(a) Note that ψ_0 is even, and ψ_1 is odd. In either case \left|\psi \right| ^2 is even, so \langle x\rangle=\int x|\psi|^{2} d x= 0 , Therefore \langle p\rangle=m d\langle x\rangle / d t= 0 . (These results hold for any stationary state of the harmonic oscillator.)
From Eqs. 2.60 and 2.63,
\psi_{0}(x)=\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} e^{-\frac{m \omega}{2\hbar} x^{2}} (2.60).
\psi_{1}(x)=A_{1} \hat{a}_{+} \psi_{0}=\frac{A_{1}}{\sqrt{2 \hbar m \omega}}\left(-\hbar \frac{d}{d x}+m \omega x\right)\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} e^{-\frac{m \omega}{2 \hbar} x^{2}} =A_{1}\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} \sqrt{\frac{2 m \omega}{\hbar}} x e^{-\frac{m \omega}{2 \hbar} x^{2}} (2.63).
\psi_{0}=\alpha e^{-\xi^{2} / 2}, \psi_{1}=\sqrt{2} \alpha \xi e^{-\xi^{2} / 2} . So
n = 0:
\left\langle x^{2}\right\rangle=\alpha^{2} \int_{-\infty}^{\infty} x^{2} e^{-\xi^{2}} d x=\alpha^{2}\left(\frac{\hbar}{m \omega}\right)^{3 / 2} \int_{-\infty}^{\infty} \xi^{2} e^{-\xi^{2}} d \xi=\frac{1}{\sqrt{\pi}}\left(\frac{\hbar}{m \omega}\right) \frac{\sqrt{\pi}}{2} = \frac{\hbar}{2 m \omega} .
\left\langle p^{2}\right\rangle=\int \psi_{0}\left(\frac{\hbar}{i} \frac{d}{d x}\right)^{2} \psi_{0} d x=-\hbar^{2} \alpha^{2} \sqrt{\frac{m \omega}{\hbar}} \int_{-\infty}^{\infty} e^{-\xi^{2} / 2}\left(\frac{d^{2}}{d \xi^{2}} e^{-\xi^{2} / 2}\right) d \xi .
=-\frac{m \hbar \omega}{\sqrt{\pi}} \int_{-\infty}^{\infty}\left(\xi^{2}-1\right) e^{-\xi^{2}} d \xi=-\frac{m \hbar \omega}{\sqrt{\pi}}\left(\frac{\sqrt{\pi}}{2}-\sqrt{\pi}\right) = \frac{m \hbar \omega}{2} .
n = 1:
\left\langle x^{2}\right\rangle=2 \alpha^{2} \int_{-\infty}^{\infty} x^{2} \xi^{2} e^{-\xi^{2}} d x=2 \alpha^{2}\left(\frac{\hbar}{m \omega}\right)^{3 / 2} \int_{-\infty}^{\infty} \xi^{4} e^{-\xi^{2}} d \xi=\frac{2 \hbar}{\sqrt{\pi} m \omega} \frac{3 \sqrt{\pi}}{4}= \frac{3 \hbar}{2 m \omega} .
\left\langle p^{2}\right\rangle=-\hbar^{2} 2 \alpha^{2} \sqrt{\frac{m \omega}{\hbar}} \int_{-\infty}^{\infty} \xi e^{-\xi^{2} / 2}\left[\frac{d^{2}}{d \xi^{2}}\left(\xi e^{-\xi^{2} / 2}\right)\right] d \xi .
=-\frac{2 m \omega \hbar}{\sqrt{\pi}} \int_{-\infty}^{\infty}\left(\xi^{4}-3 \xi^{2}\right) e^{-\xi^{2}} d \xi=-\frac{2 m \omega \hbar}{\sqrt{\pi}}\left(\frac{3}{4} \sqrt{\pi}-3 \frac{\sqrt{\pi}}{2}\right)= \frac{3 m \hbar \omega}{2} .
(b) n = 0:
\sigma_{x}=\sqrt{\left\langle x^{2}\right\rangle-\langle x\rangle^{2}}=\sqrt{\frac{\hbar}{2 m \omega}} ; \sigma_{p}=\sqrt{\left\langle p^{2}\right\rangle-\langle p\rangle^{2}}=\sqrt{\frac{m \hbar \omega}{2}} .
\sigma_{x} \sigma_{p}=\sqrt{\frac{\hbar}{2 m \omega}} \sqrt{\frac{m \omega \hbar}{2}}=\frac{\hbar}{2} (Right at the uncertainty limit.)
n = 1:
\sigma_{x}=\sqrt{\frac{3 \hbar}{2 m \omega}} ; \sigma_{p}=\sqrt{\frac{3 m \hbar \omega}{2}} ; \sigma_{x} \sigma_{p}=3 \frac{\hbar}{2}>\frac{\hbar}{2} .
(c)
\langle T\rangle=\frac{1}{2 m}\left\langle p^{2}\right\rangle = \left\{\begin{array}{l} \frac{1}{4} \hbar \omega(n=0) \\ \frac{3}{4} \hbar \omega(n=1) \end{array}\right\} ;
\langle V\rangle=\frac{1}{2} m \omega^{2}\left\langle x^{2}\right\rangle = \left\{\begin{array}{l} \frac{1}{4} \hbar \omega(n=0) \\ \frac{3}{4} \hbar \omega(n=1) \end{array}\right\} ;
\langle T\rangle+\langle V\rangle=\langle H\rangle= \left\{\begin{array}{l} \frac{1}{2} \hbar \omega(n=0)=E_{0} \\ \frac{3}{2} \hbar \omega(n=1)=E_{1} \end{array}\right\} , as expected.