Identical Particles: Symmetry & Antisymmetry of Wave Function

[Euclid]

When discussing identical particles, it is said that a wave function must be either symmetric or antisymmetric wrt exchange of particles. Could we not have, in general,
\psi(r_1,r_2)=e^{i\phi}\psi(r_2,r_1)
since the overall factor is physically meaningless?

 

[cesiumfrog]

Is this like bosons, fermions and anyons?

 

[Euclid]

I've never heard of an anyon. I found a brief section on it on wiki. Can you provide any details?

 

[StatMechGuy]

The symmetry or antisymmetry drops out of field theory in three dimensions. If you restrict yourself to two dimensions, you no longer have a strict exclusion principle, and flipping the two particles yields whatever phase you want with respect to the original state vector. These are anyons.

 

[masudr]

Let the operator for swapping the particles be \hat{S} in the position representation. Now apply it twice to our wavefunction:

<br /> \begin{array}{cll}<br /> \hat{S}^2\psi(r_1,r_2) = \hat{S}\psi(r_2,r_1)=\psi(r_1,r_2) &amp;\Rightarrow \hat{S}^2 = 1 \\<br /> &amp;\Rightarrow \hat{S} = \pm 1 \\<br /> &amp;\Rightarrow \psi(r_1,r_2)=\pm\psi(r_2,r_1)<br /> \end{array}<br />

 

[Euclid]

S^2 =1 does not imply S=\pm 1. All this implies is S=S^{-1}.

And I believe you just proved that all states are either symmetric or antisymmetric. What about the state |0>|1> ?

In fact, S must have some zero's on the diagonal in the position basis:
&lt;x,y|S|x,y&gt;=&lt;x,y|y,x&gt; = \delta_{xy}

Am I mistaken?

 

[masudr]

Can you think of an example where

\hat{S}^2 = 1 \mbox { but } \hat{S} \ne \pm 1

 

[selfAdjoint]

Can you think of an example where

\hat{S}^2 = 1 \mbox { but } \hat{S} \ne \pm 1

e^{i\frac{\pi}{4}}?

 

[masudr]

In fact, S must have some zero's on the diagonal in the position basis:
&lt;x,y|S|x,y&gt;=&lt;x,y|y,x&gt; = \delta_{xy}

And do you mean

<br /> \langle x_1&#039;,x_2&#039; | \hat{S} | x_1,x_2 \rangle =<br /> \langle x_1&#039;,x_2&#039; | x_2,x_1 \rangle = <br /> \delta_{x_1&#039;x_2,x_2&#039;x_1}<br />

 

[masudr]

e^{i\frac{\pi}{4}}?

\hat{S}=e^{i\frac{\pi}{4}} \Rightarrow \hat{S}^2 = e^{i\frac{\pi}{2}} = i \ne 1

 

[Galileo]

You can conclude from S^2=I (and that S is Hermitian) that the eigenvalues are \pm 1 (take the Pauli spin operators for example). If the particles are identical, then H must treat them on equal footing. It just means that [H,S]=0, so you can find a complete set of eigenstates common to H which are either symmetric or antisymmetric.

 

[Euclid]

Can you think of an example where

\hat{S}^2 = 1 \mbox { but } \hat{S} \ne \pm 1

Yes, a diagonal matrix with 1 and -1's on the diagonal.

 

[Euclid]

You can conclude from S^2=I (and that S is Hermitian) that the eigenvalues are \pm 1 (take the Pauli spin operators for example). If the particles are identical, then H must treat them on equal footing. It just means that [H,S]=0, so you can find a complete set of eigenstates common to H which are either symmetric or antisymmetric.

Where do anyons come into all this? What you say here shows that all particles can be thought as a superposition of fermionic and bosonic states.

Also, why would we expect S to be hermitian?

 

[Galileo]

Where do anyons come into all this?

They don't, since that wasn't my intention.

What you say here shows that all particles can be thought as a superposition of fermionic and bosonic states.

Also, why would we expect S to be hermitian?

You can prove S is Hermitian quite easily.

