7

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?

Qmechanic
  • 201,751
Gilberto
  • 305
  • 3
    "As a result of interchanging the two particles, the wave function can change only by an unimportant phase factor" is a postulate (for me) – user26143 Aug 09 '13 at 16:01

2 Answers2

5

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).

Kevinismus
  • 316
  • 1
    Can we get anyon from this approach? – user26143 Aug 09 '13 at 23:54
  • How does he explain anyons then ? – gatsu Aug 09 '13 at 23:55
  • 7
    -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. – joshphysics Aug 10 '13 at 06:58
  • 2
    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 $P_{1,2}$ – Gilberto Aug 10 '13 at 15:09
  • @joshphysics: thanks for the comment; I've edited my answer to incorporate it. I must admit that I didn't know anything about anyons prior to this (above my pay-grade as an experimental particle dunce). – Kevinismus Aug 12 '13 at 23:12
  • 2
    @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. – Ján Lalinský Dec 06 '13 at 21:08
  • @user26143 No. The bosons and fermions come from the one-dimensional irreducible representations (IRs) of the permutation group. The generalization to higher dimensional IRs is known as para-statistics. Anyons, on the other hand, loosely speaking, comes from the IRs of the pure braid group, not permutation group. – Isidore Seville Dec 07 '13 at 05:05
2

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.

Brandon Enright
  • 11,937
user35388
  • 21