(a)
\text { Let } \quad\left(m v x^{2}-2 E_{n}^{i} a t\right) / 2 \hbar w=\phi(x, t) . \quad \Phi_{n}=\sqrt{\frac{2}{w}} \sin \left(\frac{n \pi}{w} x\right) e^{i \phi} , so
\frac{\partial \Phi_{n}}{\partial t}=\sqrt{2}\left(-\frac{1}{2} \frac{1}{w^{3 / 2}} v\right) \sin \left(\frac{n \pi}{w} x\right) e^{i \phi}+\sqrt{\frac{2}{w}}\left[-\frac{n \pi x}{w^{2}} v \cos \left(\frac{n \pi}{w} x\right)\right] e^{i \phi}+\sqrt{\frac{2}{w}} \sin \left(\frac{n \pi}{w} x\right)\left(i \frac{\partial \phi}{\partial t}\right) e^{i \phi} .
=\left[-\frac{v}{2 w}-\frac{n \pi x v}{w^{2}} \cot \left(\frac{n \pi}{w} x\right)+i \frac{\partial \phi}{\partial t}\right] \Phi_{n} . \quad \frac{\partial \phi}{\partial t}=\frac{1}{2 \hbar}\left[-\frac{2 E_{n}^{i} a}{w}-\frac{v}{w^{2}}\left(m v x^{2}-2 E_{n}^{i} a t\right)\right]=-\frac{E_{n}^{i} a}{\hbar w}-\frac{v}{w} \phi .
i \hbar \frac{\partial \Phi_{n}}{\partial t}=-i \hbar\left[\frac{v}{2 w}+\frac{n \pi x v}{w^{2}} \cot \left(\frac{n \pi}{w} x\right)+i \frac{E_{n}^{i} a}{\hbar w}+i \frac{v}{w} \phi\right] \Phi_{n} .
H \Phi_{n}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Phi_{n}}{\partial x^{2}} . \quad \frac{\partial \Phi_{n}}{\partial x}=\sqrt{\frac{2}{w}}\left[\frac{n \pi}{w} \cos \left(\frac{n \pi}{w} x\right)\right] e^{i \phi}+\sqrt{\frac{2}{w}} \sin \left(\frac{n \pi}{w} x\right) e^{i \phi}\left(i \frac{\partial \phi}{\partial x}\right) .
\frac{\partial \phi}{\partial x}=\frac{m v x}{\hbar w} . \quad \frac{\partial \Phi_{n}}{\partial x}=\left[\frac{n \pi}{w} \cot \left(\frac{n \pi}{w} x\right)+i \frac{m v x}{\hbar w}\right] \Phi_{n} .
\frac{\partial^{2} \Phi_{n}}{\partial x^{2}}=\left[-\left(\frac{n \pi}{w}\right)^{2} \csc ^{2}\left(\frac{n \pi}{w} x\right)+\frac{i m v}{\hbar w}\right] \Phi_{n}+\left[\frac{n \pi}{w} \cot \left(\frac{n \pi}{w} x\right)+i \frac{m v x}{\hbar w}\right]^{2} \Phi_{n} .
So the Schrödinger equation \left(i \hbar \partial \Phi_{n} / \partial t=H \Phi_{n}\right) is satisfied ⇔
-i \hbar\left[\frac{v}{2 w}+\frac{n \pi x v}{w^{2}} \cot \left(\frac{n \pi}{w} x\right)+i \frac{E_{n}^{i} a}{\hbar w}+i \frac{v}{w} \phi\right] .
=-\frac{\hbar^{2}}{2 m}\left\{-\left(\frac{n \pi}{w}\right)^{2} \csc ^{2}\left(\frac{n \pi}{w} x\right)+\frac{i m v}{\hbar w}+\left[\frac{n \pi}{w} \cot \left(\frac{n \pi}{w} x\right)+i \frac{m v x}{\hbar w}\right]^{2}\right\}.
