Expectation Values ⟨A⟩

When you measure an observable $A$ many times on a system always prepared in the same state $|\psi\rangle$, the average result converges to the expectation value

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

This is one of the central formulas of quantum mechanics: it turns the abstract operator $A$ into a single, physically measurable number.

Where the formula comes from

Suppose $A$ has eigenvalues $a_n$ and orthonormal eigenstates $|a_n\rangle$, so $A|a_n\rangle = a_n|a_n\rangle$. Any state can be expanded in this basis:

$|\psi\rangle = \sum_n c_n |a_n\rangle, \qquad c_n = \langle a_n|\psi\rangle.$

The Born rule says that a single measurement yields $a_n$ with probability $|c_n|^2$. The statistical mean of many measurements is therefore

$\langle A\rangle = \sum_n a_n\, |c_n|^2.$

We can rewrite this in a basis-free way. Using $A|a_n\rangle = a_n|a_n\rangle$ and the completeness relation $\sum_n |a_n\rangle\langle a_n| = \mathbf{1}$:

$\langle\psi|A|\psi\rangle = \langle\psi|\left(\sum_n a_n|a_n\rangle\langle a_n|\right)|\psi\rangle = \sum_n a_n |\langle a_n|\psi\rangle|^2 = \sum_n a_n |c_n|^2.$

The two expressions match, so $\langle A\rangle = \langle\psi|A|\psi\rangle$ is not merely a definition — it is a consequence of the Born rule combined with linearity.

Computing it in matrix form

In any finite-dimensional basis, $|\psi\rangle$ is a column vector and $A$ is a square matrix. The bra $\langle\psi|$ is the conjugate-transpose row vector. The expectation value is then the ordinary matrix product

$\langle A\rangle = \psi^\dagger A\, \psi = \sum_{i,j} \psi_i^* A_{ij}\, \psi_j.$

Example — Pauli-$Z$ in the $|+\rangle$ state. The Pauli-$Z$ operator is

$\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$

with eigenstates $|0\rangle$ (eigenvalue $+1$) and $|1\rangle$ (eigenvalue $-1$). The equal-superposition state is

$|+\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle \;\leftrightarrow\; \psi = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}.$

Step 1 — apply $\sigma_z$ to $|\psi\rangle$:

$\sigma_z\psi = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}.$

Step 2 — inner-product with $\langle\psi|$:

$\langle\sigma_z\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix} = \frac{1}{2}(1) + \frac{1}{2}(-1) = 0.$

The result is exactly $0$, which makes sense: $|+\rangle$ is an equal mixture of the $+1$ and $-1$ eigenstates of $\sigma_z$, so the average $z$-spin vanishes.

Try it

This is a numerical exercise — your code should return a number. Compute $\langle\psi|\sigma_z|\psi\rangle$ for the equal-superposition state $|\psi\rangle = |+\rangle$.

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