# QM: Find expectation value of measurements

We have an observable $$O \mapsto \begin{pmatrix} 0 & 2 \\ 2 & 4 \end{pmatrix}$$

We find the eigenvalues and eigenvectors by O$\psi$ = o$\psi$. The eigenvalues will give us the possible results of the measurement $O$: $o_1 = 2(1 + \sqrt2)$ , $o_2 = -2(\sqrt2 - 1)$.

The state of the system after each measurement is the superposition of the eigenvectors $\psi_n$. The state is then $$\psi = a\psi_1 + b\psi_2 = a\begin{pmatrix} -1 + \sqrt2 \\ 1 \end{pmatrix} + b\begin{pmatrix}-1 - \sqrt2\\ 1\end{pmatrix}$$

What will the expectation value of the measurements of $O$ be?

If ${\displaystyle O}$ has a complete set of eigenvectors ${\displaystyle \phi _{j}}$, with eigenvalues ${\displaystyle o_{j}}$

${\displaystyle \langle O\rangle _{\psi_k }=\sum _{j}o_{j}|\langle \psi_k |\phi _{j}\rangle |^{2}}$ .

or

$\langle O \rangle_\psi = \langle \psi | O | \psi \rangle$

• Indeed if the measurement returns $o_k$ then the state "collapses" to an eigenstate of $O$ with eigenvalue $o_k$. In your case the eigenvalues are distinct to your initial state will collapse to a single eigenstate - call it $\psi_k$. The evaluation of the average value of $A$ is otherwise correct if the OP properly changes $\psi\to \psi_k.$ – ZeroTheHero Mar 23 '17 at 0:51
• From Wikipedia:"Any quantum state can be represented as a superposition of the eigenstates of an observable.". Why isn't this statement valid in my case? How can I find this single eigenstate? – JimiChango Mar 23 '17 at 8:41
• @JimiChango 1) The state after measurement is NOT the superposition of eigenstate. 2) you cannot find the eigenstate because you have not specified the outcome of the measurement. This could be useful: en.m.wikipedia.org/wiki/… – ZeroTheHero Mar 24 '17 at 2:55

• Thanks, yang, that had completely slipped my mind. But just to clarify, we're still talking about a density matrix representing an epistemic mixture (the measured system is in a pure state, but our knowledge of which one is incomplete, modulo real weights $w_i$), not in an ontological superposition (that can, e.g., exhibit interference between the pure states comprising the superposition). – John Forkosh Mar 24 '17 at 5:22