# Adding quantum numbers breaks exclusion principle?

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)?

• "What if I add additional quantum numbers (e.g. without any physical interpretation, just constants)?" What do you mean by this? Commented May 4, 2021 at 11:31
• Yes, this is the generalized Pauli principle, which you must have learned about in your introductory particle physics course. Any quantum numbers distinguishing particles allow for more non-vanishing antisymmetric combinations of them. Commented May 4, 2021 at 12:55
• It may appear problematic in mathematical/theoretical reasoning, but in the nature it is hard to add quantum numbers. If you can add them, it may be an indication that your initial model was not correct/complete. Commented May 4, 2021 at 17:16

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!

• The main point of my question has to do with this line of your answer: "" Note that you cannot 'neglect' the spin degree of freedom in the first place, as you would've not specified the single-particle states completely. "" If I start with a given H then I can always lift it to a H' on a bigger Hilbert space where the H' does not care about the extra DOF's. These extra DOF's then allow me to construct more ground states in such a way that the original ground state for a 2 Fermion system has higher energy than the new one. (Continued in next comment)
– Gert
Commented May 4, 2021 at 13:46
• Is it correct if I summarize your answer as: this is indeed true, and you can verify, by measuring the energy, whether you had an incomplete description of the system?
– Gert
Commented May 4, 2021 at 13:47
• @Gert I don't know if I can follow. Could you rephrase your question? Is your $H$ a hamiltonian? What I intended to say with the quoted line is that if you want to describe a particle with say spin $s=1/2$, then of course you have to include this in your description of the physical system, i.e. in the construction of the Hilbert space $\mathscr{H}_1$. Commented May 4, 2021 at 14:22
• @Gert I just don't get why we can't just add dummy internal DOF's (mathematically) to cheat on the exclusion principle. Well, I guess this would lead to contradictions with experiments? If your particles do not have certain properties, you shouldn't include them in the mathematical description, no?! Commented May 4, 2021 at 16:25
• @Gert I don't know if I understand you. I agree that if the theoretical predictions are disproved in experiment, then it seems reasonable that the model could be incomplete. But I do not see a connection to the question why you cannot arbitrarily add or omitt things for no reason. Commented May 6, 2021 at 16:06