I don't think you can prove the wavefunction must be (anti)-symmetric from QM. The fact that the states are required to be either completely symmetric or antisymmetric under the exchange of two identical particles is a postulate.

 

[quetzalcoatl9]

I don't think you can prove the wavefunction must be (anti)-symmetric from QM. The fact that the states are required to be either completely symmetric or antisymmetric under the exchange of two identical particles is a postulate.

Is that true?

I know that if you derive the (relativistic) Dirac equation the Pauli spin matrices fall right out. So I can see how the antisymmetric nature comes about for fermions. As for bosons...i have no idea.

 

[quetzalcoatl9]

Also, why would we expect S to be hermitian?

Because S commutes with respect to the Hamiltonian, and since the wavefuction is an eigenfunction of H it will be an eigenfunction of S also - the eigenvalue being the (experimentally measurable) spin. Whichever way to want to look at it, S must be hermitian.

proof:

If H is hermitian then:

H^\dagger = H

and if S and H commute:

[H,S] = 0

HS - SH = 0

HS = SH

(HS)^\dagger = (SH)^\dagger

S^\dagger H^\dagger = H^\dagger S^\dagger

S^\dagger H = H S^\daggerbut since [H,S] = [S,H] = 0 then SH=HS and so it must be true that

S^\dagger = S

and so S must be hermitian

 

[tomkeus]

First: Requierment that permuting particles leaves physical state intact leads to conclusion that representation of permutational grouop must act as multiplication with e^{i\phi}. Even, if you take that in form of some general unitary operator you will finally end with reducible operator whose irreducible subspaces are 1D, and hence his action is reduced to simple scalar multiplication by phase factor. Then, application of same permutation again must be identical operator, thus giving only two possible 1D representations, +1 and -1.
Second: take a system which is in state |\phi&gt;=|s&gt;+|a&gt;, where |s&gt; is from symmetric subspace of permutational group representation and |a&gt; is from antisymmetric part. Attack this state with some permutation and you will have D(S)|\phi&gt;=|s&gt;-|a&gt;, which is not colinear with original state, and this is not allowed by assumption of identical particles. Thus all systems must be in either symmetric or antisymmetric state.

 

[Euclid]

I think I am only becoming more confused.

Anyons do not exist?

 

[zbyszek]

When discussing identical particles, it is said that a wave function must be either symmetric or antisymmetric wrt exchange of particles. Could we not have, in general,
\psi(r_1,r_2)=e^{i\phi}\psi(r_2,r_1)
since the overall factor is physically meaningless?

The answer is here: M. D. Girardeau "Permutation Symmetry of Many-Particle Wave Functions", Phys. Rev. 139, 500 (1965).

In short, the phase \phi can depend on positions in 1D, but in 3D the only
possibilities are 0 and \pi.

Beautiful paper, anyway.

Cheers!

 

[Careful]

The answer is here: M. D. Girardeau "Permutation Symmetry of Many-Particle Wave Functions", Phys. Rev. 139, 500 (1965).

In short, the phase \phi can depend on positions in 1D, but in 3D the only
possibilities are 0 and \pi.

Beautiful paper, anyway.

Cheers!

As a general note, you have to be VERY careful when applying the credo that two particles can be considered to be indistinguishable. For example, when I have two electrons 1 and 2, each flying off in two different directions, say to the left and to the right ; then clearly the probability that the first electron is at my left and the second at my right is much bigger than the first being at my right and the second at my left. In other words flipping the positions might change the physical state !

Careful

 

[masudr]

As a general note, you have to be VERY careful when applying the credo that two particles can be considered to be indistinguishable. For example, when I have two electrons 1 and 2, each flying off in two different directions, say to the left and to the right ; then clearly the probability that the first electron is at my left and the second at my right is much bigger than the first being at my right and the second at my left. In other words flipping the positions might change the physical state !

Careful

Well, it's not that difficult. If they are flying in two different directions, then their momenta will have opposite signs. Therefore they will be in different eigenstates of momentum, and then they are obviously in different states.

 

[Careful]

Well, it's not that difficult. If they are flying in two different directions, then their momenta will have opposite signs. Therefore they will be in different eigenstates of momentum, and then they are obviously in different states.

