# Quantum mechanics/Heisenberg principle

Suppose that we have prepared a system in a state φ and this state is an eigenfunction of the operator Â corresponding to a classical quantity A (Âφ = aφ). If we measure A, we will always get a, i.e. the value of A is precisely defined. Can you also have a precise value for a different quantity B? The answer is generally no, because the eigenfunctions of ${\hat {A}}$ are generally not eigenfunctions of ${\hat {B}}$. So if the system is in eigenstate ${\hat {A}}$ then the repeated measure of a quantity B will give different results each time (so we can only calculate an expectation value of B). If A is precisely determined, B is not precisely determined. One of the most shocking ideas of quantum mechanics is that it is not possible to measure with arbitrary accuracy any desired observable.
The only way to measure precisely both A and B is to prepare a system in a state which is at the same time eigenfunctions of ${\hat {A}}$ and ${\hat {B}}$.

## Commutators

Given two operators ${\hat {A}}$ and ${\hat {B}}$, they are normally non-commutative, i.e. ${\hat {A}}{\hat {B}}\neq {\hat {B}}{\hat {A}}$. The commutator between them is defined as $\left[{\hat {A}},{\hat {B}}\right]\equiv {\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}$ and, when different than zero, it indicates the order of application of the two operators is important.
The commutator of the operator ${\hat {x}}$ and ${\hat {p}}_{x}$ is iħ (one of the postulates of quantum mechanics was ${\hat {x}}{\hat {p}}_{x}-{\hat {p}}_{x}{\hat {x}}=i\hbar$) so:
$\left[{\hat {x}},{\hat {p}}_{x}\right]=i\hbar$ (Eq. 1)
The commutator of the operators ${\hat {p}}_{x}$ and ${\hat {p}}_{x}^{2}$ is zero (see Exercise) so:
$\left[{\hat {p}}_{x},{\hat {p}}_{x}^{2}\right]=0$ (Eq. 2)
It can be proven (see Exercises) that if two operators commute they have common eigenfunctions. So we can know precisely the value of A and B at the same time only if the corresponding operators ${\hat {A}}$ and ${\hat {B}}$ commute.

## Heisenberg's uncertainty principle

Since the position and momentum operators do not commute we cannot measure at the same time with arbitrary accuracy the position and the momentum of a particle. This is known as the uncertainty principle.
From the postulates seen so far you can prove the following relation:
$\Delta x\Delta p_{x}\geq {\begin{matrix}{\frac {1}{2}}\end{matrix}}\hbar$ (Eq. 3)
where Δx is the uncertainty in the position, and Δp$_x$ the uncertainty in the momentum.

## Illustration of he Heisenberg principle for the particle in the box

A possible way to estimate the error of a quantity q whose measure changes each time is $\Delta q={\sqrt {\left\langle q^{2}\right\rangle -\left\langle q\right\rangle ^{2}}}$ (called standard deviation). We can use this formula to calculate Δx and Δp$_x$ for the particle in the box in the ground state. We have already calculated in the previous lesson <x> = L/2 amd <p> = 0. We now calculate in the usual way <x$^2$> and <p$^2$>.
$\left\langle x^{2}\right\rangle ={\frac {\int \psi ^{*}{\hat {x}}^{2}\psi \,d\tau }{\int \psi ^{*}\psi \,d\tau }}={\frac {2}{L}}\int _{0}^{L}\sin \left({\frac {\pi x}{L}}\right)x^{2}\sin \left({\frac {\pi x}{L}}\right)\,dx={\frac {2}{L}}\int _{0}^{L}x^{2}\sin ^{2}\left({\frac {\pi x}{L}}\right)\,dx={\frac {L^{2}}{3}}$ (Eq. 4)
$\left\langle p^{2}\right\rangle ={\frac {\int \psi ^{*}{\hat {p}}^{2}\psi \,d\tau }{\int \psi ^{*}\psi \,d\tau }}={\frac {2}{L}}\int _{0}^{L}\sin \left({\frac {\pi x}{L}}\right)\left(-\hbar ^{2}{\frac {\partial ^{2}}{\partial x^{2}}}\right)\sin \left({\frac {\pi x}{L}}\right)\,dx={\frac {\hbar ^{2}\pi ^{2}}{L^{2}}}{\frac {2}{L}}\int _{0}^{L}\sin ^{2}\left({\frac {\pi x}{L}}\right)\,dx={\frac {\hbar ^{2}\pi ^{2}}{L^{2}}}$ (Eq. 5)
so we get
$\Delta p={\sqrt {\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}}={\frac {\hbar \pi }{L}}$ (Eq. 6)
$\Delta x={\sqrt {\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}}={\sqrt {{\frac {L^{2}}{3}}-{\frac {L^{2}}{4}}}}={\frac {L}{2{\sqrt {3}}}}$ (Eq. 7)
and therefore $\Delta x\Delta p={\frac {\hbar \pi }{2{\sqrt {3}}}}\approx \hbar$ (Eq. 8)

