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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 ♦
    Commented Dec 4, 2017 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
    Commented Dec 4, 2017 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. $\endgroup$
    – user137289
    Commented Dec 4, 2017 at 13:28
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    $\begingroup$ see also physics.stackexchange.com/q/86116 $\endgroup$
    – Adam
    Commented Dec 5, 2017 at 6:56
  • $\begingroup$ Thanks Adam. This is very close to what I was looking for. I see the problem is far from simple. $\endgroup$
    – Benoit Sanchez
    Commented Dec 8, 2017 at 9:03

3 Answers 3

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Yes. If you study the representation theory of the permutation group, you will learn there are only two representations with the property that the $n$-particle states comes back to any multiple of itself.

For the symmetric representation, the state comes back to + itself, and for the antisymmetric to - itself.

The requirement that the state come back to $\pm$ itself is an essential consequence of the indistinguishability of particles. For any other representations (for $n\ge 3$ particles there are states of mixed symmetry transforming as linear combination of other states) there are problems in the computation of average values if the state is not fully symmetric or antisymmetric. You can find references to literature on this in this related answer.

Note that anyonic states can come back themselves up to some (in principle) generic phase, but anyons are quasi-particles rather than fundamental particles. Moreover, they live in 2d rather than 3d. There are multiple posts on this site on anyons (see for instance this one) in addition to those already cited above.

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$$ \left| \phi(x,y) \right| = \left| \phi(y,x) \right| \implies \phi(y,x) = e^{i \alpha} \phi(x,y) $$

Making exchange twice, we get

$$\phi(x,y) = e^{2 i \alpha} \phi(x,y)$$

Hence,

$$e^{2 i \alpha} = 1$$

so $\alpha = n \pi$ with $n$ integer. If it's an odd integer then $\phi(x,y) = -\phi(y,x)$. If it's an even integer then $\phi(y,x) = \phi(x,y)$ (assuming $\alpha$ is a constant)

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Oct 3, 2022 at 19:41
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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.

Edit

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
    Commented Dec 4, 2017 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
    Commented Dec 4, 2017 at 14:11
  • $\begingroup$ I am not sure, because this is a tensor product space and things get more complicated. $\endgroup$
    – eranreches
    Commented Dec 4, 2017 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
    Commented Dec 5, 2017 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
    Commented Dec 5, 2017 at 2:37

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