta_2} \rangle = \int_{\Theta_1 \cup \Theta_2} q |\Psi^{a_1, a_2}(q; \Theta_1, \Theta_2)|^2 dq. \quad (C.10) $$

**C.V. 2.** The probability of getting a result $q$ with an accuracy $\delta q$ such that $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q) \in \text{supp}(c_{\psi_1}(q))$ or $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q) \in \text{supp}(c_{\psi_2}(q))$ given by

$$ \int_{|q-q'| \le \delta q/2} \left[ (|\langle q' || \Psi_1^{a_1} [\Theta_1] \rangle|^2) * (|\langle q' || \Psi_2^{a_2} [\Theta_2] \rangle|^2) \right] dq'. \quad (C.11) $$

**Remark C.8.** Note that C.IV.3 immediately follows from C.III.3.

**C.V. 3.** Assume that the system is initially in the state $|\Psi^{a_1, a_2}[\Theta_1, \Theta_2]\rangle$. If on performing a measurement of $Q_{\Theta_1, \Theta_2}$ with an accuracy $\delta q$, the result is obtained in the range $(q - \frac{1}{2}\delta q, q + \frac{1}{2}\delta q)$, then the state of the system immediately after measurement given by

$$ \frac{\hat{P}(q_i, \delta q) |\Psi^{a_1, a_2}[\Theta_1, \Theta_2] \rangle}{\sqrt{\langle \psi | \hat{P}(q_i, \delta q) | \psi \rangle}} = \frac{\int_{|q_i - q'| \le \delta q/2} (|q' \rangle \langle q' || \psi_1^{a_1} [\Theta_1] \rangle + |q' \rangle \langle q' || \psi_2^{a_2} [\Theta_2] \rangle) dq'}{\sqrt{\int_{|q_i - q'| \le \delta q/2} [|\langle q' || \psi_1 [\Theta_1] \rangle|^2 + |\langle q' || \psi_2 [\Theta_2] \rangle|^2] dq'}} = \quad (C.12) $$

$$ \frac{\int_{|q_i - q'| \le \delta q/2} (a_1 |q'\rangle \langle q' || \psi_1 [\Theta_1] \rangle + a_2 |q'\rangle \langle q' || \psi_2 [\Theta_2] \rangle) dq'}{\sqrt{\int_{|q_i - q'| \le \delta q/2} [|\langle q' || \psi_1 [\Theta_1] \rangle|^2 + |\langle q' || \psi_2 [\Theta_2] \rangle|^2] dq'}} \in \mathbf{H}_{\Theta_i}, \\ q_i \in \Theta_i, i = 1, 2. $$

**Definition C.6.** Let $H_{1,2}$ be $H_{1,2} \triangleq H_{\Theta_1} \oplus H_{\Theta_2}$.

**Definition C.7.** Let $|\psi^a\rangle$ be a state $|\psi^a\rangle = a|\psi\rangle$, where $|\psi\rangle \in S^\infty, |a| \neq 1$ and $|\psi\rangle = \int_{\Theta_1}^{\Theta_2} c_\psi(q)|q\rangle dq$. Let $|\psi_a\rangle$ be an state such that $|\psi_a\rangle \in S^\infty$. States $|\psi^a\rangle$ and $|\psi_a\rangle$ is a $Q$-equivalent: $|\psi^a\rangle \sim_Q |\psi_a\rangle$ iff

$$ P(q, q + dq; |\psi^a\rangle) = |a|^{-2} |c_\psi(q|a|^{-2})|^2 dq = P(qq + dq; |\psi_a\rangle) dq \quad (C.13) $$

**C.V.** For any state $|\psi^a\rangle = a|\psi\rangle$, where $|\psi\rangle \in S^\infty, |a| \neq 1$ and $|\psi\rangle = \int_{\Theta_1}^{\Theta_2} c_\psi(q)|q\rangle dq$ there exist an state $|\psi_a\rangle \in S^\infty$ such that: $|\psi^a\rangle \sim_Q |\psi_a\rangle$.

**Definition C.8.** Let $|\psi^a\rangle$ be a state $|\psi^a\rangle = a|\psi\rangle$, where $|\psi\rangle \in S^\infty, |a| \neq 1$ and $|\psi\rangle = \int_{\Theta_1}^{\Theta_2} c_\psi(q)|q\rangle dq$. Let $|\psi_a\rangle$ be an state such that $|\psi_a\rangle \in S^\infty$. States $|\psi^a\rangle$ and $|\psi_a\rangle$ is a $\hat{Q}$-equivalent $(|\psi^a\rangle \sim_{\hat{Q}} |\psi_a\rangle)$ iff: $\langle \psi^a | \hat{Q} | \psi^a \rangle = \langle \psi_a | \hat{Q} | \psi_a \rangle$.

**C.VI.** For any state $|\psi^a\rangle = a|\psi\rangle$, where $|\psi\rangle \in S^\infty, |a| \neq 1$ and $|\psi\rangle = \int_{\Theta_1}^{\Theta_2} c_\psi(q)|q\rangle dq$ there exist an state $|\psi_a\rangle \in S^\infty$ such that: $|\psi^a\rangle \sim_{\hat{Q}} |\psi_a\rangle$.
# Appendix I.1. Derivation of the Path Integral for the non-Hermitian Hamiltonian.

Here we derive Path Integral for the case of a generalized non-Hermitian Hamiltonian

$$ (H_{\varepsilon}(p,x))_{\varepsilon} = \frac{p^2}{2m} - \frac{im\omega x^2}{2} + (V_{\varepsilon}(x))_{\varepsilon}. \qquad (1.1.1) $$

Here $cl[(V_{\varepsilon}(x))_{\varepsilon}] \in G(\mathbb{R}^n)$.

The quantity we want to calculate is

$$ (K_{\varepsilon}(x_1, t + \Delta t|x_0, t))_{\varepsilon} = (\langle x_1, t + \Delta t|x_0, t \rangle)_{\varepsilon} = \\ \left( \left\langle x_1 \middle| \exp\left(-\frac{iH_{\varepsilon}\Delta t}{\varepsilon}\right) \middle| x_0, t \right\rangle \right)_{\varepsilon} = \left( \int dp \langle x_1|p \rangle \langle p|\exp\left(-\frac{iH_{\varepsilon}\Delta t}{\varepsilon}\right) \middle| x_0 \rangle \right)_{\varepsilon} \qquad (1.1.2) $$

the last factor in Eq.(1.1.2) can be estimated as

$$ \left(\left\langle p \middle| \exp\left(-\frac{iH_{\varepsilon}\Delta t}{\varepsilon}\right) \middle| x_0 \right\rangle\right)_{\varepsilon} = \left(\left\langle p \middle| 1 - \frac{iH_{\varepsilon}\Delta t}{\varepsilon} + O(\Delta t)^2 \middle| x_0 \right\rangle\right)_{\varepsilon} = \\ \left(1 - \frac{i}{\varepsilon} \frac{p^2 \Delta t}{2m} - \frac{1}{\varepsilon} \frac{x_0^2 m \omega \Delta t}{2} - \frac{i}{\varepsilon} (V_{\varepsilon}(x_0, t))_{\varepsilon} \Delta t + O(\Delta t)^2\right) \frac{\exp(-ipx_0/\varepsilon)}{\sqrt{2\pi\varepsilon}} = \qquad (1.1.3) \\ \frac{1}{\sqrt{2\pi\varepsilon}} \exp\left[-\frac{i}{\varepsilon} \left(p x_0 + \frac{p^2}{2m} \Delta t - \frac{x_0^2 m \omega}{2} \Delta t + (V_{\varepsilon}(x_0, t))_{\varepsilon} \Delta t + O(\Delta t)^2\right)\right]. $$

Substitution Eq.(1.1.3) into Eq.(1.1.2) gives
# Appendix I.1. Derivation of the Path Integral for the non-Hermitian Hamiltonian.

Here we derive Path Integral for the case of a generalized non-Hermitian Hamiltonian

$$ (H_{\varepsilon}(p,x))_{\varepsilon} = \frac{p^2}{2m} - \frac{im\omega x^2}{2} + (V_{\varepsilon}(x))_{\varepsilon}. \qquad (1.1.1) $$

Here $cl[(V_{\varepsilon}(x))_{\varepsilon}] \in G(\mathbb{R}^n)$.

The quantity we want to calculate is

$$ \begin{gathered} (K_{\varepsilon}(x_1, t + \Delta t|x_0, t))_{\varepsilon} = (\langle x_1, t + \Delta t|x_0, t \rangle)_{\varepsilon} = \\ \left(\langle x_1 | \exp(-\frac{iH_{\varepsilon}\Delta t}{\varepsilon}) | x_0, t \rangle\right)_{\varepsilon} = \left(\int dp \langle x_1 | p \rangle \langle p | \exp(-\frac{iH_{\varepsilon}\Delta t}{\varepsilon}) | x_0 \rangle\right)_{\varepsilon} \end{gathered} \qquad (1.1.2) $$

the last factor in Eq.(1.1.2) can be estimated as

$$ \begin{align} \left(\langle p | \exp(-\frac{iH_{\varepsilon}\Delta t}{\varepsilon}) | x_0 \rangle\right)_{\varepsilon} &= \left(\langle p | 1 - \frac{iH_{\varepsilon}\Delta t}{\varepsilon} + O(\Delta t)^2 | x_0 \rangle\right)_{\varepsilon} \nonumber \\ & \quad \left(1 - \frac{i}{\varepsilon} \frac{p^2 \Delta t}{2m} - \frac{1}{\varepsilon} \frac{x_0^2 m \omega \Delta t}{2} - \frac{i}{\varepsilon} (V_{\varepsilon}(x_0, t))_{\varepsilon} \Delta t + O(\Delta t)^2\right) \frac{\exp(-ipx_0/\varepsilon)}{\sqrt{2\pi\varepsilon}} \tag{1.1.3} \\ & \quad \frac{1}{\sqrt{2\pi\varepsilon}} \exp\left[-\frac{i}{\varepsilon} \left(p x_0 + \frac{p^2}{2m}\Delta t - \frac{x_0^2 m \omega}{2}\Delta t + (V_{\varepsilon}(x_0, t))_{\varepsilon}\Delta t + O(\Delta t)^2\right)\right]. \nonumber \end{align} $$

Substitution Eq.(1.1.3) into Eq.(1.1.2) gives
$$
\begin{align}
\langle x_1, t + \Delta t | x_0, t \rangle &= \nonumber \\
&\quad \frac{1}{\sqrt{2\pi\epsilon}} \left( \int dp \exp\left(-\frac{ipx_1}{\epsilon}\right) \times \right. \nonumber \\
&\qquad \left. \exp\left[ -\frac{i}{\epsilon} \left( px_0 + \frac{p^2}{2m} \Delta t - \frac{i\omega x_0^2}{2} \Delta t + (V_\epsilon(x_0, t))_\epsilon \Delta t + O(\Delta t)^2 \right) \right] \right)_\epsilon = \tag{1.1.4} \\
&\quad \sqrt{\frac{m}{2\pi i \epsilon \Delta t}} \exp\left[ \frac{i}{\epsilon} \left( \frac{m}{2} \frac{(x_1 - x_0)^2}{\Delta t} + \frac{im\omega x_0^2}{2} \Delta t - (V_\epsilon(x_0, t))_\epsilon \Delta t + O(\Delta t)^2 \right) \right] \nonumber
\end{align}
$$

