Consider a system of $N$ spin $1/2$ particles. Assume the spin is the only degree of freedom and hence there is no spatial component. Then the dimension of the Hilbert space in this case is $2^N$. This follows since in this case we have $j_1 = j_2 = \ldots = j_N = 1/2. $ And the dimension of the product space is $$ (2j_1 + 1) (2j_2 + 1) \ldots (2j_N + 1). $$

Now suppose these particles obey Fermi-Dirac statistics because they are identical fermions. I'm now being asked to determine the dimension of the vectorspace.

I don't really understand the answer. The answer my professor gave is: According to Pauli exclusion principle, no two particles can occupy the same quantum state. In this case we have only two states (spin up, spin down). So there are only two options:

$$ N=1 : \qquad \mid + \rangle, \quad \mid - \rangle \qquad \dim = 2 $$ $$ N=2 : \qquad \frac{1}{\sqrt{2}} \bigg( \mid + \rangle \mid - \rangle \ - \ \mid - \rangle \mid + \rangle \bigg) \qquad \dim = 1. $$

I don't understand how one derives these dimensions. I don't see how Pauli principle can be important to determine the dimension. The particles will just fill the lowest energy levels (with no more than two occupying the same state). That doesn't mean there are only two states? They would also be characterized by the quantum number $n$ for example?

  • 2
    $\begingroup$ "Fermi-Dirac statistics" is the physical way of saying "project down to the antisymmetric subspace of the total Hilbert space". In this case, the total Hilbert space is $\mathbb C^2\otimes\cdots\otimes\mathbb C^2$ $N$ times. The projection onto the antisymmetric subspace is the same as considering the exterior product $\mathbb C^2\wedge\cdots\wedge\mathbb C^2$ $N$ times. Hence the dimensionality follows: for $N=1$ it is just $\mathbb C^2$, hence dimension 2; for $N=2$ the exterior product is just $\mathbb C$, hence dimension 1; for $N>2=\operatorname{dim}(\mathbb C^2)$ the subspace is $\{0\}$. $\endgroup$
    – Phoenix87
    Jun 6, 2016 at 10:25

1 Answer 1


If spin is the only degree of freedom, that also means all particles in your system will have the same quantum numbers for any property that is not spin. This makes your argument about occupying the lowest energy levels somewhat redundant, as there is only one energy level. Therefore, you can have no more than two particles in your system, since there are only two possibilities for the spin of a spin $\frac{1}{2}$ particle. A third particle would necessarily have the same spin as one of the other two, and therefore the Pauli exclusion principle would be violated.

Now for the dimension of the Hilbert space in both cases. In the case $N = 1$, it is obvious the particle has either spin up or spin down. In the case $N = 2$, however, you should have a Hilbert space with dimension $4$, since $2^2 = 4$. But because of the Pauli exclusion principle, both particles cannot have the same spin state. They can't be both 'spin up' or 'spin down' because that would violate the Pauli exclusion principle.

However, that still leaves you with two possible states, namely $|+\rangle|-\rangle$ and $|-\rangle|+\rangle$. I don't see how exactly your professor finds dimension one for this Hilbert space, as the sum of these states $\frac{1}{\sqrt{2}}(|+\rangle|-\rangle + |-\rangle|+\rangle)$ is also an acceptable solution (the linear cominations arise because you want eigenstates of different operators). The difference between these two is that when you take the difference you will find a total spin of $S = 0$, and when you take the sum you have a total spin of $S = 1$ (total spin is not the same as 'spin up' or 'spin down'). You might want to ask your professor about this possibility.

TL; DR: The Pauli exclusion principle rules out certain states, which reduces the dimension of your vector space.

EDITED: The $\frac{1}{\sqrt{2}}(|+\rangle|-\rangle + |-\rangle|+\rangle)$ state is symmetric and therefore ruled out because of the Fermi-Dirac statistics. Thanks to fqq and Phoenix87 for noticing this

  • 1
    $\begingroup$ "This makes your argument about occupying the lowest energy levels somewhat redundant, as there is only one energy level" Not necessarily, as no Hamiltonian is specified. $\endgroup$
    – fqq
    Jun 6, 2016 at 10:26
  • $\begingroup$ The two-electron state must be antisymmetric, so the symmetric state is not "an acceptable solution" $\endgroup$
    – fqq
    Jun 6, 2016 at 10:27
  • $\begingroup$ @Bart W. Thank you for the reply. It is a bit more clear to me now. $\endgroup$
    – Kamil
    Jun 6, 2016 at 10:52

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