(a) \langle x\rangle=\langle\alpha \mid x \alpha\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left\langle\alpha \mid\left(a_{+}+a_{-}\right) \alpha\right\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left(\left\langle a_{-} \alpha \mid \alpha\right\rangle+\left\langle\alpha \mid a_{-} \alpha\right\rangle\right)= \sqrt{\frac{\hbar}{2 m \omega}}\left(\alpha+\alpha^{*}\right) .
x^{2}=\frac{\hbar}{2 m \omega}\left(a_{+}^{2}+a_{+} a_{-}+a_{-} a_{+}+a_{-}^{2}\right) . But a_{-} a_{+}=\left[a_{-}, a_{+}\right]+a_{+} a_{-}=1+a_{+} a_{-} (Eq. 2.55):
\hat{a}_{+} \hat{a}_{-}=\frac{1}{\hbar \omega} \hat{H}-\frac{1}{2} (2.55).
=\frac{\hbar}{2 m \omega}\left(a_{+}^{2}+2 a_{+} a_{-}+1+a_{-}^{2}\right) .
\left\langle x^{2}\right\rangle=\frac{\hbar}{2 m \omega}\left\langle\alpha \mid\left(a_{+}^{2}+2 a_{+} a_{-}+1+a_{-}^{2}\right) \alpha\right\rangle=\frac{\hbar}{2 m \omega}\left(\left\langle a_{-}^{2} \alpha \mid \alpha\right\rangle+2\left\langle a_{-} \alpha \mid a_{-} \alpha\right\rangle+\langle\alpha \mid \alpha\rangle+\left\langle\alpha \mid a_{-}^{2} \alpha\right\rangle\right) .
=\frac{\hbar}{2 m\omega}\left[\left(\alpha^{*}\right)^{2}+2\left(\alpha^{*}\right) \alpha+1+\alpha^{2}\right]= \frac{\hbar}{2 m \omega}\left[1+\left(\alpha+\alpha^{*}\right)^{2}\right] .
\langle p\rangle=\langle\alpha \mid p \alpha\rangle=i \sqrt{\frac{\hbar m \omega}{2}}\left\langle\alpha \mid\left(a_{+}-a_{-}\right) \alpha\right\rangle=i \sqrt{\frac{\hbar m \omega}{2}}\left(\left\langle a_{-} \alpha \mid \alpha\right\rangle-\left\langle\alpha \mid a_{-} \alpha\right\rangle\right)=-i \sqrt{\frac{\hbar m \omega}{2}}\left(\alpha-\alpha^{*}\right).
p^{2}=-\frac{\hbar m \omega}{2}\left(a_{+}^{2}-a_{+} a_{-}-a_{-} a_{+}+a_{-}^{2}\right)=-\frac{\hbar m \omega}{2}\left(a_{+}^{2}-2 a_{+} a_{-}-1+a_{-}^{2}\right) .
=-\frac{\hbar m \omega}{2}\left[\left(\alpha^{*}\right)^{2}-2\left(\alpha^{*}\right) \alpha-1+\alpha^{2}\right]=\frac{\hbar m \omega}{2}\left[1-\left(\alpha-\alpha^{*}\right)^{2}\right] .
(b)
\sigma_{x}^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2}=\frac{\hbar}{2 m \omega}\left[1+\left(\alpha+\alpha^{*}\right)^{2}-\left(\alpha+\alpha^{*}\right)^{2}\right]=\frac{\hbar}{2 m \omega} .
\sigma_{p}^{2}=\left\langle p^{2}\right\rangle-\langle p\rangle^{2}=\frac{\hbar m \omega}{2}\left[1-\left(\alpha-\alpha^{*}\right)^{2}+\left(\alpha-\alpha^{*}\right)^{2}\right]=\frac{\hbar m \omega}{2} . \quad \sigma_{x} \sigma_{p}=\sqrt{\frac{\hbar}{2 m \omega}} \sqrt{\frac{\hbar m \omega}{2}}=\frac{\hbar}{2} . QED
(c) Using Eq. 2.68 for \psi_{n} :
\psi_{n}=\frac{1}{\sqrt{n !}}\left(\hat{a}_{+}\right)^{n} \psi_{0} (2.68).
c_{n}=\left\langle\psi_{n} \mid \alpha\right\rangle=\frac{1}{\sqrt{n !}}\left\langle\left(a_{+}\right)^{n} \psi_{0} \mid \alpha\right\rangle=\frac{1}{\sqrt{n !}}\left\langle\psi_{0} \mid\left(a_{-}\right)^{n} \alpha\right\rangle=\frac{1}{\sqrt{n !}} \alpha^{n}\left\langle\psi_{0} \mid \alpha\right\rangle=\frac{\alpha^{n}}{\sqrt{n !}} c_{0} .
(d) 1=\sum_{n=0}^{\infty}\left|c_{n}\right|^{2}=\left|c_{0}\right|^{2} \sum_{n=0}^{\infty} \frac{|\alpha|^{2 n}}{n !}=\left|c_{0}\right|^{2} e^{|\alpha|^{2}} \quad \Rightarrow c_{0}=e^{-|\alpha|^{2} / 2} .
(e) |\alpha(t)\rangle=\sum_{n=0}^{\infty} c_{n} e^{-i E_{n} t / \hbar}|n\rangle=\sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}} e^{-|\alpha|^{2} / 2} e^{-i\left(n+\frac{1}{2}\right) \omega t}|n\rangle=e^{-i \omega t / 2} \sum_{n=0}^{\infty} \frac{\left(\alpha e^{-i \omega t}\right)^{n}}{\sqrt{n !}} e^{-|\alpha|^{2} / 2}|n\rangle .
Apart from the overall phase factor e^{-i \omega t / 2} (which doesn’t affect its status as an eigenfunction of a_- , or its eigenvalue), |\alpha(t)\rangle \text { is the same as }|\alpha\rangle \text {, but with eigenvalue } \alpha(t)=e^{-i \omega t} \alpha .
(f) From (a), \langle x\rangle=\sqrt{\hbar / 2 m \omega}\left[\alpha(t)+\alpha^{*}(t)\right] . \text { From (e), } \alpha(t)=e^{-i \omega t} \alpha . So
\langle x\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left(\alpha e^{-i \omega t}+\alpha^{*} e^{i \omega t}\right)=\sqrt{\frac{\hbar}{2 m \omega}}\left(C \sqrt{\frac{m \omega}{2 \hbar}} e^{i \phi} e^{-i \omega t}+C \sqrt{\frac{m \omega}{2 \hbar}} e^{-i \phi} e^{i \omega t}\right) .
=\frac{1}{2} C\left(e^{-i(\omega t-\phi}+e^{i(\omega t-\phi)}\right)=C \cos (\omega t-\phi).
From (b), \sigma_{x}=\sqrt{\frac{\hbar}{2 m \omega}} . So the expectation value of x oscillates at the classical frequency (but this is true for all states of the harmonic oscillator-see Problem 3.40) and the wave packet maintains a constant width.
(g) Equation 2.59 says a_{-}\left|\psi_{0}\right\rangle=0 , so yes, it is a coherent state, with eigenvalue α = 0.
\hat{a}_{-} \psi_{0}=0 (2.59).