 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/34c/c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

∣ \pm ⟩_{1} = (∣0 ⟩_{1} \pm ∣1 ⟩_{1}) / 2

. Since

B_{1} + B_{1}^{†}

is the Pauli

X

operator on the first oscillator, the exact eigenvectors are

∣ \pm ⟩_{1} \otimes ∣ n_{2} ⟩_{2},

E_{\pm, n_{2}} = \pm g + E_{2} n_{2}, n_{2} = 0, 1.

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