By formally identifying $\dot{x}^2(t_1) = (x_1 - x_0)^2/(\Delta t)^2$ from Eq.(1.1.4) we obtain

$$
\langle x_1, t + \Delta t | x_0, t \rangle = \sqrt{\frac{m}{2\pi i \epsilon \Delta t}} \left( \exp\left[ \frac{i}{\epsilon} \mathcal{L}_{\epsilon}(\dot{x}(t), x(t), t) \Delta t \right] \right)_{\epsilon}, \quad (1.1.5)
$$

Here

$$
\mathcal{L}_{\epsilon}(\dot{x}(t), x(t), t) = \frac{m}{2} \dot{x}^2(t) + \frac{im\omega x_0^2}{2} - (V_{\epsilon}(x, t))_{\epsilon} \quad (1.1.6)
$$

is the complex classical Lagrangian. Using the completeness relation $N$ times one
obtain

$$
\begin{equation}
\begin{split}
\langle (x_f, t_f | x_i, t_i \rangle)_\epsilon = \int & \left( \left\langle (x_f, t_f | x_{N-1}, t_{N-1}) \right\rangle_\epsilon dx \left( \left\langle (x_{N-1}, t_{N-1} | x_{N-2}, t_{N-2}) \right\rangle_\epsilon \right) dx_{N-2} \dots \\
& \dots dx_2 \left( \left\langle (x_2, t_2 | x_1, t_1) \right\rangle_\epsilon \right) dx_1 \left( \left\langle (x_1, t_1 | x_i, t_i) \right\rangle_\epsilon \right).
\end{split}
\tag{1.1.7}
\end{equation}
$$
$$
\begin{align}
\langle x_1, t + \Delta t | x_0, t \rangle &= \nonumber \\
&\quad \frac{1}{\sqrt{2\pi\epsilon}} \left( \int dp \exp\left(-\frac{ipx_1}{\epsilon}\right) \times \right. \nonumber \\
&\qquad \left. \exp\left[ -\frac{i}{\epsilon} \left( px_0 + \frac{p^2}{2m} \Delta t - \frac{i\omega x_0^2}{2} \Delta t + (V_\epsilon(x_0, t))_\epsilon \Delta t + O(\Delta t)^2 \right) \right] \right)_\epsilon = \tag{1.1.4} \\
&\quad \sqrt{\frac{m}{2\pi i \epsilon \Delta t}} \exp\left[ \frac{i}{\epsilon} \left( \frac{m}{2} \frac{(x_1 - x_0)^2}{\Delta t} + \frac{im\omega x_0^2}{2} \Delta t - (V_\epsilon(x_0, t))_\epsilon \Delta t + O(\Delta t)^2 \right) \right] \nonumber
\end{align}
$$

By formally identifying $\dot{x}^2(t_1) = (x_1 - x_0)^2/(\Delta t)^2$ from Eq.(1.1.4) we obtain

$$
\langle x_1, t + \Delta t | x_0, t \rangle = \sqrt{\frac{m}{2\pi i \epsilon \Delta t}} \left( \exp\left[ \frac{i}{\epsilon} \mathcal{L}_{\epsilon}(\dot{x}(t), x(t), t) \Delta t \right] \right)_{\epsilon}, \quad (1.1.5)
$$

Here

$$
\mathcal{L}_{\epsilon}(\dot{x}(t), x(t), t) = \frac{m}{2} \dot{x}^2(t) + \frac{im\omega x_0^2}{2} - (V_{\epsilon}(x, t))_{\epsilon} \quad (1.1.6)
$$

is the complex classical Lagrangian. Using the completeness relation $N$ times one
obtain

$$
\begin{equation}
\begin{split}
\langle (x_f, t_f | x_i, t_i \rangle)_\epsilon = \int & \left( \left\langle (x_f, t_f | x_{N-1}, t_{N-1}) \right\rangle_\epsilon dx \left( \left\langle (x_{N-1}, t_{N-1} | x_{N-2}, t_{N-2}) \right\rangle_\epsilon \right) dx_{N-2} \dots \\
& \dots dx_2 \left( \left\langle (x_2, t_2 | x_1, t_1) \right\rangle_\epsilon \right) dx_1 \left( \left\langle (x_1, t_1 | x_i, t_i) \right\rangle_\epsilon \right).
\end{split}
\tag{1.1.7}
\end{equation}
$$
The time interval for each factor is $(\Delta t_{\epsilon})_{\epsilon} = (t_f - t_i)/(N_{\epsilon})_{\epsilon}$. Substitution Eq.(1.1.5) into Eq.(1.1.6) by taking the limit $N_{\epsilon} \to \infty$, $\epsilon \in (0,1]$ gives:

$$ \left\langle \langle x_f, t_f | x_i, t_i \rangle \right\rangle_{\epsilon} = \left( \lim_{N_{\epsilon} \to \infty} \int \prod_{i=1}^{N_{\epsilon}-1} dx_i \exp \left[ \frac{i}{\epsilon} \Delta t_{\epsilon} \sum_{j=0}^{N_{\epsilon}-1} \mathcal{L}_{\epsilon}(\dot{x}(t_j), x(t_j), t_j) \right] \right)_{\epsilon} . \quad (1.1.8) $$

Here $t_0 = t_i$. The exponent becomes to Colombeau time-integral of the generalized Lagrangian, namely the generalized action $(S_{\epsilon}[x(t), t_i, t_f])_{\epsilon}$ for each path. A change of variable: $x(t) \to x(t) + \delta x(t), \delta x(t_f) = \delta x(t_i) = 0$ does not change the result of the Feynman-Colombeau path integral, and we find

$$ \left( \begin{array}{c} \displaystyle \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp \left\{ \frac{i}{\hbar} S_{\epsilon}[x(t) + \delta x(t), t_i, t_f] \right\} \\ \\ \displaystyle \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp \left\{ \frac{i}{\hbar} (S_{\epsilon}[x(t) + \delta x(t), t_i, t_f])_{\epsilon} \right\} = \\ \\ \displaystyle \left( \begin{array}{c} \displaystyle \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp \left\{ \frac{i}{\hbar} S_{\epsilon}[x(t), t_i, t_f] \right\} \\ \\ \displaystyle \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp \left\{ \frac{i}{\hbar} (S_{\epsilon}[x(t), t_i, t_f])_{\epsilon} \right\}. \end{array} \right) \end{array} \right. \quad (1.1.9) $$

Therefore
The time interval for each factor is $(\Delta t_{\epsilon})_{\epsilon} = (t_f - t_i)/(N_{\epsilon})_{\epsilon}$. Substitution Eq.(1.1.5) into Eq.(1.1.6) by taking the limit $N_{\epsilon} \to \infty$, $\epsilon \in (0,1]$ gives:

$$ \left\langle \langle x_f, t_f | x_i, t_i \rangle \right\rangle_{\epsilon} = \left( \lim_{N_{\epsilon} \to \infty} \int \prod_{i=1}^{N_{\epsilon}-1} dx_i \exp \left[ \frac{i}{\epsilon} \Delta t_{\epsilon} \sum_{j=0}^{N_{\epsilon}-1} \mathcal{L}_{\epsilon}(\dot{x}(t_j), x(t_j), t_j) \right] \right)_{\epsilon} . \quad (1.1.8) $$

Here $t_0 = t_i$. The exponent becomes to Colombeau time-integral of the generalized Lagrangian, namely the generalized action $(S_{\epsilon}[x(t), t_i, t_f])_{\epsilon}$ for each path. A change of variable: $x(t) \to x(t) + \delta x(t), \delta x(t_f) = \delta x(t_i) = 0$ does not change the result of the Feynman-Colombeau path integral, and we find

$$ \left( \begin{array}{l} \displaystyle \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp \left\{ \frac{i}{\hbar} S_{\epsilon}[x(t) + \delta x(t), t_i, t_f] \right\} \\ \\ \displaystyle \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp \left\{ \frac{i}{\hbar} (S_{\epsilon}[x(t) + \delta x(t), t_i, t_f])_{\epsilon} \right\} = \\ \\ \displaystyle \left( \begin{array}{l} \displaystyle \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp \left\{ \frac{i}{\hbar} S_{\epsilon}[x(t), t_i, t_f] \right\} \\ \\ \displaystyle \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp \left\{ \frac{i}{\hbar} (S_{\epsilon}[x(t), t_i, t_f])_{\epsilon} \right\}. \end{array} \right. \end{array} \right) \quad (1.1.9) $$

Therefore
$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t) + \delta x(t), t_i, t_f]\right)_{\varepsilon}\right\} - $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t), t_i, t_f]\right)_{\varepsilon}\right\} = \qquad (1.1.10) $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \frac{i}{\hbar} \left( \left( \delta S_{\varepsilon}[x(t)] \right)_{\varepsilon} \right) \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t)]\right)_{\varepsilon}\right\} = 0. $$

From Eq.(1.1.10) we obtain

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \frac{i}{\hbar} \left( \left( \delta S_{\varepsilon}[x(t), t_i, t_f] \right)_{\varepsilon} \right) \times $$

$$ \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t), t_i, t_f]\right)_{\varepsilon}\right\} = \qquad (1.1.11) $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \frac{i}{\hbar} \left[ \int_{t_i}^{t_f} \left( \frac{\partial \mathcal{L}_{\varepsilon}}{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \right)_{\varepsilon} \delta x(t) dt \right] \times $$

$$ \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t), t_i, t_f]\right)_{\varepsilon}\right\} = 0. $$

Because $\delta x(t)$ is an arbitrary change of variable, the expression must be zero at all $t$ independently:
$$ \int_{\begin{subarray}{l} x(t_f)=x_f \\ x(t_i)=x_i \end{subarray}} D[x(t)] \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t)+\delta x(t), t_i, t_f]\right)_{\varepsilon}\right\} - $$

$$ \int_{\begin{subarray}{l} x(t_f)=x_f \\ x(t_i)=x_i \end{subarray}} D[x(t)] \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t), t_i, t_f]\right)_{\varepsilon}\right\} = \qquad (1.1.10) $$

$$ \int_{\begin{subarray}{l} x(t_f)=x_f \\ x(t_i)=x_i \end{subarray}} D[x(t)] \frac{i}{\hbar} \left( \left( \delta S_{\varepsilon}[x(t)] \right)_{\varepsilon} \right) \exp\left\{ \frac{i}{\hbar} \left( S_{\varepsilon}[x(t)] \right)_{\varepsilon} \right\} = 0. $$

From Eq.(1.1.10) we obtain

$$ \int_{\begin{subarray}{l} x(t_f)=x_f \\ x(t_i)=x_i \end{subarray}} D[x(t)] \frac{i}{\hbar} \left( \left( \delta S_{\varepsilon}[x(t), t_i, t_f] \right)_{\varepsilon} \right) \times $$

$$ \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t), t_i, t_f]\right)_{\varepsilon}\right\} = \qquad (1.1.11) $$

$$ \int_{\begin{subarray}{l} x(t_f)=x_f \\ x(t_i)=x_i \end{subarray}} D[x(t)] \frac{i}{\hbar} \left[ \int_{t_i}^{t_f} \left( \frac{\partial \mathcal{L}_{\varepsilon}}{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \right)_{\varepsilon} \delta x(t) dt \right] \times $$

$$ \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t), t_i, t_f]\right)_{\varepsilon}\right\} = 0. $$

