Expectation Values of Components

A measurement of an angular-momentum component yields one of its eigenvalues, with probabilities set by the state. The average over many identical measurements is the expectation value, the most directly comparable prediction between theory and experiment.

Definition

For an observable $\hat{A}$ and a normalized state $|\psi\rangle$, the expectation value is

It is the mean of the eigenvalues weighted by their measurement probabilities. For $\hat{A} = \hat{L}_z$ acting on $|\psi\rangle = \sum_m c_m |\ell, m\rangle$,

since each $|\ell, m\rangle$ contributes its eigenvalue $\hbar m$ with probability $|c_m|^2$.

The spin-1/2 case

For a single spin-$\tfrac12$ (a qubit), identify $|{\uparrow}\rangle = |0\rangle$ and $|{\downarrow}\rangle = |1\rangle$, with $\hat{S}_z = \tfrac{\hbar}{2}\,\sigma_z$. The two eigenvalues are $\pm\tfrac{\hbar}{2}$, so

where $p_0 = |\langle 0|\psi\rangle|^2$ and $p_1 = |\langle 1|\psi\rangle|^2$. The expectation value ranges continuously from $+\tfrac{\hbar}{2}$ (pure $|0\rangle$) to $-\tfrac{\hbar}{2}$ (pure $|1\rangle$), even though each individual measurement returns only one of the two discrete values.

A parametrized state

Apply a $Y$-rotation to $|0\rangle$:

Then $p_0 = \cos^2(\theta/2)$ and $p_1 = \sin^2(\theta/2)$, so using the double-angle identity $\cos^2(\theta/2) - \sin^2(\theta/2) = \cos\theta$,

This is exactly the vector-model result: $\theta$ is the polar angle of the Bloch vector, and the $z$-projection of a unit spin is $\tfrac{\hbar}{2}\cos\theta$.

Try it

Use $R_y(\theta)$ with $\theta = \pi/3$ to prepare the state, then compute and return $\langle \hat{S}_z\rangle = \tfrac{1}{2}(p_0 - p_1)$ (with $\hbar = 1$). It should equal $0.25$.

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