Expectation Values of Spin

A spin measurement along $z$ returns $\pm\tfrac{\hbar}{2}$ at random, but the average of many such measurements on identically prepared states is a definite number — the expectation value. For an observable $A$ and a state $|\psi\rangle$ ,

\langle A\rangle = \langle\psi|A|\psi\rangle.

Spin expectation values

For the spin- $z$ operator $S_z = \tfrac{\hbar}{2}\sigma_z$ ,

\langle S_z\rangle = \frac{\hbar}{2}\langle\sigma_z\rangle = \frac{\hbar}{2}\bigl(P_{\uparrow} - P_{\downarrow}\bigr),

where $P_\uparrow = |\alpha|^2$ and $P_\downarrow = |\beta|^2$ are the probabilities of measuring up and down. The expectation value is the difference of the two outcome values, weighted by their probabilities — exactly the classical notion of a mean.

The Bloch-vector shortcut

For a general state $|\psi\rangle = \cos\tfrac{\theta}{2}|0\rangle + e^{i\phi}\sin\tfrac{\theta}{2}|1\rangle$ , the three Pauli expectation values are the Cartesian components of the Bloch vector:

\langle\sigma_x\rangle = \sin\theta\cos\phi, \qquad \langle\sigma_y\rangle = \sin\theta\sin\phi, \qquad \langle\sigma_z\rangle = \cos\theta.

So computing a spin expectation value is the same as reading off a component of the Bloch arrow:

\langle\mathbf{S}\rangle = \frac{\hbar}{2}\,\langle\boldsymbol{\sigma}\rangle = \frac{\hbar}{2}\,\hat{n},

pointing along the same direction $\hat{n}$ the state was prepared in, with length $\tfrac{\hbar}{2}$ .

Worked example

Take $\theta = \tfrac{\pi}{3}$ , $\phi = 0$ . Then

\langle\sigma_z\rangle = \cos\frac{\pi}{3} = \frac{1}{2}, \qquad \langle S_z\rangle = \frac{\hbar}{2}\cdot\frac{1}{2} = \frac{\hbar}{4}.

The probabilities are $P_\uparrow = \cos^2\tfrac{\theta}{2} = \cos^2\tfrac{\pi}{6} = \tfrac{3}{4}$ and $P_\downarrow = \sin^2\tfrac{\pi}{6} = \tfrac{1}{4}$ , so $\langle S_z\rangle = \tfrac{\hbar}{2}(\tfrac{3}{4} - \tfrac{1}{4}) = \tfrac{\hbar}{4}$ — the same answer by the probability route.

Try it

Prepare the state at $\theta = \pi/3$ , $\phi = 0$ and let the grader read its Bloch vector. The expectation value $\langle S_z\rangle = \tfrac{\hbar}{2}\langle\sigma_z\rangle$ should equal $\tfrac{1}{4}$ in units where $\hbar = 1$ .

Run your code to see the quantum state.