Because $\delta x(t)$ is an arbitrary change of variable, the expression must be zero at all $t$ independently:
$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left[ \left( \frac{\partial \mathcal{L}_\varepsilon}{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L}_\varepsilon}{\partial \dot{x}} \right)_\varepsilon \right] \exp \left\{ \frac{i}{\hbar} (S_\varepsilon[x(t), t_i, t_f])_\varepsilon \right\} = $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( \frac{\partial \mathcal{L}_\varepsilon}{\partial x} \right)_\varepsilon \exp \left\{ \frac{i}{\hbar} (S_\varepsilon[x(t), t_i, t_f])_\varepsilon \right\} - $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( \left( \frac{d}{dt} \frac{\partial \mathcal{L}_\varepsilon}{\partial \dot{x}} \right)_\varepsilon \right) \exp \left\{ \frac{i}{\hbar} (S_\varepsilon[x(t), t_i, t_f])_\varepsilon \right\} = \qquad (1.1.12) $$

$$ \left( \left\langle \frac{\partial \mathcal{L}_\varepsilon}{\partial x} \right\rangle \right)_\varepsilon - \left( \left\langle \frac{d}{dt} \frac{\partial \mathcal{L}_\varepsilon}{\partial \dot{x}} \right\rangle \right)_\varepsilon = $$

$$ \left( \left\langle \frac{\partial \mathcal{L}_\varepsilon}{\partial x} \right\rangle \right)_\varepsilon - \frac{d}{dt} \left( \left\langle \frac{\partial \mathcal{L}_\varepsilon}{\partial \dot{x}} \right\rangle \right)_\varepsilon = 0. $$

Therefore, the Euler-Lagrange equation must hold as a generalized expectation value, nothing but the Ehrenfest's theorem.

## Appendix I.2. Colombeau-Schrödinger Equation from Feynman-Colombeau Path Integral.

Here we derive Colombeau-Schrödinger equation from the Feynman-Colombeau path integral. Let us first see that the momentum is given by a derivative. Starting from the Feynman-Colombeau path integral
$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left[ \left( \frac{\partial \mathcal{L}_\varepsilon}{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L}_\varepsilon}{\partial \dot{x}} \right)_\varepsilon \right] \exp \left\{ \frac{i}{\hbar} \langle S_\varepsilon[x(t), t_i, t_f] \rangle_\varepsilon \right\} = $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( \frac{\partial \mathcal{L}_\varepsilon}{\partial x} \right)_\varepsilon \exp \left\{ \frac{i}{\hbar} \langle S_\varepsilon[x(t), t_i, t_f] \rangle_\varepsilon \right\} - $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( \left( \frac{d}{dt} \frac{\partial \mathcal{L}_\varepsilon}{\partial \dot{x}} \right)_\varepsilon \right) \exp \left\{ \frac{i}{\hbar} \langle S_\varepsilon[x(t), t_i, t_f] \rangle_\varepsilon \right\} = \qquad (1.1.12) $$

$$ \left( \left\langle \frac{\partial \mathcal{L}_\varepsilon}{\partial x} \right\rangle \right)_\varepsilon - \left( \left\langle \frac{d}{dt} \frac{\partial \mathcal{L}_\varepsilon}{\partial \dot{x}} \right\rangle \right)_\varepsilon = $$

$$ \left( \left\langle \frac{\partial \mathcal{L}_\varepsilon}{\partial x} \right\rangle \right)_\varepsilon - \frac{d}{dt} \left( \left\langle \frac{\partial \mathcal{L}_\varepsilon}{\partial \dot{x}} \right\rangle \right)_\varepsilon = 0. $$

Therefore, the Euler-Lagrange equation must hold as a generalized expectation value, nothing but the Ehrenfest's theorem.

## Appendix I.2. Colombeau-Schrödinger Equation from Feynman-Colombeau Path Integral.

Here we derive Colombeau-Schrödinger equation from the Feynman-Colombeau path integral. Let us first see that the momentum is given by a derivative. Starting from the Feynman-Colombeau path integral
$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} S_{\varepsilon}[x(t), t_i, t_f]\right\} \right)_{\varepsilon} = \qquad (1.2.1) $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} (S_{\varepsilon}[x(t), t_i, t_f])_{\varepsilon}\right\}. $$

we shift the trajectory $x(t)$ by a small amount $x(t) + \delta x(t)$ with the boundary condition that $x_i$ is held fixed ($\delta x(t_i) = 0$) while $x_f$ is varied ($x(t_f) \neq 0$). Under this variation, the propagator changes by

$$ \left\langle \left(x_f + \delta x(t_f), t_f | x_i, t_i \right) \right\rangle_{\varepsilon} - \left\langle \left(x_f, t_f | x_i, t_i \right) \right\rangle_{\varepsilon} = \left( \left( \frac{\partial}{\partial x_f} \langle x_f, t_f | x_i, t_i \rangle_{\varepsilon} \right) \right)_{\varepsilon} (\delta x(t_f)) \quad (1.2.2) $$

On the other hand, the Feynman-Colombeau path integral changes by

$$ \int_{\substack{x(t_f)=x_f+\delta x(t_f) \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t) + \delta x(t), t_i, t_f + \delta x(t_f)]\right)_{\varepsilon}\right\} - \\ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t), t_i, t_f]\right)_{\varepsilon}\right\} = \qquad (1.2.3) $$

$$ \frac{i}{\hbar} \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( \left( \delta S_{\varepsilon}[x(t), t_i, t_f] \right)_{\varepsilon} \right) \exp \left\{ \frac{i}{\hbar} \left( S_{\varepsilon}[x(t), t_i, t_f] \right)_{\varepsilon} \right\} $$
$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} S_{\varepsilon}[x(t), t_i, t_f]\right\} \right)_{\varepsilon} = \qquad (1.2.1) $$

$$ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} (S_{\varepsilon}[x(t), t_i, t_f])_{\varepsilon}\right\}. $$

we shift the trajectory $x(t)$ by a small amount $x(t) + \delta x(t)$ with the boundary condition that $x_i$ is held fixed ($\delta x(t_i) = 0$) while $x_f$ is varied ($x(t_f) \neq 0$). Under this variation, the propagator changes by

$$ \left\langle \left(x_f + \delta x(t_f), t_f | x_i, t_i \right) \right\rangle_{\varepsilon} - \left\langle \left(x_f, t_f | x_i, t_i \right) \right\rangle_{\varepsilon} = \left( \left( \frac{\partial}{\partial x_f} \left\langle x_f, t_f | x_i, t_i \right\rangle_{\varepsilon} \right) \right)_{\varepsilon} (\delta x(t_f)) \quad (1.2.2) $$

On the other hand, the Feynman-Colombeau path integral changes by

$$ \int_{\substack{x(t_f)=x_f+\delta x(t_f) \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t) + \delta x(t), t_i, t_f + \delta x(t_f)]\right)_{\varepsilon}\right\} - \\ \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \exp\left\{\frac{i}{\hbar} \left(S_{\varepsilon}[x(t), t_i, t_f]\right)_{\varepsilon}\right\} = \qquad (1.2.3) $$

$$ \frac{i}{\hbar} \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( \left( \delta S_{\varepsilon}[x(t), t_i, t_f] \right)_{\varepsilon} \right) \exp \left\{ \frac{i}{\hbar} \left( S_{\varepsilon}[x(t), t_i, t_f] \right)_{\varepsilon} \right\} $$
and the action changes by

$$ (\delta S_{\varepsilon}[x(t), t_i, t_f])_{\varepsilon} = (S_{\varepsilon}[x(t) + \delta x(t), t_i, t_f])_{\varepsilon} - (S_{\varepsilon}[x(t), t_i, t_f])_{\varepsilon} = \\ \int_{t_i}^{t_f} dt \left( \frac{\partial \mathcal{L}_{\varepsilon}}{\partial x} \delta x(t) + \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \delta \dot{x}(t) \right)_{\varepsilon} = \qquad (1.2.4) $$

$$ \left( \left. \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \delta x(t) \right|_{t_i}^{t_f} + \int_{t_i}^{t_f} \left( \left( \left. \frac{\partial \mathcal{L}_{\varepsilon}}{\partial x} - \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \right)_{\varepsilon} \right) \delta x(t) dt. $$

The last terms vanishes because of the equation of motion (which holds as an expectation value, as we saw in the previous **Appendix I.1.1**), and we are left with

$$ (\delta S_{\varepsilon}[x(t), t_i, t_f])_{\varepsilon} = \left( \left( \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \right)_{\varepsilon} \right) \delta x(t_f) = \left( (p_{\varepsilon}(t_f))_{\varepsilon} \right) \delta x(t_f). \qquad (1.2.5) $$

By putting them together, and dropping $\delta x(t_f)$, we obtain

$$ \left( \left. \frac{\partial}{\partial x_f} \langle x_f, t_f | x_i, t_i \rangle_{\varepsilon} \right|_{\varepsilon} = \frac{i}{\hbar} \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( (p_{\varepsilon}(t_f))_{\varepsilon} \right) \exp \left\{ \frac{i}{\hbar} \left( S_{\varepsilon}[x(t), t_i, t_f] \right)_{\varepsilon} \right\} \qquad (1.2.6) $$

This is precisely how the momentum operator is represented in the position space. Now the Colombeau-Schrödinger equation can be derived by taking a variation with respect to $t_f$. Note that

$$ \left( \frac{\partial S_{\varepsilon}}{\partial t_f} \right)_{\varepsilon} = -(H_{\varepsilon}(t_f))_{\varepsilon} \qquad (1.2.7) $$

by using the equation of motion. Therefore,
and the action changes by

$$ (\delta S_{\varepsilon}[x(t), t_i, t_f])_{\varepsilon} = (S_{\varepsilon}[x(t) + \delta x(t), t_i, t_f])_{\varepsilon} - (S_{\varepsilon}[x(t), t_i, t_f])_{\varepsilon} = \\ \int_{t_i}^{t_f} dt \left( \frac{\partial \mathcal{L}_{\varepsilon}}{\partial x} \delta x(t) + \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \delta \dot{x}(t) \right)_{\varepsilon} = \qquad (1.2.4) $$

$$ \left( \left. \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \delta x(t) \right|_{t_i}^{t_f} + \int_{t_i}^{t_f} \left( \left( \left. \frac{\partial \mathcal{L}_{\varepsilon}}{\partial x} - \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \right)_{\varepsilon} \right) \delta x(t) dt. $$

The last terms vanishes because of the equation of motion (which holds as an expectation value, as we saw in the previous **Appendix I.1.1**), and we are left with

$$ (\delta S_{\varepsilon}[x(t), t_i, t_f])_{\varepsilon} = \left( \left( \frac{\partial \mathcal{L}_{\varepsilon}}{\partial \dot{x}} \right)_{\varepsilon} \right) \delta x(t_f) = \left( (p_{\varepsilon}(t_f))_{\varepsilon} \right) \delta x(t_f). \qquad (1.2.5) $$

