The measurement postulates · Quantum Mechanics

Intuition

Quantum mechanics does not say what a single measurement will give. It says with what probability each result will appear: expand the state in the eigenvectors of the observable, and the probability of each eigenvalue is the modulus squared of its coefficient. And the measurement leaves its mark: right afterwards, the system is in the eigenvector that was found.

A state is a set of odds, not a hidden answer. Measuring is rolling the die those odds describe — and once it has landed, the die shows one face and the odds are gone.

A photon polarised at $30^{\circ}$ above the horizontal, $\ket{\psi}=\cos30^{\circ}\ket{H}+\sin30^{\circ}\ket{V}$ . Its components along the two outcomes of a horizontal–vertical measurement are 0.866 and 0.5, so the probabilities are $0.866^{2}=0.75$ and $0.5^{2}=0.25$ .

The Born rule and the state after

Measuring an observable $\hat{A}$ with non-degenerate eigenvalues $a_{n}$ and normalised eigenvectors $\ket{a_{n}}$ on a system in the normalised state gives the result with probability equal to the modulus squared of the overlap. Immediately after a measurement that gave , the system is in the state .

P(a_{n})=\bigl|\braket{a_{n}|\psi}\bigr|^{2},\qquad \ket{\psi}\ \longrightarrow\ \ket{a_{n}}

Reading the postulates

The probabilities are those of the first chapter's expansion: in the eigenbasis of $\hat{A}$ , $P(a_{n})=|c_{n}|^{2}$ .

The probabilities add up to one

\sum_{n}P(a_{n})=\sum_{n}\bigl|\braket{a_{n}|\psi}\bigr|^{2}=1

The eigenvectors of a Hermitian operator form an orthonormal basis, so the completeness relation holds for them. Inserting it into the squared norm of the normalised state turns that norm, which is 1, into the sum of the probabilities.

Proof steps

\sum_{n}\ket{a_{n}}\bra{a_{n}}=\hat{1}

The eigenvectors of the observable form an orthonormal basis: the spectral theorem.

1=\braket{\psi|\psi}=\bra{\psi}\hat{1}\ket{\psi}

The state is normalised.

=\sum_{n}\braket{\psi|a_{n}}\braket{a_{n}|\psi}

Insert the completeness relation.

=\sum_{n}\bigl|\braket{a_{n}|\psi}\bigr|^{2}=\sum_{n}P(a_{n})

Each term is an overlap times its conjugate: a probability.

Applications

Practice

Modulus Squared of the Overlap

The probability of a result is the modulus squared of the overlap between the state and the eigenvector of that result.

P(a_{n})=\bigl|\braket{a_{n}|\psi}\bigr|^{2}

Try it

A photon is in the state $\tfrac{\sqrt3}{2}\ket{H}+\tfrac12\ket{V}$ and its polarisation is measured as horizontal or vertical. What is the probability of the result horizontal?