# I 2-particle Wavefunction and the Pauli Exclusion Principle

1. Nov 21, 2017

### Isaac0427

Say you have two particles a and b with respective positions $x_a$ and $x_b$. Particle a is in the state $\psi_a$, and particle b is in the state $\psi_b$. If they are distinguishable, the wavefunction is
$$\psi=\psi_a(x_a)\psi_b(x_b)$$
However, if they are identical fermions, the wavefunction is
$$\psi(a,b)=A\left(\psi_a(x_a)\psi_b(x_b)-\psi_a(x_b)\psi_b(x_a)\right)$$
If $\psi_a=\psi_b$ then $\psi=0$. But, if $\psi_a$ is arbitrarily close to $\psi_b$ (i.e. $\psi_a=\psi_1$ and $\psi_b=\sqrt{.99}\psi_1+.1\psi_2$ where 1 and 2 denote the 1st and 2nd principal quantum numbers of some potential) then the wavefunction is not zero, and can be normalized if A is arbitrarily large. So, it would seem as though you could have an infinite number of identical fermions in just two energy levels. I know, however, that is not the case. So, I came up with a thought on why this is, but it had a problem. Could tell me if I'm on the right track, and where I am going wrong?

The constant A in the identical fermion equation must be $\frac{1}{\sqrt{2}}$ (for N particles it would be $\frac{1}{\sqrt{N!}}$). This would require that $\psi_a$ and $\psi_b$ (and any other wavefunctions corresponding to more identical fermions) be orthogonal in order to keep $\psi$ normalized. But, this has two issues. First, as a and b get farther apart, the second identical fermions start to act distinguishable. So, $\psi_a(x_b)$ and $\psi_b(x_a)$ go to zero, and $\psi$ goes to $\frac{1}{\sqrt{2}}\psi_a(x_a)\psi_b(x_b)$, which is not normalized (as well as not the same as the distinguishable particle equation).

I also had a thought on that; my thinking was that even as a and b get arbitrarily far apart, in theory they are still identical, and particle a being in in state $\psi_a$ and particle b being in $\psi_a$ are the same things, as the fermions are indistinguishable. Thus, $\psi_a(x_b)\psi_b(x_a)$ does not go to zero. But, since $\psi_a$ and $\psi_b$ do not overlap, they are essentially orthogonal. It appears as though this would make the identical fermion equation yield about the same practical result as the distinguishable equation.

Again, this is just my thinking, I have no idea how far off I am.

2. Nov 21, 2017

### mikeyork

Trying to discuss the spin-statistics theorem without including spin dependence in your wave functions is a sure recipe for misunderstanding and error.

Then once you have spin in the system you need to be very careful how you define the frames of reference in which you measure the z-component as a $2\pi$ rotation on just one of the particles can change the sign of a fermion wave function.

Until you get that clear, any attempt to answer your question will create more confusion.

3. Nov 21, 2017

### Isaac0427

I understand that spin dependence is a factor too—but it still yields the same orthogonality question.

4. Nov 21, 2017

### mikeyork

If you rely on an anti-symmetrization rule you need to be clear about where it comes from and exactly how it applies.

BTW you have actually hit on a really important question which is how do we move without a sudden jump from distinguishable states to indistinguishable states.

5. Nov 21, 2017

### Isaac0427

Okay, I know that fermions, by definition, have antisymetric wave functions under exchange. They also happen to have half-integer spin.

The answer to this may also help me with my first question. Is there a good explanation for this somewhere?

6. Nov 21, 2017

### mikeyork

They are anti-symmetric only with a specific choice of spin reference frames. See

Symmetrizing The Symmetrization Postulate'', AIP Proceedings 545 (2000) Spin Statistics Connection And Commutation Relations'', ed. Hilborn and Tino, pp 104-110. (http://arxiv.org/pdf/quant-ph/0006101.pdf)

and

http://arxiv.org/pdf/quant-ph/9908078v2.pdf

where you will also find references to similar papers by other authors.

Yes, see the papers above. The point is that the conventional antisymmetry is due to a hidden permutation-dependent geometrical asymmetry that introduces a $2\pi$ rotation on one particle's spin frame when they are permuted. When that asymmetry is removed, the wavefunctions are symmetric for fermions as well as bosons and so the problem disappears. If the particle states are defined in a way that is completely symmetrical (permutation-independent) then the transition from distinguishable states to indistinguishable states is smooth.

To my knowledge, you won't find a reasonable answer to this question of a discontinuity in going from distinguishable to indistinguishable states anywhere else.

7. Nov 21, 2017

### Staff: Mentor

What makes you think this? By which I mean, what experimental evidence?

8. Nov 21, 2017

### Staff: Mentor

As @mikeyork has implicitly pointed out, these wave functions are not correct. You are assuming that the position part of the wave function must be antisymmetric, but that is not the case. Only the whole wave function has to be antisymmetric; but the wave function includes a spin part as well as a position part. So there are two possibilities: position antisymmetric and spin symmetric, or position symmetric and spin antisymmetric. (A good sidebar question: why are these the two possibilities?) It might be helpful to write out these two possibilities explicitly.

9. Nov 21, 2017

### Swamp Thing

mikeyork, are you M J York, the author of the papers you referred to?

10. Nov 21, 2017

### Isaac0427

Many textbooks state it, see Griffiths chapter 5.
If $\psi_a$ and $\psi_b$ are spinors, wouldn’t that issue be fixed?

If so, am I still correct about the one-over-root-2 term forcing the two states to be orthogonal?

Also, if I am getting this correct, separated identical fermions are always indestinguushable, just at a point thinking of them as distinguishable is a valid approximation. Is this correct or off-the-mark?