Euh that is not the point. In the above people tried to derive bose and fermi statistics from projective invariance of any two particle state under the swapping operation. The issue I tried to raise is; when do you call two particles indistinguishable and consequently when does the statistics apply ? You simply say, well both particles are in different states, but in a two particle wave, you cannot even speak about the state of a single particle; that is the message of EPR (at least in the standard QM formalism).

Moreover, if they are in true eigenstates of the momentum, then the latter probabilities are ill defined.

For example, two electrons bound to different nuclei spaced by a distance of a few micron cannot be expected to satisfy the above criterion (antisymmetrization of state). Even the wavefunction of two electrons in different shells of one atom should not anti-symmetrize by this argument. I am not even sure if it would apply exactly to two electrons in the same shell if all radiative corrections are taken into account.


Careful

 

[masudr]

Euh that is not the point. In the above people tried to derive bose and fermi statistics from projective invariance of any two particle state under the swapping operation. The issue I tried to raise is; when do you call two particles indistinguishable and consequently when does the statistics apply ? You simply say, well both particles are in different states, but in a two particle wave, you cannot even speak about the state of a single particle; that is the message of EPR (at least in the standard QM formalism).

That's fair enough: I must admit, I didn't really read the rest of the thread.

For example, two electrons bound to different nuclei spaced by a distance of a few micron cannot be expected to satisfy the above criterion.

As far as I know, two electrons bound to different nuclei do not satisfy the principle: even if they have the same n,l,j,m_j, their total state vector is specified by the position of the center of mass of each atom as well: and these will be different.

Even the wavefunction of two electrons in different shells of one atom should not anti-symmetrize by this argument.

Why should they? They have different values of n, l.

 

[Careful]

As far as I know, two electrons bound to different nuclei do not satisfy the principle: even if they have the same n,l,j,m_j, their total state vector is specified by the position of the center of mass of each atom as well: and these will be different.

Right, now prove that they even do not approximately form an antisymmetric state starting from fermionic free QFT where you know that creation operators anti-commute and as such always form antisymmetric wave functions. The issue is that I would for sure not expect such anti symmetrization properties to hold for free particles, which are clearly distinguishable.

Why should they? They have different values of n, l.

Since you seem to accept this so easily, you might now want to show us why we should believe in an entangled state where we have a left mover and a right mover ; both particles being clearly distinguishable, but the wave function still being antisymmetric (of course I am aware that you will appeal to conservation of spin , but nothing says this needs to be the case since we do not control the mechanism for creating an entangled state AFAIK (and only total angular momentum needs to be conserved).).

 

[Epicurus]

There is a theory called the spi-statistics theorem, which proves that half integer spin particles cannot occupy the same state. If you generate greens function from an intrinsic spin half theory like the Dirac equation, then if 2 particles occupy the same state then the greens functions will equal zero. Since Greens functions are related to the density which is related to wave functions then this give the antisymmetry condition on the wavefunction. This is however a function of dimensionality. In theories with even space dimensions we are able to have states which are neither totally symmetric nor antisymmetric. This is related to parastatistics wgich was first Elucidated by H. S. Green in 1953 (Phys. Rev 1953,90,270).

You can find a 'proof' of the spin statistics theorem in Scwabl, 'Advanced Quantum Mechanics'

 

[Epicurus]

As far as I know, two electrons bound to different nuclei do not satisfy the principle: even if they have the same , their total state vector is specified by the position of the center of mass of each atom as well: and these will be different.

This is a blatant misunderstanding of QM and is one usually performed by chemists. We use the single particle orbitals to approximate the full many-particle Hilbert space. They are not in the same state.

 

[Epicurus]

As a general note, you have to be VERY careful when applying the credo that two particles can be considered to be indistinguishable. For example, when I have two electrons 1 and 2, each flying off in two different directions, say to the left and to the right ; then clearly the probability that the first electron is at my left and the second at my right is much bigger than the first being at my right and the second at my left. In other words flipping the positions might change the physical state !

No this is also very wrong. YOU have to be VERY careful about labeling particles and as to what you are actually observing. If You observe a particle then make another measurement, you cannot say which particle you actually detected in the second instance.

