 Solution

Define the interaction picture by $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$. Differentiating and using the Schrödinger equation cancels the free Hamiltonian:Integrating once and replacing the state in the integral by its initial value givesSince $⟨b|a⟩=0$, the first-order transition amplitude is $-i/\hslash$ times the integral of $⟨b|V(t^{\prime })|a⟩e^{i(E_b-E_a)t^{\prime }/\hslash }$. Its modulus squared givesto second order in the perturbation.

For the two-state example, the matrix element is $v{t}^{\prime }$ and the free energies coincide. Thus $P_{1\to 2}=v^2{t}^4/(4{\hslash }^2)$ to order $v^2$. All Hamiltonians commute at different times because they are combinations of $I$ and the same Pauli matrix ${\sigma }_1$. Therefore the ordinary exponential of their time integral solves the equation: differentiating it gives $-(i/\hslash )(EI+vt{\sigma }_1)$ times itself, and its value at zero is $I$. Since ${\sigma }_1^2=I$, it isStarting from state 1, the exact result and the approximation condition are