(a) Normalization is the same as before: A=\left(\frac{2 a}{\pi}\right)^{1 / 4} .
(b) Equation 2.104 says
\phi(k)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \Psi(x, 0) e^{-i k x} d x . (2.104).
\phi(k)=\frac{1}{\sqrt{2 \pi}}\left(\frac{2 a}{\pi}\right)^{1 / 4} \int_{-\infty}^{\infty} e^{-a x^{2}} e^{i l x} e^{-i k x} d x . [same as before, only k → k – l]= \frac{1}{(2 \pi a)^{1 / 4}} e^{-(k-l)^{2} / 4 a} .
\Psi(x, t)=\frac{1}{\sqrt{2 \pi}} \frac{1}{(2 \pi a)^{1 / 4}} \int_{-\infty}^{\infty} e^{-(k-l)^{2} / 4 a} e^{i\left(k x-\hbar k^{2} t / 2 m\right)} d k .
Let u ≡ k – l , so k = u + l and dk = du:
\Psi(x, t)=\frac{1}{\sqrt{2 \pi}} \frac{1}{(2 \pi a)^{1 / 4}} \int_{-\infty}^{\infty} e^{-u^{2} / 4 a} e^{i\left[u x+l x-(\hbar t / 2 m)\left(u^{2}+2 u l+l^{2}\right)\right]} d u .
=\frac{1}{\sqrt{2 \pi}} \frac{1}{(2 \pi a)^{1 / 4}} e^{i l\left(x-\frac{\hbar l t}{2 m}\right)} \int_{-\infty}^{\infty} e^{-u^{2}\left(\frac{1}{4 a}+i \frac{\hbar t}{2 m}\right)+i u\left(x-\frac{\hbar l t}{m}\right)} d u .
Using the hint in Problem 2.21, the integral becomes
\frac{1}{\sqrt{\frac{1}{4 a}+i \frac{\hbar t}{2 m}}} e^{\left(x-\frac{\hbar l t}{m}\right)^{2} / 4\left(\frac{1}{4 a}+i \frac{\hbar t}{2 m}\right)} \int_{-\infty}^{\infty} e^{-y^{2}} d y=\frac{2 \sqrt{a}}{\gamma} e^{-a\left(x-\frac{\hbar l t}{m}\right)^{2} / \gamma^{2}} \sqrt{\pi} .
so
Ψ (x,t) = \left(\frac{2 a}{\pi}\right)^{1 / 4} \frac{1}{\gamma} e^{-a\left(x-\frac{\hbar l t}{m}\right)^{2} / \gamma^{2}} e^{i l\left(x-\frac{\hbar l t}{2 m}\right)} .
The gaussian envelope (the first exponential) travels at speed \hbar l / m the sinusoidal wave (the second exponential) travels at speed \hbar l /2 m .
(c)
|\Psi(x, t)|^{2}=\sqrt{\frac{2 a}{\pi}} \frac{1}{|\gamma|^{2}} e^{a\left(x-\frac{\hbar l t}{m}\right)^{2}\left[\frac{1}{\gamma^{2}}+\frac{1}{\left(\gamma^{*}\right)^{2}}\right]} .
The term in square brackets simplifies:
\left[\frac{1}{\gamma^{2}}+\frac{1}{\left(\gamma^{*}\right)^{2}}\right]=\frac{1}{|\gamma|^{4}}\left[\left(\gamma^{*}\right)^{2}+\gamma^{2}\right]=\frac{1}{|\gamma|^{4}}\left(1-\frac{2 i \hbar t}{m}+1+\frac{2 i \hbar t}{m}\right)=\frac{2}{|\gamma|^{4}} .
|\gamma|^{2}=\sqrt{\left(1+\frac{2 i a \hbar t}{m}\right)\left(1-\frac{2 i a \hbar t}{m}\right)}=\sqrt{1+\theta^{2}} .
where (as before) \theta \equiv 2 \hbar a t / m . So
|\Psi(x, t)|^{2}=\sqrt{\frac{2 a}{\pi}} \frac{1}{\sqrt{1+\theta^{2}}} e^{2 a\left(x-\frac{h t}{m}\right)^{2} /\left(1+\theta^{2}\right)}= \sqrt{\frac{2}{\pi}} w e^{-2 w^{2}\left(x-\frac{\hbar l t}{m}\right)^{2}} .
where (as before) w \equiv \sqrt{a /\left(1+\theta^{2}\right)} The result is the same as in Problem 2.21, except that x \rightarrow\left(x-\frac{\hbar l}{m} t\right) so |\Psi|^{2} has the same ( flattening gaussian) shape – only this time the center moves at constant speed v=\hbar l / m .
