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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?

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up vote 4 down vote accepted

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.

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Should that be $\operatorname{Alt}(|\psi\rangle_1 \otimes |\psi\rangle_2)$ rather than $\operatorname{Alt}(|\psi\rangle_1 \otimes |\psi\rangle_1)$ in the first line? – DumpsterDoofus Apr 22 '14 at 1:20
Also, I believe the last term in the same line should be $b_1$? Edit: Never mind that it looks fine. – Constandinos Damalas Apr 22 '14 at 1:23
You are correct. I made an edit. – Robin Ekman Apr 22 '14 at 1:24
I follow the math, but I'm not sure what the consequences of this are. Could you expand a little bit? – Javier Apr 22 '14 at 1:59
@JavierBadia The consequence is that you don't have to have prepared a particular set of eigenvalues (say "up" and "down") for the occupation limit (2 electrons) to apply. This is the PEP applies no matter what games you try to play with the electron spins, as the combined wavefunction "knows" how many are allowed (in less anthromophic language, you run out of degrees of freedom to use in anti-symmetrizing it). – dmckee Apr 22 '14 at 13:30

As you said spin states consist of linear combinations of spin up and spin down. An electron can be either up or down. You cannot have an electron having a spin of quarter up or quarter down for example. An electron at any time is either up or down. That is why you cannot have more than 2 electrons sharing the same spin state. Think of it like this, if two electrons share the same spatial state, then one of them is spin up and the other is spin down. The spin superposition you are confused about just tells you that the electrons are not always spin up and spin down, but have associated probabilities with their spin states.

For example a spin state $\frac{1}{\sqrt{2}}\mid\uparrow \rangle$ + $\frac{1}{\sqrt{2}}\mid\downarrow\rangle$ tells you that the electron is spin up 50% of the times and spin down 50% of the time. Then the other electron can only be in the other state to satisfy the Pauli Exclusion Principle.

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"You cannot have an electron having a spin of quarter up or quarter down for example. An electron at any time is either up or down." This is blatantly false in QM. An electron that is spin up with respect to the $z$-axis is in a superposition of spin up and down with respect to any other axis. – Robin Ekman Apr 21 '14 at 23:40
The superposition just implies probabilities of being up and down, you cannot just break up spin into components like you do in Classical Mechanics. I feel we are saying the same thing here. – Constandinos Damalas Apr 22 '14 at 0:05
@PhotonicBoom To be in a superposition is to not be in either of the eigenstates. The probabilities implied only materialize when you measure the z-axis component of the spin. In the mean time it is not in either state along the z-axis. The statement Robin is objecting to is not correct on it's own, and the full answer requires you to anti-symmetrize the wave function to show the occupation limit does not depend on the spin axis you chose. – dmckee Apr 22 '14 at 0:58
I see what you mean now. I was referring to measurement results but as I understand now this is not the complete picture still. – Constandinos Damalas Apr 22 '14 at 1:21

The Pauli-Exclusion-Principle states that you can not have two electrons in the same quantum state. What one has to take into account is, that $ \left | \uparrow \right >$ and $ \left | \downarrow \right >$ already are a basis of the spin part of the hilbert space. This is not comparable to e.g. the z-direction of a three-dimensional space where the x and y components are independent. If you have a quantum state $\psi(\vec{r})$ its spatial component $\phi(\vec{r})$ does depend on the basis of the spin space you choose, i.e. with the basis of the spin state you are determining a direction in space which reflects in the spatial component of your quantum state.

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