# Identical Particles: Symmetry & Antisymmetry of Wave Function

• Euclid
In summary, when discussing identical particles, it is said that their wave function must be either symmetric or antisymmetric with respect to the exchange of particles. This is a postulate in quantum mechanics and is supported by the fact that permuting particles should leave the physical state intact. However, in three dimensions, the overall phase factor of the wave function is restricted to 0 and pi, while in one dimension, it can be a function of position. This is explained in M. D. Girardeau's paper "Permutation Symmetry of Many-Particle Wave Functions". Anyons do not exist in this context, as all particles can be thought of as a superposition of fermionic and bosonic states. The operator for swapping particles
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?

Is this like bosons, fermions and anyons?

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

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.

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

$$\begin{array}{cll} \hat{S}^2\psi(r_1,r_2) = \hat{S}\psi(r_2,r_1)=\psi(r_1,r_2) &\Rightarrow \hat{S}^2 = 1 \\ &\Rightarrow \hat{S} = \pm 1 \\ &\Rightarrow \psi(r_1,r_2)=\pm\psi(r_2,r_1) \end{array}$$

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$$S^2 =1$$ 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:
$$<x,y|S|x,y>=<x,y|y,x> = \delta_{xy}$$

Am I mistaken?

Can you think of an example where

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

masudr said:
Can you think of an example where

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

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

Euclid said:
In fact, S must have some zero's on the diagonal in the position basis:
$$<x,y|S|x,y>=<x,y|y,x> = \delta_{xy}$$

And do you mean

$$\langle x_1',x_2' | \hat{S} | x_1,x_2 \rangle = \langle x_1',x_2' | x_2,x_1 \rangle = \delta_{x_1'x_2,x_2'x_1}$$

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

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

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.

masudr said:
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.

Galileo said:
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?

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Euclid said:
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.

Galileo said:
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.

Euclid said:
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:

$$[H,S] = 0$$

$$HS - SH = 0$$

$$HS = SH$$

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

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

$$S^\dagger H = H S^\dagger$$but 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

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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 $$|\phi>=|s>+|a>$$, where $$|s>$$ is from symmetric subspace of permutational group representation and $$|a>$$ is from antisymmetric part. Attack this state with some permutation and you will have $$D(S)|\phi>=|s>-|a>$$, 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.

I think I am only becoming more confused.

Anyons do not exist?

Euclid said:
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!

zbyszek said:
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

Careful said:
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.

masudr said:
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

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Careful said:
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).

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 $n, l$.

masudr said:
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.

masudr said:
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).).

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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'

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.

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.

Epicurus said:
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

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Epicurus said:
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

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Careful said:
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.

Vanesch said:
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.

Sure, but this is entirely similar in classical mechanics, and there I will call the one flying to the left number one, and the one flying to the right number two. After this, I am never going to say that number one is flying to the right. Here, by definition both particles are distinguishable, since the labelling depends upon their physical properties.

Now, suppose I would color two particles in my box red and green then I have 50 percent chance (given random dynamics within the box) that the green (red) one is flying to the right (left) and 50 percent chance that it flies to the left (right) (in case the red one would always fly to the left for some reason then the following reasoning does not apply). So in this case, I have a statistical mixture which is very different from a superposition. So, now you are left with a dilemma, either you assume your wavefunction to be a description of an ensemble of particles giving rise to independent emissions and then you might proceed as usual (but then the wavefunction is extremely nonlocal in a *spatio-temporal* sense). On the other side, on the individual level, it clearly makes no sense and you are left with the conclusion I draw. In any case, the reason for the single particles building up the correlations of the symmetric/antisymmetric state has to be very different from the reasoning employed previously.

Vanesch said:
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.

Why not ?! Imagine the situation of two different nuclei (call them 1 and 2) and two different electrons red and green bound to 1 and 2 respectively. Then, clearly particles red and green are distinguishable since they are bound to different nuclei and swapping them requires work to overcome the potential barriers of both nuclei. In order to get different statistics here, I would have to imagine that although the nuclei are fixed, the electrons were somehow placed there randomly in a previous moment in time or that in a statistical ensemble of two nuclei boxes I have freedom in identifying the nuclei.

Vanesch said:
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.

Like I explained before, this depends upon your notion of distinguishability as well as your notion of what quantum mechanics is supposed to mean. It is not that simple (and was actually much debated upon) as you try to present it.

The whole disagreement boils down to (a) when to work with statistical ensembles (b) when to work with single wave equations (and this is not silly as epicurus tries to tell us) and (c) how to label particles. Classically, I could do the same : suppose I would be studying wave dynamics in which I had two wave types red and green which emerge through one of two opposite holes of the source respectively. Then I could have emissions RG and GR each with chance 1/2. Now, nobody in his right mind would say that actually RG - GR occurs with probability one, but that our detectors only allow for R or G to be observed (in a consistent way). Now in case something strange happens, then a classical physicist would say that something else occurs which is not properly taken into account yet. This is all I wanted to say.

I do not understand, as epicurus suggests, why these considerations can be confusing (actually it is rather worrying that some do not make them).

Cheers,

Careful

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Epicurus said:
masudr said:
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.

