Suppose we have an atom. It is commonly said that because of the PEP, two electrons can't be in the ground state unless they have opposite spins, because no two electrons can have the same wavefunction.
What bugs me is that spin up and spin down aren't the only possible spin states. There's a whole continuum of linear combinations of them, and as far as I can tell the PEP wouldn't exclude the possibility of having lots of electrons, all sharing the same spatial wavefunction but with different combinations of $\mid\uparrow \rangle$ and $\mid\downarrow\rangle$. Why doesn't this happen?
Answer
The general one-particle spin state for a spin 1/2 particle is $$|\psi\rangle = a\mid\uparrow\rangle + b\mid\downarrow\rangle$$ with $|a|^2 + |b|^2 = 1$. So let us try to anti-symmetrize two of these. \begin{align*}\operatorname{Alt}(|\psi\rangle_1 \otimes |\psi\rangle_2) = & (a_1 \mid\uparrow\rangle + b_1\mid\downarrow\rangle)\otimes (a_2 \mid\uparrow\rangle + b_2 \mid\downarrow\rangle) - (a_2 \mid\uparrow\rangle + b_2 \mid\downarrow\rangle)\otimes (a_1 \mid\uparrow\rangle + b_1 \mid\downarrow\rangle)\\ = & (a_1a_2 - a_1a_2) \mid\uparrow\rangle\mid\uparrow \rangle + (b_1b_2 - b_2b_1)\mid\downarrow\rangle\mid\downarrow \rangle \\ & + (a_1b_2 - a_2 b_1)\mid \uparrow\rangle\mid\downarrow \rangle + (b_1a_2 - b_2 a_1) \mid\downarrow\rangle\mid\uparrow \rangle \\ = & (a_1b_2 - a_2b_1) (\mid\uparrow \downarrow \rangle - \mid\downarrow\uparrow\rangle) \end{align*} so whatever one-particle state you start with you end with something proportional to $\mid\uparrow \downarrow \rangle - \mid\downarrow\uparrow\rangle$.
More abstractly, if $v_1, v_2, \ldots, v_n$ are a basis for a vector space $V$, a basis the anti-symmetric rank 2 tensors on $V$ is given by $$v_i \otimes v_j - v_j \otimes v_i \quad 1 \le i < j \le n.$$ In the case that $ n = 2$, this reduces to the previous result.
No comments:
Post a Comment