= Solution
The only relevant position <matrix element> is $z_{01}=\sqrt{\hbar/(2m\omega_0)}$. To first order in the radiation amplitude, the two ground–excited-state cross terms give
$$
\langle z\rangle_t=2\sqrt{\frac{\hbar}{2m\omega_0}}
\operatorname{Re}\left[e^{-i\omega_0t}b_1(t)\right].
$$
Multiplying by the charge $-e$ gives the induced <electric dipole moment>. Since $2\sqrt{\hbar/(2m\omega_0)}C=eA_0/m$, the coefficient derived in part (b) yields
$$
\mathbf P(t)=-\frac{e^2A_0}{m}\hat{\mathbf z}\operatorname{Re}\left[
e^{i\theta}\frac{e^{-i\omega t}-e^{-i\omega_0t}}{i(\omega_0-\omega)}
+e^{-i\theta}\frac{e^{i\omega t}-e^{-i\omega_0t}}{i(\omega_0+\omega)}\right]
$$
at 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. Thus
$$
\boxed{\mathbf P(t)=-\frac{e^2A_0}{m}\hat{\mathbf z}\operatorname{Re}\left[e^{iky}\left(
\frac{e^{-i\omega t}-e^{-i\omega_0t}}{i(\omega_0-\omega)}
-\frac{e^{-i\omega t}-e^{i\omega_0t}}{i(\omega_0+\omega)}\right)\right].}
$$
The 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 $t e^{-i\omega_0t}$, 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 $\dot P_z(0^+)=-2e^2A_0\cos(ky)/m$. Because the <magnetic vector potential> jumps at zero time, its time derivative contains an impulsive <electric field>. Equivalently, the mechanical velocity is $(p_z+eA_z)/m$, whose initial mean is $2eA_0\cos(ky)/m$. The transient therefore implements the specified switch-on correctly.
Back to article page
