# Symmetric/Antisymmetric states in nature?

1. Jan 2, 2009

### hendriko373

For the composite system of identical particles only symmetric and antisymmetric states in the tensor-product (from the one-particle spaces) space are allowed to represent particles in nature. Why is that?

Is it an experimental fact which is used as an input in the theory of many particle QM?

Or

Is it a consequence of the commutation relation $$\left[P,Q\right]=0$$ with P the permutation operator and Q an observable (this commution relation is just the mathematical formulation for the indistinguishability of our many particle system)? This would conclude that P and Q have a common eigenbasis (but which space would span this basis?) whereas the eigenvectors from P are (anti)symmetric so that the action of Q on the system also puts the system in a (anti)symmetric state?

Hendrik

2. Jan 2, 2009

### tim_lou

I think it's precisely due to the second one (with Q the Hamiltonian). Whenever two hermitian operator commutes, you can simultaneously diagonalize both of them. The eigenvalues for P must be ±1 since P^2=1.

3. Jan 2, 2009

### hendriko373

Alright, this means that once your system is (anti)symmetric it stays that way because every observable commutes with P. But this does not explain why a many particle state is (anti)symmetric in the first place (or does it, and if so could you please clearify this)?

4. Jan 2, 2009

### hendriko373

Ok, maybe this is the explanation: because of the very nature of indistinguishability every operator Q must commute with every element of the permutation group. But this can only be true for symmetric or antisymmetric states, which form invariant subspaces of the product space of the one particle spaces, under permutations. These subspaces are the only two one dimensional invariant subspaces of the product space (or put it another way: the only two one dimensional representations of the permutation group). Other states belong to more dimensional representations and are rotated in a more dimensional invariant subspace (which alters their physical meaning and thus the eigenvalue of Q).

Actually I'm not sure about the last sentence, because although the permutation elements rotate the states I still think they stay in the same eigenspace of Q.

Hendrik

5. Jan 2, 2009

### tim_lou

Well, there is a theorem that says whenever two hermitian operators commute, you can choose a basis that simultaneously diagonalizes both of them. Usually, it is a choice that we make. In general, we can decomposes the Hilbert space into (edit) direct sums of symmetric and antisymmetric parts. Of course, at the end of calculations, the conclusions are independent of what basis you choose.

Now, then why is the hilbert space of bosons only the symmetric part? This is a postulate in QM and cannot be proved under its framework. The fact that particles are indistinguishable (i.e. all Q commutes with with P) forbid the existence of mixed state. By completeness of the hilbert space, then it must contain only the symmetric part or the antisymmetric part. This doesn't say anything about bosons or fermions though.

You can, however, go to second quantizations (in the framework of QFT), and show the postulate comes from imposing the relations

$$[\phi(x), \pi(x')]_\pm=i\hbar \delta^{3}(x-x')$$

Where plus indicates anti-commutation (for fermions) and minus indicates commutation (for bosons). This is simply a postulate of QFT and cannot be proved. The reason why we use anticommutation instead of commutation comes from spin statistics theorem which I don't know too much myself (which comes from Lorentz invariance, Locality and a couple other things...)

Last edited: Jan 3, 2009
6. Jan 3, 2009

### hendriko373

Why is that? Because for these states not all Pij commute with eachother?

So if I understand you well: it is within the framework of QM that we can conclude that only symmetric and antisymmetric states are physical. But to know which particles belong to the symmetrical part and which to the antisymmetrical we have to get into QFT.

7. Jan 3, 2009

### tim_lou

After some thinking, it seems that the reason why mixed state doesn't exist is quite delicate..

In usual QM texts, it is simply asserted that for indistinguishably to hold, the action of P on any subspace must be that subspace itself, hence P|a>= factor* |a>. I believe that this statement is equivalent to P commuting with all possible hermitian operators...

Here is a proof I just thought of:

Suppose P|ψ> = |ψ'>, |ψ'> does not lie in the subspace spanned by |ψ>. Let
|ψ'>=a|ψ>+ b|φ>. |ψ> and |φ> are orthonormal.
Fix any unitary operator, U that maps |ψ> to |ψ> (using Gram-Schmidt process if you wish), but rotates other vectors such that U|φ>≠ |φ'> (just make U|φ>=-|φ>). It can be proved that U=exp(iH) for some hermitian operator (H≠0 since U≠1). [H,P]=0 hence [U, P]=0.

but [U,P]|ψ>=U|ψ'> - P|ψ>=U|ψ'> - |ψ'> = 0
but U|φ>≠|φ>