32C (Principles of Quantum Mechanics)

Solution

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

. Differentiating and using the Schrödinger equation cancels the free Hamiltonian:

i ℏ \frac{d}{d t} ∣ ψ_{I} (t)⟩ = V_{I} (t) ∣ ψ_{I} (t)⟩, V_{I} (t) = e^{i H_{0} t /ℏ} V (t) e^{- i H_{0} t /ℏ} .

Integrating once and replacing the state in the integral by its initial value gives

∣ ψ_{I} (t)⟩ = ∣ a ⟩ - \frac{i}{ℏ} \int_{0}^{t} V_{I} (t^{'}) ∣ a ⟩ d t^{'} + O (V^{2}) .

Since

⟨ b ∣ a ⟩ = 0

, the first-order transition amplitude is

- i /ℏ

⟨ b ∣ V (t^{'}) ∣ a ⟩ e^{i (E_{b} - E_{a}) t^{'} /ℏ}

. Its modulus squared gives

P_{a \to b} (t) = \frac{1}{ℏ ^{2}} \int_{0}^{t} ⟨ b ∣ V (t^{'}) ∣ a ⟩ e^{i (E_{b} - E_{a}) t^{'} /ℏ} d t^{'}^{2}

For the two-state example, the matrix element is

v t^{'}

and the free energies coincide. Thus

P_{1 \to 2} = v^{2} t^{4} / (4 ℏ^{2})

to order

v^{2}

. All Hamiltonians commute at different times because they are combinations of

I

and the same Pauli matrix

σ_{1}

. Therefore the ordinary exponential of their time integral solves the equation: differentiating it gives

- (i /ℏ) (E I + v t σ_{1})

times itself, and its value at zero is

I

. Since

σ_{1}^{2} = I

, it is

e^{- i Et /ℏ} [cos (\frac{v t ^{2}}{2ℏ}) I - i sin (\frac{v t ^{2}}{2ℏ}) σ_{1}] .

Starting from state 1, the exact result and the approximation condition are

P_{1 \to 2} (t) = sin^{2} (\frac{v t ^{2}}{2ℏ}), \frac{v t ^{2}}{2ℏ} ≪ 1.

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