# 2 particle exchange operator P

1. Apr 3, 2015

### jl29488

Trying to derive two functions which are eigenfunctions of the hamiltonian of 2 identical and indistinguisable particles and also eigenfunctions of the 2-particle exchange operator P.

Need some help with my workings I think.

Have particle '1' and particle '2' in a hamiltonian given as:

$$\hat{H} = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} +V(x_1,x_2)$$

And for particles 1,2 to be indistinguisable, H must be symmetric w.r.t particle interaction, so

$$\hat{H}(1,2)=\hat{H}(2,1)$$

So the time independent S.E can be:

$$\hat{H}(1,2)\psi(1,2)=E\psi(1,2) \\ and\\ \hat{H}(2,1)\psi(2,1)=E\psi(2,1)$$

$\psi(2,1)$ is also an eigenfunction of H(1,2) belonging to the same eigenvalue E.

And for a linear hermitian exchange operator P:

$$\hat{P}f(1,2)=f(2,1)$$

Therefore, can write:

$$\hat{P}[\hat{H}(1,2)\psi(1,2)]=\hat{H}(2,1)\psi(2,1)=\hat{H}(1,2)\psi(2,1) = \hat{H}(1,2)\psi(1,2)$$

Not sure where to go from here....??

2. Apr 5, 2015

### Einj

You can't really derive the complete wave function if you don't know explicetely the expression for the potential. However, on general grounds, suppose that you are able to solve the S.E. for the Hamiltonian H(1,2) and hence find the normalized wave function, $\psi(1,2)$. Now you have two cases: (a) your particles are identical bosons and (b) they're identical fermions. In the first case the wave function must always be symmetric under the action of the exchange operator P. Now if $\psi(1,2)=\psi(2,1)$ you're lucky and that's automatically satisfied, otherwise you will have to symmetrize your wave function as $\psi_{symm}=(\psi(1,2)+\psi(2,1))/\sqrt{2}$.
For fermions, instead, the complete wave function must be anti-symmetric under P, so either $\psi(1,2)=-\psi(2,1)$ or you have to define $\psi_{anti-symm}=(\psi(1,2)-\psi(2,1))/\sqrt{2}$.

I believe that if you don't know the form of the potential you can't really go much further.