# Schrödinger equation

The Schrödinger equation is an equation that is fundamental to quantum theory. The time independent Schrödinger equation looks like
    ${\hat {H}}\psi =E\psi .$
The ${\hat {H}}$ is called the Hamiltonian Hamiltonian mechanics. $\psi \,\;$ denote the wavefunctions.

## How to construct the Schrödinger equation for a system

The operator ${\hat {H}}$ is called the Hamiltonian. It contains two parts:
* The kinetic energy, denoted by the operator ${\hat {T}}$
* The potential energy, denoted by the operator ${\hat {U}}$
These two operators represent the principal types of energies in any physical system. Putting these two parts together, we can write this as ${\hat {H}}={\hat {T}}+{\hat {U}}$. We can think of the Hamiltonian as a mathematical object that encodes how the energies in a system can be distributed.

### Kinetic energy operator

In classical mechanics, we know that the kinetic energy of a moving particle is given by
    $E={\frac {1}{2}}mv^{2}={\frac {1}{2}}m{\vec {v}}\cdot {\vec {v}}$
in which $m$ corresponds to the mass of the particle, and ${\vec {v}}$ is its velocity. Noting that classically the linear momentum ${\vec {p}}$ is given by ${\vec {p}}=m{\vec {v}}$, so the kinetic energy can also be written as
    $E={\frac {p^{2}}{2m}}$.
In order to get the quantum mechanical version of this, we substitute all occurrences of the momentum ${\vec {p}}$ with
    ${\vec {p}}\to -i\hbar \nabla$,
where $\nabla$ is the vector analogue of the partial differential operator ${\frac {\partial }{\partial x}}$.
Hence, the kinetic energy operator is given by
    ${\hat {T}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}$.

### Total energy operator and the time dependent Schrödinger equation

The total energy operator introduces the time propagation to the wavefunctions so the time dependent Schrödinger equation tells how the quantum system developes is time while ist was found is some state at the original time. It is basically constructed from the condition that its operation on time dependent oscillatory factors characteristic to the plane waves obtained from the classical wave equation describing the mechanical waves like sound or the waves on the water surface should produce the energy quanta $\hbar \omega$ as the eigenvalues from the eigen-equation namely
    ${\hat {E}}_{TOT}e^{-i\omega t}=\hbar \omega e^{-i\omega t}$
One therefore may readily guess it as
    ${\hat {E}}_{TOT}=i\hbar {\frac {\partial }{\partial t}}$
While te total energy in the classical mechanics is normally equal to the Hamiltonian the time dependent Schrödinger equation is obtained by equalizing the two:
    ${\hat {H}}\psi =i\hbar {\dot {\psi }}.$

## How to solve the Schrödinger equation
### Example 1: The ground state of the hydrogen atom

The time independent Schrödinger equation for the electron in the Coulomb potential in the spherical coordinates takes the form:
    $-{\frac {\hbar ^{2}}{2m}}\left[{1 \over r^{2}}{\partial \over \partial r}\left(r^{2}{\partial \over \partial r}\right)+{1 \over r^{2}\sin \theta }{\partial \over \partial \theta }\left(\sin \theta {\partial \over \partial \theta }\right)+{1 \over r^{2}\sin ^{2}\theta }{\partial ^{2} \over \partial \varphi ^{2}}\right]\psi (r,\theta ,\varphi )-{\frac {e^{2}}{4\pi \epsilon _{0}r}}\psi (r,\theta ,\varphi )=E\psi (r,\theta ,\varphi ).$
For the ground state which is the state with the lowest possible $E$ we seek the solution in the simplest possible form not dependent on the spherical angles $\theta$ and $\varphi$. The Schrödinger equation greatly simplifies to the second order ordinary partial differential equation of the one radial variable:
    $-{\frac {\hbar ^{2}}{2m}}\left[{1 \over r^{2}}{\partial \over \partial r}\left(r^{2}{\partial \over \partial r}\right)\right]\psi _{0}(r)-{\frac {e^{2}}{4\pi \epsilon _{0}r}}\psi _{0}(r)=E_{0}\psi _{0}(r).$
The part of it contains derivatives multiplied by the various powers of the $r$ variable and we guess the solution in the exponential form since the exponential function is the eigenfunction of the differentiation operator while the derivative of the exponential function is proportional to it:
    $\psi _{0}(r)=Ne^{-ar}$
Substituting it to the Schrödinger equation we obtain the following equation:
    $-{\frac {\hbar ^{2}}{2m}}\left[-{\frac {2a}{r}}+a^{2}\right]\psi _{0}(r)-{\frac {e^{2}}{4\pi \epsilon _{0}r}}\psi _{0}(r)=E_{0}\psi _{0}(r)$
The only way this equation is valid for all values of $r$ is that the coefficient multiplying the $1/r$ is equal to $0$:
    ${\frac {a\hbar ^{2}}{mr}}={\frac {e^{2}}{4\pi \epsilon _{0}r}}$
that gives the value of the Bohr radius $a_{0}$ (the range how the ground state mostly extents radially):
    $a_{0}=1/a={\frac {4\pi \varepsilon _{0}\hbar ^{2}}{me^{2}}}$
and the ground state energy
    $E_{0}=-{\frac {a^{2}\hbar ^{2}}{2m}}=-{\frac {me^{4}}{8{\varepsilon _{0}}^{2}\pi ^{2}\hbar ^{2}}}$.

