(a) Check the answer given: x_{c}=\omega \int_{0}^{t} f\left(t^{\prime}\right) \sin \left[\omega\left(t-t^{\prime}\right)\right] d t^{\prime} \Longrightarrow x_{c}(0)=0 .
\dot{x}_{c}=\omega f(t) \sin [\omega(t-t)]+\omega^{2} \int_{0}^{t} f\left(t^{\prime}\right) \cos \left[\omega\left(t-t^{\prime}\right)\right] d t^{\prime}=\omega^{2} \int_{0}^{t} f\left(t^{\prime}\right) \cos \left[\omega\left(t-t^{\prime}\right)\right] d t^{\prime} \Rightarrow \dot{x}_{c}(0)=0 .
\ddot{x}_{c}=\omega^{2} f(t) \cos [\omega(t-t)]-\omega^{3} \int_{0}^{t} f\left(t^{\prime}\right) \sin \left[\omega\left(t-t^{\prime}\right)\right] d t^{\prime}=\omega^{2} f(t)-\omega^{2} x_{c} .
Now the classical equation of motion is m\left(d^{2} x / d t^{2}\right)=-m \omega^{2} x+m \omega^{2} f . For the proposed solution, m\left(d^{2} x_{c} / d t^{2}\right)=m \omega^{2} f-m \omega^{2} x_{c} , so it does satisfy the equation of motion, with the appropriate boundary conditions.
(b) Let z \equiv x-x_{c} \quad\left(\text { so } \psi_{n}\left(x-x_{c}\right)=\psi_{n}(z)\right. , and z depends on t as well as x).
\frac{\partial \Psi}{\partial t}=\frac{d \psi_{n}}{d z}\left(-\dot{x}_{c}\right) e^{i\{\}}+\psi_{n} e^{i\{\}} \frac{i}{\hbar}\left[-\left(n+\frac{1}{2}\right) \hbar \omega+m \ddot{x}_{c}\left(x-\frac{x_{c}}{2}\right)-\frac{m}{2} \dot{x}_{c}^{2}+\frac{m \omega^{2}}{2} f x_{c}\right] .
[]=-\left(n+\frac{1}{2}\right) \hbar \omega+\frac{m \omega^{2}}{2}\left[2 x\left(f-x_{c}\right)+x_{c}^{2}-\frac{\dot{x}_{c}^{2}}{\omega^{2}}\right] .
\frac{\partial \Psi}{\partial t}=-\dot{x}_{c} \frac{d \psi_{n}}{d z} e^{i\{\}}+i \Psi\left\{-\left(n+\frac{1}{2}\right) \omega+\frac{m \omega^{2}}{2 \hbar}\left[2 x\left(f-x_{c}\right)+x_{c}^{2}-\frac{\dot{x}_{c}^{2}}{\omega^{2}}\right]\right\} .
\frac{\partial \Psi}{\partial x}=\frac{d \psi_{n}}{d z} e^{i\{\}}+\psi_{n} e^{i\{\}} \frac{i}{\hbar}\left(m \dot{x}_{c}\right) ; \quad \frac{\partial^{2} \Psi}{\partial x^{2}}=\frac{d^{2} \psi_{n}}{d z^{2}} e^{i\{\}}+2 \frac{d \psi_{n}}{d z} e^{i\{\}} \frac{i}{\hbar}\left(m \dot{x}_{c}\right)-\left(\frac{m \dot{x}_{c}}{\hbar}\right)^{2} \psi_{n} e^{i\{\}} .
H \Psi=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi}{\partial x^{2}}+\frac{1}{2} m \omega^{2} x^{2} \Psi-m \omega^{2} f x \Psi .
=-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi_{n}}{d z^{2}} e^{i\{\}}-\frac{\hbar^{2}}{2 m} 2 \frac{d \psi_{n}}{d z} e^{i\{\}} \frac{i m \dot{x}_{c}}{\hbar}+\frac{\hbar^{2}}{2 m}\left(\frac{m \dot{x}_{c}}{\hbar}\right)^{2} \Psi+\frac{1}{2} m \omega^{2} x^{2} \Psi-m \omega^{2} f x \Psi .
But -\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi_{n}}{d z^{2}}+\frac{1}{2} m \omega^{2} z^{2} \psi_{n}=\left(n+\frac{1}{2}\right) \hbar \omega \psi_{n} , so
H \Psi=\cancel {\left(n+\frac{1}{2}\right) \hbar \omega \Psi}-\frac{1}{2} m \omega^{2} z^{2} \Psi-\cancel {i \hbar \dot{x}_{c} \frac{d \psi n}{d z} e^{i\{\}}}+\frac{m}{2} \dot{x}_{c}^{2} \Psi+\frac{1}{2} m \omega^{2} x^{2} \Psi-m \omega^{2} f x \Psi .
