Symmetrization Postulate

1. Feb 3, 2016

zonde

I was reading this paper http://arxiv.org/abs/quant-ph/0511002 but I did not get clear answer to the question: is there some mathematical reason (given other postulates of QM) why we require that description of two (several) indistinguishable particles has to have any symmetry?
Basically why description of two indistinguishable particles can't be probabilistic combination of two descriptions of single particles? Is this just empirical observation (that they can't) or is it related to other postulates of QM?

2. Feb 3, 2016

strangerep

In relativistic QFT, one has the spin-statistics theorem.

Nonrelativistic QM lacks such a theorem, hence symmetrization must be introduced as a postulate.
Ballentine covers this. (Sorry, I don't have the book handy to give a chapter reference).

3. Feb 3, 2016

stevendaryl

Staff Emeritus
If you think of particles as excitations of a field then it's completely natural to understand the symmetry requirements. The state in which "Particle number one is at position $x=a$ and particle number two is at position $x=b$" is exactly the same state as the state in which
"Particle number two is at position $x=a$ and particle number one is at position $x=b$". They both are different ways of describing the state in which there are excitations at $a$ and $b$. So those two different descriptions refer to the same physical state. So the wave functions corresponding to those descriptions must be the same up to nonphysical differences such as a phase difference.

4. Feb 3, 2016

zonde

It does not seem that spin-statistics theorem answers my question.
Say look at this simple summary of spin-statistics theorem (http://math.ucr.edu/home/baez/spin_stat.html):
1) Quantum mechanics says that if you turn a particle around 360°, its wavefunction changes by a phase of either +1 (that is, not at all) or -1. It also says that if you interchange two particles of the same type, their joint wavefunction changes by a phase of +1 or -1.
2) The spin-statistics theorem says that these are not independent choices: you get the same phase in both cases! The phase you get by rotating a particle is related to its spin, while the phase you get by switching two goes by the funny name of "statistics". The spin-statistics theorem says how these are related.

I looked at Chapter 17 "Identical Particles" from Ballentine.
He identifies three expressions of particle identity:
(1) permutation symmetry of the Hamiltonian
(2) permutation symmetry of all observables
(3) the symmetrization postulate

Looking at (1) he says:
Two physical situations that differ only by the interchange of identical particles are indistinguishable. One of the consequences of this fact is that any physical Hamiltonian must be invariant under permutation of identical particles.
Following that he combines two particles in single Hilbert space and introduces permutation operator. This idea that combination of two particles can be described by single Hilbert space seems to imply that two particles don't have independent existence.

5. Feb 3, 2016

zonde

As I said in my previous post the idea that two particles can be described by the same wavefunction seems to have certain consequences.
Say we look at both particles at a later time. If we use combined Hilbert space description of both particles then both particles contribute to the later state of each particle (loosely speaking). This is somewhat along the lines how Feynman explains this in his book QED: The strange theory of light and matter.

Basically this is quite non-classical reasoning.

6. Feb 3, 2016

PeroK

Do you know basic probability theory? When you get dealt two cards, your hand is the same regardless of what order you get dealt the cards. This essentially means that the cards are indistinguishable. There's no difference between $2D, 5H$ and $5H, 2D$. It's the same hand. The same state.

If particles are indistinguishable, it means (by definition) that there is only one state where there is a particle at postion $x_1$ and a particle at position $x_2$. There is no state where "particle A" is at position $x_1$ and particle "B" at at position $x_2$ and then a different state (with a different probability) for the particles the other way round. There is only one state, hence symmetry of the wavefunction.

7. Feb 3, 2016

stevendaryl

Staff Emeritus
I don't think it's so terribly nonclassical. It's not a perfect analogy, but if you have a rope and flick it to send impulses down the rope, there is no identity to the impulses. If two impulses meet each other (one going one way, and the other going the other) there is no distinction between "the impulses passed through each other" and "the impulses bounced off each other".

