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When discussing identical particles books often use that the states are eigenstates of the permutation operator:

$P_{ij}|\psi\rangle = \lambda |\psi\rangle$

for bosons this is easy to see if I use the commutation relation: $[P_{ij}, H ] =0$

$P_{ij} |\psi\rangle = P_{ij}\frac{E}{E}|\psi\rangle = \frac{1}{E}P_{ij}H|\psi\rangle = \frac{1}{E} ( H P_{ij}|\psi\rangle) = \frac{E}{E}|\psi\rangle = |\psi\rangle$

I used that the eigenenergy is (as an observable) invariant:

$ H P_{ij}|\psi\rangle= E|\psi\rangle $

Where did I go wrong? I don't see where I'd introduce a $-1$ or even a $\lambda$. Once I have $\lambda$ the $\pm 1 $ is easy.

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  • $\begingroup$ Are you sure that $\frac{1}{E} ( H P_{ij}|\psi\rangle) = \frac{E}{E}|\psi\rangle$ is true for fermions? $\endgroup$
    – Bosoneando
    Jul 22, 2016 at 15:12
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    $\begingroup$ In step 4 to 5 you're assuming $P_{ij}\psi=\psi$. However, as you stated earlier, $P_{ij}\psi=\lambda\psi$ with $\lambda=1$ or $-1$. In particular, $-1$ for fermions. $\endgroup$
    – Daniel
    Jul 22, 2016 at 15:12

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In one of your hang you are already assuming that $P_{ij}|\psi\rangle = |\psi\rangle$, which not true. One way to see that the permutation operator has possible eigenvalues $\pm 1$ is by using that exchanging two identical particles and then exchanging them back should give back the original state,

$$P_{ij}P_{ij}|\psi\rangle = P_{ij}\lambda|\psi\rangle = \lambda P_{ij}|\psi\rangle = \lambda \lambda |\psi\rangle = \lambda^2|\psi\rangle = |\psi\rangle.$$

In the first line we just apply two swap operators to an eigenstate with eigenvalue $\lambda$. We see that we get back original eigenstate multiplied by $\lambda^2$, but we required that we would also get our original state back. This gives $\lambda^2 = 1$, which has solutions $\lambda \pm1$.

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  • $\begingroup$ I'm not assuming this. I use $H(P_{ij}|\psi>) = H|\psi>$ i.e that the permutation does not change the eigenenergy (which is an observable) $\endgroup$
    – Stein
    Jul 22, 2016 at 17:54
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    $\begingroup$ Indeed, permutation does not change the eigenenergy. But this only means that $H(P_{ij})|\psi\rangle = E (P_{ij})|\psi\rangle$, but not necessarily that $H(P_{ij})|\psi\rangle = H |\psi\rangle = E |\psi \rangle$! $\endgroup$ Jul 22, 2016 at 18:33
  • $\begingroup$ Ok you're right. $\endgroup$
    – Stein
    Jul 22, 2016 at 18:44

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