Past exam of the advanced quantum physics course of the University of Cambridge / 2021 / Section B / 5

The components of the momentum operator do not generally commute with a position-dependent magnetic vector potential. Expanding minimal coupling with the ordering intact givesActing on a wavefunction shows that $\mathbf{p}\cdot \mathbf{A}-\mathbf{A}\cdot \mathbf{p}=-i\hslash \nabla \cdot \mathbf{A}$. In the Coulomb gauge the divergence vanishes, so the two cross terms agree. Discarding the quadratic term in the weak radiation amplitude givesThe equality of the two operator orderings follows from the gauge condition, not from treating $\mathbf{p}$ and $\mathbf{A}$ as commuting numbers component by component.

For the requested oscillator response, treat $y$ as the fixed transverse position of a one-dimensional quantum harmonic oscillator along $z$, and put $\theta =ky$. The supplied radiation is transverse and already satisfies the Coulomb gauge. Its perturbation isThe annihilation operator and creation operator giveThus the only first-order transition out of the ground state is to $|1⟩$. More generally, the oscillator electric-dipole selection rule is $\Delta n=\pm 1$. This is a rule for matrix elements; exact energy conservation is replaced by a finite-time resonance profile during a switched finite-duration interaction.

To display both the resonant and nonresonant terms without spurious singularities, defineThe interaction picture coefficient in the given expansion is thenThe full Schrödinger picture coefficient of $|1⟩$ is $e^{-3i{\omega }_{0}t/2}{b}_1(t)$, while that of $|0⟩$ is $e^{-i{\omega }_{0}t/2}$ at this order. The normalization changes only at order $A_0^2$. In particular, at $\omega ={\omega }_0$ the first term is $C{e}^{i\theta }t$, finite for every finite time, and the perturbative result requires $|b_1|\ll 1$.

The fixed-$y$ interpretation is needed for the displayed target formulae. If $p^2$ were instead a three-dimensional kinetic operator with no transverse confinement, that model would have no normalizable transverse ground state, and $e^{\pm iky}$ would transfer transverse momentum $\pm \hslash k$. A treatment of that different model needs specified transverse states and recoil energies; they are absent from the question. The following parts use the intended localized-oscillator response.

The only relevant position matrix element is $z_{01}=\sqrt{\hslash /(2m{\omega }_0)}$. To first order in the radiation amplitude, the two ground–excited-state cross terms giveMultiplying by the charge $-e$ gives the induced electric dipole moment. Since $2\sqrt{\hslash /(2m{\omega }_0)}C=e{A}_0/m$, the coefficient derived in part (b) yieldsat linear order. Taking the complex conjugate inside the real part of the second term changes it to minus the same term with $e^{i\theta }$ and reversed frequency signs. ThusThe terms at ${\omega }_0$ are switch-on transients, while those at $\omega$ describe the driven response. Both are needed to make $\mathbf{P}(0)=0$. The expression is understood by continuity at $\omega ={\omega }_0$; the first quotient then tends to $t{e}^{-i{\omega }_{0}t}$, producing resonant secular growth rather than an infinite dipole at a finite time.

There is also an initial-velocity check on this harmonic-oscillator dipole response to switched radiation. The formula gives ${\dot{P}}_z(0^{+})=-2e^2{A}_0\cos (ky)/m$. Because the magnetic vector potential jumps at zero time, its time derivative contains an impulsive electric field. Equivalently, the mechanical velocity is $(p_z+e{A}_z)/m$, whose initial mean is $2e{A}_0\cos (ky)/m$. The transient therefore implements the specified switch-on correctly.

For $\omega \ne {\omega }_0$, isolate the terms proportional to $e^{-i\omega t}$ in the previous answer. Their coefficient isHence the driven electric dipole moment component isFor $t>0$, away from the instantaneous switch, the radiation electric field isThus the dynamic electric polarizability isBelow resonance the driven electric dipole moment is in phase with the electric field; above resonance its sign reverses. The same susceptibility follows from the classical Lorentz oscillator model without damping. At exact resonance the separation into bounded driven and transient terms fails: their apparent poles combine into the finite-time secular term identified in part (c). Without damping there is no bounded steady resonant response, and the first-order state expansion eventually ceases to apply.