## Summary of the postulates of quantum mechanics

* Postulate 1. The state of a quantum mechanical system is completely specified by a function Ψ(r,t) (called the wavefunction) that depends on the coordinates of the particle(s) and on time. |Ψ(r,t)|$^2$ is proportional to the probability of finding the particle in a region of space around r.
* Postulate 2. To every observable in classical mechanics, there corresponds a linear Hermitian operator in quantum mechanics. All operators can be built from their classical expression identifying ${\hat {x}}\to x,{\hat {p}}_{x}\to -i\hbar {\frac {\partial }{\partial x}}$ (and similar expressions for the other coordinates). The operator of the total energy is called the Hamiltonian (${\hat {H}}$).
* Postulate 3. In any measurement of the observable associated with operator ${\hat {A}}$, the only values that will ever be observed are the eigenvalues λ$_i$ which satisfy the eigenvalue equation ${\hat {A}}\varphi _{i}=\lambda _{i}\varphi _{i}$.
* Postulate 4. If a system is in a state described by a normalised wavefunction Ψ(r,t) then the average value of the observable corresponding to the operator ${\hat {A}}$ is given by $\left\langle A\right\rangle =\int \psi ^{*}{\hat {A}}\psi d,d\tau$.
* Postulate 5. The wavefunction of a system evolves with time according to the time dependent Schrödinger equation: $i\hbar {\frac {\partial \Psi (r,t)}{\partial t}}={\hat {H}}\Psi (r,t)$

