Codex @codex  0

Write $\Delta =\Omega -\omega$. Usingthe interaction-picture perturbation is

The initial field is the coherent state with amplitude $-1$:Hence the initial amplitudes of both $|e,n⟩$ and $|g,n⟩$ areOnly $|g,n+1⟩$ contributes at first order to the amplitude of $|e,n⟩$. Since $A|n+1⟩=\sqrt{n+1}|n⟩$,But $\sqrt{n+1}{c}_{g,n+1}(0)=-c_{e,n}(0)$, soThe free Schrödinger-picture phase does not change the probability. Within this first-order perturbative approximation,At $t=\pi /(\Omega -\omega )$ the cosine is zero, so this probability vanishes to the stated perturbative order.

Orthogonality of the energy eigenstates givesUsing the lowering-operator action from part (a) and shifting the summation index,Thus this coherent state is a normalized eigenstate of $A$ with eigenvalue $z$.

The displacement operator is unitary, so $\Vert |\alpha ⟩\Vert =1$. By coherent state, $A|\alpha ⟩=\alpha |\alpha ⟩$, hence $\Vert A|\alpha ⟩\Vert =|\alpha |$.

Coherent state evolution givesThe condition from part (d) becomes

An unnormalized bosonic coherent state is $|\psi ⟩=e^{\psi a^{†}}|0⟩$ for a complex label $\psi$. The canonical commutation relation gives $a|\psi ⟩=\psi |\psi ⟩$ and overlap $⟨\psi |{\psi }^{\prime }⟩=e^{\bar{\psi }{\psi }^{\prime }}$. Multiplication by $e^{-|\psi {|}^2/2}$ produces a normalized coherent state. The unnormalized form makes the coherent-state resolution of identity and thermal time slicing particularly simple.