# Does measuring the operator of a wave function collapse the wave function to the measured eigenstate?

Suppose you have a state described by the wave function $\psi(x) = \phi_1(x)+2\phi_2(x)+3\phi_3(x)$ , where the $\phi$s are normalised eigenfunctions of a Hermitian operator $\hat{O}$ with eigenvalues $\lambda_1 = 1, \lambda_2 = 5, \lambda_3 = 9$.

What happens to the wave function if you measure $\hat{O}$ and you obtain the result $\lambda_2=5$?

My answer is that the wave function immediately collapses to $\psi(x) = 5\phi_2$

EDIT: My answer should be $\psi(x) = 10\phi_2$, I think

Is that correct? Or should I state the probability with which the wave function will collapse to that state:

$<\hat{O}> = \sum|c_n|^2\lambda_n$ where $|c_n|^2$ is the probability of the $n^{th}$ state.

• If the $\phi$ are normalized, then your $\psi$ is not normalized, making this question difficult to understand. – Mark Eichenlaub Nov 6 '13 at 16:59
• @MarkEichenlaub The normalised wave function is $\psi(x) = \sqrt{1/14}(\phi_1(x)+2\phi_2(x)+3\phi_3(x))$ – turnip Nov 6 '13 at 17:06
• Your final wave function, whatever it is, should be normalized. This will tell you what coefficient to put out in front (if any). – BMS Nov 6 '13 at 17:58

## 2 Answers

Well wavefunctions are technically rays, so what you wrote is acceptable, although usually one would normalize the wavefunction again. Once you have measured $\hat O$ and obtained $5$, you are past the point of asking about probabilities. Before measurement you have probabilities but it's like rolling dice. Once the dice is rolled and you got $6$, you don't ask what the probability is of having a $4$ because it is clearly zero.

Put another way, probability in quantum mechanics means that if you have $n$ identically prepared systems, and you perform the measurement on each one of them, then the fraction of times you get a result $x$ is given by some probability function $p(x)$ as $n\to \infty$. If you only perform the measurement once, or have already performed the measurement, then the probability is simply not defined.

Btw, all this wavefunction collapse business is just an idealization, because all our experiments point to the fact that everything is ruled by quantum mechanics, and unitarity is a central tenet of quantum mechanics. Wave function collapse is not a unitary process, decoherence and einselection is. Cheers.

I think there is a little confusion here: the thing you wrote in the last line is the expectation value of a measurement on the state $\psi$: $\langle\psi|\hat O|\psi\rangle \equiv \langle \hat O \rangle$. However, the expectation value of finding your system in the state $|\phi_n\rangle$ after a measurement $\hat O$ is $\langle\phi_m|\hat O|\psi\rangle$, but I get a different result from yours: \begin{align} \langle\phi_m|\hat O|\psi\rangle &= \langle\phi_m|\sum_n |\phi_n\rangle\lambda_n \langle\phi_n|\sum_i c_i |\phi_i\rangle\\ &= \sum_n\sum_i c_i\lambda_n\langle\phi_m|\phi_n\rangle\langle\phi_n|\phi_i\rangle \\ &= \sum_n\sum_i c_i\lambda_n\delta_{mn}\delta_{in}\\ &= c_m\lambda_m \end{align}

Thus, the expectation value of finding your system in eigenstate 2 is 5*2 = 10.

• "the expectation value of finding your system in eigenstate 2" I'm not sure that's really a thing. – lionelbrits Nov 6 '13 at 17:20