Yes that is exactly the point I was making: that two electrons belonging to the different atoms are not in the same state. Where is the misunderstanding?

masudr said:
Yes that is exactly the point I was making: that two electrons belonging to the different atoms are not in the same state. Where is the misunderstanding?
You said in post 23 that a two electron wave of electrons in different orbitals does not antisymmetrize while in the textbook treatment this is assumed to be the case (see the Hartree Fock approximation). As I said in my previous post, the entire subtlety is in one's interpretation of distinguishability and QM. Similarly, you agreed that two electrons around different nuclei spaced a micron apart do not obey spin statistics, while many people believe spin statistics also to apply in this case.

Careful

Careful said:
Sure, but this is entirely similar in classical mechanics, and there I will call the one flying to the left number one, and the one flying to the right number two. After this, I am never going to say that number one is flying to the right. Here, by definition both particles are distinguishable, since the labelling depends upon their physical properties.

The notion of "distinguishable" or not is indeed a physical notion which is largely independent of quantum theory per se. As you do somewhere else in your post, you "color" the particles to keep them distinguishable in principle. Of course, if in quantum mechanics, I have an internal degree of freedom which colors them, then the entanglement with this degree of freedom will ALSO make them behave as if they are distinguishable. So these tricks are not allowed (no coloring, painting, numbering them).
So what's left ? What's left is that we use the dynamical state itself to distinguish them.

Note that if two electrons are (postulated to be) physically indistinguishable (no colors, paint, numbers... on them), then in classical physics too we've made an "error" in setting up the configuration (or phase) space, because the point corresponding to "electron 1 to the left and electron 2 to the right" (where the 1 and 2 simply come from the order in which we label the coordinates in phase space) is a different point from the point in classical phase space where "electron 1 to the right and electron 2 to the right". So our "phase space" contains different points which correspond to identical physical situations, which shouldn't be the case. To be entirely correct we should have a phase space "modulo physical equivalence" also in classical phase space.

However, this doesn't affect, because of its structure, the results of dynamics in classical physics. This didn't have to be the case, but it is. In other words, the dynamics of classical physics allows you to be sloppy with the phase space (that is, having different points which correspond to the same physical situation) and nevertheless the results always come out all right. So never anybody bothered about this error in setting up a phase space of identical particles, because it doesn't affect the results.

In quantum theory, there's no difference in approach: to each dimension in hilbert space must correspond a different configuration, and we have exactly the same error as in classical phase space when we compose the hilbert space of two identical particles by their tensor product.
We should have enumerated the dimensions of hilbert space by different configurations, and we didn't: we have introduced an unwanted degeneracy by having different dimensions which correspond to the same physical situation.
And, rats, quantum mechanics doesn't have the same forgiving property that classical physics had: it doesn't produce the same results with and without the error (while classical physics did).

We can of course look for the "origin" of the classical forgivingness: it is probably due to the fact that, apart from some pathological cases, the hamiltonian flow in phase space will never "confuse" the trajectory of the (r1,p1,r2,p2) point with the (r2,p2,r1,p1) point - there are no collisions between trajectories in classical phase space. So, dynamically, you can use one of both, and never bother about the other one, and this will not bring you dynamical problems: whether you've explicitly "cut away" the other trajectory of classical phase space or not, it doesn't affect the dynamics of the other one. But this is not true in quantum theory, where the dynamics of the state |r1,r2> is potentially influenced by the existence or not of the |r2,r1> dimension.

So, the point I wanted to make, is this:
if we take the physical assumption that the state (r1,r2) is physically identical (is the same physical state) as (r2,r1), then, keeping the two different points in state space (whether it is classical phase space, or quantum hilbert space) is an error. But in the classical case, the error has no consequences, while in the quantum case, it does.

Careful said:
Similarly, you agreed that two electrons around different nuclei spaced a micron apart do not obey spin statistics, while many people believe spin statistics also to apply in this case.

Well, of course they obey spin statistics. That is:

We shouldn't write the state of the two electrons as:
|atom1,orb1>|atom2,orb2> but rather:
|atom1,orb1>|atom2,orb2> - |atom2,orb2>|atom1,orb1>

However, nobody does this, because in this case, even quantum mechanics will (in most cases) forgive you the error by considering the two different hilbert dimensions ||atom1,orb1>|atom2,orb2> and |atom2,orb2>|atom1,orb1>, and most results you could potentially be interested in will come out all right (just as in classical mechanics), no matter what "representative" you pick from the "physical equivalence" class.

So the poor guy/gal who applies his principles entirely correctly will hit himself simply with more involved algebra, to find the same results (in most cases). That's why people don't bother symmetrising between electrons in different atoms... unless they are solid-state physicists, where it does make a difference again, when the atoms get close together!

The rule when to be allowed to make the error (and hence simplify the algebra), is this:

when, during the entire dynamics, there's never a serious non-orthogonality of the evolutions of the "erroneous" different representatives of the equivalence class, then you'll be allowed to the error (and you don't have to symmetrise).

So as long as "atom1" stays away from "atom2", then the states will remain essentially orthogonal, and you can do your thing with or without symmetrising, as you like.

This is also why in classical mechanics, you can (apart from very pathological cases) always get away with "making the error".

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