If I was to have a list of the theories which I believe people thought they understood but actually didn't so they confused the hell out of everyone else it would be

1) Evolution and natural selection
2)Angular momentum
3)Indistinguishability of particles.

What you have said is contradictory.

 

[Careful]

There is a theory called the spi-statistics theorem, which proves that half integer spin particles cannot occupy the same state. If you generate greens function from an intrinsic spin half theory like the Dirac equation, then if 2 particles occupy the same state then the greens functions will equal zero. Since Greens functions are related to the density which is related to wave functions then this give the antisymmetry condition on the wavefunction. This is however a function of dimensionality. In theories with even space dimensions we are able to have states which are neither totally symmetric nor antisymmetric. This is related to parastatistics wgich was first Elucidated by H. S. Green in 1953 (Phys. Rev 1953,90,270).

You can find a 'proof' of the spin statistics theorem in Scwabl, 'Advanced Quantum Mechanics'

Sigh, I would expect an epicurus to be able to read, but alas. The discussion went about the derivation of statistics from the presumed projective invariance of any state containing two ``identical'' particles under the swapping operation. Given this assumption, you arrive at either bose or fermi statistics. So, our discussion went about the plausibility of the latter assumption, under what circumstances does it apply? Now, here you come with the TRIVIAL remark that in FREE quantum field theory, integer spin particle wave functions are symmetric while half integer spin wave functions are anti symmetric. First, I said this myself a few posts ago and second I was asking wheter one could expect these results to be reasonable given the arguments about distinguishability/indistinguishability which lead to our previous derivation of statistics (without any connection to spin whatsoever). Later on I basically said I expect such result to be wrong for free particles : in other words I would expect it to be a simple product state, no matter what the ``spin'' is.

Careful

 

[Careful]

No this is also very wrong. YOU have to be VERY careful about labeling particles and as to what you are actually observing. If You observe a particle then make another measurement, you cannot say which particle you actually detected in the second instance.

If I was to have a list of the theories which I believe people thought they understood but actually didn't so they confused the hell out of everyone else it would be

1) Evolution and natural selection
2)Angular momentum
3)Indistinguishability of particles.

What you have said is contradictory.

Absolutely not, there is no contradiction in what I said whatsoever and if you think there is one, YOU are welcome to point it out (instead of pointing to a reference which everyone knows).

Careful

 

[vanesch]

As a general note, you have to be VERY careful when applying the credo that two particles can be considered to be indistinguishable. For example, when I have two electrons 1 and 2, each flying off in two different directions, say to the left and to the right ; then clearly the probability that the first electron is at my left and the second at my right is much bigger than the first being at my right and the second at my left. In other words flipping the positions might change the physical state !

Uh

The idea is simply that we have "two electrons, one in the state "left to me and flying off to the left" and one in the state "right to me and flying off to the right".

The problem is that the way we write the (non-relativistic) wavefunction, we have to assign certain degrees of freedom as "belonging to electron 1", and other degrees of freedom as "belonging to electron 2".
But clearly, the physical description:

(A) we have "two electrons, one in the state "left to me and flying off to the left" and one in the state "right to me and flying off to the right"

and

(B) we have "two electrons, one in the state "right to me and flying off to the right" and one in the state "left to me and flying off to the left"

are equivalent physical descriptions ; in other words, they should not correspond to *different* physical states. But the way we write the wavefunction, there are a priori two different wavefunctions corresponding to (A) and to (B), simply because when making a cartesian product of hilbert spaces (for "electron 1" and for "electron 2") there's a difference between H1 x H2 and H2 x H1. So we have a whole lot of states in H1 x H2 which are different from those in H2 x H1, and that shouldn't be.

So we have to work "modulo the flip" of H1 and H2, because in H1 x H2, there are different states which correspond to identical physical situations (A) and (B).
Doing this "modulo" thing would give us a lousy space, so another way to do the "modulo" thing is to find in each equivalence class a "preferred representative". This is the state that is "symmetrical" in the equivalence class. It can be the symmetrical, or the anti-symmetrical one.