# Eigenvalues of Exchange Operator

$$\hat P$$ 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.

• Have you heard of anyons? Jun 23 at 14:07

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.

• 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 Jun 23 at 21:53
• Particles are labelled by their coordinate (ignoring spin as per your example) so how do you differentiate between exchanging coordinates and exchanging particles? Jun 23 at 22:03
• you can’t, which is why you need properly symmetrized states. This way none of the measurable quantities depend on the assignment of labels. Jun 23 at 22:35
• 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 $\hat P \hat P\psi(1,2)=e^{i \lambda} \psi(1,2)$ Jun 23 at 22:38
• so maybe these are both true.. interesting Jun 23 at 22:40