Exclusion principle in Helium 3

[EternalStudent]

TL;DR

Asking whether there are multiple Helium 3 states with the same spin.

I was wondering about the following question. On the one hand, Helium 3 is spin 1/2. But, on the other hand, there are multiple combinations of internal spins that would lead to total spin being spin-up or spin-down. So what would happen if we insist that all helium 3 atoms are at lowest energies. Would we be allowed to have more than one spin-up atom, as long as these atoms have different internal spin combinations? Please let me know.

 

[PeterDonis]

On the one hand, Helium 3 is spin 1/2.

Are you talking about a neutral Helium-3 atom in its ground state, or a Helium-3 nucleus in its ground state? I'm assuming it's the first (the atom).

there are multiple combinations of internal spins that would lead to total spin being spin-up or spin-down

Please give an example.

what would happen if we insist that all helium 3 atoms are at lowest energies. Would we be allowed to have more than one spin-up atom

Why do you think spin up atoms and spin down atoms would have different energies?

 

[EternalStudent]

Are you talking about a neutral Helium-3 atom in its ground state, or a Helium-3 nucleus in its ground state? I'm assuming it's the first (the atom).

I am asking about the neutral Helium-3 atom

Please give an example.

Okay I see your point. I was thinking I could alter spins of constituent particles, but I can't, since exclusion principle demands that electrons have two opposite spins and protons have two opposite spins and spin of a neutron will determine it.

But here is a different question Suppose I have an atom that has 3 electrons, 3 protons and 3 neutrons. Then the lower energy of each will have opposite spins. But then we are left with one higher energy electron, one higher energy proton, one higher energy neutron. Then this is spin 3/2 system. But, since 1/2<3/2, we can legitimately ask about spin 1/2 state that this spin 3/2 system might attain. And then we can get spin-up in three different ways:

a) Higher energy electron and higher energy proton are both spin up, while higher energy neutron is spin down

b) Higher energy electron and higher energy neutron are both spin up while higher energy proton is spin down

c) Higher energy proton and higher energy neutron are spin up, while higher energy electron is spin down.

So the option a has different energy, but options b and c have the same energy. So the question is: are we allowed to have one of those atoms in state b and the other one in state c, despite the fact that they have the same spin and the same energy?

Why do you think spin up atoms and spin down atoms would have different energies?

I didn't say they do. On the contrary, my point is that they have *the same* energy -- yet they can be treated as two separate states due to different internal structure. So the question is: if they have the same total spin and the same energy, can we still treat them as two separate states when it comes to fermi exclusion principle?

 

[Vanadium 50]

You can't look at a multibody system and say this constituent (e.g. electron) but not that constituent. You can only talk about the total wavefunction.

 

[PeterDonis]

Suppose I have an atom that has 3 electrons, 3 protons and 3 neutrons. Then the lower energy of each will have opposite spins. But then we are left with one higher energy electron, one higher energy proton, one higher energy neutron.

This is a Lithium-6 atom in its ground state.

Then this is spin 3/2 system.

Not in the ground state; in the ground state it is a spin 1/2 system.

First, there are two possibilities for the nucleus: spin 0 (unpaired proton and neutron with opposite spins) and spin 1 (unpaired proton and neutron with the same spin). It turns out that spin 1 has slightly lower energy because of the coupling between the magnetic moments of the proton and neutron. So the nucleus in its ground state is spin 1.

Then, there are two possibilities for the atom as a whole, given a spin 1 nucleus: spin 3/2 (nucleus and electron spins aligned) or spin 1/2 (nucleus and electron spins opposite). It turns out that the latter case has slightly lower energy, so the atom as a whole in its ground state is spin 1/2.

we can get spin-up in three different ways

Not with the same energy. There is only one state of lowest energy (ground state), and it is the one I described above (which seems to be the third of the states you are describing).

if they have the same total spin and the same energy, can we still treat them as two separate states when it comes to fermi exclusion principle?

Yes, because they don't have the same position. The exclusion principle says that no two fermions can have the exact same values for all degrees of freedom; spin degrees of freedom are not the only ones.

More precisely (to take into account the valid point @Vanadium 50 makes), suppose we have a system consisting of two Li-6 atoms, both in the ground state as described above. The proper application of the exclusion principle to this two-fermion system is that if we exchange the two atoms, the wave function changes sign. That means that the wave function must be such that changing its sign does not affect the values of any observables (since we have no way of distinguishing which atom is which, so both wave functions, the original one and the "exchanged" sign-changed one, must describe the same physical state, i.e., the same values for all observables).

