Suppose I have a state $|\psi\rangle$ and I want to estimate the probability of obtaining a computational basis state $|x\rangle$. Then by Born rule:

$$ p(x) = |\langle x|\psi\rangle|^2 = Tr[|x\rangle \langle x|\psi\rangle \langle \psi|]. $$

However, I could alternatively achieve the same by defining an observable $O = |x\rangle\langle x|$. This satisfies the definition of observable - $O$ is Hermitian; and the only projector $|x\rangle\langle x|$ is Hermitian and $(|x\rangle\langle x|)^2 = |x\rangle\langle x|$. So now I can calculate expectation value of the observable:

$$ \langle O \rangle = Tr[O|\psi \rangle \langle \psi|] = Tr[|x\rangle \langle x|\psi\rangle \langle \psi|] = p(x). $$

Is this reasoning correct?


1 Answer 1


Yes, the reasoning is correct.

In fact, it can be generalized beyond pure states. By definition, every mixed quantum state $\rho$ is a positive semidefinite operator with unit trace. Since every positive semidefinite operator is Hermitian, we may interpret $\rho$ as an observable.

In this case, the expectation of observable $\rho$ in state $\sigma$

$$ \langle\rho\rangle_\sigma = \mathrm{tr}(\rho\sigma) = \langle\sigma\rangle_\rho $$

is the Hilbert-Schmidt inner product of the two operators. Moreover, if one of the states is pure then $\mathrm{tr}(\rho\sigma)$ is the fidelity $F(\rho, \sigma)$ between $\rho$ and $\sigma$.


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