By putting them together, and dropping $\delta x(t_f)$, we obtain

$$ \left( \left. \frac{\partial}{\partial x_f} \langle x_f, t_f | x_i, t_i \rangle_{\varepsilon} \right|_{\varepsilon} = \frac{i}{\hbar} \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( (p_{\varepsilon}(t_f))_{\varepsilon} \right) \exp \left\{ \frac{i}{\hbar} \left( S_{\varepsilon}[x(t), t_i, t_f] \right)_{\varepsilon} \right\} \qquad (1.2.6) $$

This is precisely how the momentum operator is represented in the position space. Now the Colombeau-Schrödinger equation can be derived by taking a variation with respect to $t_f$. Note that

$$ \left( \frac{\partial S_{\varepsilon}}{\partial t_f} \right)_{\varepsilon} = -(H_{\varepsilon}(t_f))_{\varepsilon} \qquad (1.2.7) $$

by using the equation of motion. Therefore,
$$ \left( \frac{\partial}{\partial t_f} \langle x_f, t_f | x_i, t_i \rangle_{\epsilon} \right)_{\epsilon} = - \frac{i}{\hbar} \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( (H_{\epsilon}(t_f))_{\epsilon} \right) \exp \left\{ \frac{i}{\hbar} (S_{\epsilon}[x(t), t_i, t_f])_{\epsilon} \right\} \quad (1.2.8) $$

If

$$ (H_{\epsilon})_{\epsilon} = \frac{p^2}{2m} + (V_{\epsilon}(x,t))_{\epsilon} \quad (1.2.9) $$

the momentum can be rewritten using Eq. (1.2.6), and we recover the Colombeau-Schrödinger equation:

$$ i\hbar \frac{\partial}{\partial t} (\langle x, t | x_i, t_i \rangle_{\epsilon}) = \left[ \frac{1}{2m} \left( \frac{\hbar}{i} \frac{\partial}{\partial x} \right)^2 + (V_{\epsilon}(x,t))_{\epsilon} \right] (\langle x, t | x_i, t_i \rangle_{\epsilon}). \quad (1.2.10) $$

## Appendix I.3. Colombeau-Feynman oscillatory path integral.

Let us calculate a simple Colombeau-Feynman oscillatory path integral:
$$ \left( \frac{\partial}{\partial t_f} \langle x_f, t_f | x_i, t_i \rangle_{\epsilon} \right)_{\epsilon} = - \frac{i}{\hbar} \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t)] \left( (H_{\epsilon}(t_f))_{\epsilon} \right) \exp \left\{ \frac{i}{\hbar} \left( S_{\epsilon}[x(t), t_i, t_f] \right)_{\epsilon} \right\} \quad (1.2.8) $$

If

$$ (H_{\epsilon})_{\epsilon} = \frac{p^2}{2m} + (V_{\epsilon}(x,t))_{\epsilon} \quad (1.2.9) $$

the momentum can be rewritten using Eq. (1.2.6), and we recover the Colombeau-Schrödinger equation:

$$ i\hbar \frac{\partial}{\partial t} (\langle x, t | x_i, t_i \rangle_{\epsilon}) = \left[ \frac{1}{2m} \left( \frac{\hbar}{i} \frac{\partial}{\partial x} \right)^2 + (V_{\epsilon}(x,t))_{\epsilon} \right] (\langle x, t | x_i, t_i \rangle_{\epsilon}). \quad (1.2.10) $$

## Appendix I.3. Colombeau-Feynman oscillatory path integral.

Let us calculate a simple Colombeau-Feynman oscillatory path integral:
$$ (\langle x_f, t_f | x_i, t_i \rangle)_\epsilon \triangleq \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \epsilon] \exp \left[ \frac{i}{\hbar_\epsilon} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \left[ \frac{d\tilde{\mathbf{R}},\epsilon}{d\tilde{\mathbf{R}},\epsilon} t \right] x(t) \right]^2 dt \right) \Bigg|_{\epsilon \in (0,1]} = $$

$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \epsilon] \exp \left[ \frac{i}{\hbar_\epsilon} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \left[ \dot{x}_{\tilde{\mathbf{R}},\epsilon}(t) \right]^2 dt \right] \right) \Bigg|_{\epsilon \in (0,1]} = $$

$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \epsilon] \exp \left[ \frac{i}{\hbar_\epsilon} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \dot{x}_\epsilon^2(t) dt \right] \right) \Bigg|_{\epsilon \in (0,1]} = \qquad (1.3.1) $$

$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \epsilon] \exp \left[ \frac{i}{\hbar_\epsilon} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \dot{x}^2(t) dt \right] \right)_{\epsilon \in (0,1]} , $$

$$ \mathbf{m} = m_1 + im_2, m_2 \geq 0, $$

$$ x(t) \in \mathbf{C}[t_i, t_f], $$

$$ (x_{\epsilon}(t))_{\epsilon} \in G(\mathbb{R}). $$

Here operator $\frac{d\tilde{\mathbf{R}},\epsilon}{d\tilde{\mathbf{R}},\epsilon}t: \mathbf{C}[t_i, t_f] \mapsto G(\mathbb{R})$ and Colombeau-Feynman measure $(D[x(t), \epsilon])_\epsilon$ will be fixed later.

**Remark.1.3.1.** We let for short abbreviation $\frac{d\tilde{\mathbf{R}},\epsilon}{d\tilde{\mathbf{R}},\epsilon}t x(t) = \dot{x}_{\tilde{\mathbf{R}},\epsilon}(t) = \dot{x}_\epsilon(t)$ or even simply $\dot{x}(t)$.
$$ (\langle x_f, t_f | x_i, t_i \rangle)_\epsilon \triangleq \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \epsilon] \exp \left[ \frac{i}{\hbar_\epsilon} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \left[ \frac{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon}}{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon}t} x(t) \right]^2 dt \right] \right)_{\epsilon \in (0,1]} = $$

$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \epsilon] \exp \left[ \frac{i}{\hbar_\epsilon} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \left[ \dot{x}_{\tilde{\mathbf{r}}_{\mathbb{R},\epsilon}}(t) \right]^2 dt \right] \right)_{\epsilon \in (0,1]} = $$

$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \epsilon] \exp \left[ \frac{i}{\hbar_\epsilon} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \dot{x}_\epsilon^2(t) dt \right] \right)_{\epsilon \in (0,1]} = \qquad (1.3.1) $$

$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \epsilon] \exp \left[ \frac{i}{\hbar_\epsilon} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \dot{x}^2(t) dt \right] \right)_{\epsilon \in (0,1]} , $$

$$ \mathbf{m} = m_1 + im_2, m_2 \geq 0, $$

$$ x(t) \in \mathbf{C}[t_i, t_f], $$

$$ (x_{\epsilon}(t))_{\epsilon} \in G(\mathbb{R}). $$

Here operator $\frac{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon}}{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon}t} : \mathbf{C}[t_i, t_f] \mapsto G(\mathbb{R})$ and Colombeau-Feynman measure $(D[x(t), \epsilon])_\epsilon$ will be fixed later.

**Remark.1.3.1.** We let for short abbreviation $\frac{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon}}{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon}t}x(t) = \dot{x}_{\tilde{\mathbf{r}}_{\mathbb{R},\epsilon}}(t) = \dot{x}_\epsilon(t)$ or even simply $\dot{x}(t)$.
The classical path is

$$x_{\text{cl}}(t) = x_i + \frac{x_f - x_i}{t_f - t_i}(t - t_i). \qquad (1.3.2)$$

Assume that $x(t) \in C[t_i, t_f]$. Hence one can write

$$x(t) = \tilde{x}_{\text{cl}}(t) + (\delta_{\varepsilon}(t))_{\varepsilon}, \varepsilon \in (0,1], \qquad (1.3.3)$$

where the left-over piece must vanish at the initial and the final time. Therefore, we can expand $(\delta_{\varepsilon}(t))_{\varepsilon}$ without loss of generality, in Colombeau-Fourier series such that

$$(\delta_{\varepsilon}(t))_{\varepsilon} = \left( \sum_{n=1}^{m(\varepsilon)} a_{n,\varepsilon} \sin \left( \frac{n\pi(t-t_i)}{t_f-t_i} \right) \right)_{\varepsilon \in (0,1]} \qquad (1.3.4)$$

with arbitraly function $m(\varepsilon)$ such that $m(\varepsilon) \to \infty$ iff $\varepsilon \to 0$. Then the integral over all paths $x(t) \in C[t_i, t_f]$ such that $a_{n,\varepsilon} = 0$ for $n > m(\varepsilon)$, can then be viewed as integrals over all $a_{n,\varepsilon}$, with $n \le m(\varepsilon)$, i.e.

$$\int D[x(t), \varepsilon](\cdot) = \int D[\{a_{n,\varepsilon}\}, \varepsilon](\cdot) = N_{\varepsilon} \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_n(\cdot) \qquad (1.3.5)$$

and therefore the integral over all paths $x(t) \in C[t_i, t_f]$ can be viewed as Colombeau over all $a_{n,\varepsilon} \in (0,1]$, i.e.

$$(\int D[x(t), \varepsilon](\cdot))_{\varepsilon} = (\int D[\{a_n\}, \varepsilon](\cdot))_{\varepsilon} = ((N_{\varepsilon})_{\varepsilon}) \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_n(\cdot) \right)_{\varepsilon} \qquad (1.3.6)$$

The overall normalization factor $(N_{\varepsilon})_{\varepsilon}$ that depends on $m$ and $t_f - t_i$ will be fixed later.

The action, on the other hand, can be calculated. The starting point is
The classical path is

$$x_{cl}(t) = x_i + \frac{x_f - x_i}{t_f - t_i} (t - t_i). \qquad (1.3.2)$$

Assume that $x(t) \in C[t_i, t_f]$. Hence one can write

$$x(t) = \tilde{x}_{cl}(t) + (\delta_{\varepsilon}(t))_{\varepsilon}, \varepsilon \in (0,1], \qquad (1.3.3)$$

where the left-over piece must vanish at the initial and the final time. Therefore, we can expand $(\delta_{\varepsilon}(t))_{\varepsilon}$ without loss of generality, in Colombeau-Fourier series such that

$$(\delta_{\varepsilon}(t))_{\varepsilon} = \left( \sum_{n=1}^{m(\varepsilon)} a_{n,\varepsilon} \sin \left( \frac{n\pi(t-t_i)}{t_f-t_i} \right) \right)_{\varepsilon \in (0,1]} \qquad (1.3.4)$$

with arbitraly function $m(\varepsilon)$ such that $m(\varepsilon) \to \infty$ iff $\varepsilon \to 0$. Then the integral over all paths $x(t) \in C[t_i, t_f]$ such that $a_{n,\varepsilon} = 0$ for $n > m(\varepsilon)$, can then be viewed as integrals over all $a_{n,\varepsilon}$, with $n \le m(\varepsilon)$, i.e.

$$\int D[x(t), \varepsilon](\cdot) = \int D[\{a_{n,\varepsilon}\}, \varepsilon](\cdot) = N_{\varepsilon} \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_n(\cdot) \qquad (1.3.5)$$

and therefore the integral over all paths $x(t) \in C[t_i, t_f]$ can be viewed as Colombeau over all $a_{n,\varepsilon} \in (0,1]$, i.e.

$$(\int D[x(t), \varepsilon](\cdot))_{\varepsilon} = (\int D[\{a_n\}, \varepsilon](\cdot))_{\varepsilon} = ((N_{\varepsilon})_{\varepsilon}) \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_n(\cdot) \right)_{\varepsilon} \qquad (1.3.6)$$

The overall normalization factor $(N_{\varepsilon})_{\varepsilon}$ that depends on $m$ and $t_f - t_i$ will be fixed later.

