Expectation Value Dynamics

We have followed states and probabilities through time. Often the most physical question is how an observable's average moves — the position of a wavepacket, the magnetisation of a spin. This lesson tracks $\langle \hat A\rangle(t)$ directly.

Two routes to the same curve

There are two equivalent ways to compute $\langle \hat A\rangle(t)$:

1. Evolve the state, then sandwich: $\langle \hat A\rangle(t) = \langle\psi(t)|\hat A|\psi(t)\rangle$ with $|\psi(t)\rangle = U(t)|\psi(0)\rangle$.
2. Evolve the operator (Heisenberg form): $\langle \hat A\rangle(t) = \langle\psi(0)|\,U^\dagger \hat A\,U\,|\psi(0)\rangle = \langle\psi(0)|\hat A_H(t)|\psi(0)\rangle$.

For our worked example the operator route is cleanest because Pauli operators rotate into each other.

A precessing spin

Take $\hat H = \tfrac{\hbar\omega}{2}\sigma_z$ and watch the observable $\sigma_x$. A standard identity for conjugating a Pauli by a $Z$-rotation gives the Heisenberg-evolved operator

and similarly $\sigma_y(t) = \cos(\omega t)\,\sigma_y + \sin(\omega t)\,\sigma_x$, while $\sigma_z(t) = \sigma_z$. Now choose the initial state $|+\rangle$, the $+1$ eigenstate of $\sigma_x$, for which $\langle\sigma_x\rangle_0 = 1$, $\langle\sigma_y\rangle_0 = 0$, $\langle\sigma_z\rangle_0 = 0$. Taking expectation values:

The Bloch picture

These three averages are the components of the Bloch vector $\vec r(t) = (\langle\sigma_x\rangle, \langle\sigma_y\rangle, \langle\sigma_z\rangle)$. Here

a unit vector sweeping counterclockwise around the $z$-axis in the equatorial plane at angular frequency $\omega$. The expectation values are not abstract — they are literally the coordinates of the spin's arrow on the Bloch sphere, precessing like a gyroscope.

Consistency with Ehrenfest

You can also get $\langle\sigma_x\rangle(t) = \cos\omega t$ from Ehrenfest's theorem: $\tfrac{d}{dt}\langle\sigma_x\rangle = \tfrac{i}{\hbar}\langle[\hat H,\sigma_x]\rangle$. Since $[\sigma_z,\sigma_x] = 2i\sigma_y$, this gives $\dot{\langle\sigma_x\rangle} = -\omega\langle\sigma_y\rangle$, and likewise $\dot{\langle\sigma_y\rangle} = \omega\langle\sigma_x\rangle$. Solving this coupled pair with the given initial conditions reproduces $\langle\sigma_x\rangle(t)=\cos\omega t$ and $\langle\sigma_y\rangle(t)=+\sin\omega t$ — consistent with the operator route above — and the next lesson formalises this Ehrenfest route.

Try it

This is a numerical exercise — your code should return a number. Compute $\langle\sigma_x\rangle(t) = \cos(\omega t)$ at $\omega t = \pi/3$.

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