Measurement Statistics

A single quantum measurement gives one eigenvalue, chosen at random. The interesting physics lies in the spread of those outcomes when you repeat the measurement on many identically prepared copies of the same state. That spread is captured by the standard deviation $\Delta A$ , sometimes called the uncertainty in $A$ .

Variance from the Born rule

Let $\hat{A}$ be an observable with eigenvalues $a_n$ and corresponding probabilities $p_n = |\langle a_n|\psi\rangle|^2$ when the system is in state $|\psi\rangle$ . The expectation value is

$\langle \hat{A}\rangle = \sum_n a_n\, p_n.$

The variance measures how far individual outcomes typically stray from that mean:

$(\Delta A)^2 = \sum_n (a_n - \langle \hat{A}\rangle)^2\, p_n.$

Expanding the square and using $\sum_n p_n = 1$ gives the computationally convenient identity

$(\Delta A)^2 = \langle \hat{A}^2\rangle - \langle \hat{A}\rangle^2.$

This is exactly the classical formula for variance, now applied to quantum probabilities. The standard deviation is $\Delta A = \sqrt{(\Delta A)^2}$ .

Operator form

In Dirac notation the same result reads

$(\Delta A)^2 = \langle\psi|\hat{A}^2|\psi\rangle - \bigl(\langle\psi|\hat{A}|\psi\rangle\bigr)^2.$

An equivalent, often-used form defines the shifted operator $\widehat{\delta A} = \hat{A} - \langle\hat{A}\rangle\,\hat{I}$ , so that

$(\Delta A)^2 = \langle\psi|(\widehat{\delta A})^2|\psi\rangle.$

Both expressions are identical; the shift form makes it manifest that $\Delta A \geq 0$ , with equality only when $|\psi\rangle$ is an eigenstate of $\hat{A}$ .

Worked example: Pauli-Z in a weighted superposition

Consider the state

$|\psi\rangle = \sqrt{\frac{3}{4}}\,|0\rangle + \sqrt{\frac{1}{4}}\,|1\rangle,$

measured in the $\sigma_z$ basis. The eigenvalues of $\sigma_z$ are $+1$ (for $|0\rangle$ ) and $-1$ (for $|1\rangle$ ), so the Born-rule probabilities are $p_0 = 3/4$ and $p_1 = 1/4$ .

Step 1 — expectation value:

$\langle\sigma_z\rangle = (+1)\cdot\frac{3}{4} + (-1)\cdot\frac{1}{4} = \frac{1}{2}.$

Step 2 — expectation value of the square. Because $\sigma_z^2 = \hat{I}$ (squaring a $\pm 1$ Pauli gives the identity), every eigenvalue squared equals $1$ :

$\langle\sigma_z^2\rangle = (+1)^2\cdot\frac{3}{4} + (-1)^2\cdot\frac{1}{4} = 1.$

Step 3 — variance and standard deviation:

\qquad \Delta\sigma_z = \frac{\sqrt{3}}{2} \approx 0.866.$$ The outcome is not certain ($\Delta\sigma_z \neq 0$), but it is not maximally uncertain either. Compare with the $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ state, where $\langle\sigma_z\rangle = 0$ and $\Delta\sigma_z = 1$ — the largest possible spread for a $\pm 1$ observable. <Callout type="tip"> For any two-outcome observable with eigenvalues $+1$ and $-1$ and probabilities $p$ and $1-p$, the variance simplifies to $1 - (2p-1)^2 = 4p(1-p)$, which is maximised at $p = 1/2$ and equals zero at $p = 0$ or $p = 1$. </Callout> ## Zero variance: the eigenstate case When $|\psi\rangle$ is already an eigenstate of $\hat{A}$ with eigenvalue $a$, all probability sits at a single outcome. Then $\langle\hat{A}\rangle = a$, $\langle\hat{A}^2\rangle = a^2$, and $(\Delta A)^2 = a^2 - a^2 = 0$. No measurement of $\hat{A}$ on this state will ever surprise you — the spread is literally zero. This is why eigenstates are sometimes called **states with definite values** of the corresponding observable. ## Try it This is a numerical exercise — your code should `return` a number. For the state $|\psi\rangle = \sqrt{3/4}\,|0\rangle + \sqrt{1/4}\,|1\rangle$ and the observable $\sigma_z$, compute the standard deviation $\Delta\sigma_z$. <RunnableExample />