# The collapse of one wave function is the creation of another

1. Jan 20, 2009

### thenewmans

Poor title. Actually I have a whole bunch of wave function questions. I don’t know the boundaries of this concept. Assuming a correct wave function, can a particle have more than one? Can 2 observers each have their own wave function? The moment a particle encounters another particle, does one wave function collapse and a new one pop up? (I’m not necessarily thinking of a wave function as an actual thing. It just comes out that way.) If you don’t know a particle is entangled can your wave function still be correct? When an entangled particle encounters some other particle, does that make it no longer entangled?

2. Jan 20, 2009

### muppet

No. The wave function is a unique description of a state, and tells you everything you could possibly find out about the results of measurements you haven't performed yet. However, what you can have are different representations of the same wave function, which superficially look different.
Sure :tongue: but I think what you're really asking is whether or not it's possible for the two observers to describe a system accurately by two different wave functions, which is really the same question as the one above, so no. (*Relativity messes up observers' common perception of space and time, so in relativistic QM this answer might not hold, but I'm not the person to ask about that I'm afraid.)
No. The "collapse of the wavefunction" usually refers to something specific. In QM there is a special class of wave functions called eigenfunctions, or eigenstates. After any precise measurement of a physical observable, the wavefunction of the system is said to be in an eigenstate of that observable. The maths of QM says that any wave function -at all- can be represented by a sum of eigenstates; the idea of wavefunction collapse is that when you measure some quantity, all the eigenstates in the sum vanish apart from the one corresponding to the result of the measurement.
You're right, however, in that the wave function of an entangled particle is different from that of a "lone" one. What happens is that you need to start talking about the wavefunction for the whole system. This is usually based on products of the wavefunctions of the individual wavefunctions that comprise the system, but there's a few technical adjustments that need to be made, relating to the fact that "quantum particles" are intrinsically indistinguishable. For example, the pauli exclusion principle says that two fermions must be described by a wavefunction that yields zero probability of them being in the same state.
In the sense that you should really be talking about the wavefunction of the whole system, the answer is no. But the other particle might not affect the results of your measurements on the particle you know about. For example, say you're measuring the spin of the particle about the z-axis. The entanglement may be such that the spins of the two particles are anti-correlated (so if you measure spin up on one, you know that the other will be spin down). But it doesn't affect the probability of you measuring either spin up or spin down on the particle that you do know about (which, all other things being equal, is exactly 50%).
Not sure. But I think the answer depends on what kind of interaction you're talking about. It might be that the new particle becomes entangled with the old ones to become a part of the new system; but some interactions (like measurement) can destroy entanglement.

Hope that helps!

3. Jan 20, 2009

### Dmitry67

A perfect example of what we discuss here:

4. Jan 20, 2009

### thenewmans

muppet,

Thank you so much. That's just what I was looking for.

5. Jan 20, 2009

### muppet

You're welcome
Dmitry67: I deliberately didn't mention MWI because it's only something you're in a position to understand once you understand entanglement as an independent concept, although I appreciate it has very direct relevance to the final question in particular.