Expectation Values with ρ

The density operator turns expectation values into a single, basis-independent formula that works for pure and mixed states alike.

The master formula

For any observable $A$ (a Hermitian operator) and any state $\rho$,

That one trace replaces the pure-state expression $\langle\psi|A|\psi\rangle$ and the ensemble average $\sum_i p_i \langle\psi_i|A|\psi_i\rangle$ — both are special cases.

Why it works

Start from the ensemble average over the pure states making up $\rho$:

Use the cyclic property of the trace, $\langle\psi_i|A|\psi_i\rangle = \operatorname{Tr}\big(A|\psi_i\rangle\langle\psi_i|\big)$. Pull the sum inside the (linear) trace:

The last equality again uses cyclicity, $\operatorname{Tr}(A\rho) = \operatorname{Tr}(\rho A)$. The result depends only on $\rho$, never on which ensemble produced it — exactly the ambiguity-free behavior we wanted.

Reading off probabilities

Measurement probabilities are a special case. The probability of outcome $k$ with projector $P_k = |k\rangle\langle k|$ is

So the diagonal entries of $\rho$ are precisely the computational-basis outcome probabilities — which is why we called them populations.

Try it

For the mixed state $\rho = \begin{pmatrix}3/4 & 1/4\\ 1/4 & 1/4\end{pmatrix}$, compute $\langle Z\rangle = \operatorname{Tr}(\rho Z)$ with $Z = \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$. The answer is the difference of the two populations.

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