But the requirement that the wave function must change sign on exchange only applies to the entire wave function, including all degrees of freedom, not just spin. There is no requirement that both atoms cannot have the same spin--just that if they have the same spin (so the spin part of the wave function is symmetric, not changing sign on exchange), then the spatial part of the wave function must be antisymmetric (i.e., changing sign on exchange).

 

[EternalStudent]

Yes, because they don't have the same position

But they are still small enough to be quantum particles. And, as such, they can both look like particle in the box and have the same exact wave function in a position space.

 

[PeterDonis]

they are still small enough to be quantum particles

So what? That does not affect anything I said.

they can both look like particle in the box and have the same exact wave function in a position space.

You are missing the point. Saying that the two particles "have the same exact wave function" is meaningless. If you have two particles, you do not have two wave functions, one for each particle. You have one wave function, for the two-particle system. That one wave function has to satisfy the properties I described if both particles are fermions.

 

[EternalStudent]

You are missing the point. Saying that the two particles "have the same exact wave function" is meaningless. If you have two particles, you do not have two wave functions, one for each particle. You have one wave function, for the two-particle system. That one wave function has to satisfy the properties I described if both particles are fermions.

I guess "different wave functions" is a sloppy way of saying that \psi_1 and \psi_2 have to be different in equation \psi (x,y) = \psi_1 (x) \psi_2 (y) - \psi_2 (x) \psi_1 (y)

In any case, the question is: should two Lithium 3 atoms in the box be treated the same way as two electrons in an atom (meaning they should have different wave functions in the above sense) or should they be allowed to have the same "external" wave function due to different internal wave functions?

I guess strictly speaking we have to look at the configuration space. So we don't just have \psi_1 and \psi_2 but rather we have \psi_{11}, ... , \psi_{19} and \psi_{21}, ..., \psi_{29} since each of them have internal structure, Then \psi_1 and \psi_2 are some sort of integrals of those things. So maybe its possible to have \psi_1=\psi_2 as long as \psi_{1k} is not equal to \psi_{2k}. That is, maybe we can have an antisymmetric function that is really non-zero yet its integral over internal degrees of freedom produces zero?

 

[PeterDonis]

I guess "different wave functions" is a sloppy way of saying that \psi_1 and \psi_2 have to be different in equation \psi (x,y) = \psi_1 (x) \psi_2 (y) - \psi_2 (x) \psi_1 (y)

Which is the wrong way to look at it to begin with. What does "different" even mean when the two functions are functions of different arguments?

You have two sets of degrees of freedom. We informally describe these as those corresponding to "atom number 1" and "atom number 2", but those labels are misleading. What we really have is one single wave function whose arguments include two atoms' worth of degrees of freedom: two positions and two spins. We could express this as a function $\Psi(x_1, x_2, s_1, s_2)$.

What "exchange" means is that we switch arguments in the wave function; so for fermions, changing sign under exchange means $\Psi(x_1, x_2, s_1, s_2) = - \Psi(x_2, x_1, s_2, s_1)$, where the "1" and "2" arguments of the wave function, corresponding to the degrees of freedom, are switched.

Notice that none of this has anything at all to do with how, or even whether, the function $\Psi(x_1, x_2, s_1, s_2)$ can be expressed as a sum of products of other functions that each only cover some of the arguments. For simplicity, wave functions like $\Psi(x_1, x_2, s_1, s_2)$ are often written in terms of such sums of products. But it has to be done more carefully than you are doing it.

 

[PeterDonis]

I guess "different wave functions" is a sloppy way of saying that \psi_1 and \psi_2 have to be different in equation \psi (x,y) = \psi_1 (x) \psi_2 (y) - \psi_2 (x) \psi_1 (y)

To follow up on this, going on from my previous post: having the two atoms "in the same state" does not mean they are "described by the same function". It means that, in the wave function $\Psi(x_1, x_2, s_1, s_2)$ that I described in my last post, we have $x_1 = x_2$ and $s_1 = s_2$. And it should be obvious from what I wrote in my last post that if that is the case for two fermions, we must have $\Psi = 0$, i.e., zero probability. That is how the Pauli exclusion principle is realized in this case.