The action, on the other hand, can be calculated. The starting point is
$$ \dot{x}_{\varepsilon}(t) = \dot{x}_{cl}(t) + \dot{\delta}_{\tilde{\mathbb{R}},\varepsilon}(t) = \frac{x_f - x_i}{t_f - t_i} + \sum_{n=1}^{m(\varepsilon)} a_{n,\varepsilon} \frac{n\pi}{t_f - t_i} \cos\left(\frac{n\pi(t-t_i)}{t_f - t_i}\right). \quad (1.3.7) $$

Because different modes are orthogonal upon t-integral, the action $(S_{\varepsilon})_{\varepsilon}$ in coordinates $\{a_{n,\varepsilon}\}_{n=1}^{m(\varepsilon)}$ is

$$ (S_{\varepsilon})_{\varepsilon} = \frac{\mathbf{m}}{2} \left( \int_{t_i}^{t_f} \dot{x}_{\tilde{\mathbb{R}},\varepsilon}^2 (t) dt \right)_{\varepsilon} = \frac{\mathbf{m}}{2} \frac{(x_f - x_i)^2}{t_f - t_i} + \frac{\mathbf{m}}{2} \left( \sum_{n=1}^{m(\varepsilon)} \frac{1}{2} \frac{n^2 \pi^2}{t_f - t_i} a_{n,\varepsilon}^2 \right)_{\varepsilon}. \quad (1.3.8) $$

The first term is nothing but the classical action and thus the path integral (1.3.1) reduces to Colombeau oscillatory integral:

$$ \left( \int_{x(t_f)=x_f}^{x(t_i)=x_i} D[x(t), \varepsilon] \exp \left[ \frac{i}{\hbar_{\varepsilon}} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \dot{x}_{\tilde{\mathbb{R}},\varepsilon}^2 (t) dt \right] \right)_{\varepsilon} = ((N_{\varepsilon})_{\varepsilon}) \exp \left[ \frac{i}{(\hbar_{\varepsilon})_{\varepsilon}} \frac{\mathbf{m}}{2} \frac{(x_f - x_i)^2}{t_f - t_i} \right] \times \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_{n,\varepsilon} \exp \left[ \left( \frac{i}{\hbar_{\varepsilon}} \frac{\mathbf{m}}{4} \sum_{n=1}^{m(\varepsilon)} \frac{1}{2} \frac{n^2 \pi^2}{t_f - t_i} \right) a_{n,\varepsilon}^2 \right] \right)_{\varepsilon}. \quad (1.3.7) $$

Thus we obtain
$$ \dot{x}_{\varepsilon}(t) = \dot{x}_{cl}(t) + \dot{\delta}_{\tilde{\mathbb{R}},\varepsilon}(t) = \frac{x_f - x_i}{t_f - t_i} + \sum_{n=1}^{m(\varepsilon)} a_{n,\varepsilon} \frac{n\pi}{t_f - t_i} \cos\left(\frac{n\pi(t-t_i)}{t_f - t_i}\right). \quad (1.3.7) $$

Because different modes are orthogonal upon t-integral, the action $(S_\varepsilon)_\varepsilon$ in coordinates $\{a_{n,\varepsilon}\}_{n=1}^{m(\varepsilon)}$ is

$$ (S_{\varepsilon})_{\varepsilon} = \frac{\mathbf{m}}{2} \left( \int_{t_i}^{t_f} \dot{x}_{\tilde{\mathbb{R}},\varepsilon}^2 (t) dt \right)_{\varepsilon} = \frac{\mathbf{m}}{2} \frac{(x_f - x_i)^2}{t_f - t_i} + \frac{\mathbf{m}}{2} \left( \sum_{n=1}^{m(\varepsilon)} \frac{1}{2} \frac{n^2 \pi^2}{t_f - t_i} a_{n,\varepsilon}^2 \right)_{\varepsilon}. \quad (1.3.8) $$

The first term is nothing but the classical action and thus the path integral (1.3.1) reduces to Colombeau oscillatory integral:

$$ \left( \int_{\substack{x(t_f)=x_f \\ x(t_i)=x_i}} D[x(t), \varepsilon] \exp \left[ \frac{i}{\hbar_{\varepsilon}} \int_{t_i}^{t_f} \frac{\mathbf{m}}{2} \dot{x}_{\tilde{\mathbb{R}},\varepsilon}^2 (t) dt \right] \right)_{\varepsilon} = ((N_{\varepsilon})_{\varepsilon}) \exp \left[ \frac{i}{(\hbar_{\varepsilon})_{\varepsilon}} \frac{\mathbf{m}}{2} \frac{(x_f - x_i)^2}{t_f - t_i} \right] \times \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_{n,\varepsilon} \exp \left[ \left( \frac{i}{\hbar_{\varepsilon}} \frac{\mathbf{m}}{4} \sum_{n=1}^{m(\varepsilon)} \frac{1}{2} \frac{n^2 \pi^2}{t_f - t_i} \right) a_{n,\varepsilon}^2 \right] \right)_{\varepsilon}. \quad (1.3.7) $$

Thus we obtain
$$ (\langle x_f, t_f | x_i, t_i \rangle_\epsilon)_\epsilon = ( (N_\epsilon)_\epsilon ) \exp \left[ \frac{i}{(\hbar_\epsilon)_\epsilon} \frac{\mathbf{m}}{2} \frac{(x_f - x_i)^2}{t_f - t_i} \right] \times \\ \left( \prod_{n=1}^{m(\epsilon)} \left( -\frac{i}{\pi \hbar_\epsilon} \frac{\mathbf{m}}{2} \frac{1}{2} \frac{n^2 \pi^2}{t_f - t_i} \right)^{-1/2} \right)_\epsilon \quad (1.3.9) $$

Therefore, the result is simply

$$ (\langle x_f, t_f | x_i, t_i \rangle_\epsilon)_\epsilon = (f_\epsilon(t_f - t_i))_\epsilon \exp \left[ \frac{i}{(\hbar_\epsilon)_\epsilon} \frac{\mathbf{m}}{2} \frac{(x_f - x_i)^2}{t_f - t_i} \right] \quad (1.3.10) $$

The normalization Colombeau constant $(f_\epsilon(t_f - t_i))_\epsilon$ can depend only on the time interval $t_f - t_i$ and is determined by the requirement that

$$ \left( \int dx (\langle x_f, t_f | x, t \rangle_\epsilon) (\langle x, t | x_i, t_i \rangle_\epsilon) \right)_\epsilon = (\langle x_f, t_f | x_i, t_i \rangle_\epsilon)_\epsilon \quad (1.3.11) $$

And hence one obtain

$$ ((f_\epsilon(t_f - t))_\epsilon)((f_\epsilon(t - t_i))_\epsilon) \sqrt{\frac{2\pi i (\hbar_\epsilon)_\epsilon (t_f - t)(t - t_i)}{\mathbf{m}(t_f - t_i)}} = (f_\epsilon(t_f - t_i))_\epsilon. \quad (1.3.12) $$

Therefore we find

$$ (f_\epsilon(t))_\epsilon = \sqrt{\frac{\mathbf{m}}{2\pi i ((\hbar_\epsilon)_\epsilon)t}}. \quad (1.3.13) $$

Finally we obtain
$$ ((x_f, t_f|x_i, t_i)_\epsilon)_\epsilon = ((N_\epsilon)_\epsilon) \exp\left[ \frac{i}{(\hbar_\epsilon)_\epsilon} \frac{\mathbf{m}}{2} \frac{(x_f - x_i)^2}{t_f - t_i} \right] \times \\ \left( \prod_{n=1}^{m(\epsilon)} \left( -\frac{i}{\pi\hbar_\epsilon} \frac{\mathbf{m}}{2} \frac{1}{2} \frac{n^2\pi^2}{t_f - t_i} \right)^{-1/2} \right)_\epsilon \quad (1.3.9) $$

Therefore, the result is simply

$$ ((x_f, t_f|x_i, t_i)_\epsilon) = (f_\epsilon(t_f - t_i))_\epsilon \exp\left[ \frac{i}{(\hbar_\epsilon)_\epsilon} \frac{\mathbf{m}}{2} \frac{(x_f - x_i)^2}{t_f - t_i} \right] \quad (1.3.10) $$

The normalization Colombeau constant $(f_\epsilon(t_f - t_i))_\epsilon$ can depend only on the time interval $t_f - t_i$ and is determined by the requirement that

$$ \left( \int dx \left( \langle x_f, t_f | x, t \rangle_\epsilon \right) \left( \langle x, t | x_i, t_i \rangle_\epsilon \right) \right)_\epsilon = \left( \langle x_f, t_f | x_i, t_i \rangle_\epsilon \right)_\epsilon \quad (1.3.11) $$

And hence one obtain

$$ ((f_\epsilon(t_f - t))_\epsilon)((f_\epsilon(t - t_i))_\epsilon) \sqrt{\frac{2\pi i (\hbar_\epsilon)_\epsilon (t_f - t)(t - t_i)}{\mathbf{m}(t_f - t_i)}} = (f_\epsilon(t_f - t_i))_\epsilon. \quad (1.3.12) $$

Therefore we find

$$ (f_\epsilon(t))_\epsilon = \sqrt{\frac{\mathbf{m}}{2\pi i ((\hbar_\epsilon)_\epsilon)t}}. \quad (1.3.13) $$

Finally we obtain
$$ \langle (x_f, t_f | x_i, t_i \rangle)_\varepsilon \rangle_\varepsilon = \sqrt{\frac{m}{2\pi i ((h_\varepsilon)_\varepsilon)(t_f - t_i)}} \exp \left[ \frac{i}{((h_\varepsilon)_\varepsilon)} \frac{m}{2} \frac{(x_f - x_i)^2}{t_f - t_i} \right]. \quad (1.3.14) $$

In order to obtain this result directly from the Colombeau-Feynman path integral, we should have chosen the normalization of the complex Colombeau-Feynman measure $(D[x(t), \varepsilon])_\varepsilon$ to be

$$ (D[x(t), \varepsilon])_\varepsilon = \sqrt{\frac{m}{2\pi T i ((h_\varepsilon)_\varepsilon}}} \left( \left( \prod_{n=1}^{m(\varepsilon)} \left( \frac{1}{i\pi h_\varepsilon} \frac{m}{2} \frac{n^2 \pi^2}{T} \right)^{1/2} \right) \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_{n,\varepsilon} \right) \right)_\varepsilon, \quad (1.3.15) $$

$$ T = t_f - t_i. $$

**Definition 1.3.1.** (i) If $m_2 = 0$, we define positive Colombeau-Feynman measure $(D^+[x(t), \varepsilon])_\varepsilon$ by formula

$$ (D^+[x(t), \varepsilon])_\varepsilon = \left( \left( \left( \prod_{n=1}^{8m(\varepsilon)} \left( \frac{1}{i\pi h_\varepsilon} \frac{m_1}{2} \frac{n^2 \pi^2}{T} \right)^{1/2} \right) \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{8m(\varepsilon)} da_{n,\varepsilon}(\cdot) \right) \right)_\varepsilon \right) = \quad (1.3.16.a) $$

$$ \left( \left( \prod_{n=1}^{8m(\varepsilon)} \left( \frac{1}{\pi h_\varepsilon} \frac{m_1}{2} \frac{n^2 \pi^2}{T} \right)^{1/2} \right) \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{8m(\varepsilon)} da_{n,\varepsilon}(\cdot) \right) \right)_\varepsilon $$

