Adding quantum numbers breaks exclusion principle?

Question

Consider 2 non-interacting Fermions with spin $s\ne 0$ trapped in a 1D harmonic potential with ground state $\left|\phi_0\right\rangle$. Due the Pauli exclusion principle I would expect that only 1 electron could occupy the ground state. Now if we add the spin of the Fermions as a quantum number (the spins not interacting with the potential) suddenly we could have 2 different states for the Fermions: $\left|\phi_0\right\rangle\otimes\left|s_1\right\rangle$ and $\left|\phi_0\right\rangle\otimes\left|s_2\right\rangle$ with $s_1 \ne s_2$.

So do the Fermions both have ground state-energy or do they need to have different energy? What if I add additional quantum numbers (e.g. some internal degrees of freedom with no effect on the Hamiltonian beside increasing the size of Hilbert space)?

Answer 1

I don't know if this really addresses your question, but here we go:

To start, let us consider a single-particle Hilbert space $\mathscr{H}_1$. If, for example, we have $\mathscr{H}_1 = L^2(\mathbb{R}) \otimes \mathbb{C}^2$, then $$\{|n\rangle \otimes |s\rangle\} $$ is a basis for $\mathscr{H}_1$ if $\{|n\rangle\}_{n\in\mathbb{N}}$ is a basis for $L^2(\mathbb{R})$ and $\{|s\rangle\}_{s=\pm1/2}$ a basis for $\mathbb{C}^2$. We will write $|ns\rangle \equiv |n\rangle \otimes |s\rangle$. Note that you cannot 'neglect' the spin degree of freedom in the first place, as you would've not specified the single-particle states in $\mathscr{H}_1$ completely.

Now consider some Hamiltonian $h$ with eigenfunctions $|ns\rangle$: $$h\, |n s\rangle = E_{ns} \, |ns\rangle \quad. $$

It could for instance take the following form: $$h = \frac{p^2}{2m} + v(x) \otimes \mathbb{I} \quad , \tag{1}$$

where $\mathbb{I}$ denotes the identity operator.

---

The Hilbert space of two indistinguishable fermions is the anti-symmetrized product of the single-particle Hilbert space:

$$\mathscr{H}^F_2 \equiv \mathscr{H}_1 \wedge \mathscr{H}_1 \quad .$$ A basis for this space is given by the anti-symmetrized products of the bases states of the single-particle Hilbert spaces. In other words, vectors of the form $$ |ns\rangle \otimes |n^\prime s^\prime\rangle - |n^\prime s^\prime\rangle \otimes |ns\rangle \tag{2} $$ constitute a basis in $\mathscr{H}^F_2$. A non-interacting Hamiltonian for such a system could be given as

$$H = h \otimes \mathbb{I} + \mathbb{I} \otimes h \quad , \tag{3} $$

for which we see that the aforementioned basis vectors are eigenstates: $$H\, \left( |ns\rangle \otimes |n^\prime s^\prime\rangle - |n^\prime s^\prime\rangle \otimes |ns\rangle\right) = \left(E_{ns}+ E_{n^\prime s^\prime}\right) \, \left(|ns\rangle \otimes |n^\prime s^\prime\rangle - |n^\prime s^\prime\rangle \otimes |ns\rangle\right) $$

---

To answer your questions: First of all, if the single-particle Hamiltonian is of the form of equation $(1)$, then there is some degeneracy, i.e. states with $s=\pm 1/2$ yield the same eigenvalue. In the following let's assume that $E_{0s}\leq E_{ns} $ for $n\geq0$.

Second, note that if in equation $(2)$ we have $n=n^\prime$ and $s=s^\prime$, then the state reduces to the zero vector: The two fermions cannot be in the same single-particle states (Pauli-Principle), i.e. cannot have both the same quantum numbers.

Consequently, when considering the eigenvalue problem of equation $(3)$, we see that the lowest possible energy we could obtain such that the eigenvector is different from the zero vector is $E_{0s=+1/2} + E_{0s=-1/2}$. So, yes, it is possible that both fermions (here electrons) are in single-particle states which yield the ground state energy of the Hamiltonian $(1)$ and thus of the Hamiltonian $(3)$ too. Nevertheless, they are not in the same single-particle state, as their spin quantum numbers differ!