# Permutation operator

1. Nov 16, 2007

### cks

I can't really imagine how this was approached.

Let $$P_{\alpha0}$$ fixed

$$P_{a0}A=\frac{1}{N!}\sum_{\alpha}\epsilon_{\alpha}P_{\alpha0}P_{\alpha}=\frac{1}{N!}\epsilon_{\alpha0}\sum_{\alpha}\epsilon_{\beta}P_{\beta}=\epsilon_{\alpha0}A$$

I can understand that $$P_{\alpha0}P_{\alpha} = P_{\beta}$$ is a new permutation operator.

$$P_{a0}A=\frac{1}{N!}\sum_{\alpha}\epsilon_{\alpha}P_{\alpha0}P_{\alpha}=\frac{1}{N!}\sum_{\alpha}\epsilon_{\alpha}P_{\beta}$$

2. Nov 18, 2007

### Vazier

can you tell me where you got this from or what is this about?

what i think happened was that

Ea = EB x Ea0 (sorry i dont know how to put in the greek words and the subscripts here)

so since Ea0 is a constant it was brought out of the equation, then the summation divided by the N! was equal to A.... and hence we get the answer

3. Nov 20, 2007

### cks

Actually, P is the permutation operator that we frequently come across from a chapter of identical particles of any quantum textbook.