(ii) If $m_2 > 0$ we define modulus $(|D[x(t), \varepsilon]|)_\varepsilon$ of the complex Colombeau-Feynman measure $(D[x(t), \varepsilon])_\varepsilon$ by formula
$$ (\langle x_f, t_f | x_i, t_i \rangle)_\epsilon = \sqrt{\frac{m}{2\pi i ((h_\epsilon)_\epsilon)(t_f - t_i)}} \exp\left[ \frac{i}{((h_\epsilon)_\epsilon)} \frac{m}{2} \frac{(x_f - x_i)^2}{t_f - t_i} \right]. \quad (1.3.14) $$

In order to obtain this result directly from the Colombeau-Feynman path integral, we should have chosen the normalization of the complex Colombeau-Feynman measure $(D[x(t), \epsilon])_\epsilon$ to be

$$ \begin{gathered} (D[x(t), \epsilon])_\epsilon = \\ \sqrt{\frac{m}{2\pi T i ((h_\epsilon)_\epsilon}}} \left( \left( \prod_{n=1}^{m(\epsilon)} \left( \frac{1}{i\pi h_\epsilon} \frac{m}{2} \frac{n^2 \pi^2}{T} \right)^{1/2} \right) \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\epsilon)} da_{n,\epsilon} \right) \right)_\epsilon, \quad (1.3.15) \\ T = t_f - t_i. \end{gathered} $$

**Definition 1.3.1.** (i) If $m_2 = 0$, we define positive Colombeau-Feynman measure $(D^+[x(t), \epsilon])_\epsilon$ by formula

$$ \begin{aligned} (D^+[x(t), \epsilon])_\epsilon = & \left( \left( \prod_{n=1}^{8m(\epsilon)} \left( \frac{1}{i\pi h_\epsilon} \frac{m_1}{2} \frac{n^2 \pi^2}{T} \right)^{1/2} \right) \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{8m(\epsilon)} da_{n,\epsilon}(\cdot) \right) \right)_\epsilon \\ & \left( \left( \prod_{n=1}^{8m(\epsilon)} \left( \frac{1}{\pi h_\epsilon} \frac{m_1}{2} \frac{n^2 \pi^2}{T} \right)^{1/2} \right) \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{8m(\epsilon)} da_{n,\epsilon}(\cdot) \right) \right)_\epsilon \end{aligned} \quad (1.3.16.a) $$

(ii) If $m_2 > 0$ we define modulus $|(D[x(t), \epsilon])_\epsilon|$ of the complex Colombeau-Feynman measure $(D[x(t), \epsilon])_\epsilon$ by formula
$$ (|D[x(t), \varepsilon]|)_{\varepsilon} = $$

$$ \left( \left( \left| \prod_{n=1}^{m(\varepsilon)} \left( \frac{1}{i\pi\hbar_{\varepsilon}} \frac{\mathbf{m}}{2} \frac{1}{2} \frac{n^2\pi^2}{T} \right)^{1/2} \right| \right)_{\varepsilon} \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_{n,\varepsilon}(\cdot) \right)_{\varepsilon} \right) = \qquad (1.3.16.b) $$

$$ \left( \left( \prod_{n=1}^{m(\varepsilon)} \left( \frac{1}{\pi \hbar_{\varepsilon}} \frac{\sqrt{m_1^2 + m_2^2}}{2} \frac{1}{2} \frac{n^2 \pi^2}{T} \right)^{1/2} \right)_{\varepsilon} \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_{n,\varepsilon}(\cdot) \right)_{\varepsilon} \right) $$

Let us calculate now Colombeau oscillatory path integral
$$ (|D[x(t), \varepsilon]|)_{\varepsilon} = $$

$$ \left( \left( \left| \prod_{n=1}^{m(\varepsilon)} \left( \frac{1}{i\pi\hbar_{\varepsilon}} \frac{\mathbf{m}}{2} \frac{1}{2} \frac{n^2\pi^2}{T} \right)^{1/2} \right| \right)_{\varepsilon} \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_{n,\varepsilon}(\cdot) \right)_{\varepsilon} \right) = \qquad (1.3.16.b) $$

$$ \left( \left( \prod_{n=1}^{m(\varepsilon)} \left( \frac{1}{\pi \hbar_{\varepsilon}} \frac{\sqrt{m_1^2 + m_2^2}}{2} \frac{1}{2} \frac{n^2 \pi^2}{T} \right)^{1/2} \right)_{\varepsilon} \right) \left( \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\varepsilon)} da_{n,\varepsilon}(\cdot) \right)_{\varepsilon} \right) $$

Let us calculate now Colombeau oscillatory path integral
$$ (\langle x_f, t_f | x_i, t_i \rangle)_\epsilon = \left( \int_{\substack{x(T)=x_f \\ x(0)=x_i}} D[x(t), \epsilon] \times \exp\left\{ \frac{i}{\hbar_\epsilon} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon} x(t)}{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon} t} \right)^2_\epsilon - \frac{\omega_1^2}{\hbar_\epsilon} \frac{\mathbf{m}}{2} x^2(t) \right] - \frac{\omega_2^2}{\hbar_\epsilon} \int_0^T dt \left[ \frac{\mathbf{m}}{2} x^2(t) \right] \right) \right\}_\epsilon = \tag{1.3.17} $$

$$ \exp\left\{ \frac{i}{\hbar_{\epsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon} x(t)}{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon} t} \right)^2_{\epsilon} - \frac{\omega^2}{\hbar_{\epsilon}} \frac{\mathbf{m}}{2} x^2(t) \right] \right) \right\}_{\epsilon}, $$ $$ \omega^2 = \omega_1^2 - i\omega_2^2. $$

The classical path is

$$ x_{\text{cl}}(t) = x_i \frac{\sin[\omega(t_f - t)]}{\sin[\omega(t_f - t_i)]} + x_f \frac{\sin[\omega(t - t_i)]}{\sin[\omega(t_f - t_i)]}. \tag{1.3.18} $$

The action $S_{\text{cl}}$ along the classical path is
$$ (\langle x_f, t_f | x_i, t_i \rangle)_\epsilon = \left( \int_{\substack{x(T)=x_f \\ x(0)=x_i}} D[x(t), \epsilon] \times \exp\left\{ \frac{i}{\hbar_\epsilon} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon} x(t)}{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon} t} \right)^2_\epsilon - \frac{\omega_1^2}{\hbar_\epsilon} \frac{\mathbf{m}}{2} x^2(t) \right] - \frac{\omega_2^2}{\hbar_\epsilon} \int_0^T dt \left[ \frac{\mathbf{m}}{2} x^2(t) \right] \right) \right\}_\epsilon = \tag{1.3.17} $$

$$ \exp\left\{ \frac{i}{\hbar_{\epsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon} x(t)}{d\tilde{\mathbf{r}}_{\mathbb{R},\epsilon} t} \right)^2_{\epsilon} - \frac{\omega^2}{\hbar_{\epsilon}} \frac{\mathbf{m}}{2} x^2(t) \right] \right) \right\}_{\epsilon}, $$ $$ \omega^2 = \omega_1^2 - i\omega_2^2. $$

The classical path is

$$ x_{\text{cl}}(t) = x_i \frac{\sin[\omega(t_f - t)]}{\sin[\omega(t_f - t_i)]} + x_f \frac{\sin[\omega(t - t_i)]}{\sin[\omega(t_f - t_i)]}. \tag{1.3.18} $$

The action $S_{\text{cl}}$ along the classical path is
$$
S_{\text{cl}} = \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{dx_{\text{cl}}(t)}{dt} \right)^2 - \omega^2 \frac{\mathbf{m}}{2} x_{\text{cl}}^2(t) \right] = \\
\frac{\mathbf{m}\omega}{2} \frac{(x_i^2 + x_f^2) \cos[\omega(t_f - t_i)] - 2x_i x_f}{\sin[\omega(t_f - t_i)]} \tag{1.3.19}
$$

One can write

$$
x(t) = \tilde{\mathcal{R}} x_{\text{cl}}(t) + (\delta_{\varepsilon}(t))_{\varepsilon}, \quad (1.3.20)
$$

where the left-over piece must vanish at the initial and the final time. Therefore, we
can expand $(\delta_{\epsilon}(t))_{\epsilon}$ without loss of generality, in Fourier-Colombeau series such
that

$$
(\delta_{\epsilon}(t))_{\epsilon} = \left( \sum_{n=1}^{m(\epsilon)} a_{n,\epsilon} \sin\left( \frac{n\pi(t-t_i)}{t_f-t_i} \right) \right)_{\epsilon} \quad (1.3.21)
$$

with arbitraly function $m(\varepsilon)$ such that $m(\varepsilon) \to \infty$ iff $\varepsilon \to 0$. Then the integral over all paths $x(t) \in C[t_i, t_f]$ such that $a_{n,\varepsilon} = 0$ for $n > m(\varepsilon)$, can then be viewed as integrals over all $a_{n,\varepsilon}$, with $n \le m(\varepsilon)$, i.e.

$$
\int D[x(t), \varepsilon](\cdot) = \int D[\{a_{n\varepsilon}\}, \tilde{\varepsilon}](\cdot) = N_{\varepsilon} \int \prod_{n=1}^{\infty} m(\varepsilon) da_{n\varepsilon}(\cdot) \quad (1.3.22)
$$

and therefore the integral over all paths $x(t) \in C[t_i, t_f]$ can be viewed as Colombeau
over all $a_{n,\epsilon}$, i.e.
$$
S_{\text{cl}} = \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{dx_{\text{cl}}(t)}{dt} \right)^2 - \omega^2 \frac{\mathbf{m}}{2} x_{\text{cl}}^2(t) \right] = \\
\frac{\mathbf{m}\omega}{2} \frac{(x_i^2 + x_f^2) \cos[\omega(t_f - t_i)] - 2x_i x_f}{\sin[\omega(t_f - t_i)]} \tag{1.3.19}
$$