### Example 2: Time dependent perturbation theory of weak field ionization

The first thing one can think to do above finding the ground state or the other allowed energies of the system is to solve the time dependent Schrödinger equation approximately i.e. in the perturbative manner. Let us apply it to the small destruction effects on the ground state of the quantum system and for the simplicity in one spatial dimension which reduces the kinetic energy operator to the second derivative with the proper coefficients. Let the ground state of the atom, for example the atom of Hydrogen will be such that $E_{0}$ is the smallest of possible values for the eigen-equation:
    $H_{0}\Psi _{0}=E_{0}\Psi _{0}$
Let us add the external electric field effect to the Hamiltonian $H_{0}$ in the simplest linear manner so the Schrödinger equation is
    $H_{0}\Psi (t)-exf(t)E_{0}\sin(\omega t)\Psi (t)=i\hbar {\dot {\Psi }}(t)$.
We will assume $\Psi (0)=\Psi _{0}$ for the initial condition and will check what happens to that state under the influence of the perturbing $-exf(t)E_{0}\sin(\omega t)$ term, more specifically where it will go in terms of the expansion in the basis of the unbound states of the system which in very good approximation are simply plane waves. For the convenience of the calculations we added the $f(\tau )$ envelope function normalized to $1$ in time so the electric field is turned on and off in the smooth and easy to Fourier transform Gaussian manner:
    $f(\tau )=N_{1}e^{-(\tau -\tau _{0})^{2}/T_{0}^{2}}$.
it is now convenient to absorb the $H_{0}$ term in the equation in the so-called interaction picture introducing the transformed solution:
    $\Psi (t)=e^{-iH_{0}t/\hbar }\Phi (t)$
The original time-dependent equation greatly simplifies to
    $-ex(t)f(t)E_{0}\sin(\omega t)\Phi (t)=i\hbar {\dot {\Phi }}(t)$.
where the position operator becomes time dependent namely
    $x(t)=e^{iH_{0}t/\hbar }xe^{-iH_{0}t/\hbar }$
With the zero iteration condition $\Phi (t)=\Psi _{0}$ we get in the first approximation
    $\Phi (t)=\Psi _{0}-{\frac {1}{i\hbar }}\int _{0}^{t}ef(\tau )E_{0}\sin(\omega \tau )e^{iH_{0}\tau /\hbar }xe^{-iH_{0}\tau /\hbar }\Psi _{0}d\tau$
we are now interested in the unbound continuum state component in the final solution which tells us about the energy spectrum of the ionized electrons. We can get that by projecting the solution onto the continuum state, simply:
    $E(k)=|\int _{-\infty }^{\infty }<\Psi (x,+\infty )|e^{ikx}>dx|^{2}$
where we extended the integration in time from the minus infinity to infinity since the electric field was localized in time with the Gaussian envelope. The only significant term which remains after the integration is:
    $E(k)={\frac {e^{2}E_{0}^{2}}{4\hbar ^{2}}}|<\Psi _{0}|x]e^{ikx}>{\tilde {f}}(\hbar ^{2}k^{2}/2m-\hbar \omega -E_{0})|^{2}$
Where the ${\tilde {f}}$ is the Fourier transform of the field envelope function $f$ and while $f$ is broadly localized the ${\tilde {f}}$ is sharply loalized around the point with the condition
    $\hbar ^{2}k^{2}/2m-\hbar \omega -E_{0}=0$
expressing the energy conservation condition
    $E(k)=\hbar ^{2}k^{2}/2m=E_{0}+\hbar \omega$
stating that while the ground state energy is negative there is no ionization below the threshold (for the photon or field frequencies that low that the kinetic energy would be impossibly negative) and the ejected electron kinetic energy is equal the energy of the absorbed photon lowered by the ionization or the binding energy.

