 Solution

ID: past-exam-of-the-field-theories-and-collective-phenomena-in-condensed-matter-course-of-the-university-of-oxford/2025/3/a/solution

a^{2} = (a^{†})^{2} = 0

. Thus the number operator satisfies

N_{a}^{2} = a^{†} a a^{†} a = a^{†} (1 - a^{†} a) a = N_{a} .

n_{a}^{2} = n_{a}

and are

n_{a} \in {0, 1}

. The same argument gives

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 ⟩ = a^{†} b^{†} ∣0 ⟩

state ∣00 ⟩ = ∣0 ⟩ ∣10 ⟩ = a^{†} ∣0 ⟩ ∣01 ⟩ = b^{†} ∣0 ⟩ ∣11 ⟩ = a^{†} b^{†} ∣0 ⟩ n_{a} 0101 n_{b} 0011

Each individual number operator has two-dimensional eigenspaces: for

N_{a}

these are spanned by

{∣00 ⟩, ∣01 ⟩}

and

{∣10 ⟩, ∣11 ⟩}

; for

N_{b}

they are spanned by

{∣00 ⟩, ∣10 ⟩}

and

{∣01 ⟩, ∣11 ⟩}

. 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 ⟩

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