One can write

$$
x(t) = \tilde{\mathcal{R}} x_{\text{cl}}(t) + (\delta_{\varepsilon}(t))_{\varepsilon}, \quad (1.3.20)
$$

where the left-over piece must vanish at the initial and the final time. Therefore, we
can expand $(\delta_{\epsilon}(t))_{\epsilon}$ without loss of generality, in Fourier-Colombeau series such
that

$$
(\delta_{\epsilon}(t))_{\epsilon} = \left( \sum_{n=1}^{m(\epsilon)} a_{n,\epsilon} \sin\left( \frac{n\pi(t-t_i)}{t_f-t_i} \right) \right)_{\epsilon} \quad (1.3.21)
$$

with arbitraly function $m(\varepsilon)$ such that $m(\varepsilon) \to \infty$ iff $\varepsilon \to 0$. Then the integral over all paths $x(t) \in C[t_i, t_f]$ such that $a_{n,\varepsilon} = 0$ for $n > m(\varepsilon)$, can then be viewed as integrals over all $a_{n,\varepsilon}$, with $n \le m(\varepsilon)$, i.e.

$$
\int D[x(t), \varepsilon](\cdot) = \int D[\{a_{n\varepsilon}\}, \tilde{\varepsilon}](\cdot) = N_{\varepsilon} \int \prod_{n=1}^{\infty} da_{n\varepsilon}(\cdot) \quad (1.3.22)
$$

and therefore the integral over all paths $x(t) \in C[t_i, t_f]$ can be viewed as Colombeau
over all $a_{n,\epsilon}$, i.e.
$$ \left( \int D[x(t), \epsilon](\cdot) \right)_{\epsilon} = \left( \int D[\{a_{n,\epsilon}\}, \epsilon](\cdot) \right)_{\epsilon} = \left( (N_{\epsilon})_{\epsilon} \right) \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\epsilon)} da_{n,\epsilon}(\cdot) \right)_{\epsilon} \quad (1.3.23) $$

The overall normalization factor $(f_{\epsilon}t_f - t_i)_{\epsilon}$ that depends on $m$ and $t_f - t_i$ will be fixed later. The action, on the other hand, can be calculated. The starting point is

$$ \left( \dot{x}_{\tilde{\mathbb{R}},\epsilon}(t) \right)_{\epsilon} = \dot{x}_{\mathrm{cl}}(t) + \dot{\delta}_{\epsilon}(t) = \dot{x}_{\mathrm{cl}}(t) + \left( \sum_{n=1}^{m(\epsilon)} a_{n,\epsilon} \frac{n\pi}{t_f - t_i} \cos\left(\frac{n\pi(t-t_i)}{t_f - t_i}\right) \right)_{\epsilon}. \quad (1.3.24) $$

Because different modes are orthogonal upon $t$-integral, the action $(S_{\epsilon})_{\epsilon}$ in coordinates $\{a_{n,\epsilon}\}_{n=1}^{m(\epsilon)}$ is

$$ (S_{\epsilon})_{\epsilon} = S_{\mathrm{cl}} + \frac{\mathbf{m}}{4} \left( \sum_{n=1}^{m(\epsilon)} \left[ \frac{n^2 \pi^2}{t_f - t_i} - \omega^2 (t_f - t_i) \right] a_{n,\epsilon}^2 \right)_{\epsilon}. \quad (1.3.25) $$

Therefore the Colombeau-Feynman path integral (1.3.17) is Colombeau-Fresnel integral over $\{a_{n,\epsilon}\}_{n=1}^{\infty}$. Using the Colombeau-Feynman measure given by Eq.(1.3.15)

we obtain

$$ \begin{aligned} \left(\langle x_f, t_f | x_i, t_i \rangle_{\epsilon}\right)_{\epsilon} = & \left[\exp\left(\frac{iS_{\mathrm{cl}}}{(\hbar_{\epsilon})_{\epsilon}}\right)\right] \times \sqrt{\frac{\mathbf{m}}{2\pi T i (\hbar_{\epsilon})_{\epsilon}}} \left(\left(\prod_{n=1}^{m(\epsilon)}\left(\frac{1}{i\pi\hbar_{\epsilon}}\frac{\mathbf{m}}{2}\frac{1}{2}\frac{n^2\pi^2}{T}\right)^{1/2}\right)_{\epsilon}\right) \times \\ & \left(\int_{-\infty}^{\infty} da_{n,\epsilon} \exp\left[\frac{i}{\hbar_{\epsilon}}\frac{\mathbf{m}}{2}\left(\frac{n^2\pi^2}{t_f - t_i} - \omega^2(t_f - t_i)\right)\frac{1}{2}a_{n,\epsilon}^2\right]\right)_{\epsilon}. \end{aligned} \quad (1.3.26) $$
$$ \left( \int D[x(t), \epsilon](\cdot) \right)_{\epsilon} = \left( \int D[\{a_{n,\epsilon}\}, \epsilon](\cdot) \right)_{\epsilon} = \left( (N_{\epsilon})_{\epsilon} \right) \left( \int_{-\infty}^{\infty} \prod_{n=1}^{m(\epsilon)} da_{n,\epsilon}(\cdot) \right)_{\epsilon} \quad (1.3.23) $$

The overall normalization factor $(f_{\epsilon}t_f - t_i)_{\epsilon}$ that depends on $m$ and $t_f - t_i$ will be fixed later. The action, on the other hand, can be calculated. The starting point is

$$ \left( \dot{x}_{\tilde{\mathbb{R}}, \epsilon}(t) \right)_{\epsilon} = \dot{x}_{\text{cl}}(t) + \dot{\delta}_{\epsilon}(t) = \dot{x}_{\text{cl}}(t) + \left( \sum_{n=1}^{m(\epsilon)} a_{n, \epsilon} \frac{n\pi}{t_f - t_i} \cos\left(\frac{n\pi(t - t_i)}{t_f - t_i}\right) \right)_{\epsilon}. \quad (1.3.24) $$

Because different modes are orthogonal upon $t$-integral, the action $(S_{\epsilon})_{\epsilon}$ in coordinates $\{a_{n,\epsilon}\}_{n=1}^{m(\epsilon)}$ is

$$ (S_{\epsilon})_{\epsilon} = S_{\text{cl}} + \frac{\mathbf{m}}{4} \left( \sum_{n=1}^{m(\epsilon)} \left[ \frac{n^2 \pi^2}{t_f - t_i} - \omega^2 (t_f - t_i) \right] a_{n, \epsilon}^2 \right)_{\epsilon}. \quad (1.3.25) $$

Therefore the Colombeau-Feynman path integral (1.3.17) is Colombeau-Fresnel integral over $\{a_{n,\epsilon}\}_{n=1}^{\infty}$. Using the Colombeau-Feynman measure given by Eq.(1.3.15)

we obtain

$$ \begin{aligned} \left( \langle x_f, t_f | x_i, t_i \rangle_{\epsilon} \right)_{\epsilon} = & \left[ \exp\left( \frac{iS_{\text{cl}}}{(\hbar_{\epsilon})_{\epsilon}} \right) \right] \times \sqrt{\frac{\mathbf{m}}{2\pi T i (\hbar_{\epsilon})_{\epsilon}}} \left( \left( \prod_{n=1}^{m(\epsilon)} \left( \frac{1}{i\pi\hbar_{\epsilon}} \frac{\mathbf{m}}{2} \frac{1}{2} \frac{n^2\pi^2}{T} \right)^{1/2} \right)_{\epsilon} \right) \times \\ & \left( \int_{-\infty}^{\infty} da_{n,\epsilon} \exp\left[ \frac{i}{\hbar_{\epsilon}} \frac{\mathbf{m}}{2} \left( \frac{n^2\pi^2}{t_f - t_i} - \omega^2(t_f - t_i) \right) \frac{1}{2} a_{n,\epsilon}^2 \right] \right)_{\epsilon}. \end{aligned} \quad (1.3.26) $$
Now we resort to the following Colombeau representation of the sine function

$$ \left( \prod_{n=1}^{m(\varepsilon)} \left( 1 - \frac{x^2}{n^2} \right) \right)_{\varepsilon} = \tilde{\mathbb{R}} \frac{\sin \pi x}{\pi x}. \qquad (1.3.27) $$

Thus we find

$$ \langle x_f, t_f | x_i, t_i \rangle_{\varepsilon} = \tilde{\mathcal{C}} \left[ \exp \left( \frac{i S_{cl}}{(\hbar_{\varepsilon})_{\varepsilon}} \right) \right] \times \sqrt{\frac{m\omega}{2\pi i (\hbar_{\varepsilon})_{\varepsilon} \sin \omega(t_f - t_i)}}. \qquad (1.3.28) $$

This agrees with the result from the more conventional method as given in literature.

**Definition 1.3.2.** Assume that $(V_{\varepsilon}(x))_{\varepsilon} \in G(\mathbb{R}^n)$. We define Colombeau-Feinman propagator $(\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon})_{\varepsilon \in (0,1]}$ by formula

$$ \langle x_f, t_f | x_i, t_i \rangle_{\varepsilon} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbb{R}}_{\varepsilon} x(t)}{d\tilde{\mathbb{R}}_{\varepsilon} t} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon \in (0,1]} \qquad (1.3.29) $$

And in a more general case we define Colombeau-Feinman propagator $(\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon, \delta})_{\varepsilon, \delta \in (0,1]}$ by formula
Now we resort to the following Colombeau representation of the sine function

$$ \left( \prod_{n=1}^{m(\varepsilon)} \left( 1 - \frac{x^2}{n^2} \right) \right)_{\varepsilon} = \tilde{\mathbb{R}} \frac{\sin \pi x}{\pi x}. \qquad (1.3.27) $$

Thus we find

$$ \langle x_f, t_f | x_i, t_i \rangle_{\varepsilon} = \tilde{\mathcal{C}} \left[ \exp \left( \frac{i S_{cl}}{(\hbar_{\varepsilon})_{\varepsilon}} \right) \right] \times \sqrt{\frac{m\omega}{2\pi i (\hbar_{\varepsilon})_{\varepsilon} \sin \omega(t_f - t_i)}}. \qquad (1.3.28) $$

This agrees with the result from the more conventional method as given in literature.

