Conserved Quantities

Some quantities never change as a system evolves — energy, and often momentum, charge, or spin along an axis. Quantum mechanics gives a crisp, operator-level criterion for which observables are conserved.

The conservation criterion

Ehrenfest's theorem, for an observable $\hat A$ with no explicit time dependence, reads

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

If $\hat A$ commutes with the Hamiltonian, $[\hat H, \hat A] = 0$ , then the right-hand side vanishes for every state, so

[\hat H, \hat A] = 0 \quad\Longrightarrow\quad \frac{d}{dt}\langle \hat A\rangle = 0 \;\;\text{for all }|\psi\rangle.

Such an $\hat A$ is a constant of the motion. The statement is stronger than "the average is constant in one state" — commuting with $\hat H$ makes the expectation value (and, as shown below, the full probability distribution of $\hat A$ ) constant in all states.

Energy is always conserved (for time-independent H)

The Hamiltonian commutes with itself, $[\hat H, \hat H] = 0$ , so $\langle \hat H\rangle$ is conserved whenever $\hat H$ does not depend explicitly on time. Energy conservation is the most basic instance of the criterion, and it follows with no extra assumptions.

Conserved means more than a steady average

When $[\hat H, \hat A] = 0$ , the two operators can be simultaneously diagonalised: there is a common eigenbasis $\{|E_n, a\rangle\}$ labelled by both energy and the value of $\hat A$ . Time evolution only multiplies each such basis state by a phase $e^{-iE_n t/\hbar}$ , which does not move probability between different $\hat A$ -eigenvalues. So the entire probability distribution of the measurement of $\hat A$ — not merely its mean — is frozen in time.

A simple check

For a spin with $\hat H = \tfrac{\hbar\omega}{2}\sigma_z$ , the observable $\sigma_z$ commutes with $\hat H$ (they are the same operator up to a constant), so $\langle\sigma_z\rangle$ is conserved — consistent with the previous lesson's $\langle\sigma_z\rangle(t) = 0$ staying flat. By contrast $\sigma_x$ does not commute with $\hat H$ ( $[\sigma_z,\sigma_x] = 2i\sigma_y \neq 0$ ), so $\langle\sigma_x\rangle$ is free to oscillate — and it does, as $\cos\omega t$ .

Try it

This is a numerical exercise — your code should return a number. Verify that $\sigma_z$ is conserved under $\hat H \propto \sigma_z$ by computing the relevant commutator entry of $[\hat H, \sigma_z]$ , which must be $0$ (hence $d\langle\sigma_z\rangle/dt = 0$ ).

Run your code to see the quantum state.