Cotangent terms: -i \hbar\left(\frac{n \pi x v}{w^{2}}\right) \stackrel{?}{=}-\frac{\hbar^{2}}{2 m}\left(2 \frac{n \pi}{w} i \frac{m v x}{\hbar w}\right)=-i \hbar \frac{n \pi v x}{w^{2}} .
Remaining trig terms on right:
-\left(\frac{n \pi}{w}\right)^{2} \csc ^{2}\left(\frac{n \pi}{w} x\right)+\left(\frac{n \pi}{w}\right)^{2} \cot ^{2}\left(\frac{n \pi}{w} x\right)=-\left(\frac{n \pi}{w}\right)^{2}\left[\frac{1-\cos ^{2}(n \pi x / w)}{\sin ^{2}(n \pi x / w)}\right]=-\left(\frac{n \pi}{w}\right)^{2} .
This leaves:
i\left[\frac{v}{2 w}+i \frac{E_{n}^{i} a}{\hbar w}+i \frac{v}{w}\left(\frac{m v x^{2}-2 E_{n}^{i} a t}{2 \hbar w}\right)\right] \stackrel{?}{=} \frac{\hbar}{2 m}\left[-\left(\frac{n \pi}{w}\right)^{2}+\frac{i m v}{\hbar w}-\frac{m^{2} v^{2} x^{2}}{\hbar^{2} w^{2}}\right] .
\cancel {\frac{i v}{ 2}}-\frac{E_{n}^{i} a}{\hbar}-\cancel {\frac{m v^{2} x^{2}}{2 \hbar w}}+\frac{v E_{n}^{i} a t}{\hbar w} \stackrel{?}{=}-\frac{\hbar n^{2} \pi^{2}}{2 m w}+\cancel {\frac{i v}{ 2}}-\cancel {\frac{m v^{2} x^{2}}{2 \hbar w}}.
-\frac{E_{n}^{i} a}{\hbar w}(w-v t)=-\frac{E_{n}^{i} a^{2}}{\hbar w} \stackrel{?}{=}-\frac{\hbar n^{2} \pi^{2}}{2 m w} \Leftrightarrow-\frac{n^{2} \pi^{2} \hbar^{2}}{2 m a^{2}} \frac{a^{2}}{\hbar w}=-\frac{\hbar n^{2} \pi^{2}}{2 m w}=\text { r.h.s. } .
So \Phi_{n} does satisfy the Schrödinger equation, and since \Phi_{n}(x, t)=(\cdots) \sin (n \pi x / w) , it fits the boundary conditions: \Phi_{n}(0, t)=\Phi_{n}(w, t)=0 .
(b)
Equation 11.137 \Longrightarrow \Psi(x, 0)=\sum c_{n} \Phi_{n}(x, 0)=\sum c_{n} \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi}{a} x\right) e^{i m v x^{2} / 2 \hbar a} .
Multiply by \sqrt{\frac{2}{a}} \sin \left(\frac{n^{\prime} \pi}{a} x\right) e^{-i m v x^{2} / 2 \hbar a} and integrate:
\sqrt{\frac{2}{a}} \int_{0}^{a} \Psi(x, 0) \sin \left(\frac{n^{\prime} \pi}{a} x\right) e^{-i m v x^{2} / 2 \hbar a} d x=\sum c_{n}[\underbrace{\frac{2}{a} \int_{0}^{a} \sin \left(\frac{n \pi}{a} x\right) \sin \left(\frac{n^{\prime} \pi}{a} x\right) d x}_{\delta_{n n^{\prime}}}]=c_{n^{\prime}} .
So, in general: c_{n}=\sqrt{\frac{2}{a}} \int_{0}^{a} e^{-i m v x^{2} / 2 \hbar a} \sin \left(\frac{n \pi}{a} x\right) \Psi(x, 0) d x . In this particular case,
c_{n}=\frac{2}{a} \int_{0}^{a} e^{-i m v x^{2} / 2 \hbar a} \sin \left(\frac{n \pi x}{a}\right) \sin \left(\frac{\pi}{a} x\right) d x . \quad \text { Let } \quad \frac{\pi}{a} x \equiv z ; \quad d x=\frac{a}{\pi} d z ; \quad \frac{m v x^{2}}{2 \hbar a}=\frac{m v z^{2}}{2 \hbar a} \frac{a^{2}}{\pi^{2}}=\frac{m v a}{2 \pi^{2} \hbar} z^{2} .
