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In a system with two indistinguishable particles, the eigenvalue to the particle exchange operator $\hat{P_{ij}}$ is $+1$ if the two particles are exchange symmetric, ie. bosons, and $-1$ if they are exchange antisymmetric, ie fermions. In addition, the particle exchange operator $\hat{P_{ij}}$ is Hermitian, and hence is an observable. It has also been claimed in our lecture notes that the exchange symmetry of such a pair of particles can be experimentally determined. How can such exchange symmetry be experimentally measured?

The question here, Eigenvalues of the exchange operator determined by the particle type(boson or fermion) in a two particle system, seems to have something remotely connected, but is not very detailed about how can the exchange symmetry be experimentally measured.

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    $\begingroup$ As far as my understanding goes, people do not measure $P$ directly, but infer the eigenvalues of it from its consequences (i.e. the Pauli exclusion principle, Fermi-Dirac and Bose-Einstein distributions). $\endgroup$ – Bob Knighton Apr 15 '18 at 14:48
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I think the experiment should have something to do with a rotation by $2\pi$ of one of the particles while leaving the other unrotated (it doesn't matter which one is rotated).

Because topologically rotation by $2\pi$ is not equivalent to doing nothing (i.e. not rotating), but, rotating by $4\pi$ is the same thing as doing nothing. So, I guess this is the basis of the experiment. At least the rotation can be made to a magnetic field, instead of the particles themselves.

Fermions, indeed, would get a minus sign in front of their state vectors when rotated by $2\pi$ however bosons get a plus sign (i.e., no phase change). $$ \tag{fermions@2$\pi$} | \psi \rangle \rightarrow -| \psi \rangle $$ $$ \tag{bosons@2$\pi$} | \phi \rangle \rightarrow + | \phi \rangle $$ Individually for the states, it does not matter if there is a phase factor on the state because the probability of the individual state would eliminate the phase factor. However, if there are two states interfering each other, then one of them getting a minus sign will imply a shift on the interference pattern (or the probability distribution).

So, in order to check if the indistinguishable 2-particle system is made of bosons or fermions, you just do an experiment that applies a full rotation to one of them (actually rotating a magnetic field is what could be done, instead of rotating the particle itself) while leaving the other as it is. The result of the experiment has two outcomes:

  1. The interference of two particles would change, i.e., they will cancel each other if they have the same wave function. So, they are fermions

  2. The interference of two particles would not change. So, they are bosons.

where, in fact, these outcomes correspond to the eigenvalues of the operator and the rotation in the experiment corresponds to the operator itself.


Please check the proposal for an experiment of a particle in two possible states

  • Y. Aharanov and L. Susskind, Observability of the Sign Change of Spinors under 2Pi Rotations, Phys. Rev. 158, 1237 (1967).

  • H. J. Bernstein, Phys. Rev. Lett. 18, 1102 (1967).

And the experimental confirmation:

  • H. Rauch, A. Zeilinger, G. Badurek, A. Wilfing, W. Bauspiess, and U. Bonse, Phys. Lett. 54A, 425 (1975).

  • A. G. Klein and G. I. Opat, Phys. Rev. D 11, 523 (1975).

  • A. G. Klein and G. I. Opat, Phys. Rev. Lett. 37, 238 (1976).

  • S. A. Werner, R. Colella, A. W. Overhauser, and C. F. Eagen, Phys. Rev. Lett. 35, 1053 (1975).

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