### Example 3: Theory of Above Threshold Ionization (ATI)

Above Threshold Ionization may be explained by solving the time dependent Schrödinger equation
in the approximate manner. The Schrödinger equation for the free electron in the field of the electromagnetic wave in one dimension and in the radiation gauge is given by
    ${\frac {1}{2m}}\left({\frac {\hbar }{i}}{\frac {\partial }{\partial x}}-eA(t)\right)^{2}\Psi =i\hbar {\frac {\partial \Psi }{\partial t}}$,
where
    $A(t)={\frac {E_{0}}{\omega }}\cos(\omega t)$
so the electric field is given by
    $E(t)=-{\frac {\partial A}{\partial t}}=E_{0}\sin(\omega t)$
Substituting
    ${\frac}\Psi (t)=e^{iC(t)+ikx}$
we are obtaining the equation for $C(t)$
    ${\dot {C}}(t)=\left(\hbar k-eA(t)\right)^{2}/2m$
with the solution
    $C(t)=-{\frac {\hbar ^{2}k^{2}}{2m}}t+\alpha k\sin(\omega t)-\beta \sin(2\omega t)-\gamma t$,
where
    $\alpha ={\frac {(eE_{0})\hbar }{m\omega ^{2}}}$
    $\beta ={\frac {(eE_{0})^{2}}{8m\omega ^{3}}}$
    $\gamma ={\frac {(eE_{0})^{2}}{4m\omega ^{2}}}$.
The Schrödinger equation for the electron in the wave field and in the atomic potential will be given by
    $\left[H_{0}(t)+V(x)\right]\Psi =i\hbar {\frac {\partial \Psi }{\partial t}}$,
where, $H_{0}$ is the Hamiltonian of the free electron in the field. Adding and subtracting the energy of the ground state from which the electron is to be ionized we obtain
    $\left[H_{0}(t)+V(x)+E_{0}-E_{0}\right]\Psi =i\hbar {\frac {\partial \Psi }{\partial t}}$
Because in the ground state the electron kinetic energy is equall the total energy but with the opposite sign (virial theorem) and only this energy will be left after the fast removal of the electron, we neglect in this equation the sum $v(x)+E_{0}$ for all $x$ and obtain the approximate equation
    $\left[H_{0}(t)-E_{0}\right]\Psi =i\hbar {\frac {\partial \Psi }{\partial t}}$,
where the only remaining of the atomic potential is the constant.
This equation may now be solved using the previous result for the free electron in the field and expanding the ground state into its Fourier components
    $\Psi (0)=Ne^{-|x|/a_{0}}=\int c_{k}e^{ikx}dk$,
with
    $c_{k}={\frac {N}{a_{0}}}{\frac {1}{k^{2}a_{0}^{2}+1}}$
This equation has therefore the solution
    $\Psi (t)=\int c_{k}e^{ikx}e^{iE_{0}t}e^{iC(t)}dk$
We obtain the ionization spectrum from the formula
    $E(k)=|\int _{-\infty }^{\infty }<\Psi (\tau )|e^{ikx}>g(\tau )d\tau |^{2}$,
telling how much of the plane wave of the free electron with the given kinetic energy is at the and of the ionization process, where $g(\tau )$ is the averaging function of the measuring detector for example
    $g(\tau )=N_{1}e^{-(\tau -\tau _{0})^{2}/T_{0}^{2}}$.
Expanding the factor
    ${\frac}e^{i\alpha k\sin(\omega t)-i\beta \sin(2\omega t)}=\sum _{n}{\mathcal {J}}_{n}(-\alpha k,\beta )e^{-in\omega t}$
with the generalized Bessel functions ${\mathcal {J}}_{n}$ defined by the inverse transform, we obtain
    $E(k)=|\sum _{n}c_{k}{\mathcal {J}}_{n}(-\alpha k,\beta ){\tilde {g}}(\hbar ^{2}k^{2}/2m-n\hbar \omega -E_{0}+\gamma )|^{2}$
(${\tilde {g}}(\omega )$ is the Fourier transform of the detector function) therefore the sum of sharp or fuzzy maxima localized around the energy condition of emitted electrons $\hbar ^{2}k^{2}/2m=n\hbar \omega +E_{0}-\gamma$ depending on the speed i.e the averaging parameter of the detector $T_{0}$.