c_{n}=\frac{2}{\pi} \int_{0}^{\pi} e^{-i \alpha z^{2}} \sin (n z) \sin (z) d z<br />
. QED
(c)
w\left(T_{e}\right)=2 a \Rightarrow a+v T_{e}=2 a \Rightarrow v T_{e}=a \Rightarrow T_{e}=a / v ; \quad e^{-i E_{1} t / \hbar} \Rightarrow \omega=\frac{E_{1}}{\hbar} \Rightarrow T_{i}=\frac{2 \pi}{\omega}=2 \pi \frac{\hbar}{E_{1}}, \text { or } ,
T_{i}=\frac{2 \pi \hbar}{\pi^{2} \hbar^{2}} 2 m a^{2}=\frac{4}{\pi} \frac{m a^{2}}{\hbar} . T_{i}=\frac{4 m a^{2}}{\pi \hbar} . Adiabatic \Rightarrow T_{e} \gg T_{i} \Rightarrow \frac{a}{v} \gg \frac{4 m a^{2}}{\pi \hbar} \Rightarrow \frac{4}{\pi} \frac{m a v}{\hbar} \ll 1, \text { or } ,
8 \pi\left(\frac{m a v}{2 \pi^{2} \hbar}\right)=8 \pi \alpha \ll 1, \quad \text { so } \quad \alpha \ll 1 . \text { Then } c_{n}=\frac{2}{\pi} \int_{0}^{\pi} \sin (n z) \sin (z) d z=\delta_{n 1} . Therefore
\Psi(x, t)=\sqrt{\frac{2}{w}} \sin \left(\frac{\pi x}{w}\right) e^{i\left(m v x^{2}-2 E_{1}^{i} a t\right) / 2 \hbar w} .
which (apart from a phase factor) is the ground state of the instantaneous well, of width w, as required by the adiabatic theorem. (Actually, the first term in the exponent, which is at most \frac{m v(2 a)^{2}}{2 \hbar(2 a)}=\frac{m v a}{\hbar} \ll 1 and could be dropped, in the adiabatic regime.)
(d)
\theta(t)=-\frac{1}{\hbar}\left(\frac{\pi^{2} \hbar^{2}}{2 m}\right) \int_{0}^{t} \frac{1}{\left(a+v t^{\prime}\right)^{2}} d t^{\prime}=-\left.\frac{\pi^{2} \hbar}{2 m}\left[-\frac{1}{v}\left(\frac{1}{a+v t^{\prime}}\right)\right]\right|_{0} ^{t}
=-\frac{\pi^{2} \hbar}{2 m v}\left(\frac{1}{a}-\frac{1}{w}\right)=-\frac{\pi^{2} \hbar}{2 m v}\left(\frac{v t}{a w}\right)=-\frac{\pi^{2} \hbar t}{2 m a w} .
So (dropping the \frac{m v x^{2}}{2 \hbar w} term, as explained in (c)) \Psi(x, t)=\sqrt{\frac{2}{w}} \sin \left(\frac{\pi x}{w}\right) e^{-i E_{1}^{i} a t / \hbar w} can be written (since -\frac{E_{1}^{i} a t}{\hbar w}=-\frac{\pi^{2} \hbar^{2}}{2 m a^{2}} \frac{a t}{\hbar w}=-\frac{\pi^{2} \hbar t}{2 m a w}=\theta ) : \Psi(x, t)=\sqrt{\frac{2}{w}} \sin \left(\frac{\pi x}{w}\right) e^{i \theta} .
This is exactly what one would naively expect: For a fixed well (of width a) we’d have \Psi(x, t)= \Psi_{1}(x) e^{-i E_{1} t / \hbar} ; for the (adiabatically) expanding well, simply replace a by the (time-dependent) width w, and integrate to get the accumulated phase factor, noting that E_1 is now a function of t.