# Does collapse violate exchange symmetry?

1. Nov 27, 2008

### Demystifier

Consider a system of two bosons with spin, say photons. The wave function must be symmetric under the exchange of the two particles. For example,
|psi> = |up> |down> + |down> |up>
So far, so good. Now let us measure the spin of the particles, and let the outcome of the measurement be that the first particle is in the state |up> and the second particle is in the state |down>. If measurement induces the collapse, then after the measurement the wave function is
|psi'> = |up> |down>
However, |psi'> is NOT symmetric under the exchange of the two particles.
Does it mean that the collapse violates the exchange symmetry?

2. Nov 27, 2008

Just a guess

It should not be possible to violate the exchange principle. Since bosons seem to live in symmetrized spaces, which don't allow for non symmetric functions. I think the key lies with the fact that the principle only holds for indistinguishable particles.
So either you don't know which particle you measure the spin of, or the knowledge of the spin makes the particles distinguishable. I tend to the latter explanation since it should be possible to distinguish two bosons which display some hyperfine splitting, but I have to give it some more thought.

3. Nov 27, 2008

From Wikipedia
Spin is an observable which changes dependent on the particles' exchange, so the exchange principle does not apply after the measurement anymore.

4. Nov 27, 2008

### Count Iblis

Include a label for the position in the kets:

|up, x1>|down,x2>

This has to be symmetrized w.r.t. interchange of the two particles:

|up, x1>|down, x2> + |down, x2>|up, x1>

Then we measure the spin of whatever particle there happens to be at position x1. In this case the measurement will yield up with 100% probability. You can write down other possible states in which the meurement outcome is not certain, then these will collapse into a state of the above form.

5. Nov 27, 2008

### malawi_glenn

Demystifier, please state what operator you are using to make that collapse.

6. Nov 27, 2008

### Count Iblis

Another way to look at this problem is to consider two electrons both with spin up. The question would then be how you can have two electrons in the spin up state as such a state would necessarily be symmetric.

Of course, the answer is that you have to write down the complete wave function and that complete wavefunction is anti-symmetric. So, it could be a state like:

|up, x1> |up, x2> - |up,x2>|up,x1>

If x1 = x2, then it becomes impossible to have an anti-symmetric state with identical spins.

7. Nov 28, 2008

### Demystifier

OK, I knew the right answer all the time, but I wanted to see how other people will respond to this apparent "paradox".

The answers provided by Count Iblis is essentially correct, but let me make it slightly more precise.

I was talking about "first" particle and "second" particle, but these notions do not make sense when the particles are indistinguishable. What does make sense, however, is, e.g., a particle localized on the left (with the wavefunction $$\psi_L(x)$$) and a particle localized on the right (with the wavefunction $$\psi_R(x)$$). Thus, instead of the wave function |up>|down>, the correctly written wave function after the measurement is
$$\psi_L(x_1) |up> \otimes \psi_R(x_2) |down> + \psi_L(x_2) |up> \otimes \psi_R(x_1) |down>$$
This wave function is indeed symmetric under the exchange of $$x_1$$ and $$x_2$$.