Measurement

Measuring an observable does not reveal a value that already existed before the measurement: quantum mechanics postulates what happens, building on the eigenbasis decomposition seen earlier.

The postulate

Let $\hat A$ be a Hermitian operator (an observable), with eigenbasis $| \psi_n \rangle$ and eigenvalues $a_n$, and let $| \psi \rangle = \sum_n c_n | \psi_n \rangle$ be the state of the particle, with $c_n = \langle \psi_n | \psi \rangle$. Measuring $\hat A$:

• can only give one of the eigenvalues $a_n$;
• gives the result $a_n$ with probability $|c_n|^2$ — the Born rule;
• immediately projects the state onto $| \psi_n \rangle$: this is the collapse of the wavefunction.

Repeating the measurement on identically prepared systems therefore gives, on average,

Why Hermitian

None of this holds if $\hat A$ is not Hermitian. A Hermitian operator has two properties measurement relies on directly: its eigenvalues $a_n$ are real — a measurement outcome is a real number, not a complex one — and its eigenstates form a complete orthonormal basis, which guarantees $\sum_n |c_n|^2 = 1$: the probabilities of the different outcomes sum to 1, as they must. A non-Hermitian operator offers neither guarantee: its eigenvalues can be complex, and its eigenstates generally form neither an orthogonal basis, nor even a basis at all.

The degenerate case

If several eigenstates $| \psi_n^{(1)} \rangle, \dots, | \psi_n^{(d_n)} \rangle$ share the same eigenvalue $a_n$ (an eigenspace of dimension $d_n$), nothing distinguishes which one the measurement selects. The probability of $a_n$ is then summed over the whole eigenspace, and the state collapses onto that entire eigenspace rather than onto any single eigenvector. Writing $\hat P_n = \sum_k | \psi_n^{(k)} \rangle \langle \psi_n^{(k)} |$ for the projector onto it,

When $d_n = 1$, this reduces exactly to the non-degenerate case above.