Wave function of multiple particles

1. Nov 21, 2012

randomafk

When dealing with n-particle systems that are identical, is the superposition of them just a mathematical construct, or is it similar to how the state of a single particle can be in multiple eigenstates until its measured.

For instance, if I have two fermions: $\Psi = \Psi_a(x_1)\Psi_b(x_2) - \Psi_b(x_1)\Psi_a(x_2)$ then are we describing it in this way only because we don't know which state it's in? Not necessarily because the state is in both configurations prior to measure?

And moreover, what does subtraction mean here? How can you subtract two states?

2. Nov 21, 2012

lugita15

Attached is an excerpt from Townsend's "A Modern Introduction to Quantum Mechanics", which explains how symmetric and anti-symmetric states work, assuming you already know the bra-ket mathematical formalism used in quantum mechanics. If the equations in the PDF look foreign to you, then I can provide you with a simpler explanation, but it would help to know how much quantum mechanics you already know.

Attached Files:

• Townsend Identical Particles Excerpt.pdf
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3. Nov 22, 2012

andrien

A factor of 1/√2 is missing for normalization,For fermions which obey exclusion principle the exchanged state must be negative of the original one i.e. when 1 and 2 are interchanged as 2 and 1,then the latter must form an anti-symmetric state together with first.

4. Nov 22, 2012

Staff: Mentor

The system is in the state you posted.
There are no particles with the labels "1" and "2", but our wave functions use those labels. As a result, you have to build a state where those labels do not matter (exchange of them just changes the sign for fermions).

5. Nov 22, 2012

randomafk

Hmm, okay.

So then, just like in a single particle system, if I measure the system then I have a 1/2 probability of collapsing the wave function into one of those two possible configurations? I'm not quite sure what you mean that the labels don't matter. I thought the problem lies in the fact that the particles are completely indistinguishable, but in truth there are still are two particles with labels 1 and 2. We just can't track them.

It just seems a little odd that if the particles are distinguishable (with no coupling), none of this applies and we can treat them classically as the total state being the product of the two individual states.

6. Nov 23, 2012

Staff: Mentor

You cannot measure $\Psi_a(x_1)\Psi_b(x_2)$ or $\Psi_b(x_1)\Psi_a(x_2)$.

Right.

Define "truth" - but no.

Well, if they are distinguishable we can assign labels 1 and 2 to them.