Math Infinitum

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Intuition

A quantity is conserved when nothing about it changes as the system evolves: not its average, not its spread, not the probability of any of its values. The test is a single commutator — an observable that commutes with the Hamiltonian is conserved, in every state. Energy always is. And the energy–time relation says how fast anything can change when the energy is spread by a given amount.

An ice skater spinning with no friction keeps her angular momentum however she moves her arms. In quantum mechanics the promise is stronger: the whole probability distribution of a conserved quantity stays exactly as it was.

Conservation laws

The rate of change of any expectation value is the average of the commutator of the Hamiltonian with the observable. If the commutator vanishes the average is constant in every state, and so is the probability of each eigenvalue: the observable is conserved.

\begin{aligned}&\frac{d}{dt}\langle\hat{A}\rangle=\frac{i}{\hbar}\bigl\langle[\hat{H},\hat{A}]\bigr\rangle\\ &[\hat{H},\hat{A}]=0\ \Rightarrow\ \hat{A}\ \text{conserved}\end{aligned}

What conservation means

Energy is conserved for a time-independent $\hat{H}$ , since $[\hat{H},\hat{H}]=0$ .

How averages change

\frac{d}{dt}\bra{\psi(t)}\hat{A}\ket{\psi(t)}=\frac{i}{\hbar}\bra{\psi(t)}[\hat{H},\hat{A}]\ket{\psi(t)}

Differentiate the bracket with the product rule and use the Schrödinger equation for the ket and its adjoint for the bra. The two terms are the Hamiltonian acting before and after the observable, and together they make the commutator.

Proof steps

\frac{d}{dt}\bra{\psi}\hat{A}\ket{\psi}=\Bigl(\frac{d}{dt}\bra{\psi}\Bigr)\hat{A}\ket{\psi}+\bra{\psi}\hat{A}\Bigl(\frac{d}{dt}\ket{\psi}\Bigr)

The product rule, for an observable that does not depend on time itself.

=\frac{i}{\hbar}\bra{\psi}\hat{H}\hat{A}\ket{\psi}-\frac{i}{\hbar}\bra{\psi}\hat{A}\hat{H}\ket{\psi}

$\frac{d}{dt}\ket{\psi}=\frac{1}{i\hbar}\hat{H}\ket{\psi}=-\frac{i}{\hbar}\hat{H}\ket{\psi}$ and its adjoint.

=\frac{i}{\hbar}\bra{\psi}[\hat{H},\hat{A}]\ket{\psi}

Collect the two terms into a commutator.

[\hat{H},\hat{A}]=0\ \Rightarrow\ \frac{d}{dt}\langle\hat{A}\rangle=0

A vanishing commutator makes the average constant in every state.

Applications

Practice

The Commutator Test

The average of an observable changes at a rate given by the average of its commutator with the Hamiltonian. If that commutator is zero, the observable is conserved.

\frac{d}{dt}\langle\hat{A}\rangle=\frac{i}{\hbar}\langle[\hat{H},\hat{A}]\rangle

Try it

For $\hat{H}=\tfrac{\hbar\omega}{2}Z$ , which observable is conserved?