Collapse means that any future measurements/interactions of the wave function can be assumed to have evolved from the measured state. Most measurements themselves normally only tell you with high probability information about state, but that's fine.

If you have a wave function which can be in two different states X1 and X2, then it's full state is going to be some linear combination of the states X= a X1+b X2. It can, over time, evolve deterministically, with a and b potentially becoming larger and smaller over time.

When a measurement occurs, such as the detection of a photon of a certain frequency which indicates that the particle made a state transition, you then assign all but one of the coefficients to zero (or approximately zero depending on the measurement). If the particle is in a certain compatible potential (one in which the measured state has a specific energy), the coefficients will stay at zero.

To answer your question: " Isn't the measuring instrument itself a wave function?" Yes it is. Unfortunately, if you have two particles, to describe the system with both particles in it, as opposed to just one, you don't just add the particle wave functions together, you have to take the cross (well tensor, but it's similar) product of the two vector spaces of the particles (remember, function spaces are vector spaces). This is true _even if they do not interact with each other_.

Yudkowsky has a pretty good description of this http://lesswrong.com/lw/pp/decoherence/

Thanks for your good answer! Quantum theory is really strange and amazing.