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...
 A: 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 $).
A: 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