### Example 4: Basic theory of the Trojan Wave Packet

Because the Trojan wave packet is an approximate eigenstate of the Hamiltonian its search starts from the Schrödinger´s equation for the Hydrogen atom in the field of the circularly polarized wave in the coordinate system rotating with the frequency of the field
    $H\Psi =E{\tfrac}\Psi$,
where
    $H={\frac {\mathbf {p} }{2}}^{2}-{\frac {1}{r}}-{\epsilon }x-\omega L_{z}$,
and $L_{z}$ is the operator of z-component of the angular momentum.
For the simplification one can restrict himself to only two spacial dimensions. This equation after neglecting the curvature terms of the Laplacian and assumption that some terms are slowly varying or even constant takes in the polar coordinate system the simplified and separated form
    $\left[-{\frac {\partial ^{2}}{2\partial r^{2}}}+{\frac {l^{2}}{2r^{2}}}-{\frac {1}{r}}-{\epsilon }r-{\frac {\partial ^{2}}{2r_{0}^{2}\partial \phi ^{2}}}-{\epsilon }r_{0}\cos \phi +\epsilon r_{0}\right]\chi =E\chi$,
    ${\frac}\chi =e^{il\phi }\Psi ,l=\omega ^{2}r_{0}$
where $\hbar =1,m_{e}=1$, and $r_{0}$ is the point of the potential of the part of the Hamiltonian dependent only on the radial coordinate and therefore also the radius of the classical orbit. Because the radial potential has the minimum the Trojan wavepacket is in this theory the product of a well localized radial Gaussian function and a well localized excited state of the quantum mathematical pendulum, corresponding to the pendulum totally reversed upside down which is also the Mathieu function with the characteristic value (and the energy) equal with a good approximation to $r_{0}\epsilon$. The more precise theory shows that it is the pendulum of a negative and fractional electron mass equal -1/3 (of the electron) and that is why it is on the contrary the ground state with approximately the same energy but in the reversed spectrum.
To notice that let us more keep the correction from the expansion of the centrifugal term
    $\Delta H=-2i{\frac {l}{r_{0}^{3}}}{\frac {\partial }{\partial \phi }}(r-r_{0})$
and let us consider its action on the Fourier component of $\chi$, e.g. $e^{ik}$.
The additional centrifugal potential will be
    $\Delta V=2i{\frac {l}{r_{0}^{3}}}k(r-r_{0})$
It will cause the shift of the minimum and the energy lowering of the radial harmonic oscillator equal to
    $\Delta E=-{\frac {2}{r_{0}^{2}}}k^{2}$
Replacing back $k^{2}$ with the operator of the z component of the angular momentum we get
    $\Delta H={\frac {2\partial ^{2}}{r_{0}^{2}\partial \phi ^{2}}}$
Adding this correction in the Schródinger equation we get the equation with the negative and fractional electron mass -1/3 ($m_{e}$):
    $\left[-{\frac {\partial ^{2}}{2\partial r^{2}}}+{\frac {l^{2}}{2r^{2}}}-{\frac {1}{r}}-{\epsilon }r+{\frac {3\partial ^{2}}{2r_{0}^{2}\partial \phi ^{2}}}-{\epsilon }r_{0}\cos \phi +\epsilon r_{0}\right]\chi =E\chi$,
which allows the Gaussian localization of the electron in the maximum but not in the minimum of the potential in the angular direction.