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$\hat P$ is the exchange operator, the standard derivation of its eigenvalues, $\pm 1$, takes advantage of the fact that exchanging the particles two times changes nothing. Mathematically: $$\hat P \hat P\psi(1,2)=\psi(1,2)$$

However, I'm not convinced of this condition because two wave functions that differ just by a phase factor are the same quantum state. So the equation that tells that changing two times the particles changes nothing should be: $$\hat P \hat P\psi(1,2)=e^{i\lambda}\psi(1,2)$$

I suppose that, if the second equation is wrong, it means that the phase factor change something in the quantum state.

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  • $\begingroup$ Have you heard of anyons? $\endgroup$ Jun 23 at 14:07

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If the exchange operator is viewed as a permutation operator (here on two particles), then $\hat P^2=\mathbb{I}$ follows because permuting twice two objects is equivalent to no permutation.

Within the context of spin statistics, exchange operators are viewed as permutation operators so, for two particles $\hat P^2=\mathbb{I}$ will always hold. For $n$ particles one will have $\hat P_\sigma^n=\mathbb{I}$ for any permutation $\sigma$, although there might be some smaller exponent $k$ for which some permutations satisfy $P_\mu^k=\mathbb{I}$. (Transpositions, which always permute only two particles, always satisfy $\hat P_k^2=\mathbb{I}$.)

It is possible to replace the permutation groups by braid groups. In the simplest case of two particles then indeed exchanging two particles my give the same state multiplied by a non-trivial phase, but the braiding is non-commutative so that braiding 1 and 2 is not the same as braiding 2 and 1. Braid groups are much more complicated than permutations.

However, there are restrictions: if average values are to be independent of the labelling of the particles in a quantum state, only those representations of the permutation group which are one-dimensional can appear when the particles are indistinguishable. This also imposes restrictions on the possible braidings since complicated sequences of braiding involving multiple particles must transform not as a linear combinations of basis states but as a phase multiple of the initial state. For instance, braiding operations acting on the symmetric state $$ \psi_a(x_1)\psi_b(x_2)+\psi_b(x_1)\psi_a(x_2) $$ must produce the same global phase when braiding the states, whereas braiding $x_1$ and $x_2$ will produce the conjugate phase to braiding $x_2$ and $x_1$. The notion of indistinguishability is not so clearly defined because there is a "direction" to the braiding in the sense that one must identify which particle braids "on top" of the other.

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Edit:

Maybe an example beyond 2 particles (which is somewhat trivial) would help.

Suppose you have a 3-particle system and you define \begin{align} P_{12}\psi(x_1,x_2,x_3)&=e^{i\lambda}\psi(x_2,x_1,x_3)\, ,\\ P_{23}\psi((x_1,x_2,c_3)&=e^{i\lambda}\psi(x_1,x_3,x_2) \end{align} Then what of $P_{12}P_{23}\psi(x_1,x_2,x_3)$? You would then pick up $e^{2i\lambda}$ (one phase for every permutation). But what to do with $P_{13}=P_{12}P_{23}P_{13}$? It’s a permutation of two particles but it’s written as a product of three permutations. Do you pick up $e^{i\lambda}$ or $e^{3i\lambda}$?

It only gets worse. Try inserting phases in a linear combination of product states and see if you can construct a fully symmetric or fully antisymmetric state up to a phase, i.e. explicitly construct a state $\psi(x_1,x_2,x_3)$ as a sum of products $\psi_a\psi_b\psi_c$ so that $P_{12}\psi(x_1,x_2,x_3)$ comes back to $e^{i\lambda}$ time itself, and $P_{12}P_{23}\psi(x_1,x_2,x_3)$ also comes back to a multiple of itself.

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  • $\begingroup$ I don't understand why this should answer to the question, if $\hat P$ is seen as coordinate exchange operator i can see $\hat P \hat P=I$ but if you interpret it as operator that exchanges particles then it's not ture $\endgroup$
    – SimoBartz
    Jun 23 at 21:53
  • $\begingroup$ Particles are labelled by their coordinate (ignoring spin as per your example) so how do you differentiate between exchanging coordinates and exchanging particles? $\endgroup$ Jun 23 at 22:03
  • $\begingroup$ you can’t, which is why you need properly symmetrized states. This way none of the measurable quantities depend on the assignment of labels. $\endgroup$ Jun 23 at 22:35
  • $\begingroup$ Sorry i deleted the comment before you posted. I agree, but consider this $\hat P \hat P\psi(1,2)=\psi(1,2)$ then since $\psi(1,2)=e^{i \lambda} \psi(1,2)$ we have $\hat P \hat P\psi(1,2)=\psi(1,2)=e^{i \lambda} \psi(1,2)$ namely $\hat P \hat P\psi(1,2)=e^{i \lambda} \psi(1,2)$ $\endgroup$
    – SimoBartz
    Jun 23 at 22:38
  • $\begingroup$ so maybe these are both true.. interesting $\endgroup$
    – SimoBartz
    Jun 23 at 22:40

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