2

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?

Qmechanic
  • 201,751
Benoit Sanchez
  • 121
  • 2
    The buzzword is "spin-statistics theorem". Possible duplicate: https://physics.stackexchange.com/q/354625/50583 – ACuriousMind Dec 04 '17 at 12:10
  • 1
    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. – J0KerSpin Dec 04 '17 at 13:10
  • 4
    This is in 3D - two times the exchange operator must give the same state. But in 2D one can have anyons. –  Dec 04 '17 at 13:28
  • 1
    see also https://physics.stackexchange.com/q/86116/ – Adam Dec 05 '17 at 06:56
  • Thanks Adam. This is very close to what I was looking for. I see the problem is far from simple. – Benoit Sanchez Dec 08 '17 at 09:03

3 Answers3

2

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.

ZeroTheHero
  • 45,515
2

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

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)

ZaellixA
  • 1,779
JtX
  • 21
  • 3
    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. – Community Oct 03 '22 at 19:41
-1

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.

Kishore
  • 27
  • 2
    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 https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem. – eranreches Dec 04 '17 at 13:07
  • @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 $\pm 1$. – Kishore Dec 04 '17 at 14:11
  • I am not sure, because this is a tensor product space and things get more complicated. – eranreches Dec 04 '17 at 14:21
  • @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. – Prahar Dec 05 '17 at 02:28
  • @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. – Kishore Dec 05 '17 at 02:37
  • @Kishore - That's wrong. $\psi$ is an eigenstate, but you don't yet know its eigenvalue. Suppose $P \psi = \eta \psi $ for some number $\eta$. Then, we know that $P^2 \psi = \eta^2 \psi$. Now, since $P^2$ returns the system to itself, the state must also return to itself, but only up to a phase since if two states differ only by a phase they describe the same system. This implies that we must have $\eta^2 = e^{i \theta_\psi}$ for some choice of $\theta_\psi \in {\mathbb R}$. In fact, typically this phase is non-zero and is called the "intrinsic parity" of the state. – Prahar Dec 05 '17 at 02:45
  • @Kishore - This discussion is more complicated than a comment section can handle. I suggest you read the discussion of the $P$ operator in Weinberg Vol. 1 (page 74 and 124) – Prahar Dec 05 '17 at 02:46
  • Thanks for the answer. That is interesting. Maybe the idea holds for the splinless assumption since formally $P^2=I$. The idea that $H$ and $P$ commute also implies that if $\psi(t_0)$ is symmetric then $\frac{d\psi}{dt}(t_0)$ is and thus $\psi(t)$ stays symmetric. (same for antisymmetric). – Benoit Sanchez Dec 08 '17 at 09:14
  • Your argument only works for transpositions but fails for types of permutations involving permutations of $3$ particles or coordinates. It could fail as soon as you have $3$ or more particles. – ZeroTheHero Feb 26 '21 at 00:05
  • @Prahar 1. $e^{i \theta_\phi}|\phi\rangle$ doesn't distribute properly over superpositions unless $\theta_\phi$ is constant rather than indexed, e.g. if $P^2~|01\rangle=i~|01\rangle$ and you try to soak the inconsistency with global phase ambiguity, the problem is if you choose something like $P|10\rangle=|10\rangle$ or some nonsense, then $P^2(|01\rangle +|10\rangle)$ is no longer multiplying that by a simple phase factor. So $\theta$ has to be globally shared for the global phase ambiguity to apply. Ar which point 2. WLOG define $\hat P=e^{-i\theta/2} P.$ – CR Drost Oct 04 '22 at 07:17