**Definition 1.3.2.** Assume that $(V_{\varepsilon}(x))_{\varepsilon \in G(\mathbb{R}^n)}$. We define Colombeau-Feinman propagator $(\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon})_{\varepsilon \in (0,1]}$ by formula

$$ \langle x_f, t_f | x_i, t_i \rangle_{\varepsilon} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbb{R}}_{\varepsilon} x(t)}{d\tilde{\mathbb{R}}_{\varepsilon} t} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon \in (0,1]} \qquad (1.3.29) $$

And in a more general case we define Colombeau-Feinman propagator $(\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon, \delta})_{\varepsilon, \delta \in (0,1]}$ by formula
$$ (\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon, \delta})_{\varepsilon, \delta \in (0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon,\delta}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{R}}_{\varepsilon,x}(t)}{d\tilde{\mathbf{R}}_{\varepsilon,t}} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon, \delta \in (0,1]} . \quad (1.3.30) $$

**Remark.1.3.2.** We let for short the abbreviations

$$ (\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon})_{\varepsilon(0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{R}}_{\varepsilon,x}(t)}{d\tilde{\mathbf{R}}_{\varepsilon,t}} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon \in (0,1]} \quad (1.3.31) $$

and

$$ (\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon, \delta})_{\varepsilon, \delta \in (0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{R}}_{\varepsilon,x}(t)}{d\tilde{\mathbf{R}}_{\varepsilon,t}} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon, \delta \in (0,1]} \quad (1.3.32) $$

or even simply
$$ (\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon, \delta})_{\varepsilon, \delta \in (0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon,\delta}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{R}}_{\varepsilon,x}(t)}{d\tilde{\mathbf{R}}_{\varepsilon,t}} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon, \delta \in (0,1]} . \quad (1.3.30) $$

**Remark.1.3.2.** We let for short the abbreviations

$$ (\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon})_{\varepsilon(0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{R}}_{\varepsilon,x}(t)}{d\tilde{\mathbf{R}}_{\varepsilon,t}} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon \in (0,1]} \quad (1.3.31) $$

and

$$ (\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon, \delta})_{\varepsilon, \delta \in (0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{d\tilde{\mathbf{R}}_{\varepsilon,x}(t)}{d\tilde{\mathbf{R}}_{\varepsilon,t}} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon, \delta \in (0,1]} \quad (1.3.32) $$

or even simply
$$ \langle \langle x_f, t_f | x_i, t_i \rangle \rangle_{\varepsilon \in (0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar \varepsilon} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{dx(t)}{dt} \right)^2 - V_{\varepsilon}(x(t)) \right] \right) \right\} \right)_{\varepsilon \in (0,1]} \quad (1.3.33) $$

and

$$ \langle \langle x_f, t_f | x_i, t_i \rangle \rangle_{\varepsilon, \delta \in (0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar \varepsilon} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{dx(t)}{dt} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon, \delta \in (0,1]} . \quad (1.3.34) $$

## Apendix II.1.1.Upper and Lower Infinite-Dimensional Subintegrals.Colombeau-Feynman Measure.

Let $\mathbb{R}^\infty$ be the space
$$ (\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon})_{\varepsilon \in (0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{dx(t)}{dt} \right)^2 - V_{\varepsilon}(x(t)) \right] \right) \right\} \right)_{\varepsilon \in (0,1]} \quad (1.3.33) $$

and

$$ (\langle x_f, t_f | x_i, t_i \rangle_{\varepsilon, \delta})_{\varepsilon, \delta \in (0,1]} = $$

$$ \left( \int_{\substack{q(t_f)=x_f \\ q(t_i)=x_i}} D[x(t), \varepsilon] \times \exp \left\{ \frac{i}{\hbar_{\varepsilon}} \left( \int_{t_i}^{t_f} dt \left[ \frac{\mathbf{m}}{2} \left( \frac{dx(t)}{dt} \right)^2 - V_{\delta}(x(t)) \right] \right) \right\} \right)_{\varepsilon, \delta \in (0,1]} . \quad (1.3.34) $$

## Apendix II.1.1. Upper and Lower Infinite-Dimensional Subintegrals. Colombeau-Feynman Measure.

Let $\mathbb{R}^{\infty}$ be the space
$$ \mathbb{R}^{\infty} = \bigoplus_{n=1}^{\infty} \mathbb{R}^{n} \qquad (1.1.1) $$

**Definition 1.1.1.** For $f \in (\mathbb{R}^{\infty})^*$ we define the infinite dimensional normed by sequence $\{c_n\}$, Feynman integral $\mathfrak{I}_{\infty}(f)$ as limit

$$ \mathfrak{I}_{\infty}(f) = \int_{\mathbb{R}^{\infty}} f(\mathbf{x}) D[\mathbf{x}] = \lim_{n \to \infty} \int_{\mathbb{R}^n} f_n(\mathbf{x}) c_n d^n \mathbf{x} = \qquad (1.1.2) $$

$$ \lim_{n \to \infty} \int \dots \int f_n(x_1, x_2, \dots, x_n) c_n dx_1 dx_2 \dots dx_n $$

iff limit in RHS of Eq.(1.1.2) exists and $f_n(x) = f(\mathbf{x}) \upharpoonright \mathbb{R}^n$.

**Remark 1.1.1.** We note that one can write Definition 1.1.1 using completely Colombeau form.

**Definition 1.1.2.** For $f \in (\mathbb{R}^{\infty})^*$ we define the infinite dimensional integral $\mathfrak{I}_{\infty}(f)$ as Colombeau quantity such that
$$ \mathbb{R}^{\infty} = \bigoplus_{n=1}^{\infty} \mathbb{R}^n \qquad (1.1.1) $$

**Definition 1.1.1.** For $f \in (\mathbb{R}^{\infty})^*$ we define the infinite dimensional normed by sequence $\{c_n\}$, Feynman integral $\mathfrak{I}_{\infty}(f)$ as limit

$$ \mathfrak{I}_{\infty}(f) = \int_{\mathbb{R}^{\infty}} f(\mathbf{x}) D[\mathbf{x}] = \lim_{n \to \infty} \int_{\mathbb{R}^n} f_n(\mathbf{x}) c_n d^n \mathbf{x} = \qquad (1.1.2) $$

$$ \lim_{n \to \infty} \int \dots \int f_n(x_1, x_2, \dots, x_n) c_n dx_1 dx_2 \dots dx_n $$

iff limit in RHS of Eq.(1.1.2) exists and $f_n(x) = f(\mathbf{x}) \upharpoonright \mathbb{R}^n$.

**Remark 1.1.1.** We note that one can write Definition 1.1.1 using completely Colombeau form.

**Definition 1.1.2.** For $f \in (\mathbb{R}^{\infty})^*$ we define the infinite dimensional integral $\mathfrak{I}_{\infty}(f)$ as Colombeau quantity such that
$$ \mathfrak{I}_{\infty}(f) = \int_{\mathbb{R}^{\infty}} f(\mathbf{x})D[\mathbf{x}] = \tilde{\zeta} \int_{\mathbb{R}^{\infty}} f(\mathbf{x})D[\mathbf{x}, \varepsilon] = \tilde{\zeta} $$

$$ \left( \int_{\mathbb{R}^{n_{\varepsilon}}} f_{n_{\varepsilon}}(x) c_{n_{\varepsilon}} d^{n_{\varepsilon}} x \right)_{\varepsilon} = \tilde{\zeta} $$

$$ \left( \int \dots \int f(x_1, x_2, \dots, x_{n_{\varepsilon}}) c_{n_{\varepsilon}} dx_1 dx_2 \dots dx_{n_{\varepsilon}} \right)_{\varepsilon} = \tilde{\zeta} \quad (1.1.3) $$

$$ ((c_{n_{\varepsilon}})_{\varepsilon}) \left( \int_{\mathbb{R}^{n_{\varepsilon}}} f_{n_{\varepsilon}}(x) d^{n_{\varepsilon}} x \right)_{\varepsilon} = \tilde{\zeta} $$

$$ ((c_{n_{\varepsilon}})_{\varepsilon}) \left( \int \dots \int f(x_1, x_2, \dots, x_{n_{\varepsilon}}) dx_1 dx_2 \dots dx_{n_{\varepsilon}} \right)_{\varepsilon}, $$

where $n_{\varepsilon} \to \infty$ if $\varepsilon \to 0$.

**Remark 1.1.2.** We note that Definition 1.1.2 work if we limit in RHS of Eq.(1.1.2) does not exists.

**Definition 1.1.3.** We define Colombeau-Feynman measure $(D[\mathbf{x}, \varepsilon])_{\varepsilon}$ by formula

$$ \begin{gathered} (D[\mathbf{x}, \varepsilon])_{\varepsilon} = (c_{n_{\varepsilon}} dx_1 dx_2 \dots dx_{n_{\varepsilon}})_{\varepsilon} = \\ ((c_{n_{\varepsilon}})_{\varepsilon}) (dx_1 dx_2 \dots dx_{n_{\varepsilon}})_{\varepsilon}. \end{gathered} \quad (1.1.4) $$

**Definition 1.1.4.** Let $\Sigma(\mathbb{R}^{\infty})$ be the measurable space $(\mathbb{R}^{\infty}, \Sigma)$ and Let $\mathbf{M}(\mathbb{R}^{\infty}, \Sigma)$ be algebra with operations: (i) $\forall c \in \mathbb{C}, \forall A \in \Sigma : (c\mu)(A) = c\mu(A)$ (ii) $\forall A \in \Sigma : (\mu + \nu)(A) = \mu(A) + \nu(A)$. Set $E(\mathbf{M}(\mathbb{R}^{\infty}, \Sigma)) = (\mathbf{M}(\mathbb{R}^{\infty}, \Sigma))^I, I = (0, 1]$,
$$ \mathfrak{I}_{\infty}(f) = \int_{\mathbb{R}^{\infty}} f(\mathbf{x})D[\mathbf{x}] = \tilde{\zeta} \int_{\mathbb{R}^{\infty}} f(\mathbf{x})D[\mathbf{x}, \varepsilon] = \tilde{\zeta} $$

$$ \left( \int_{\mathbb{R}^{n_{\varepsilon}}} f_{n_{\varepsilon}}(x) c_{n_{\varepsilon}} d^{n_{\varepsilon}} x \right)_{\varepsilon} = \tilde{\zeta} $$

$$ \left( \int \dots \int f(x_1, x_2, \dots, x_{n_{\varepsilon}}) c_{n_{\varepsilon}} dx_1 dx_2 \dots dx_{n_{\varepsilon}} \right)_{\varepsilon} = \tilde{\zeta} \quad (1.1.3) $$

$$ ((c_{n_{\varepsilon}})_{\varepsilon}) \left( \int_{\mathbb{R}^{n_{\varepsilon}}} f_{n_{\varepsilon}}(x) d^{n_{\varepsilon}} x \right)_{\varepsilon} = \tilde{\zeta} $$

$$ ((c_{n_{\varepsilon}})_{\varepsilon}) \left( \int \dots \int f(x_1, x_2, \dots, x_{n_{\varepsilon}}) dx_1 dx_2 \dots dx_{n_{\varepsilon}} \right)_{\varepsilon}, $$

where $n_{\varepsilon} \to \infty$ if $\varepsilon \to 0$.

**Remark 1.1.2.** We note that Definition 1.1.2 work if we limit in RHS of Eq.(1.1.2) does not exists.

**Definition 1.1.3.** We define Colombeau-Feynman measure $(D[\mathbf{x}, \varepsilon])_{\varepsilon}$ by formula

$$ \begin{gathered} (D[\mathbf{x}, \varepsilon])_{\varepsilon} = (c_{n_{\varepsilon}} dx_1 dx_2 \dots dx_{n_{\varepsilon}})_{\varepsilon} = \\ ((c_{n_{\varepsilon}})_{\varepsilon}) (dx_1 dx_2 \dots dx_{n_{\varepsilon}})_{\varepsilon}. \end{gathered} \quad (1.1.4) $$

**Definition 1.1.4.** Let $\Sigma(\mathbb{R}^{\infty})$ be the measurable space $(\mathbb{R}^{\infty}, \Sigma)$ and Let $\mathbf{M}(\mathbb{R}^{\infty}, \Sigma)$ be algebra with operations: (i) $\forall c \in \mathbb{C}, \forall A \in \Sigma : (c\mu)(A) = c\mu(A)$ (ii) $\forall A \in \Sigma : (\mu + \nu)(A) = \mu(A) + \nu(A)$. Set $E(\mathbf{M}(\mathbb{R}^{\infty}, \Sigma)) = (\mathbf{M}(\mathbb{R}^{\infty}, \Sigma))^I, I = (0, 1]$,