Clarification on measurement in QM

Supppose we are given a quantum state that isn't pure state, such that it is a linear combination of the eigenstates of a Hermitian operator $\hat O$. $$|\psi\rangle=N\sum \alpha_i |i\rangle$$ where $N$ is just a normalization constant. I just want to make something clear about measuring. If we measure the corresponding eigenvalue of $\hat O$ and find it to be in the $|1\rangle$ state then immediately measure again, would I be right in thinking that the probability of getting the $|2\rangle$ state is $0$?

I think that because I believe the wavefunction collapses to the $|1\rangle$ state upon the first measurement...

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Yes, it "collapses", so when you get 1, you can't immediately get a different result 2. However, the collapse isn't material in any sense. You should interpret it in terms of subjective knowledge, the validity of propositions, and conditional propositions. The statement that we can't get 2 immediately after the measurement when we got 1 is the statement that the conditional probability of the result's being 2 given the assumption that it was 1 a split second earlier is zero. –  Luboš Motl May 22 '13 at 10:34
Thanks, @LubošMotl ! –  valerie v. May 22 '13 at 10:44

If that operator provides a basis for the Hilbert space of your problem, and you actually can is do the linear combination, the wave function collapse is the following statement

$$|\Psi\rangle = N \sum_{i}\alpha_{i}|i\rangle \Rightarrow |\Psi\rangle_{>}=M \alpha_{j}|j\rangle$$

where $\Rightarrow$ means measurement of the observable linked to $\hat{O}$ and the subscript $>$ means the wave function after that measure. You can see that in the state

$$|\Psi\rangle=M \alpha_{j}|j\rangle$$

the probability of measuring any eigenvalue different from $\alpha_{j}$ is zero, this is, the probability of measuring $\alpha_{j}$ is one. However, due to the nature of the Schrödinger equation, the wave function will evolve into other linear combination after a certain amount of time

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Thank you, Nivalth! –  valerie v. May 22 '13 at 10:42

For simplicity, we could write your state as:

$$|\psi> = \alpha_1\,|1> + \alpha_2\,|2>$$ with $\alpha_1^2 + \alpha_2^2 = 1$ for correct normalisation.

Here $|1>$ and $|2>$ are eingenvectors of your hermitian operator $\hat O$

Be careful, this state $|\psi>$ is a pure state. A pure state is any linear combination of states. The general case, however, is mixed states. You will have to consider the density matrix : if the density matrix is factorisable, that is : if it exists $\psi$ such that $$\rho = |\psi> <\psi|$$ then the density matrix corresponds to a pure state. But it is only a very particular case.

Going back to your question, you are right: after your measurement, the state will be :

$$|\psi> = |1>$$

So, now the probabily of getting the $|2>$ state is :

$$|<1|2>|^2$$

But this probability is zero, because 2 eingenvectors corresponding to 2 different eigenvalues of an Hermitian operator are orthogonal ($<1|2> = 0$).

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Thanks, Trimok! –  valerie v. May 22 '13 at 10:43