Wherever I have read about exchange operator(P), it is stated that for two identical bosons it introduces a plus sign after exchange and minus sign for fermions. P and Hamiltonian(H) commute for two identical particles.Eigenvalues of P are $\pm1$. I think it commutes also for non-identical particles.

This is what I think.

Let the two non-identical particles are in state $\psi(\vec r_1,\vec r_2)$ and after exchange they are in the state $\phi(\vec r_1,\vec r_2)$(I believe even the wave function changes but I don't have any proof) and E is the total energy of the system which remains unchanged even after the exchange.

$H\psi(\vec r_1,\vec r_2)=E\psi(\vec r_1,\vec r_2),~~H\phi(\vec r_1,\vec r_2)=E\phi(\vec r_1,\vec r_2)$

$P\psi(\vec r_1,\vec r_2)=\phi(\vec r_1,\vec r_2),~~P\phi(\vec r_1,\vec r_2)=\psi(\vec r_1,\vec r_2)$

$PH\psi(\vec r_1,\vec r_2)=PE\phi(\vec r_1,\vec r_2)=E\psi(\vec r_1,\vec r_2)$

$HP\psi(\vec r_1,\vec r_2)=H\phi(\vec r_1,\vec r_2)=E\psi(\vec r_1,\vec r_2)$


  • $\begingroup$ Let's say my non-identical particles have different masses and I place them in a uniform (classical) gravitational field with the heavier particle higher up than the lighter one. If I exchange the particles is the energy unchanged? $\endgroup$ – By Symmetry Aug 22 '18 at 10:01
  • $\begingroup$ Nope, the energy changes. $\endgroup$ – Asit Srivastava Aug 22 '18 at 16:35

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