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I am currently reading Schwartz' book on QFT, Section 12.2 on Spin and statistics. He shows, that in 3D there are only two inequivalent ways to exchange two indistinguishable particles. More formally, this means that the fundamental group of the configuration space (6D, identify (x1, x2) with (x2,x1)) is $Z_2$.

Now my question: Why can we deduce from the above that there are only two possible statistics/ ways the wavefunction can transform under this particle exchange? I think my question boils down to: If we tranform the state along two homotopic paths in the configuration space, why must the resulting state be the same for both paths?

sika_98
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1 Answers1

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The wavefunction $\psi(x_1,x_2)$ assigns a complex number to each point in configuration space and is continuous, but possibly has many branches as far as phase is concerned. Consider a path $\gamma_0(\tau)$ in configuration space such that $$\gamma_0(0)=(x_1,x_2),\quad \gamma_0(1)=(x_2,x_1).$$ Then the value of the complex number $\psi(\gamma_0(\tau))$ will change continuously from $\psi(\gamma_0(0))$ to $\psi(\gamma_0(1))=\eta_0\psi(\gamma_0(0))$, where $\eta_0$ is some phase. This last equality up to a phase follows from the indistinguishability of the particles.

We can repeat this with another path $\gamma_1$ with the same endpoints and get $\psi(\gamma_1(1))=\eta_1\psi(\gamma_1(0))$, where $\eta_1$ can be different from $\eta_0$ if the two paths go on different branches.

But if the paths are homotopic, i.e. there is a continuous function $\gamma_\sigma(\tau)$ of $\sigma,\tau$, which interpolates between $\gamma_0$ and $\gamma_1$ then we must have $\eta_0=\eta_1$ by continuity.

Moreover we can construct a loop which starts and ends on $(x_1,x_2)$ by traversing the path $\gamma_0$ and then the reverse of the path $\gamma_1$, and in general the endpoint of this loop would lead to $\eta_0\eta_1^*\psi(x_1,x_2)$, but if the two paths are homotopic this loop can be contracted to the trivial loop which never moves from $(x_1,x_2)$ so $\eta_0\eta_1^*=\eta_0\eta_0^*=1$, from which we see $\eta_0=\pm 1$.

octonion
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  • Thanks for the help, now I know that it follows from the continuity of the transformation between the paths. However, I can't quite figure out how this carries over to the trasformation of the state.. Could you elaborate on that i.e. explicitly show how this leads to the phases to be equal? – sika_98 Oct 29 '21 at 09:42
  • The wavefunction $\psi$ is the state. What I'm trying to say in my answer is that the phases $\eta_0$ and $\eta_1$ can only be different if $\psi$ is a multivalued function on configuration space and the two paths go over a branch cut. But if they are homotopic they can't go over a branch cut. – octonion Oct 29 '21 at 09:55
  • Hmm, I am not very fit in that region of maths. For now I would be fine if I understood the simple case, i.e. why for a single-valued wavefunction the phases must be equal. Does that make sense? – sika_98 Oct 29 '21 at 10:00
  • Well, if the wavefunction is single-valued then $\eta_0$ must be equal to $\eta_1$ by the definition of being single-valued. $\psi(\gamma(\tau))$ is just a complex number. If it is single valued and the argument is the same then it must be the exact same complex number. Maybe the notion of the exchange as a transformation of states (which is how this is often explained) is what is confusing. My answer is very similar to this longer answer here: https://physics.stackexchange.com/a/168728/59281, and that is how I'm explaining it rather than transformation. – octonion Oct 29 '21 at 10:18
  • I see, stupid question then. Thanks for the link, I'll have a look at it. – sika_98 Oct 29 '21 at 10:27