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It is written in my quantum physics book that the K shell contains only 2 electrons due to the Pauli principle.

I know that if $n = 1, l = 0, m = 0$, then the Hilbert space associated to the spin is of dimension $2$. I also know that Pauli principle says that if we have a vector $\lvert\psi\rangle$ which represents the state of N electrons, then $\lvert\psi\rangle$ must be antisymmetric by transposition. But how can we conclude please ?

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up vote 7 down vote accepted

Suppose you had three electrons, with individual wavefunctions $\lvert \psi_1 \rangle$, $\lvert \psi_2 \rangle$, and $\lvert \psi_3 \rangle$. Let them all have the same $\vec{x}$, $l$, and $m$, so they can only differ in intrinsic spin. Since spin is a two-dimensional Hilbert space, as you noted, then three vectors must be linearly dependent. That is, there exist complex numbers $\alpha$ and $\beta$ such that $$ \lvert \psi_3 \rangle = \alpha \lvert \psi_1 \rangle + \beta \lvert \psi_2 \rangle. $$

Now the state of all three electrons is given by the tensor product $$ \lvert 1,2,3 \rangle \equiv \lvert \psi_1 \rangle \otimes \lvert \psi_2 \rangle \otimes \lvert \psi_3 \rangle. $$ The tensor product respects the structure of the underlying vector spaces, which is a fancy way of saying we can write $$ \lvert 1,2,3 \rangle = \alpha \lvert 1,2,1 \rangle + \beta \lvert 1,2,2 \rangle. $$ But $\lvert 1,2,1 \rangle$, as a product state of identical fermions, is antisymmetric under interchange, in particular under interchance of the $\lvert \psi_1 \rangle$ components: $$ \lvert 1,2,1 \rangle = - \lvert 1,2,1 \rangle. $$ Thus $\lvert 1,2,1 \rangle \equiv 0$. Similarly, $\lvert 1,2,2 \rangle$ vanishes. We have just shown that our 3-electron wavefunction is a linear combination of $0$-vectors, and so it too vanishes.

This argument easily generalizes to more than three electrons by appending the combined wavefunction of all the others to the end of our state and just carrying it through the computations; that is, just do the same thing splitting up $\lvert \psi_3 \rangle$, but apply it to $$ \lvert 1,2,3 \rangle \otimes \lvert 4,\ldots,n \rangle. $$

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Thank you very much ! – Arnaud Apr 20 '13 at 6:13

Statement of the Pauli principle is that two fermion can not be at the same quantum state, otherwise anti-symmetric state vector would give zero. So in K shell, $ l = 0, m = 0$ . Therefore there's only two different quantum state which is spin up and spin down states.

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Why we couldn't add a third electron, which could be in a linear superposition of $|+>$ and $|->$ ? I'm sorry, I assume that this question is stupid but I haven't the answer... – Arnaud Apr 19 '13 at 18:47
Linear superposition of up&down state is not an independent state vector from what we've got already. This superposition state will collapse to either spin-up or spin-down state upon measurement. – Hikmet Apr 19 '13 at 18:55
Yes, I agree, but I don't see how can we conclude from this statement that the K shell contains only 2 electrons. – Arnaud Apr 19 '13 at 18:57
You can think of in terms of probabilities. Suppose that you have an anti-symmetric state vector describing the behavior of two fermions. If these two fermions happens to be at the same quantum number, then state vector would be zero, which means that probability of finding these two fermions in the same quantum number is just zero. – Hikmet Apr 19 '13 at 19:04
I don't understand, sorry. – Arnaud Apr 19 '13 at 19:11

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