8. Feb 3, 2016

zonde

Nice example.
But let's consider description of distinguishable particles. There is no symmetrization requirement. So how does it appears with indistinguishability?
And I think I know the way how to narrow down the question.
Say we take two description side by side: one where we have two indistinguishable particles and another one where we have two distinguishable particles. Now we try to claim there has to be symmetrization in both cases. As we proceed with our reasoning the second description has to fail at some point while the first one goes through.
So we can ask why for the first description we proceed differently.

Say can we combine two distinguishable particles in single Hilbert space?
Can we introduce permutation operator that swaps both particles?

9. Feb 4, 2016

zonde

Your example is not very good. Let me rewrite it.
Let's compare two cases. I get dealt two cards. In one case it's $2D, 5H$ but in other case it's $5H, 5H$. In classical world I can't tell the two cases apart without looking at the front of the cards but in quantum world I can do that.

10. Feb 4, 2016

PeroK

You're missing something fundamental about basic probability theory that is making it hard to understand QM: which is the symmetry of probabilities where "order doesn't matter".

11. Feb 4, 2016

zonde

Probably.
My understanding is that we can get probability where "order doesn't matter" by summing up probabilities that we get by imagining that "order matters". Is this understanding wrong?

12. Feb 4, 2016

DrDu

Note that beside Fermi and Bose statistics, there are also para-Fermi and para-Bose statistics. If they are of infinite dimension, you get Boltzmann statistics.

13. Feb 4, 2016

PeroK

Yes, exactly. The idea of symmetry is that there is therefore only one outcome, not two (or more) outcomes that are distinguishable. In QM, you are dealing with probability amplitudes, not directly wih probabilities. But, the same symmetry principle applies: there is only one outcome, not two distinguishable outcomes. Also, with probability amplitudes you can subtract as well as add. In general, for indistinguishable particles, you have:

$\psi(x_1, x_2) = C[\psi_a(x_1) \psi_b(x_2) \pm \psi_b(x_1) \psi_a(x_2)]$ (i)

What this means is: the probability amplitude of finding one particle at $x_1$ and one particle at $x_2$.

If you compare this with the simple product of wavefunctions for distinguishable particles:

$\psi(x_1, x_2) = \psi_a(x_1) \psi_b(x_2)$ (ii)

What this means is: the probability amplitude of finding the first particle (a) at $x_1$ and the second particle (b) at $x_2$.

Note that if you try to apply this for indistinguishable particles, then you could have different probability amplitudes for the two indistinguishable outcomes. You would also have:

$\psi(x_2, x_1) = \psi_a(x_2) \psi_b(x_1)$ (iii)

But, (ii) and (iii) give the probability amplitude for the same outcome, so they cannot be different. It's the lack of symmetry in the two spatial variables that is the problem. You need to sort that out by taking a linear combination of these two terms, hence the equation (i).

Last edited: Feb 4, 2016
14. Feb 4, 2016

stevendaryl

Staff Emeritus
Let $\Psi$ be a two-particle state, and let $S(\Psi)$ be the state that is obtained from $\Psi$ by swapping the two particles. Then $\Psi$ and $S(\Psi)$ are physically the same state. There are no observable differences. So $\Psi$ and $S(\Psi)$ can differ by at most a phase factor: $S(\Psi) = e^{i \theta} \Psi$. The argument that $\theta = 0$ or $\theta = \pi$ is a little subtle. If you assume that (1) $S$ is a linear operator, and (2) that $S(S(\Phi)) = \Phi$ (for every $\Phi$), then it follows that $e^{i \theta} = \pm 1$, but there's more to the argument as to why those two facts must be true.

If the two particles are distinguishable, then $S(\Phi)$ is not physically the same state as $\Phi$. So there is no requirement that $S(\Phi)$ be the same, up to a phase factor.

But why would you claim that, in the case of distinguishable particles?

