Expectation values · Quantum Mechanics

Intuition

Measure the same quantity on many identically prepared systems and average the results. That average, the expectation value, is predicted without listing the probabilities: it is the bracket of the state with the observable acting on it. It need not be a value any single measurement can give — the average roll of a die is three and a half.

The expected score of a game is the sum of each score times its chance. Nobody ever scores the expected value in one game; it is where a long run of games settles.

Results $-1$ with probability $0.25$ and $+1$ with probability $0.75$ average to $\langle\hat{A}\rangle=0.5$ , the hollow mark. Every single measurement gives $\pm1$ ; only the average sits at $0.5$ .

The average of many measurements

The expectation value of an observable in a normalised state is the probability-weighted average of its eigenvalues. It can be computed from the state and the operator directly, without first finding the eigenvalues.

\langle\hat{A}\rangle=\sum_{n}a_{n}P(a_{n})=\bra{\psi}\hat{A}\ket{\psi}

Properties

It is real, because $\hat{A}$ is Hermitian, and it lies between the smallest and the largest eigenvalue.
It is linear in the observable: $\langle\alpha\hat{A}+\beta\hat{B}\rangle=\alpha\langle\hat{A}\rangle+\beta\langle\hat{B}\rangle$ for real .

The expectation value is a bracket

\sum_{n}a_{n}\bigl|\braket{a_{n}|\psi}\bigr|^{2}=\bra{\psi}\hat{A}\ket{\psi}

Write the observable through its spectral decomposition and sandwich it between the state's bra and ket. Each projector turns into the probability of its eigenvalue, so the bracket is exactly the probability-weighted average of the eigenvalues.

Proof steps

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

The spectral decomposition of the observable.

\bra{\psi}\hat{A}\ket{\psi}=\sum_{n}a_{n}\braket{\psi|a_{n}}\braket{a_{n}|\psi}

Put it between the bra and the ket, and take the numbers out of the sum's terms.

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

Each product is an overlap times its conjugate: the Born probability.

\bra{\psi}\hat{A}\ket{\psi}=\sum_{n}a_{n}P(a_{n})=\langle\hat{A}\rangle

The bracket is the average of the eigenvalues weighted by their probabilities.

Applications

Practice

A Weighted Average

The expectation value is each eigenvalue times its probability, added up. For $Z$ , whose eigenvalues are $\pm1$ , that is the difference of the two probabilities.

\langle Z\rangle=(+1)|\alpha|^{2}+(-1)|\beta|^{2}

Try it

What is $\langle Z\rangle$ in the state $0.6\ket{0}+0.8\ket{1}$ , where $Z=\ket{0}\bra{0}-\ket{1}\bra{1}$ ?