Deriving the Commutator of Exchange Operator and Hamiltonian

[Samama Fahim]

In the boxed equation, how would you get the right hand side from the left hand side? We know that $H(1,2) = H(2,1)$, but we first have to apply $H(1,2)$ to $\psi(1,2)$, and then we would apply $\hat{P}_{12}$; the result would not be $H(2,1) \psi(2,1)$. $\hat{P}_{12}$ is the exchange operator and $H(1,2)$ is the hamiltonian.

Source: https://books.google.com.pk/books?i...inction as to which particle is which&f=false

 

[HomogenousCow]

$$P_{12}H(1,2)\psi(12) = P_{12}H(1,2){P_{12}}^\dagger P_{12}\psi(1,2) = H(2,1)\psi(2,1)$$
using $P_{12}^2 = I$ and ${P_{12}}^\dagger = P_{12}$.