The operator

∣ e ⟩ ⟨ g ∣ \otimes A

annihilates one field quantum and excites the atom. Its adjoint

∣ g ⟩ ⟨ e ∣ \otimes A^{†}

de-excites the atom and creates one field quantum. These are the two exchange processes in the Jaynes-Cummings model.

Unitary time evolution requires the Hamiltonian to be Hermitian. Since the common coefficient

\frac{1}{2} ℏ (Ω - ω)

is real, the two terms must be adjoints with equal coefficients. Therefore

β = 1 .

ii

Solution

Write

Δ = Ω - ω

. Using

e^{i H_{0} t /ℏ} (∣ e ⟩ ⟨ g ∣ \otimes A) e^{- i H_{0} t /ℏ} = e^{i Δ t} ∣ e ⟩ ⟨ g ∣ \otimes A,

the interaction-picture perturbation is

δ H_{I} (t) = \frac{ℏΔ}{2} (e^{i Δ t} ∣ e ⟩ ⟨ g ∣ \otimes A + e^{- i Δ t} ∣ g ⟩ ⟨ e ∣ \otimes A^{†}) .

- 1

e^{- 1/2} e^{- A^{†}} ∣0 ⟩ = e^{- 1/2} \sum_{m = 0}^{\infty} \frac{( - 1 ) ^{m}}{m !} ∣ m ⟩ .

Hence the initial amplitudes of both

∣ e, n ⟩

and

∣ g, n ⟩

are

c_{e, n} (0) = c_{g, n} (0) = \frac{e ^{- 1/2} ( - 1 ) ^{n}}{2 n !} .

Only

∣ g, n + 1 ⟩

contributes at first order to the amplitude of

∣ e, n ⟩

. Since

A ∣ n + 1 ⟩ = n + 1 ∣ n ⟩

c_{e, n}^{I} (t) = c_{e, n} (0) - \frac{i Δ}{2} n + 1 c_{g, n + 1} (0) \int_{0}^{t} e^{i Δ t^{'}} d t^{'} = c_{e, n} (0) - \frac{1}{2} n + 1 c_{g, n + 1} (0) (e^{i Δ t} - 1) .

But

n + 1 c_{g, n + 1} (0) = - c_{e, n} (0)

, so

c_{e, n}^{I} (t) = \frac{c _{e, n} ( 0 )}{2} (1 + e^{i Δ t}) .

The free Schrödinger-picture phase does not change the probability. Within this first-order perturbative approximation,

P {atom e, field n} = \frac{e ^{- 1}}{2 n !} cos^{2} (\frac{( Ω - ω ) t}{2}) .

t = π / (Ω - ω)

the cosine is zero, so this probability vanishes to the stated perturbative order.

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