 Solution

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.