Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 32D (Principles of Quantum Mechanics)

For a time-independent reference Hamiltonian, define interaction-picture states and observables byThe transformation is unitary, so matrix elements and expectation values are unchanged; the two pictures make the same physical predictions. Differentiating the state and using the Schrödinger equation cancels the $H_0$ term and gives the interaction picture evolutionIntegrating once and replacing the evolving state in the integral by its initial state gives $|{\psi }_I(t)⟩=|n⟩-(i/\hslash ){\int }_0^t{V}_I(t^{\prime })|n⟩d{t}^{\prime }+O(V^2)$. For $m\ne n$, the first-order transition amplitude therefore yields

For the oscillator, $a|n⟩=\sqrt{n}|n-1⟩$ and $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$. Only neighboring levels have a first-order amplitude. Write $\Delta =\nu -\omega$. The two phase integrals are ${\int }_0^t{e}^{i\Delta t^{\prime }}d{t}^{\prime }$ and its conjugate, whose squared modulus is $4{\sin }^2(\Delta t/2)/{\Delta }^2$. To order ${\lambda }^2$,At resonance take their continuous limits, ${\lambda }^{2}n{t}^2$ and ${\lambda }^2(n+1)t^2$. All other distinct-level probabilities vanish to this order. Normalization gives $P_{n\to n}=1-4{\lambda }^2(2n+1){\sin }^2(\Delta t/2)/{\Delta }^2$ to this order, and the downward probability is zero when $n=0$.

The allowed transition rate is proportional to $t{\mathrm{sinc}}^2(\Delta t/2)$, so its central peak has width of order $1/t$ about $\nu =\omega$. Distributionally, $4{\sin }^2(\Delta t/2)/(t{\Delta }^2)\to 2\pi \delta (\Delta )$. Thus both allowed transitions become sharply resonant at the oscillator frequency. Near the peak, perturbation theory still requires ${\lambda }^2(n+1)t^2\ll 1$; its coherent $t^2$ growth cannot be extrapolated indefinitely at fixed coupling.