Permutation operator and Hamiltonian

  • #1
367
13
The permutation operator commutes with the Hamiltonian when considering identical particles, which implies:

$$ [\hat{P}_{21}, \hat{H}] = 0 \tag{1}$$

Now given a general eigenvector ##{\lvert} {\psi} {\rangle}##, where

$$ \hat{P}_{21} (\hat{H}{\lvert}{\psi}){\rangle} = (\hat{P}_{21} \hat{H}) {\lvert} {\psi}{\rangle} + \hat{H}(\hat{P}_{21}{\lvert}{\psi}{\rangle}) $$

Using (1):
$$
(\hat{P}_{21} \hat{H}){\lvert} {\psi}{\rangle} = 0
$$

But how exactly does this last relation follow? Why does acting the permutation operator on the Hamiltonian result in 0? In this case, if the Hamiltonian is symmetric with respect to permutation, how does this term going to 0 indicate that?
 

Answers and Replies

  • #2
DrClaude
Mentor
7,428
3,702
Now given a general eigenvector ##{\lvert} {\psi} {\rangle}##, where

$$ \hat{P}_{21} (\hat{H}{\lvert}{\psi}){\rangle} = (\hat{P}_{21} \hat{H}) {\lvert} {\psi}{\rangle} + \hat{H}(\hat{P}_{21}{\lvert}{\psi}{\rangle}) $$
I don't understand how you get that equation.
 
  • #3
367
13
I don't understand how you get that equation.
Yikes. Looking back at it, I don't quite understand my step either. I was under the assumption the operator ##\hat{P}_{21}## was acting on the total state determined by ##(\hat{H}{\lvert}{\psi}{\rangle})## and thus expanded it out to act on each term in that state, but switching the order of operation here seems very wrong.
 

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