Question 3.41: A harmonic oscillator is in a state such that a measurement ...

Evidently \Psi(x, t)=c_{0} \psi_{0}(x) e^{-i E_{0} t / \hbar}+c_{1} \psi_{1}(x) e^{-i E_{1} t / \hbar}, \text { with }\left|c_{0}\right|^{2}=\left|c_{1}\right|^{2}=1 / 2, \text { so } c_{0}=e^{i \theta_{0}} / \sqrt{2}, c_{1}=e^{i \theta_{1}} / \sqrt{2} , for some real \theta_{0}, \theta_{1}.

\langle p\rangle=\left|c_{0}\right|^{2}\left\langle\psi_{0} \mid p \psi_{0}\right\rangle+\left|c_{1}\right|^{2}\left\langle\psi_{1} \mid p \psi_{1}\right\rangle+c_{0}^{*} c_{1} e^{i\left(E_{0}-E_{1}\right) t / \hbar}\left\langle\psi_{0} \mid p \psi_{1}\right\rangle+c_{1}^{*} c_{0} e^{i\left(E_{1}-E_{0}\right) t / \hbar}\left\langle\psi_{1} \mid p \psi_{0}\right\rangle .

But E_{1}-E_{0}=\left(\frac{3}{2} \hbar \omega\right)-\left(\frac{1}{2} \hbar \omega\right)=\hbar \omega , and (Problem 2.11) \left\langle\psi_{0} \mid p \psi_{0}\right\rangle=\left\langle\psi_{1} \mid p \psi_{1}\right\rangle=0 while (Eqs. 2.70 and 2.67).

x=\sqrt{\frac{\hbar}{2 m \omega}}\left(\hat{a}_{+}+\hat{a}_{-}\right) ; \quad \hat{p}=i \sqrt{\frac{\hbar m \omega}{2}}\left(\hat{a}_{+}-\hat{a}_{-}\right)       (2.70).

\hat{a}_{+} \psi_{n}=\sqrt{n+1} \psi_{n+1}, \quad \hat{a}_{-} \psi_{n}=\sqrt{n} \psi_{n-1}       (2.67).

\left\langle\psi_{0} \mid p \psi_{1}\right\rangle=i \sqrt{\frac{\hbar m \omega}{2}}\left\langle\psi_{0} \mid\left(a_{+}-a_{-}\right) \psi_{1}\right\rangle=i \sqrt{\frac{\hbar m \omega}{2}}\left[\left\langle\psi_{0} \mid \sqrt{2} \psi_{2}\right\rangle-\left\langle\psi_{0} \mid \sqrt{1} \psi_{0}\right\rangle\right]=-i \sqrt{\frac{\hbar m \omega}{2}} ;\left\langle\psi_{1} \mid p \psi_{0}\right\rangle=i \sqrt{\frac{\hbar m \omega}{2}} .

\langle p\rangle=\frac{1}{\sqrt{2}} e^{-i \theta_{0}} \frac{1}{\sqrt{2}} e^{i \theta_{1}} e^{-i \omega t}\left(-i \sqrt{\frac{\hbar m \omega}{2}}\right)+\frac{1}{\sqrt{2}} e^{-i \theta_{1}} \frac{1}{\sqrt{2}} e^{i \theta_{0}} e^{i \omega t}\left(i \sqrt{\frac{\hbar m \omega}{2}}\right) .

=\frac{i}{2} \sqrt{\frac{\hbar m \omega}{2}}\left[-e^{-i\left(\omega t-\theta_{1}+\theta_{0}\right)}+e^{i\left(\omega t-\theta_{1}+\theta_{0}\right)}\right]=-\sqrt{\frac{\hbar m \omega}{2}} \sin \left(\omega t+\theta_{0}-\theta_{1}\right) .

The maximum is \sqrt{\hbar m \omega / 2} it occurs at t = 0 \Leftrightarrow \sin \left(\theta_{0}-\theta_{1}\right)=-1, \text { or } \theta_{1}=\theta_{0}+\pi / 2 . We might as well pick \theta_{0}=0, \theta_{1}=\pi / 2 ; then

\Psi(x, t)=\frac{1}{\sqrt{2}}\left[\psi_{0} e^{-i \omega t / 2}+\psi_{1} e^{i \pi / 2} e^{-3 i \omega t / 2}\right]= \frac{1}{\sqrt{2}} e^{-i \omega t / 2}\left(\psi_{0}+i \psi_{1} e^{-i \omega t}\right) .