\stackrel{?}{=} i \hbar \frac{\partial \Psi}{\partial t}=\cancel {-i \hbar \dot{x}_{c} \frac{d ψ_{n}}{d z} e^{i\{\}}}-\hbar \Psi\left[-\cancel {\left(n+\frac{1}{2}\right) \omega}+\frac{m \omega^{2}}{2 \hbar}\left(2 x f-2 x x_{c}+x_{c}^{2}-\frac{1}{\omega^{2}} \dot{x}_{c}^{2}\right)\right] .
-\frac{1}{2} m \omega^{2} z^{2}+\cancel {\frac{m}{ 2 } {x_{c}}^{2}}+\frac{1}{2} m \omega^{2} x^{2}-\cancel {m ω^2 f x} \stackrel{?}{=}-\frac{m \omega^{2}}{2}\left(\cancel {2 x f}-2 x x_{c}+x_{c}^{2}-\cancel {\frac{1}{ω^{2}} \dot{x}_{c}^{2}}\right) .
z^{2}-x^{2} \stackrel{?}{=}-2 x x_{c}+x_{c}^{2} ; \quad z^{2} \stackrel{?}{=}\left(x^{2}-2 x x_{c}+x_{c}^{2}\right)=\left(x-x_{c}\right)^{2} .
(c)
Eq. 11.140 ⇒ H=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+\frac{1}{2} m \omega^{2}\left(x^{2}-2 x f+f^{2}\right)-\frac{1}{2} m \omega^{2} f^{2} . \quad \text { Shift origin: } \quad u \equiv x-f .
H=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial u^{2}}+\frac{1}{2} m \omega^{2} u^{2}\right]-\left[\frac{1}{2} m \omega^{2} f^{2}\right] .
The first term is a simple harmonic oscillator in the variable u; the second is a constant (with respect to position). So the eigenfunctions are \psi_{n}(u) , and the eigenvalues are harmonic oscillator ones, ( n + \left.\frac{1}{2}\right) \hbar \omega , E_{n}=\left(n+\frac{1}{2}\right) \hbar \omega-\frac{1}{2} m \omega^{2} f^{2} .
(d) Note that \sin \left[\omega\left(t-t^{\prime}\right)\right]=\frac{1}{\omega} \frac{d}{d t^{\prime}} \cos \left[\omega\left(t-t^{\prime}\right)\right], \quad \text { so } \quad x_{c}(t)=\int_{0}^{t} f\left(t^{\prime}\right) \frac{d}{d t^{\prime}} \cos \left[\omega\left(t-t^{\prime}\right)\right] d t^{\prime}, , or
x_{c}(t)=\left.f\left(t^{\prime}\right) \cos \left[\omega\left(t-t^{\prime}\right)\right]\right|_{0} ^{t}-\int_{0}^{t}\left(\frac{d f}{d t^{\prime}}\right) \cos \left[\omega\left(t-t^{\prime}\right)\right] d t^{\prime}=f(t)-\int_{0}^{t}\left(\frac{d f}{d t^{\prime}}\right) \cos \left[\omega\left(t-t^{\prime}\right)\right] d t^{\prime} .
(since f(0) = 0). Now, for an adiabatic process we want df/dt very small; specifically: \frac{d f}{d t^{\prime}} \ll \omega f(t) .
\left(0<t^{\prime} \leq t\right) . Then the integral is negligible compared to f(t), and we have x_{c}(t) \approx f(t) . (Physically, this says that if you pull on the spring very gently, no fancy oscillations will occur; the mass just moves along as though attached to a string of fixed length.)
\text { (e) Put } x_{c} \approx f \text { into Eq. 11.142, using Eq. 11.143: }
\Psi(x, t)=\psi_{n}(x, t) e^{\frac{1}{\hbar}\left[-\left(n+\frac{1}{2}\right) \hbar \omega t+m \dot{f}(x-f / 2)+\frac{m \omega^{2}}{2} \int_{0}^{t} f^{2}\left(t^{\prime}\right) d t^{\prime}\right]} .
The dynamic phase (Eq. 11.92) is
e^{i \theta_{n}(t)}, \quad \text { where } \theta_{n}(t) \equiv-\frac{1}{\hbar} \int_{0}^{t} E_{n}\left(t^{\prime}\right) d t^{\prime} (11.92).
\theta_{n}(t)=-\frac{1}{\hbar} \int_{0}^{t} E_{n}\left(t^{\prime}\right) d t^{\prime}=-\left(n+\frac{1}{2}\right) \omega t+\frac{m \omega^{2}}{2 \hbar} \int_{0}^{t} f^{2}\left(t^{\prime}\right) d t^{\prime}, \quad \text { so } \quad \Psi(x, t)=\psi_{n}(x, t) e^{i \theta_{n}(t)} e^{i \gamma_{n}(t)} .
confirming Eq. 11.144, with the geometric phase given (ostensibly) by \gamma_{n}(t)=\frac{m}{\hbar} \dot{f}(x-f / 2) . But the eigenfunctions here are real, and this means the geometric phase should be zero. The point is that (in the adiabatic approximation) \dot{f} is extremely small (see above), and hence in this limit \frac{m}{\hbar} \dot{f}(x-f / 2) \approx 0 (at least, in the only region of x where \psi_{n}(x, t) is nonzero).