Pauli exclusion and superposition states

[thegreenlaser]

In basic chemistry, we "fill up" the energy levels of an atom by putting two electrons in each energy level. The justification for this (that I've seen) is that the Pauli exclusion principle only allows one electron per state and there are two states in each energy level (spin up and spin down). My question is, why can't we use superposition states? My understanding is that the Pauli exclusion principle says that two fermions can't occupy the same state, but it does not require that those states be orthogonal.

As a simplified example, let's ignore spin and say that we have eigenstates of the Hamiltonian $\left| 1 \right\rangle, \left|2\right\rangle, \left|3\right\rangle, \ldots$. Let's say we have one electron in state $\left| 1 \right\rangle$. Could we then have a second electron in state $2^{-1/2} \big( \left|1\right\rangle + \left|2\right\rangle \big)$? It seems to me that there is no violation of Pauli exclusion here, yet this is a lower energy configuration than having one particle in $\left| 1 \right\rangle$ and one in $\left| 2 \right\rangle$ (which is what we seem to usually assume).

Am I making any mistakes in my thinking here? If not, then I guess my question is why do we seem to always assume that electrons occupy eigenstates of the Hamiltonian? Why do we expect that particles should be in states of definite energy?

 

[Orodruin]

The real statement behind the Pauli exclusion principle is that the state of the system must be anti-symmetric under exchange of two identical fermions. You really are not assigning one specific electron to a specific orbital, but the ground state would be an asymmetric linear combination of $|1,2\rangle$ and $|2,1\rangle$ (where the first number corresponds to the state of electron 1 and the second of electron 2). You cannot have electrons in the same state as it is impossible to make the state $|1,1\rangle$ (or similar) antisymmetric. Consequently, any such state is also not allowed in a linear combination as this would also not be possible to make asymmetric.

 

[thegreenlaser]

I realize that the actual statement has to do with symmetry of the state vector, but isn't $|\psi\rangle = C( |\alpha\rangle |\beta\rangle - |\beta\rangle |\alpha\rangle )$ always an antisymmetric state vector, even if $|\alpha\rangle$ and $|\beta\rangle$ are not orthogonal, as in my example in the original post? IThe only requirement is that $|\alpha\rangle\neq |\beta\rangle$: the Pauli exclusion principle.

If we set $|\alpha\rangle = |1\rangle$ and $|\beta\rangle = 2^{-1/2} \left( |1\rangle + |2\rangle \right)$, then wouldn't $C ( |\alpha\rangle |\beta\rangle - |\alpha\rangle |\beta\rangle )$ be the antisymmetric state representing the situation where one particle is in state $|1\rangle$ and the other particle is in state $2^{-1/2} \left(|1\rangle + |2\rangle\right)$?

Maybe as an answer to my own question, it's actually not hard to show that
$C ( |\alpha\rangle |\beta\rangle - |\alpha\rangle |\beta\rangle ) = 2^{-1/2} ( |1\rangle |2\rangle - |2\rangle |1\rangle )$
because the overlap between $|\alpha\rangle$ and $|\beta\rangle$ cancels out. So even though we seem to have one particle in $|\alpha\rangle$ and one particle in $|\beta\rangle$, we actually have one particle in $|1\rangle$ and one particle in $|2\rangle$.

So maybe the answer is that there's a stronger version of the Pauli exclusion principle: antisymmetry of the state vector not only implies that no two particles can have the same state, but it also implies that the states of any two particles must be orthogonal to each other. Is that correct?

 

[Orodruin]

That would rather be a consequence of any symmetric part of the wave function cancelling out when anti-symmetrising the state. It is not really that the particles cannot occupy the same state - after all, the actual state is a linear combination of $|1,2\rangle$ and $|2,1\rangle$ so you cannot really say that the first particle is in state 1.

 

[thegreenlaser]

That would rather be a consequence of any symmetric part of the wave function cancelling out when anti-symmetrising the state. It is not really that the particles cannot occupy the same state - after all, the actual state is a linear combination of $|1,2\rangle$ and $|2,1\rangle$ so you cannot really say that the first particle is in state 1.

But can't you say that you found one particle in state 1 and the other particle in state 2, without specifying which particle is in which state?

 

[Orodruin]

Well, you have to say it like that - the particles are identical and there is no way of telling which you found in which state.

 

[thegreenlaser]

Okay, that makes sense. But I guess what's still unclear to me is, do the particles have to be found in orthogonal states? If I say that I found one particle in state $\alpha$ and one particle in state $\beta$ (without specifying which is in which), do states $\alpha$ and $\beta$ have to be orthogonal? If so, why?

 

[VantagePoint72]

Okay, that makes sense. But I guess what's still unclear to me is, do the particles have to be found in orthogonal states? If I say that I found one particle in state α\alpha and one particle in state β\beta (without specifying which is in which), do states α\alpha and betabeta have to be orthogonal? If so, why?

Yes, and the reason was answered above:

That would rather be a consequence of any symmetric part of the wave function cancelling out when anti-symmetrising the state.

And it's pretty easy to see mathematically why anti-symmetrizing cancels out any non-orthogonal components—it's just taking a determinant, and determinants are invariant when you add a scalar multiple of one column to another. So, if you take non-orthogonal states and apply the Slater determinant, the non-orthogonal parts will always be eliminated.

 

[VantagePoint72]

Also worth noting is that completely anti-symmetric states have the same form in any orthogonal basis for the same reason. Going from one orthogonal basis to another just amounts to multiplying the matrix inside the Slater determinant by some unitary, which, again, does not change the determinant. Hence, the completely anti-symmetric state spanned by some particular set of states is unique up to phase.

 

[thegreenlaser]

Yes, and the reason was answered above:
And it's pretty easy to see mathematically why anti-symmetrizing cancels out any non-orthogonal components—it's just taking a determinant, and determinants are invariant when you add a scalar multiple of one column to another. So, if you take non-orthogonal states and apply the Slater determinant, the non-orthogonal parts will always be eliminated.

Okay, sorry this is taking me so long. So I think the math makes sense... any antisymmetric wavefunction made up of non-orthogonal states is going to reduce (i.e., be mathematically equivalent to) to a wavefunction made of orthogonal states.

But then what's the physical interpretation of that? I see two possibilities:
(1) Having particles in non-equal but non-orthogonal states is physically impossible. (I.e., fermions must be in orthogonal states to each other)
(2) A system of particles in non-equal but non-orthogonal states is always exactly equivalent to some system of particles in orthogonal states, so we might as well assume that they're in orthogonal states.

I'm thinking (2) is correct?