(d)
\langle x\rangle=\int_{-\infty}^{\infty} x|\Psi(x, t)|^{2} d x Let <br />
y \equiv x-v t, \text { so } x=y+v t .
=\int_{-\infty}^{\infty}(y+v t) \sqrt{\frac{2}{\pi}} w e^{-2 w^{2} y^{2}} d y=v t =\frac{\hbar l}{m} t .
(The first integral is trivially zero; the second is 1 by normalization.)
\langle p\rangle=m \frac{d\langle x\rangle}{d t} = \hbar l .
\left\langle x^{2}\right\rangle=\int_{-\infty}^{\infty}(y+v t)^{2} \sqrt{\frac{2}{\pi}} w e^{-2 w^{2} y^{2}} d y=\frac{1}{4 w^{2}}+0+(v t)^{2}=\frac{1}{4 w^{2}}+\left(\frac{\hbar l t}{m}\right)^{2} .
(The first integral is same as in Problem 2.21):
\left\langle p^{2}\right\rangle=-\hbar^{2} \int_{-\infty}^{\infty} \Psi^{*} \frac{\partial^{2} \Psi}{\partial x^{2}} d x ; \quad \frac{\partial \Psi}{\partial x}=\left[-\frac{2 a}{\gamma^{2}}\left(x-\frac{\hbar l t}{m}\right)+i l\right] \Psi ,
\frac{\partial^{2} \Psi}{\partial x^{2}}=-\frac{2 a}{\gamma^{2}} \Psi+\left[-\frac{2 a}{\gamma^{2}}\left(x-\frac{\hbar l t}{m}\right)+i l\right]^{2} \Psi=\left[A x^{2}+B x+C\right] \Psi ,
where
A \equiv\left(\frac{2 a}{\gamma^{2}}\right)^{2}, \quad B \equiv-\left(\frac{2 a}{\gamma^{2}}\right)^{2} \frac{2 \hbar l t}{m}-\frac{4 i a l}{\gamma^{2}}=-\frac{4 i a l}{\gamma^{4}} .
C \equiv-\frac{2 a}{\gamma^{2}}+\left(\frac{2 a}{\gamma^{2}}\right)^{2}\left(\frac{\hbar l t}{m}\right)^{2}+\left(\frac{4 i a l}{\gamma^{2}}\right)\left(\frac{\hbar l t}{m}\right)-l^{2}=-\frac{1}{\gamma^{4}}\left(2 a \gamma^{2}+l^{2}\right) .
\left\langle p^{2}\right\rangle=-\hbar^{2} \int_{-\infty}^{\infty} \Psi^{*}\left[A x^{2}+B x+C\right] \Psi d x=-\hbar^{2}\left[A\left\langle x^{2}\right\rangle+B\langle x\rangle+C\right] .
=-\frac{\hbar^{2}}{\gamma^{4}}\left[4 a^{2}\left(\frac{1}{4 w^{2}}+\left(\frac{\hbar l t}{m}\right)^{2}\right)-4 i a l\left(\frac{\hbar l t}{m}\right)-\left(2 a \gamma^{2}+l^{2}\right)\right] .
=-\frac{\hbar^{2}}{\gamma^{4}}\left\{\left[a+a\left(\frac{2 \hbar a t}{m}\right)^{2}-2 a-\frac{4 i a^{2} \hbar t}{m}\right]+l^{2}\left[-1-\frac{4 i a \hbar t}{m}+4\left(\frac{\hbar a t}{m}\right)^{2}\right]\right\} .
=-\frac{\hbar^{2}}{\gamma^{4}}\left(-a \gamma^{4}-l^{2} \gamma^{4}\right)= \hbar^{2}\left(a+l^{2}\right) .
\sigma_{x}^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2}=\frac{1}{4 w^{2}}+\left(\frac{\hbar l t}{m}\right)^{2}-\left(\frac{\hbar l t}{m}\right)^{2}=\frac{1}{4 w^{2}} \Rightarrow \sigma_{x}=\frac{1}{2 w} ;
\sigma_{p}^{2}=\left\langle p^{2}\right\rangle-\langle p\rangle^{2}=\hbar^{2} a+\hbar^{2} l^{2}-\hbar^{2} l^{2}=\hbar^{2} a , so \sigma_{p}=\hbar \sqrt{a} .
(e) σ_x and σ_p are same as before, so the uncertainty principle still holds.