# Does an electron-positron wavefunction have to be antisymmetric?

1. May 11, 2012

### kof9595995

My first response to the question is "No", since electron and positron are distinguishable by their charge, so it's not necessary the wavefunction remains the same(up to +/-) upon permutation of particles. However on second thought, in the quantized Dirac field electron and positron creation operators anti-commute, which suggest an electron-positron wavefunction has to be anti-symmetric.
Seems to be a paradox, which part of the arguments went wrong?

2. May 11, 2012

### Bill_K

If you have a two-particle state consisting of an electron and a positron, Fermi statistics requires it to be antisymmetric under complete interchange of the two particles. But now there's an extra degree of freedom. You must interchange the spatial coordinates (P), interchange the spins (S) and interchange the charge (C). Particle interchange is the product of all three.

Space parity is (-)L. Spin parity is +1 for triplet state and -1 for singlet state. Charge parity is ±(-)L where the upper sign applies to singlet states and the lower sign applies to triplet states. Multiplying all three parities PSC together we always get -1.

3. May 11, 2012

### geoduck

Why is space correlated with charge? Assume you have a singlet state. If you choose an even value for L for space parity, then it seems that the wavefunciton must be even on exchange of just the two charges, since L is the same.

4. May 11, 2012

### Meir Achuz

The charge only enters if you want to add it as another degree of freedom.
The e and p don't have to be antisymmetric. The symmetry will be related to whether the state is even or odd under charge conjugation.

5. May 11, 2012

### Bob S

6. May 11, 2012

### Bill_K

Yes, there are two ways of looking at it. If you regard electron and positron as different types of particle, then the interchange only involves space and spin, and the wavefunction may be either symmetric or antisymmetric. On the other hand, you may regard them as charge states of the same particle. In which case you need to include charge parity and the wavefunction under interchange of all three variables is antisymmetric.

7. May 11, 2012

### geoduck

But as kof9595995 mentioned, the creation operators for positrons and electrons anticommute. This suggests that charge is being added as another degree of freedom. If instead of charge being added as a degree of freedom, you had the electron and positron as two separate particles, then would not the creation operators of different particles commute instead of anticommute?

Also, do the tau and the electron operators anticommute? How would you reconcile this with the view that charge is a degree of freedom? Would the extra degree of freedom be charge+mass, and not just charge by itself?

8. May 12, 2012

### Bill_K

The fact that the creation operators anticommute (even for different varieties of fermion) has nothing to do with particle interchange. Sure, you can write a*b*|0> = - b*a*|0>, but particle "a" still has the same momentum, spin, etc that it did before, and so does "b". You have not interchanged them.

9. May 12, 2012

### Dickfore

Positrons and electrons are described by 2-row subcolumns in the Dirac 4-spinor.

10. May 12, 2012

### kof9595995

I see, so the pitfall is the meaning of "interchange" depends on which perspective I take. To elaborate what you said, when treating electron and postron as different particles, interchange means,e.g.:
$$a^{\dagger (s)}_{p_1}b^{\dagger (r)}_{p_2}|0\rangle\rightarrow a^{\dagger (r)}_{p_2}b^{\dagger (s)}_{p_1}|0\rangle$$
but in general $a^{\dagger (s)}_{p_1}b^{\dagger (r)}_{p_2}$ and $a^{\dagger (r)}_{p_2}b^{\dagger (s)}_{p_1}$ do not anticommute, hence no antisymmetry.
Treating them as the same particles, interchange means
$$a^{\dagger (s)}_{p_1}b^{\dagger (r)}_{p_2}|0\rangle\rightarrow b^{\dagger (r)}_{p_2}a^{\dagger (s)}_{p_1}|0\rangle$$
then the two anticommute and a minus sign is acquired.

11. May 13, 2012

### geoduck

Yeah, but why two different rules? The first rule you have swaps labels, but not operators. The second rule you have swaps operators, but not labels. The "interchange" operation shouldn't really depend on what particles you have, whether they're identical or not, should it?

12. May 16, 2012

### kof9595995

Sorry for the late reply. I think "interchange" does depend on the perspectives. The reason is we want the definition of "interchange" as natural as possible in both cases. When we treat the two as different particles, the attribute of charges offers a natural labeling. By interchange we will interchange the wavefunctions associated to the labels, thus in the first case we keep charge fixed while interchange all other properties. When treating them identical particles with different states in charge, since charges are not labels but part of the wavefunctions, we have to interchange them also.