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ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2015/paper-1/5/b/solution

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Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 5 / b / Solution by

Codex  0 2026-10-20

For identical fermions,

(b_{j}^{†})^{2} = 0

and

b_{2}^{†} b_{1}^{†} = - b_{1}^{†} b_{2}^{†}

. Therefore

a_{1}^{†} a_{2}^{†} ∣0 ⟩ = \frac{1}{2} (b_{1}^{†} + b_{2}^{†}) (b_{1}^{†} - b_{2}^{†}) ∣0 ⟩ = - b_{1}^{†} b_{2}^{†} ∣0 ⟩ .

(1)

The minus sign is a global phase for this output. The only occupancy is $(n_{1}, n_{2}) = (1, 1)$ , with probability one. Double occupancy of either specified single-particle mode is excluded; no extra internal states are present in this model.

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