(a)
1=|A|^{2} \int_{-\infty}^{\infty} e^{-2 a x^{2}} d x=|A|^{2} \sqrt{\frac{\pi}{2 a}} ; A=\left(\frac{2 a}{\pi}\right)^{1 / 4} .
(b)
\int_{-\infty}^{\infty} e^{-\left(a x^{2}+b x\right)} d x=\int_{-\infty}^{\infty} e^{-y^{2}+\left(b^{2} / 4 a\right)} \frac{1}{\sqrt{a}} d y=\frac{1}{\sqrt{a}} e^{b^{2} / 4 a} \int_{-\infty}^{\infty} e^{-y^{2}} d y=\sqrt{\frac{\pi}{a}} e^{b^{2} / 4 a} .
\phi(k)=\frac{1}{\sqrt{2 \pi}} A \int_{-\infty}^{\infty} e^{-a x^{2}} e^{-i k x} d x=\frac{1}{\sqrt{2 \pi}}\left(\frac{2 a}{\pi}\right)^{1 / 4} \sqrt{\frac{\pi}{a}} e^{-k^{2} / 4 a}=\frac{1}{(2 \pi a)^{1 / 4}} e^{-k^{2} / 4 a} .
\Psi(x, t)=\frac{1}{\sqrt{2 \pi}} \frac{1}{(2 \pi a)^{1 / 4}} \int_{-\infty}^{\infty} \underbrace{e^{-k^{2} / 4 a} e^{i\left(k x-\hbar k^{2} t / 2 m\right)}}_{e^{-\left[\left(\frac{1}{4 a}+i h t / 2 m\right) k^{2}-i x k\right]}} d k .
=\frac{1}{\sqrt{2 \pi}(2 \pi a)^{1 / 4}} \frac{\sqrt{\pi}}{\sqrt{\frac{1}{4 a}+i \hbar t / 2 m}} e^{-x^{2} / 4\left(\frac{1}{4 a}+i \hbar t / 2 m\right)} = \left(\frac{2 a}{\pi}\right)^{1 / 4} \frac{e^{-a x^{2} /(1+2 i \hbar a t / m)}}{\sqrt{1+2 i \hbar a t / m}} .
(c)
Let \theta \equiv 2 \hbar a t / m . Then |\Psi|^{2}=\sqrt{\frac{2 a}{\pi}} \frac{e^{-a x^{2} /(1+i \theta)} e^{-a x^{2} /(1-i \theta)}}{\sqrt{(1+i \theta)(1-i \theta)}} .
The exponent is
-\frac{a x^{2}}{(1+i \theta)}-\frac{a x^{2}}{(1-i \theta)}=-a x^{2} \frac{(1-i \theta+1+i \theta)}{(1+i \theta)(1-i \theta)}=\frac{-2 a x^{2}}{1+\theta^{2}} ; |\Psi|^{2}=\sqrt{\frac{2 a}{\pi}} \frac{e^{-2 a x^{2} /\left(1+\theta^{2}\right)}}{\sqrt{1+\theta^{2}}} .
Or, with w \equiv \sqrt{\frac{a}{1+\theta^{2}}} , |\Psi|^{2}=\sqrt{\frac{2}{\pi}} w e^{-2 w^{2} x^{2}} . As t increases, the graph of |\Psi|^{2} attens out and broadens.
(d)
\langle x\rangle=\int_{-\infty}^{\infty} x|\Psi|^{2} d x= 0 (odd integrand); \langle p\rangle = \frac{d \langle x\rangle}{dt} = 0 .
\left\langle x^{2}\right\rangle=\sqrt{\frac{2}{\pi}} w \int_{-\infty}^{\infty} x^{2} e^{-2 w^{2} x^{2}} d x=\sqrt{\frac{2}{\pi}} w \frac{1}{4 w^{2}} \sqrt{\frac{\pi}{2 w^{2}}} = \frac{1}{4w^2} , \left\langle p^{2}\right\rangle=-\hbar^{2} \int_{-\infty}^{\infty} \Psi^{*} \frac{\partial^{2} \Psi}{\partial x^{2}} d x .
Write \Psi=B e^{-b x^{2}} , where B \equiv\left(\frac{2 a}{\pi}\right)^{1 / 4} \frac{1}{\sqrt{1+i \theta}} and b \equiv \frac{a}{1+i \theta} .
\frac{\partial^{2} \Psi}{\partial x^{2}}=B \frac{\partial}{\partial x}\left(-2 b x e^{-b x^{2}}\right)=-2 b B\left(1-2 b x^{2}\right) e^{-b x^{2}} .
\Psi^{*} \frac{\partial^{2} \Psi}{d \partial x^{2}}=-2 b|B|^{2}\left(1-2 b x^{2}\right) e^{-\left(b+b^{*}\right) x^{2}} ; b+b^{*}=\frac{a}{1+i \theta}+\frac{a}{1-i \theta}=\frac{2 a}{1+\theta^{2}}=2 w^{2} .
|B|^{2}=\sqrt{\frac{2 a}{\pi}} \frac{1}{\sqrt{1+\theta^{2}}}=\sqrt{\frac{2}{\pi}} w . So \Psi^{*} \frac{\partial^{2} \Psi}{\partial x^{2}}=-2 b \sqrt{\frac{2}{\pi}} w\left(1-2 b x^{2}\right) e^{-2 w^{2} x^{2}} .
\left\langle p^{2}\right\rangle=2 b \hbar^{2} \sqrt{\frac{2}{\pi}} w \int_{-\infty}^{\infty}\left(1-2 b x^{2}\right) e^{-2 w^{2} x^{2}} d x .
=2 b \hbar^{2} \sqrt{\frac{2}{\pi}} w\left(\sqrt{\frac{\pi}{2 w^{2}}}-2 b \frac{1}{4 w^{2}} \sqrt{\frac{\pi}{2 w^{2}}}\right)=2 b \hbar^{2}\left(1-\frac{b}{2 w^{2}}\right) .
But 1-\frac{b}{2 w^{2}}=1-\left(\frac{a}{1+i \theta}\right)\left(\frac{1+\theta^{2}}{2 a}\right)=1-\frac{(1-i \theta)}{2}=\frac{1+i \theta}{2}=\frac{a}{2 b} , so
\left\langle p^{2}\right\rangle=2 b \hbar^{2} \frac{a}{2 b}=\hbar^{2}a , \sigma_{x}=\frac{1}{2 w} ; \sigma_{p}=\hbar \sqrt{a} .
(e)
\sigma_{x} \sigma_{p}=\frac{1}{2 w} \hbar \sqrt{a}=\frac{\hbar}{2} \sqrt{1+\theta^{2}}=\frac{\hbar}{2} \sqrt{1+(2 \hbar a t / m)^{2}} \geq \frac{\hbar}{2} .
Closest at t = 0 at which time it is right at the uncertainty limit.
