Consider a charged oscillator, of positive charge q and mass m, which is subject to an oscillating electric field E_{0}\cos \omega t; the particle’s Hamiltonian is \hat{H}=P_{2}/(2m)+k\hat{X}^{2} /2+ qE_{0} \hat{X}\cos \omega t.
(a) Calculate d〈\hat{X}〉/dt,d〈\hat{P}〉/dt,d〈\hat{H}〉/dt.
(b) Solve the equation for d〈\hat{X}〉/dt and obtain 〈\hat{X}〉(t) such that 〈\hat {X}〉(0)=x_{0}.
(a) Since the position operator \hat{X} does not depend explicitly on time (i.e.,∂\hat{X}/∂t=0), equation (3.88)
\frac{d}{dt}〈\hat{A}〉=\frac{1}{i\hbar} 〈[\hat{A},\hat{H}]〉+〈\frac{∂\hat{A}}{∂t} 〉. (3.88)
yields
\frac{d}{dt}〈\hat{X}〉=\frac{1}{i\hbar} 〈[\hat{X},\hat{H}]〉= \frac{1}{i\hbar}〈\left[\hat{X},\frac{P_{2}}{2m} \right] 〉=\frac{〈\hat {P}〉}{m}. (3.231)
Now, since [\hat{P},\hat{X}]=-i\hbar ,[\hat{P},\hat{X}^{2}]= -2i \hbar \hat{X} and ∂\hat{P}/∂t=0, we have
\frac{d}{dt}〈\hat{P}〉=\frac{1}{i\hbar} 〈[\hat{P},\hat{H}]〉= \frac {1}{i\hbar}〈\left[\hat{P},\frac{1}{2}k\hat{X}^{2}+ qE_{0} \hat {X}\cos \omega t \right] 〉=-k〈\hat{X}〉-qE_{0}\cos \omega t, (3.232)
\frac{d}{dt}〈\hat{H}〉=\frac{1}{i\hbar} 〈[\hat{H},\hat{H}]〉+〈\frac {∂\hat{H}}{∂t} 〉=〈\frac{∂\hat{H}}{∂t} 〉=-qE_{0}\omega 〈\hat{X}〉\sin \omega t. (3.233)
(b) To find 〈\hat{X}〉 we need to take a time derivative of (3.231) and then make use of (3.232):
\frac{d^{2} }{dt^{2} }〈\hat{X}〉=\frac{1}{m}\frac{d}{dt}〈\hat{P}〉=-\frac{k}{m}〈\hat{X}〉-\frac{qE_{0}}{m}\cos \omega t. (3.234)
The solution of this equation is
〈\hat{X}〉(t)=〈\hat{X}〉(0)\cos \left(\sqrt{\frac{k}{m} }t \right) -\frac{qE_{0}}{m\omega }\sin \omega t+A, (3.235)
where A is a constant which can be determined from the initial conditions; since 〈\hat {X}〉(0)=x_{0} we have A = 0, and hence
〈\hat{X}〉(t)=x_{0}\cos \left(\sqrt{\frac{k}{m} }t \right)-\frac {qE_{0}}{m\omega }\sin \omega t. (3.236)