The symmetrisation postulate is known for stating that, in nature, particles have either completely symmetric or completely antisymmetric wave functions. According to these postulate, these states are thought to be sufficient do describe all possible systems of identical particles.

However, in Landau Lifshitz Quantum Mechanics, in the first page of Chapter IX - Identity of Particles, he comes to the same conclusion without needing to state any ad-hoc postulate.

It goes like this: Let $\psi(\xi_1,\xi_2)$ be the wave function of the system, $\xi_1$ and $\xi_2$ denoting the three coordinates and spin projection for each particle. As a result of interchanging the two particles, the wave function can change only by an unimportant phase factor: $$ \psi(\xi_1,\xi_2)=e^{i\alpha}\psi(\xi_2,\xi_1) $$ By repeating the interchange, we return to the original state, while the function $\psi$ is multiplied by $e^{2i\alpha}$. Hence it follows that $e^{2i\alpha}=1$ or $e^{i\alpha}=\pm1$. Thus $$ \psi(\xi_1,\xi_2)=\pm\psi(\xi_2,\xi_1) $$ Thus there are only two possibilities: the wave function is either symmetrical or antisymmetrical.

It goes on by explaining how to generalize this concept to systems with any number of identical particles, etc.

Im summary, no symmetrisation postulate was ever stated in this rationale. Is "shifting by an unimportant phase factor" a too strong requirement for ensuring identity of particles?

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    $\begingroup$ "As a result of interchanging the two particles, the wave function can change only by an unimportant phase factor" is a postulate (for me) $\endgroup$ – user26143 Aug 9 '13 at 16:01

The way Shankar addresses the problem (pg. 278) is by introducing an "Exchange Operator" $P_{1,2}$, which would swap your two particles as follows:

$P_{1,2} |\xi_1, \xi_2 \rangle = |\xi_2, \xi_1 \rangle$

I like the operator notation because it makes it clear (to me, at least) that applying the operator twice is just the identity operator, since swapping two particles twice just gets you back to your original state:

$P_{1,2}^2 |\xi_1, \xi_2 \rangle = |\xi_1, \xi_2 \rangle \longrightarrow P_{1,2}^2 = 1$

This shows that the eigenvalues of the swap are $\pm 1$, meaning your wave function is only either symmetric or anti-symmetric, although there is an implicit assumption that the system in question is in fact an eigenvector of the exchange operator. This is true of particles in the Standard Model in three dimensions, but not generally true (see anyons, for example).

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    $\begingroup$ Can we get anyon from this approach? $\endgroup$ – user26143 Aug 9 '13 at 23:54
  • $\begingroup$ How does he explain anyons then ? $\endgroup$ – gatsu Aug 9 '13 at 23:55
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    $\begingroup$ -1: (I'll glady remove the downvote if you strengthen the answer). Your argument tacitly assumes that a vector representing the state of a system of identical particles is an eigenvector of the exchange operator. This fact, or something equivalent, cannot be derived; it is an extra physical input in the mathematical model. In other words, some postulate is necessary. $\endgroup$ – joshphysics Aug 10 '13 at 6:58
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    $\begingroup$ It seems correct to assume that the vector representing the state of two identical particles is an eigenvector of $P^2_{1,2}$ because two consecutive exchanges give back the original state. And since $P^2_{1,2}$ commutes with $P_{1,2}$ then the vector $|\xi_1,\xi_2\rangle$ is also an eigenvector of $P_{1,2}$ $\endgroup$ – Gilberto Aug 10 '13 at 15:09
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    $\begingroup$ @Gilberto: $P^2$ is operator identity per definition, so every function is its eigenfunction. Although identity commutes with $P$, this does not allow us to infer that functions to be used in physics are necessarily eigenfunctions of $P$. Joshphysics above is right on this point, assuming eigenfunction of $P$ is an additional assumption. $\endgroup$ – Ján Lalinský Dec 6 '13 at 21:08

Check out the section about Chapter 17 Identical particles in Ballentines, he not only points out why looking at the Permutation operators of two particles in a multi particle setting is misleading but also discusses some errors in previous claims.

  • $\begingroup$ Hi user35388, can you please provide a short summary of this? We expect questions to be mostly self-contained. $\endgroup$ – Brandon Enright Dec 6 '13 at 21:17

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