Sure, the two-particle, spinless Hilbert space for particles moving in one spatial dimension consists of all functions of two variables $\Psi(x_1, x_2)$ such that $\int dx_1 dx_2 \Psi^*(x_1, x_2) \Psi(x_1, x_2)$ is finite. The permutation operator $S(\Psi)$ is well-defined: $S(\Psi) = \Psi'$ where $\Psi'(x_1, x_2) = \Psi(x_2, x_1)$. But there is no restriction that $S(\Psi)$ and $\Psi$ be the same, up to a phase.

If the particles are indistinguishable, then you have a reduced Hilbert space.

[edit: changed 'distinguishable' to 'indistinguishable' in the last sentence]

Last edited: Feb 4, 2016
15. Feb 4, 2016

vanhees71

You mean, if the particles are indinstinguishable, then you have a reduced Hilbert space, i.e., in your example of spin-zero particles, the wave functions must be symmetric (bosons) under exchange.

Note that for more than 2 particles, you have more representations of the permutation group $S_N$ than just the trivial one (bosons) or the alternating one (fermions). Indinstinguishable particles whose wave function obey other permutation properties are called anyons. There's a topological argument that you cannot have anyons in $\geq 3$ space dimensions:

M. G. G. Laidlaw and C. M. DeWitt. Feynman Functional Integrals for Systems of Indistinguishable Particles. Phys. Rev. D, 3:1375, 1970.

16. Feb 4, 2016

stevendaryl

Staff Emeritus
Yeah, that's why I skipped the argument, because I couldn't reproduce it off the top of my head. The simple argument from the states being physically identical only gets you to the conclusion that swapping the particles can at most produce a phase change. Why that phase change must be $\pm 1$ is more complicated to show.

17. Feb 4, 2016

DrDu

That's a little bit vague. In 3D more than 2 particles may belong to other representations of the permutation group than the one dimensional one. Specifically the parastatistics I already mentioned.
In two dimensions, the permutation of particles is not described by the group S_N but by so-called braid groups whose representations are called anyons.
Parastatistics can be reduced to ordinary bose or fermi statistics if one introduces a hidden degree of freedom. E.g. for electrons spin is not an observable non-relativistically. Hence the spatial part of an electronic wavefunction obeys a parafermionic statistics.

The article by Laidlaw has been criticised in this respect in some comments. Strangely enough, the claim that parastatistics is excluded appears only in the abstract but is not repeated in the paper, as far as I remember.
PS: Ah, here is the reference:
arxiv.org/pdf/math-ph/0406011
and an explicit construction of the path integral
http://cds.cern.ch/record/300043/files/9603179.pdf

Last edited: Feb 4, 2016
18. Feb 4, 2016

vanhees71

Interesting. Can you summarize what's wrong with Laidlaw&deWitt?

19. Feb 4, 2016

zonde

I think I understand one way how it follows from basic postulates of QM. Let me describe it:
If we take Copenhagen interpretation which says that state in Hilbert space is the maximum physical description of single particle then the reasoning goes like that:
Let $\phi$ be a state of particle at $x_1$ and $\psi$ a state of particle at $x_2$. Then let's change $\phi$ into $\phi'$ so that it describes the same particle at $x_2$ and respectively change $\psi$ into $\psi'$ as description of the same particle at $x_1$. We can now compare description of $\phi$ & $\psi$ with $\psi'$ & $\phi'$. If two descriptions are identical we can state that two physical configurations are identical as well.
That way we arrive at conclusion using only single particle descriptions.

This is a bit less clear.
First when we combine two Hilbert spaces we make the relative phase factor fixed (seems like an assumption even if it's reasonable).
Second $S(S(\Phi)) = \Phi$ this does not have to be true. It can be $S(S(\Phi)) = e^{i \theta'}\Phi$

I think that if we postulate that state in Hilbert space is maximum physical description of many particle configuration this reasoning should go through.

I suppose that for distinguishable particle it is certain that $S(\Psi)$ and $\Psi$ are not the same, neglecting a phase factor, right?

20. Feb 4, 2016

stevendaryl

Staff Emeritus
Well, both particles could be in the same one-particle state. There is nothing preventing that. Then exchanging the particles would do nothing (physically)