Consider three (or any number bigger than 2) electrons without spatial degrees of freedom, thus the only degree of freedom is the spins. The Hilbert space is then formed by the tensor product of the space of each electron. Now according to my literature about the so-called antisymmetric tensor, that there are no antisymmetric tensor can be formed if the number of vectors to be tensor-multiplied is bigger than the dimension of each vector space. If applied to my example in the beginning, it seems that I cannot form an antisymmetric state in the system of three electrons because the number of ket to be tensor-multiplied is three (there are three electrons) while the dimension of each is 2 (due to spin 1/2). If this is true, then is three electron system without spatial degrees of freedom impossible? But this is strange.


1 Answer 1


Yes, it is not possible to construct a totally antisymmetric spin state with more than two electrons. This is just a statement of Pauli's exclusion principle.

  • $\begingroup$ Does that mean that inclusion of the spatial part of the wave function is essential in creating a 3-electron state? $\endgroup$
    – garyp
    Commented May 16, 2016 at 15:07
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    $\begingroup$ Wait, wait, wait... After some times pondering on it, if I stick on $|\pm\rangle |\pm\rangle |\pm\rangle$ basis, indeed it doesn't seem possible to construct an anti-symmetric state. But if I introduce another state, say $|\pi\rangle$ which is some linear combination of $|+\rangle$ and $|-\rangle$, I can actually form an antisymmetric state, which is $$ |A\rangle = |+\rangle |-\rangle |\pi\rangle + |\pi\rangle |-\rangle |+\rangle + |-\rangle |\pi\rangle |+\rangle - |\pi\rangle |-\rangle |+\rangle -|-\rangle |+\rangle |\pi\rangle - |+\rangle |\pi\rangle |-\rangle $$ What do you think? $\endgroup$
    – nougako
    Commented May 16, 2016 at 15:26
  • $\begingroup$ @nougako indeed it doesn't violate Pauli principle: all three electrons are in different single-particle states, so the total state can be antisymmetrized. $\endgroup$
    – Ruslan
    Commented May 16, 2016 at 16:10
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    $\begingroup$ @nougako No, you cannot. Expand $|\pi\rangle$ as an arbitrary linear combination of $|1\rangle$ and $|2\rangle$ and you will see that $|A\rangle = 0$. If you have a vector space $V$, the antisymmetric product states are a subspace of $V\otimes V$. For a 2d space, $\text{Asymm}(V\otimes V)$ is one-dimensional, and so $\text{Asymm}(V\otimes V\otimes V)$ is zero-dimensional. $\endgroup$ Commented May 16, 2016 at 16:41
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    $\begingroup$ @nougako I was sloppy and I'm not sure the implication is correct. However, $\text{Asymm}(V\otimes V\otimes V)$ is zero-dimensional if $V$ is 2-dimensional. To see this, try to construct a basis. Or, try unapologetic.wordpress.com/2008/12/23/antisymmetric-tensors $\endgroup$ Commented May 16, 2016 at 18:58

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