# Filling a System of independent Electrons and the Pauli Exclusion Principle

Consider a system of spineless electrons in the independent electron approximation. Thus we consider a one-particle Hamiltonian $H$.

Assume for simplicity that $H$ has discrete spectrum labeled by $E_n$. Also assume that the corresponding eigenstates are infinitely degenerate, with $\alpha$ eigenstates per unit area for each $n$.

Now we want to fill the many-body system up to some energy $\mu$. The Pauli exclusion principle says that no two electrons may occupy the same state. However, we may put $\alpha$ electrons in the first level and then the $(\alpha+1)$th could be a superposition of states, thus not in the same state as the previous $\alpha$ electrons.

Is the reason why this would fail the fact that such a many-body wave-function wouldn't be anti-symmetric with respect to exchange of any pair of particles?

I'm not completely sure I understand your question. But if you are imagining something like putting one electron in the state $|\alpha\rangle$ and then the next one in the state $\frac{1}{\sqrt{2}}( |\alpha\rangle+|\alpha +1\rangle)$, then yes, that won't work. A slightly more precise way to state the Pauli exclusion principle is that electrons must all occupy linearly independent states, which the two above clearly aren't. And indeed, if you tried to antisymmetrize two electrons in $|\alpha\rangle$ and $\frac{1}{\sqrt{2}}( |\alpha\rangle+|\alpha +1\rangle)$ you just find that the amplitude for both being in $|\alpha\rangle$ cancels out.

• If the first electron state $|\alpha\rangle$ lives in $\mathcal{H}_1$ and the second lives in $\mathcal{H}_2$ how can you sum $|\alpha\rangle + |\alpha + 1\rangle$? They don't belong to the same vector space. – gented Oct 6 '15 at 22:02
• Hi @GennaroTedesco- Sure, whenever we talk about constructing a many-body fermionic state by "filling up single particle states" it is not quite right, but it is a common and often useful way to get intuition for what the state ends up looking like. If you like, I am considering antisymmetrizing the state $|\alpha\rangle_1 \otimes (|\alpha \rangle_2 + |\alpha +1 \rangle_2)$, where e.g. $|\alpha\rangle_1$ is a state in the single-particle Hilbert space that you call $H_1$. – Rococo Oct 6 '15 at 23:28
• *Sorry, make that $\mathcal{H}_1$ – Rococo Oct 6 '15 at 23:36
• If $|\alpha\rangle$ is the name of the state in $\mathcal{H}_1$ and not the occupation number, what is $|\alpha + 1\rangle$? Also, if $|\alpha\rangle \in \mathcal{H}_1$ what is $|\alpha\rangle_2$? – gented Oct 7 '15 at 9:46
• $|\alpha\rangle$ is a single-particle state, labelled by some quantum number $\alpha$. $|\alpha +1\rangle$ is another single particle state, labelled by the quantum number $\alpha+1$. $|\alpha\rangle_1$, $|\alpha +1\rangle_1$ are single particle states in $\mathcal{H}_1$. $|\alpha\rangle_2$, $|\alpha +1\rangle_2$ are single particle states in $\mathcal{H}_2$. – Rococo Oct 7 '15 at 15:28

Consider a system of spineless electrons in the independent electron approximation. Thus we consider a one-particle Hamiltonian $H$.

If you have a system of $N$ (non)-interacting free particles, the total Hamiltonian will be the direct sum of each single Hamiltonian, namely $$H(q_1, p_1;\ldots;q_n, p_n) =\oplus_{j=1}^N H_j(q_j, p_j)$$ the sum being intended as $H_1\otimes 1_2\otimes\ldots\otimes 1_N \oplus 1_1\otimes H_2\otimes 1_3\otimes\ldots\otimes 1_N$ and so on and so forth. If $|i_p\rangle_k$ is a single particle state $i_p$ for the $k^{\textrm{th}}$ particle, then an $N$-particle state is written as $$|\psi\rangle = \frac{1}{\sqrt{N!}}\sum_{\sigma} (-1)^{\sigma}P\Big(|i_1\rangle\otimes|i_2\rangle\otimes\ldots\otimes|i_N\rangle\Big)$$ $P$ being all the pairwise permutations of $N$ elements, taken with the corresponding signs. The state being fermionic is included in the alternate signs $(-1)^{\sigma}$ (take $+1$ for bosons instead). Since we have to take permutations, we can equivalently specify the above state indicating how many particles $n_1$ happen to occupy the single particle state $|i_1\rangle$, how many particles $n_2$ occupy $|i_2\rangle$ and so on. Therefore, normalisation factors besides, the state $|\psi\rangle$ can equivalently be denoted as $$|\psi\rangle = c(n_1, n_2,\ldots) |n_1, n_2, \ldots\rangle.$$ As you can see from the alternate signs in the permutations, if a single particle state happens twice then the overall state vanishes because it would be the sum of similar contributions having pairwise opposite signs. As such the numbers $|n_1, n_2, \ldots\rangle$ can be only either $0$ or $1$.

Assume for simplicity that $H$ has discrete spectrum labeled by $E_n$.

Since $H=\oplus H_j$ then $E_n = \epsilon_{n_1} +\ldots+ \epsilon_{n_k}$ with $n_1 + n_2 + \ldots = N$ and each $\epsilon_{n_p}$ being the single particle energy contribution of the single particle state $|i_p\rangle$.

Since in your example the particles are spinless, we assume the only degrees of freedom to be orbital.

The Pauli exclusion principle says that no two electrons may occupy the same state. However, we may put $\alpha$ electrons in the first level.

You may not put $\alpha$ electrons in the first level, because as per the above the occupation numbers for each single particle state $n_p$ may be either $0$ or $1$. Moreover it is not clear what you exactly mean by first level. Each single particle state that makes up the overall $N$-particle state has its own energy $\epsilon_{n_p}$ and may be occupied at most once. Yet, it may happen that two different single particle states accidentally share the same energy, that is $\epsilon_{n_i} = \epsilon_{n_j}$ for some $i\neq j$. Nevertheless, the two underlying states are different, hence electrons will not be in the same state anyways.

Addendum: good reference for the above is "F. Schwabl: Advanced quantum mechanics".

• Thanks for your answer. I don't understand why I cannot put $\alpha$ electrons per unit area in the first level, if the first level is degenerate with $\alpha$ eigenstates per unit area. – PPR Oct 6 '15 at 21:49
• What do you exactly mean by "first level"? You don't put particles in a level, you put particles in a state. – gented Oct 6 '15 at 21:52
• I mean that you have many states ($\alpha$ eigenstates per unit area) all with the same energy, which I call the first level because it is the lowest one. – PPR Oct 6 '15 at 21:54
• If the states are different as element of Hilbert spaces then yes, you may. – gented Oct 6 '15 at 21:57
• I think the OP's terminology of "levels" is supposed to reference the degenerate structure of Landau levels. – Rococo Oct 6 '15 at 23:34