I've read in many text books that the indistinguishability of two identical particles at $x$ and $y$ implies:

$$|\psi(x,y)|=|\psi(y,x)|\quad (1)$$

This sounds rather natural.

Then they say there are two solutions: the symmetric and antisymmetric ones leading to bosons and fermions:

  • Symmetric: $\psi(x,y)=\psi(y,x)$
  • Antisymmetric: $\psi(x,y)=-\psi(y,x)$

I can't help to wonder why there is no other solution. Is it a theorem that can be proven from (1)? Or is it just that these two solutions are the only ones observed in nature?

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    $\begingroup$ The buzzword is "spin-statistics theorem". Possible duplicate: physics.stackexchange.com/q/354625/50583 $\endgroup$ – ACuriousMind Dec 4 '17 at 12:10
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    $\begingroup$ More than a theorem, is a principle. If you have one symeetric state and one antisymmetric state you don't need anything else to describe identical particles. I you want or need it, I can give a more detailed answer later. $\endgroup$ – J0KerSpin Dec 4 '17 at 13:10
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    $\begingroup$ This is in 3D - two times the exchange operator must give the same state. But in 2D one can have anyons. $\endgroup$ – Pieter Dec 4 '17 at 13:28
  • $\begingroup$ I really don't understand what are you telling to me $\endgroup$ – J0KerSpin Dec 4 '17 at 14:05
  • $\begingroup$ see also physics.stackexchange.com/questions/86116/… $\endgroup$ – Adam Dec 5 '17 at 6:56

The physically measurable quantity in QM is the probability density. This gives you $\left|\psi\left(x,y\right)\right|^2 = \left|\psi\left(y,x\right)\right|^2$ (since when we say two particles are indistiguishable, what we actually mean is that we cannot make any measurements to distinguish them). This would imply that $\psi\left(x,y\right) = \pm\psi\left(y,x\right)$, which tells you that the wavefunction must be odd or even.


As pointed out by eranreches, the reasoning above is flawed, since the wavefunction with particles interchanged could be multiplied by a complex phase factor, in general.

A better way of looking at the above problem would be to define a particle exchange operator $P$ such that $P\psi\left(x,y\right) = \psi\left(y,x\right)$. If the system does not change under the action of this operator, it means that the Hamiltonian is invariant under the action of this operator, which implies $\left[H,P\right] = 0$. This implies that the energy eigenstates are also eigenstates of $P$. We can also see that by definition, $P^2 = I$, where $I$ is the identity operator. Let $p$ be the eigenvalue for $P$ of some energy eigenstate $\psi\left(x,y\right)$. We can then see that $P^2\psi\left(x,y\right) = p^2\psi\left(x,y\right) = I\psi\left(x,y\right) = \psi\left(x,y\right)$. This implies $p^2 = 1$, which implies $p = \pm 1$. Thus we get $\psi\left(y,x\right) = P\psi\left(x,y\right) = \pm\psi\left(x,y\right)$.

This is not the complete story, however. If you want to delve further into why particles with half-integer spins have antisymmetric wavefunctions, and why particles with integer spin have symmetric wavefunctions, you would have to look at the Spin statistics theorem, as pointed out by ACuriousMind.

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    $\begingroup$ This actually implies that $\psi\left(x,y\right)=e^{i\varphi}\psi\left(y,x\right)$ for some $\varphi$. As @ACuriousMind said, it is the result of the "Spin-Statistics Theorem". See here en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem. $\endgroup$ – eranreches Dec 4 '17 at 13:07
  • $\begingroup$ @eranreches Could we instead argue that if the Hamiltonian is invariant under an operator $P$ which interchanges the coordinates, the energy eigenstates must be eigenstates of $P$? We know that $P^2 = 1$, by definition, so the square of the eigenvalue of P should be 1, which would imply the eigenvalue is $\pm 1$. $\endgroup$ – Kishore Dec 4 '17 at 14:11
  • $\begingroup$ I am not sure, because this is a tensor product space and things get more complicated. $\endgroup$ – eranreches Dec 4 '17 at 14:21
  • $\begingroup$ @Kishore - Even with $P$ you are having the same problem. It is true that $P^2$ should leave the system invariant. On the state this then implies that generically $P^2 |\psi\rangle = e^{i \theta_\psi} |\psi\rangle$. This is because states are identified up to phases. $\endgroup$ – Prahar Dec 5 '17 at 2:28
  • $\begingroup$ @Prahar We don't have the same problem with P, because psi is an eignenstate of P, and from that, we get that the square of the eigenvalue (not the magnitude of the eigenvalue) is one. Also, by definition, P^2 is the identity operator, so it doesn't introduce a phase factor. $\endgroup$ – Kishore Dec 5 '17 at 2:37

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