## The angular momentum operators and their commutators

The angular momentum operators are very important to understand the atomic spectra, and a rigorous textbook on quantum mechanics will spend quite some time on them. Here we simplify the matter, giving without proof the basic commutator relations for the angular momentum operator.
$\left[{\hat {L}}^{2},{\hat {L}}_{x}\right]=\left[{\hat {L}}^{2},{\hat {L}}_{y}\right]=\left[{\hat {L}}^{2},{\hat {L}}_{z}\right]=0$ (Eq. 9)
$\left[{\hat {L}}_{x},{\hat {L}}_{y}\right]=i\hbar {\hat {L}}_{z};\left[{\hat {L}}_{y},{\hat {L}}_{z}\right]=i\hbar {\hat {L}}_{x};\left[{\hat {L}}_{x},{\hat {L}}_{z}\right]=i\hbar {\hat {L}}_{y}$ (Eq. 10)
${\hat {L}}^{2}$ is the quantum operator corresponding to the square of the angular momentum and ${\hat {L}}_{x},{\hat {L}}_{y},{\hat {L}}_{z}$ are the quantum operators corresponding to the three components of angular momentum.
According to the postulates of quantum mechanics, it is not possible to measure accurately the three components of the angular momentum (they do not commute). We can only measure the absolute value (L$^2$) and its component along one direction (conventionally the z direction).
We have already studied the eigenfunctions of ${\hat {L}}^{2}$ because the Hamiltonian of the free rotor can be written as:
${\hat {H}}={\frac {{\hat {L}}^{2}}{2I}}$ (Eq. 11)
Considering that ${\hat {H}}$ and ${\hat {L}}^{2}$ differ only by a constant, the spherical harmonics are also eigenfunctions of the ${\hat {L}}^{2}$ operator:
${\hat {L}}^{2}Y_{lm}(\theta ,\phi )=\hbar ^{2}l(l+1)Y_{lm}(\theta ,\phi )$ (Eq. 12)
The total angular momentum for a free rotor is quantized and its absolute value can take the values $\hbar {\sqrt {l(l+1)}}$. ${\hat {L}}^{2}$ and ${\hat {L}}_{z}$ commute so they must have common eigenfunctions. The following relation can also be proven:
${\hat {L}}_{z}Y_{lm}(\theta ,\phi )=\hbar mY_{lm}(\theta ,\phi )$ (Eq. 13)
We can then achieve an intuitive understanding of the quantum number l and m which appears in the free rotor and the hydrogen atom eigenfunction. l is intuitively related to the total angular momentum (how quickly the electron rotates) and m is related to the direction of the motion. Remember that |m| ≤ l and so the component along z of the momentum (ħm) is always smaller than the absolute value of the momentum.
Exercise
Prove that if two operators commute, they have the same eigenfunctions.
Show that the commutators $\left[{\hat {x}},{\hat {x}}^{2}\right],\left[{\hat {p}}_{x},{\hat {p}}_{x}^{2}\right],\left[{\hat {x}},{\hat {y}}\right],\left[{\hat {x}},{\hat {p}}_{y}\right],\left[{\hat {y}},{\hat {p}}_{x}\right]$ are null.
Show that the eigenfunction ψ(x) = e$^{ikx}$ of the free particle (Exercise 3.8) is also an eigenfunction of the linear momentum operator ${\hat {p}}_{x}$. What would be the result of the measure of the linear momentum for a particle in state ψ(x) = e$^{ikx}$?
Can energy and linear momentum be known exactly for the free particle? And for the harmonic oscillator?
With the help of a famous search engine, find out what "Schrödinger's cat" is.
Solutions
Hypothesis $\left[{\hat {A}},{\hat {B}}\right]=0,{\hat {A}}\Psi =a\Psi$ , thesis ${\hat {B}}\Psi =b\Psi$ , proof: ${\begin{matrix}{\hat {A}}\Psi =a\Psi ;&{\hat {B}}{\hat {A}}\Psi =a{\hat {B}}\Psi \\\ &{\hat {A}}{\hat {B}}\Psi =a{\hat {B}}\Psi \\\ &{\hat {A}}({\hat {B}}\Psi )=a({\hat {B}}\Psi )\\\ &{\hat {B}}\Psi =b\Psi ;\end{matrix}}$
${\hat {p}}_{x}=-i\hbar {\frac {\partial }{\partial x}},{\hat {p}}_{x}^{2}=-i\hbar ^{2}{\frac {\partial ^{2}}{\partial x^{2}}},{\hat {p}}_{x}^{2}{\hat {p}}_{x}={\hat {p}}_{x}{\hat {p}}_{x}^{2}=i\hbar ^{3}{\frac {\partial ^{3}}{\partial x^{3}}},{\hat {y}}{\hat {p}}_{x}={\hat {p}}_{x}{\hat {y}}=-i\hbar y{\frac {\partial }{\partial x}}$
We have to prove ${\hat {p}}_{x}e^{ikx}=\lambda e^{ikx}$ and indeed $\left[-i\hbar {\frac {\partial }{\partial x}}\right]e^{ikx}=[\hbar k]e^{ikx}$. The result of measuring the momentum of the free particle in state ψ(x) = e$^{ikx}$ is ħk.
Yes because $[{\hat {H}},{\hat {p}}_{x}]=0$ for the free particle.
No because $[{\hat {H}},{\hat {p}}_{x}]\neq 0$ in the harmonic oscillator. :
Next: Lesson 10 - Hydrogen atom in a magnetic field and the discovery of spin