The permutation operator commutes with the Hamiltonian when considering identical particles, which implies:(adsbygoogle = window.adsbygoogle || []).push({});

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

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?

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# I Permutation operator and Hamiltonian

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