Last edited: Nov 21, 2017
11. Nov 21, 2017

### Staff: Mentor

I do have a copy of Griffiths and I could scan over chapter 5, but I can assure you are misinterpreting what he is saying. That being the case best to re-read it and post his EXACT wording if you are still in any doubts.

It's the difference between orthogonal and orthonormal. I suspect you haven't studied Linear algebra yet - that should be corrected. Once you do that the answer is obvious. Of course I could tell you - but then I would hvve to kill you . Seriously you learn and understand better what you sort out yourself.

Thanks
Bill

12. Nov 21, 2017

### Isaac0427

I apologize-I worded that question poorly. I understand a reasonable amount of linear algebra. And I understand that the one-over-root-2 keeps a superposition normalized if the super posed wavefunctions are orthogonal (I am pretty sure). My question was if the one-over-root-2 term has to be in the equation for identical fermions, which would mean that identical fermions would have to have orthogonal wavefunctions.

13. Nov 21, 2017

### Staff: Mentor

Exactly. (Pun sort of intended. )

14. Nov 21, 2017

### Isaac0427

Well, I do see where I got confused. Griffiths did say that you could pretend that the particles are distinguishable for practical matters. I’m assuming then that it is not really the limit of the identical fermion equation, it is just another way look at the situation that yields the same practical result.

15. Nov 21, 2017

### mikeyork

Yes.

16. Nov 21, 2017

### mikeyork

Identical particles, by themselves, are intrinsically indistinguishable by definition. However, their space-time or energy-momentum states may be distinguishable.

17. Nov 22, 2017

### Isaac0427

Right, but theoretically for any case saying that particle a is in the state $\psi_a$ is essentially the same as saying that particle b is in that state, as particles a and b are identical (i.e. it doesn’t make sense to talk about particle a or particle b as you need to talk about particle a and particle b because they are identical). Is that correct?

However, if $\psi_a$ and $\psi_b$ do not overlap, you can say that one particle is in the state $\psi_a$ and one particle is in the state $\psi_b$, and get the same practical results as if you had made the particles identical fermions. Is this correct as well?

18. Nov 22, 2017

### Staff: Mentor

No, that's not what "identical particles" means. What it means is that you can't say "particle a is in state $\psi_a$" at all (or particle b). All you can say is "a particle is in state $\psi_a$ and a particle is in state $\psi_b$". Or, more precisely, that you have a quantum system consisting of two identical particles, with one in state $\psi_a$ and one in state $\psi_b$. Labeling the particles as "particle a" and "particle b" is an artifact of the way we model such systems mathematically; it doesn't correspond to anything in the system itself.

If the two identical particles are fermions, you can also say something else: that it is impossible to have a state of the two-particle quantum system in which both particles are in state $\psi_a$. (Note that the state here has to include all the degrees of freedom of the system, both position and spin, and in the general case there might be others as well.) Mathematically, this implies that the wave function of the system must be antisymmetric.

19. Nov 22, 2017

### Isaac0427

I guess that is sort-of what I meant (though using very incorrect wording). I meant that you can't say anything about particle a being in a state or particle b being in a state as you can only talk about "some particle," and you cannot draw distinctions between a and b.
Again, though, do the states just have to be different or orthogonal? If they have to be orthogonal, why would that be, mathematically. If they don't, then couldn't an infinite number of identical fermions all be in the ground energy level with slightly different spin-up and spin-down superpositions?

20. Nov 22, 2017

### Staff: Mentor

The short answer is "just different". But I think it's worth expanding on that.

First, there are no "states" of the individual particles, strictly speaking. There is only the state of the 2-particle system. The Hilbert space of states for the system will have a basis of orthogonal vectors, but those are state vectors for the 2-particle system, not for either particle in isolation.

One way to construct the Hilbert space of the 2-particle system (for fermions) is to start with the Hilbert space for a one-particle system and then form an antisymmetric tensor product of two copies of that Hilbert space. (For bosons, you would form the symmetric tensor product.) Similarly, you can construct a basis for the Hilbert space of the 2-particle system by starting with a basis for the one-particle system (which would consist of a set of mutually orthogonal vectors in that Hilbert space) and forming antisymmetric combinations of those vectors. That's basically what you were getting at in the OP of this thread.

The basis states of the 2-particle Hilbert space will be mutually orthogonal, as I said above. And if you express them the way I just described, they will each look like linear combinations of terms that are each products of two 1-particle basis vectors, and in each such product, the two 1-particle basis vectors will be different (since the combinations are antisymmetric), so they will be orthogonal. However, that is not the same as saying the two "one particle states" that make up the two-particle state (i.e., the two states that appear in the description "one particle is in state $\psi_a$ and one particle is in state $\psi_b$") must be orthogonal. That is not the case. They only need to be different.

The example you picked in the OP might obscure this because you only considered two possible positions. More generally, you were considering a discrete spectrum of states (which of course you have to if you are considering bound systems). For a discrete spectrum of states, there is no real difference, for the purposes of this discussion, between "different" and "orthogonal": heuristically, there are only a finite number of "buckets" into which you can put fermions. See further comments below.

No, because you are, again, talking about a discrete spectrum of states. Consider the difference between bound fermions (electrons in an atom, say) and free fermions (e.g., electrons moving through free space). In the latter case, you could indeed have an infinite number of identical fermions all with the same energy but slightly different superpositions of other degrees of freedom. (And of course there are also a continuous infinity of possible energies.)

It's also worth noting that so far we have not considered entangled states; considering them complicates all this even further. For example, here is a description of an entangled state: "there is a fermion at position $x_a$ and a fermion at position $x_b$, and their spins are opposite". You might want to think about how this state would be written down mathematically, as compared with a non-entangled state such as "there is a fermion at position $x_a$ and a fermion at position $x_b$" (with nothing specified about their spins).