 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/34a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

A^{†} = \frac{mω X - i P}{2 m ℏ ω} .

[X, P] = i ℏ I

[A, A^{†}] = \frac{1}{2 m ℏ ω} [mω X + i P, mω X - i P] = I .

Thus

A

and

A^{†}

For part (b), write

K = \frac{1}{2} (A^{† 2} - A^{2}), S (γ) = e^{- γ K} .

Since

[A, K] = A^{†}

and

[A^{†}, K] = A

, differentiation of the unitary conjugation gives

\frac{d}{d γ} A (γ) = - S^{†} [A, K] S = - A^{†} (γ), \frac{d}{d γ} A^{†} (γ) = - A (γ) .

A (0) = A

and

A^{†} (0) = A^{†}

A (γ) = A cosh γ - A^{†} sinh γ .

This is the transformation implemented by the single-mode squeeze operator.

For part (c), invert the definition of

A

X = \frac{ℏ}{2 mω} (A + A^{†}), P = - i \frac{m ℏ ω}{2} (A - A^{†}) .

The transformation above implies

S^{†} XS = e^{- γ} X, S^{†} PS = e^{γ} P .

ℏ/ (2 mω)

and

m ℏ ω /2

∣ γ ⟩ = S (γ) ∣0 ⟩

has

(Δ X)_{γ}^{2} = e^{- 2 γ} \frac{ℏ}{2 mω}, (Δ P)_{γ}^{2} = e^{2 γ} \frac{m ℏ ω}{2} .

The reciprocal squeezing factors cancel, so

Δ X Δ P = \frac{ℏ}{2} .

The state therefore continues to saturate the Heisenberg uncertainty relation.

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