Post-measurement $\psi$ in quantum mechanics I have a question regarding the wave function after a measurement. Everything I found online says that this is the following formula: 
$\psi = \frac{M_m\psi}{\sqrt{P(m)}}$
Where $P(m)$ is the probability of observing m, the $\psi$ on the left is the wavefunction AFTER the measurement and the $\psi$ on the right is the original wavefunction. However, I cannot find a good definition on how I would go about calculating $M_m$? The Berkley lecture notes say that this is the measurement operator, but how would I go about finding that for my specific problem?
Also the probability function is $P(m) = |<\psi|\omega>|^2$, how would I find $\omega$ in this case? Is it just the eigenstate at that observable? 
 A: You have to think about what a measurement is in QM. Generally speaking, you are going to measure an eigenvalue of an observable. Lets say that you measure $A$ that follows the eigenstate equation $A|a>=a|a>$. Then if you begin with a general state $|\psi>$, after the measurement you will get another state $|\psi'>$ following the equation:
$$|\psi'>=\frac{|a><a|\psi>}{\sqrt{P(a)}}$$
Where $|a><a|$ is called the projector (or measurement operator by your lectures). It basically gives you the part of $|\psi>$ that is in the subspace of eigenstates of eigenvalue $a$.
Physically it means that after a measurement is done, you can be sure that the state of the system is in a subspace with eigenvalue $a$, so that any other measurement of the obsevable $A$ that you make on the system afterwards will spit out the eigenvalue $a$ all the time (if you keep the system unchanged).
So you if you want to calculate $M_m$, which is the projector into the subspace of some eigenvalue, just find the eigenstates of that eigenvalue (lets say $A|a_n>=a|a_n>$ for $n=0,1,...,g$). Then.
$$M_m=\sum_{n=0}^{g}|a_n><a_n|$$
A: It is instructive if you work with some simple setups. Consider the example of a spin-half system where you’re measuring the spin along the z-axis. We know that the possible outcomes are either up or down. So in general your Say your pre-measurement state is a normalised linear combination:
$$|\psi\rangle=\alpha|u\rangle+\beta|d\rangle$$

What this says is that probability of observing the state to be up when measured is given by:
$$P(u)=\big|\langle u|\psi\rangle\big|^2=|\alpha|^2$$
And the probability to observe a down state under spin measurement is similarly given by:
$$P(d)=\big|\langle d|\psi\rangle\big|^2=|\beta|^2$$
Remember that after measurement of any observable the state will always be in exactly one of the eigenstates. Like in this example, the state after measurement is either up or down. And not a linear combination of the two. 
