 Solution

Define the interaction picture byDifferentiating and using the Schrödinger equation cancels the $H_0$ terms, leavingIts integral equation, iterated once from the initial $|a⟩$, gives $|{\psi }_I(t)⟩=|a⟩-(i\lambda /\hslash ){\int }_0^t{V}_I(t^{\prime })|a⟩d{t}^{\prime }+O({\lambda }^2)$.

Since $a,b$ are orthogonal energy eigenstates with distinct energies, the zeroth-order transition amplitude is zero, whileReturning to the Schrodinger picture changes this component only by a unit-modulus phase. Squaring the first-order amplitude therefore gives

For exponential switching, put $\Delta =E_b-E_a$. The integral isBecause $\mu >0$, its exponential vanishes at late time, yieldingThis is the leading weak-perturbation result, not an exact probability formula at arbitrary coupling.