What is often called the Indistinguishability Postulate is expressed in (at least) two different ways depending on the textbook.

For any normalized composite states of N identical particles $|\psi \rangle$ in $H^{N}$, and observable O on $H^{N}$, and any permutation operator $P$ in the permutation group $S_{N}$,

  1. $\langle \psi | O | \psi \rangle = \langle \psi |P^{\dagger}OP|\psi\rangle $. (That is, $[P, O] = 0$.)


  1. $\langle \psi|O| \psi \rangle = \langle P \psi |O | P \psi \rangle$.

I take it that 1 and 2 are equivalent, that is, 1 is true if and only if 2 is true. But apparently, the are saying two different things. 1 is a restriction on which operators can represent observables, and 2 is a restriction on which vectors can represent physical states. How can we show that 1 and 2 are actually equivalent?

Furthermore, is there any reason to prefer one over the other as a better articulation of the postulate?


$$ | P \psi \rangle = P | \psi \rangle $$ by definition. Hence $$ \langle P \psi | = \langle \psi | P^\dagger . $$ So $$ \langle P\psi|O| P\psi \rangle = \langle \psi | P^\dagger O P | \psi \rangle $$ by definition. That is, the meaning of the notation on the left hand side is given by the expression on the right hand side. So your question is not looking at two different expressions, but two ways of writing the same expression.


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