(a) For the free particle, V (x) = 0, so the time-dependent Schrödinger equation reads
i \hbar \frac{\partial \Psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi}{\partial x^{2}} . \quad \Psi(x, t)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} e^{i p x / \hbar} \Phi(p, t) d p \Rightarrow
\frac{\partial \Psi}{\partial t}=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} e^{i p x / \hbar} \frac{\partial \Phi}{\partial t} d p, \quad \frac{\partial^{2} \Psi}{\partial x^{2}}=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty}\left(-\frac{p^{2}}{\hbar^{2}}\right) e^{i p x / \hbar} \Phi d p . So
\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} e^{i p x / \hbar}\left[i \hbar \frac{\partial \Phi}{\partial t}\right] d p=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} e^{i p x / \hbar}\left[\frac{p^{2}}{2 m} \Phi\right] d p .
But two functions with the same Fourier transform are equal (as you can easily prove using Plancherel’s theorem), so
i \hbar \frac{\partial \Phi}{\partial t}=\frac{p^{2}}{2 m} \Phi .
\frac{1}{\Phi} d \Phi=-\frac{i p^{2}}{2 m \hbar} d t \quad \Rightarrow \Phi(p, t)=e^{-i p^{2} t / 2 m \hbar} \Phi(p, 0) .
(b)
\Psi(x, 0)=A e^{-a x^{2}} e^{i l x}, \quad A=\left(\frac{2 a}{\pi}\right)^{1 / 4} . (Problem 2.42(a)).
\Phi(p, 0)=\frac{1}{\sqrt{2 \pi \hbar}}\left(\frac{2 a}{\pi}\right)^{1 / 4} \int_{-\infty}^{\infty} e^{-i p x / \hbar} e^{-a x^{2}} e^{i l x} d x=\frac{1}{\left(2 \pi a \hbar^{2}\right)^{1 / 4}} e^{-(l-p / \hbar)^{2} / 4 a} . (Problem 2.42(b)).
\Phi(p, t)=\frac{1}{\left(2 \pi a \hbar^{2}\right)^{1 / 4}} e^{-(l-p / \hbar)^{2} / 4 a} e^{-i p^{2} t / 2 m \hbar} ; \quad|\Phi(p, t)|^{2}=\frac{1}{\sqrt{2 \pi a} \hbar} e^{-(l-p / \hbar)^{2} / 2 a} .
(c)
\langle p\rangle=\int_{-\infty}^{\infty} p|\Phi(p, t)|^{2} d p=\frac{1}{\sqrt{2 \pi a} \hbar} \int_{-\infty}^{\infty} p e^{-(l-p / \hbar)^{2} / 2 a} d p
\text { [Let } y \equiv(p / \hbar)-l, \text { so } p=\hbar(y+l) \text { and } d p=\hbar d y \text {.] }
=\frac{\hbar}{\sqrt{2 \pi a}} \int_{-\infty}^{\infty}(y+l) e^{-y^{2} / 2 a} d y [but the first term is odd]
=\frac{2 \hbar l}{\sqrt{2 \pi a}} \int_{0}^{\infty} e^{-y^{2} / 2 a} d y=\frac{2 \hbar l}{\sqrt{2 \pi a}} \sqrt{\frac{\pi a}{2}}=\hbar l [as in Problem 2.42(d)].
\left\langle p^{2}\right\rangle=\int_{-\infty}^{\infty} p^{2}|\Phi(p, t)|^{2} d p=\frac{1}{\sqrt{2 \pi a} \hbar} \int_{-\infty}^{\infty} p^{2} e^{-(l-p / \hbar)^{2} / 2 a} d p=\frac{\hbar^{2}}{\sqrt{2 \pi a}} \int_{-\infty}^{\infty}\left(y^{2}+2 y l+l^{2}\right) e^{-y^{2} / 2 a} d y .
=\frac{2 \hbar^{2}}{\sqrt{2 \pi a}}\left[\int_{0}^{\infty} y^{2} e^{-y^{2} / 2 a} d y+l^{2} \int_{0}^{\infty} e^{-y^{2} / 2 a} d y\right]
=\frac{2 \hbar^{2}}{\sqrt{2 \pi a}}\left[2 \sqrt{\pi}\left(\sqrt{\frac{a}{2}}\right)^{3}+l^{2} \sqrt{\frac{\pi a}{2}}\right]=\left(a+l^{2}\right) \hbar^{2} . [as in Problem 2.42(d)].
(d) H=\frac{p^{2}}{2 m} ; \quad\langle H\rangle=\frac{1}{2 m}\left\langle p^{2}\right\rangle=\frac{\hbar^{2}}{2 m}\left(l^{2}+a\right)=\frac{1}{2 m}\langle p\rangle^{2}+\frac{\hbar^{2} a}{2 m} . \quad \operatorname{But}\langle H\rangle_{0}=\frac{1}{2 m}\left\langle p^{2}\right\rangle_{0}=\frac{\hbar^{2} a}{2 m} (Problem 2.21(d)).
So \langle H\rangle=\frac{1}{2 m}\langle p\rangle^{2}+\langle H\rangle_{0} . QED Comment: The energy of the traveling gaussian is the energy of the same gaussian at rest, plus the kinetic energy \left(\langle p\rangle^{2} / 2 m\right) associated with the motion of the wave packet as a whole.