2
$\begingroup$

I am reading chapter 6 of Sakurai's Modern Quantum Mechanics and have come across the 'symmetrization postulate', which tells me that for any given system of identical particles, all states must either be symmetric with respect to the exchange operator, or antisymmetric.

Why can we not have some states that are symmetric with respect to the exchange operator, and some that are antisymmetric? As far as I can tell, this question is not even referenced anywhere in the material I've been reading.

I understand that this is the symmetrization postulate, but is there any chance anyone could elucidate on why this is the case? Is this a consequence of quantum field theory?

$\endgroup$
5
  • $\begingroup$ see physics.stackexchange.com/a/614795/36194 $\endgroup$ Feb 26 at 1:26
  • 1
    $\begingroup$ You'd probably be interested in the anyon quasiparticles, note of course that these are not normal particles of course. $\endgroup$
    – Triatticus
    Feb 26 at 2:26
  • $\begingroup$ @OP: Are you asking why in the Hilbert space of $N$ identical particles contains either anti-symmetrized or symmetrized states (and not both)? Is that your question? $\endgroup$
    – Jakob
    Feb 26 at 9:19
  • $\begingroup$ Yes, that is my question. In other words, why can we not have a system of particles such that there exists a state $| \psi \rangle = | a, b; A \rangle + | a', b'; S \rangle$ that behaves under the exchange operator in a manner such that $P | \psi \rangle = -| b, a, A \rangle + | a', b'; S \rangle $? $\endgroup$ Feb 26 at 10:15
  • $\begingroup$ In fact, the more I am thinking about this, the more that it seems like identical particles are not necessarily indistinguishable at all. For instance, the EPR paradox and its resolution, we consider a Bell state consisting of two identical photons, that we can distinguish because one is provided to Alice and one is provided to Bob. $\endgroup$ Feb 26 at 10:16
-1
$\begingroup$

If particles are identical they follow exactly the same rules, therefore any operator on any of them affect any other particle in the same exact way. You can not choose or find any operator that would act on some in one way and on some others in a different way since they are identical. Symmetry or anti-symmetry comes when you think what could happen when to exchange two particles (not that you move them around, that would be a current, but rather you just swap with the other): $$\psi(r_1, r_2) \rightarrow \psi(r_2, r_1)$$ and you can always extend this to any number of particles since reorganizing any number of them can be done by pairs. In that case the state of the system should maintain $$|\psi(r_1, r_2)|^2$$ unchanged since they are identical. Then, under any reorganization of any number of identical particles any exchange of two of them would result in either symmetry or anti-symmetry of the wave-function or state.

$\endgroup$
2
  • $\begingroup$ Why can't that operation just multiplya phase factor $\exp(i\omega)$? $\endgroup$ Feb 25 at 23:23
  • $\begingroup$ The wave-function is invariant under a phase factor transformation in general, but trivial since the $w$ needs to affect all states at once and therefore it is global phase, giving you neither new information nor a different state and thus leaving nothing you can measure. $\endgroup$ Feb 26 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.