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

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.
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}