32C (Principles of Quantum Mechanics)

Solution

∣ ψ_{I} (t)⟩ = e^{i H_{0} t /ℏ} ∣ ψ_{S} (t)⟩

and

V_{I} (t) = e^{i H_{0} t /ℏ} V (t) e^{- i H_{0} t /ℏ}

. Differentiation cancels the free Hamiltonian operator and yields

i ℏ \frac{d}{d t} ∣ ψ_{I} (t)⟩ = λ V_{I} (t) ∣ ψ_{I} (t)⟩ .

Assume the interaction is switched off sufficiently fast in the remote past for the following integral to converge. Starting in

∣ n ⟩

⟨ m ∣ ψ_{I} (t)⟩ = - \frac{iλ}{ℏ} \int_{- \infty}^{t} ⟨ m ∣ V (t^{'}) ∣ n ⟩ e^{i (E_{m} - E_{n}) t^{'} /ℏ} d t^{'} + O (λ^{2})

for

m \neq = n

. The picture change only multiplies the measured amplitude by a phase. Squaring therefore gives

P_{n \to m} (t) = \frac{λ ^{2}}{ℏ ^{2}} \int_{- \infty}^{t} ⟨ m ∣ V (t^{'}) ∣ n ⟩ e^{i (E_{m} - E_{n}) t^{'} /ℏ} d t^{'}^{2} + O (λ^{3}) .

∣1 ⟩ = a^{†} ∣0 ⟩

is normalized and

⟨ 1∣ (a + a^{†}) ∣0 ⟩ = 1

ℏ ω

, and

\int_{- \infty}^{\infty} e^{- p ∣ t ∣} e^{iω t} d t = \frac{1}{p + iω} + \frac{1}{p - iω} = \frac{2 p}{p ^{2} + ω ^{2}} .

Hence the lowest nonzero transition probability is

P_{0 \to 1} (\infty) = \frac{4 λ ^{2} p ^{2}}{ℏ ^{2} ( p ^{2} + ω ^{2} ) ^{2}} + O (λ^{4}) .

The stronger remainder here follows from parity: every insertion of

a + a^{†}

reverses oscillator parity, so the amplitude between

∣0 ⟩

and

∣1 ⟩

has only odd powers of

λ

. The general displayed

O (λ^{3})

estimate remains valid.

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