(a) From the Ehrenfest equations d〈\hat{P}〉/dt=〈[\hat{P},\hat {V}(x,t)]〉/i\hbar as shown in (3.134),
\frac{d}{dt}〈\hat{\vec{P}}〉=\frac{1}{i\hbar }〈[\hat{\vec{P}}, \hat{V}(\hat{\vec{R}},t)]〉=-〈\vec{∇}\hat{V}(\hat{\vec{R}},t)〉.
and since for a free particle \hat{V}(x,t)=0, we see that d〈\hat{P}〉/dt=0. As expected this leads to 〈\hat {P}〉(t)=p_{0}, since the linear momentum of a free particle is conserved. Inserting 〈\hat{P}〉=p_{0} into Ehrenfest’s other equation d〈\hat{X}〉/dt=〈\hat{P}〉/m (see (3.132)),
\frac{d}{dt}〈\hat{\vec{R}}〉=\frac{1}{m}〈\hat{\vec{P}}〉.
we obtain
\frac{d〈\hat{X}〉}{dt}=\frac{1}{m}p_{0}. (3.237)
The solution of this equation with the initial condition 〈\hat{X}〉(0)=x_{0} is
〈\hat{X}〉(t)=\frac{p_{0}}{m}t+x_{0}. (3.238)
(b) First, the proof of d〈\hat{P}^{2}〉/dt=0 is straightforward. Since [\hat{P}^{2},\hat{H}]=[\hat{P}^{2}, \hat{P}^{2}/2m]=0 and ∂\hat{P}^{2}/∂t=0 (the momentum operator does not depend on time), (3.124)
\frac{d}{dt}〈\hat{A}〉=\frac{1}{i\hbar} 〈[\hat{A},\hat{H}]〉+〈\frac{∂\hat{A}}{∂t} 〉,
yields
\frac{d}{dt}〈\hat{P}^{2}〉=\frac{1}{i\hbar} 〈[\hat{P}^{2},\hat {H}]〉+〈\frac{∂\hat{P}^{2}}{∂t} 〉=0. (3.239)
For d〈\hat{X}^{2}〉/dt we have
\frac{d}{dt}〈\hat{X}^{2}〉=\frac{1}{i\hbar} 〈[\hat{X}^{2},\hat {H}]〉=\frac{1}{2im\hbar }〈[\hat{X}^{2},\hat{P}^{2}]〉, (3.240)
since ∂\hat{X}^{2}/∂t=0. Using [\hat{X},\hat{P}] =i\hbar, we obtain
[\hat{X}^{2},\hat{P}^{2}]=\hat{P}[\hat{X}^{2},\hat{P}]+[\hat{X}^{2},\hat{P}]\hat{P}
=\hat{P}\hat{X}[\hat{X},\hat{P}]+\hat{P}[\hat{X},\hat {P}]\hat{X}+\hat{X}[\hat{X},\hat{P}]\hat{P}+[\hat{X},\hat{P}] \hat{X}\hat{P}
=2i\hbar (\hat{P}\hat{X}+\hat{X}\hat{P})=2i\hbar (2\hat{P} \hat{X}+i\hbar ); (3.241)
hence
\frac{d}{dt}〈\hat{X}^{2}〉=\frac{2}{m}〈\hat{P}\hat{X}〉+\frac {i\hbar }{m}. (3.242)
(c) As the position fluctuation is given by (\Delta x)^{2}=〈\hat {X}^{2}〉-〈\hat{X}〉^{2}, we have
\frac{d(\Delta x)^{2}}{dt}=\frac{d〈\hat{X}^{2}〉}{dt}-2〈\hat {X}〉\frac{d〈\hat{X}〉}{dt}=\frac{2}{m}〈\hat{P}\hat{X}〉+\frac{i\hbar }{m}-\frac{2}{m}〈\hat{X}〉〈\hat{P}〉. (3.243)
In deriving this expression we have used (3.242) and d〈\hat{X}〉 /dt=〈\hat{P}〉/m. Now, since d(〈\hat{X}〉〈\hat{P}〉)/dt=〈\hat{P}〉d〈\hat{X}〉/dt=〈\hat{P}〉^{2}/m amd
\frac{d〈\hat{P}\hat{X}〉}{dt}=\frac{1}{i\hbar}〈[\hat{P}\hat{X} ,\hat{H}]〉=\frac{1}{2im\hbar }〈[\hat{P}\hat{X},\hat{P}^{2}]〉=\frac {1}{m}〈\hat{P}^{2}〉, (3.244)
we can write the second time derivative of (3.243) as follows:
\frac{d^{2}(\Delta x)^{2}}{dt^{2} }=\frac{2}{m} \left(\frac{d〈\hat{P}\hat{X} 〉}{dt}-\frac{d〈\hat{X}〉〈\hat{P}〉}{dt} \right)=\frac{2}{m^{2}} \left(〈\hat{P}^{2}〉-〈\hat{P}〉^{2}\right)=\frac{2}{m^{2}} (\Delta p)^{2}_{0}, (3.245)
where (\Delta p)^{2}_{0}=〈\hat{P}^{2}〉-〈\hat{P}〉^{2}=〈\hat {P}^{2}〉_{0}-〈\hat{P}〉^{2}_{0}; the momentum of the free particle is a constant of the motion. We can verify that the solution of the differential equation (3.245) is given by
(\Delta x)^{2}=\frac{1}{m^{2}}(\Delta p)^{2}_{0}t^{2} +(\Delta x)^{2}_{0}. (3.246)
This fluctuation is similar to the spreading of a Gaussian wave packet we derived in Chapter 1.