= Solution
The <canonical anticommutation relations> imply $a^2=(a^\dagger)^2=0$. Thus the <number operator> satisfies
$$
N_a^2=a^\dagger aa^\dagger a
=a^\dagger(1-a^\dagger a)a=N_a.
$$
It is a <orthogonal projection>, so its <eigenvalues> obey $n_a^2=n_a$ and are $\boxed{n_a\in\{0,1\}}$. The same argument gives $\boxed{n_b\in\{0,1\}}$. This is the <Pauli exclusion principle> for each orbital. The two <number operators> commute, so they have simultaneous <eigenstates>.
Take the <vacuum state> to have unit norm and fix the double-occupation convention $|11\rangle=a^\dagger b^\dagger|0\rangle$. An <orthonormal basis> of the four-dimensional <fermionic Fock space> is
$$
\boxed{\begin{array}{c|c|c}
\text{state}&n_a&n_b\\\hline
|00\rangle=|0\rangle&0&0\\
|10\rangle=a^\dagger|0\rangle&1&0\\
|01\rangle=b^\dagger|0\rangle&0&1\\
|11\rangle=a^\dagger b^\dagger|0\rangle&1&1
\end{array}}
$$
Each individual <number operator> has two-dimensional <eigenspaces>: for $N_a$ these are spanned by $\{|00\rangle,|01\rangle\}$ and $\{|10\rangle,|11\rangle\}$; for $N_b$ they are spanned by $\{|00\rangle,|10\rangle\}$ and $\{|01\rangle,|11\rangle\}$. Higher <occupation numbers> vanish because the square of a <fermionic creation operator> is zero. Reversing the order of the two <fermionic creation operators> changes only the sign convention for $|11\rangle$.
Back to article page
