(a) Schrödinger equation:
-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}=E \psi, \quad \text { or } \quad \frac{d^{2} \psi}{d x^{2}}=-k^{2} \psi \quad(k \equiv \sqrt{2 m E} / \hbar) \quad\left\{\begin{array}{l} 0<x<\frac{1}{2} a+\epsilon \\ \frac{1}{2} a+\epsilon<x<a \end{array}\right. .
Boundary conditions: \psi(0)=\psi\left(\frac{1}{2} a+\epsilon\right)=\psi(a)=0 .
Solution:
\text { (1) } 0<x<\frac{1}{2} a+\epsilon: \quad \psi(x)=A \sin k x+B \cos k x . \quad \text { But } \quad \psi(0)=0 \Rightarrow B=0 , and
\psi\left(\frac{1}{2} a+\epsilon\right)=0 \Rightarrow\left\{\begin{array}{l}k\left(\frac{1}{2} a+\epsilon\right)=n \pi \quad(n=1,2,3, \ldots) \Rightarrow E_{n}=n^{2} \pi^{2} \hbar^{2} / 2 m(a / 2+\epsilon)^{2} \\ \text { or else } A=0 \end{array}\right.,
\text { (2) } \frac{1}{2} a+\epsilon<x<a: \quad \psi(x)=F \sin k(a-x)+G \cos k(a-x) . \quad \text { But } \quad \psi(a)=0 \Rightarrow G=0 , and
\psi\left(\frac{1}{2} a+\epsilon\right)=0 \Rightarrow\left\{\begin{array}{l}k\left(\frac{1}{2} a-\epsilon\right)=n^{\prime} \pi \quad\left(n^{\prime}=1,2,3, \ldots\right) \Rightarrow E_{n^{\prime}}=\left(n^{\prime}\right)^{2} \pi^{2} \hbar^{2} / 2 m(a / 2-\epsilon)^{2} \\ \text { or else } \quad F=0 \end{array}\right. .
The ground state energy is \begin{cases}\text { either } \quad E_{1}=\frac{\pi^{2} \hbar^{2}}{2 m\left(\frac{1}{2} a+\epsilon\right)^{2}} \quad(n=1), \quad \text { with } \quad F=0 \\\text { or else } \quad E_{1^{\prime}}=\frac{\pi^{2} \hbar^{2}}{2 m\left(\frac{1}{2} a-\epsilon\right)^{2}} \quad\left(n^{\prime}=1\right), \quad \text { with } \quad A=0\end{cases} .
Both are allowed energies, but E_1 is (slightly) lower (assuming ε is positive), so the ground state is
\psi(x)= \begin{cases}\sqrt{\frac{1}{2} a+\epsilon} \sin \left(\frac{\pi x}{\frac{1}{2} a+\epsilon}\right), & 0 \leq x \leq \frac{1}{2} a+\epsilon \\ 0, & \frac{1}{2} a+\epsilon \leq x \leq a\end{cases} .
(b)
-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+f(t) \delta\left(x-\frac{1}{2} a-\epsilon\right) \psi=E \psi \quad \Rightarrow \quad \psi(x)=\left\{\begin{array}{l}A \sin k x, & 0 \leq x<\frac{1}{2} a+\epsilon, \\ F \sin k(a-x), \frac{1}{2} a+\epsilon<x \leq a, \end{array}\right\} \quad \text { where } k \equiv \frac{\sqrt{2 m E}}{\hbar} .
Continuity in \psi \text { at } x=\frac{1}{2} a+\epsilon :
A \sin k\left(\frac{1}{2} a+\epsilon\right)=F \sin k\left(a-\frac{1}{2} a-\epsilon\right)=F \sin k\left(\frac{1}{2} a-\epsilon\right) \quad \Rightarrow F=A \frac{\sin k\left(\frac{1}{2} a+\epsilon\right)}{\sin k\left(\frac{1}{2} a-\epsilon\right)} .
Discontinuity in \psi^{\prime} \text { at } x=\frac{1}{2} a+\epsilon (Eq. 2.128):
\Delta\left(\frac{d \psi}{d x}\right)=-\frac{2 m \alpha}{\hbar^{2}} \psi(0) (2.128).
-F k \cos k(a-x)-A k \cos k x=\frac{2 m f}{\hbar^{2}} A \sin k x \Rightarrow F \cos k\left(\frac{1}{2} a-\epsilon\right)+A \cos k\left(\frac{1}{2} a+\epsilon\right)=-\left(\frac{2 m f}{\hbar^{2} k}\right) A \sin k\left(\frac{1}{2} a+\epsilon\right) .
A \frac{\sin k\left(\frac{1}{2} a+\epsilon\right)}{\sin k\left(\frac{1}{2} a-\epsilon\right)} \cos k\left(\frac{1}{2} a-\epsilon\right)+A \cos k\left(\frac{1}{2} a+\epsilon\right)=-\left(\frac{2 T}{z}\right) A \sin k\left(\frac{1}{2} a+\epsilon\right) .
\sin k\left(\frac{1}{2} a+\epsilon\right) \cos k\left(\frac{1}{2} a-\epsilon\right)+\cos k\left(\frac{1}{2} a+\epsilon\right) \sin k\left(\frac{1}{2} a-\epsilon\right)=-\left(\frac{2 T}{z}\right) \sin k\left(\frac{1}{2} a+\epsilon\right) \sin k\left(\frac{1}{2} a-\epsilon\right) .
\sin k\left(\frac{1}{2} a+\epsilon+\frac{1}{2} a-\epsilon\right)=-\left(\frac{2 T}{z}\right) \frac{1}{2}\left[\cos k\left(\frac{1}{2} a+\epsilon-\frac{1}{2} a+\epsilon\right)-\cos k\left(\frac{1}{2} a+\epsilon+\frac{1}{2} a-\epsilon\right)\right] .
\sin k a=-\frac{T}{z}(\cos 2 k \epsilon-\cos k a) \Rightarrow z \sin z=T[\cos z-\cos (z \delta)] .
(c)
\sin z=\frac{T}{z}(\cos z-1) \quad \Rightarrow \quad \frac{z}{T}=\frac{\cos z-1}{\sin z}=-\tan (z / 2) \Rightarrow \quad \tan (z / 2)=-\frac{z}{T} .
\text { Plot } \tan (z / 2) \text { and }-z / 7 on the same graph, and look for intersections:
\text { As } t: 0 \rightarrow \infty, T: 0 \rightarrow \infty , and the straight line rotates counterclockwise from 6 o’clock to 3 o’clock, so the smallest z goes from π to 2π, and the ground state energy goes from k a=\pi \Rightarrow E(0)=\frac{\hbar^{2} \pi^{2}}{2 m a^{2}} .
(appropriate to a well of width a) to k a=2 \pi \Rightarrow E(\infty)=\frac{\hbar^{2} \pi^{2}}{2 m(a / 2)^{2}} . (appropriate for a well of width a/2.
(d) Mathematica yields the following table:
| 1000 |
100 |
20 |
5 |
1 |
0 |
T |
| 6.21452 |
6.13523 |
5.72036 |
4.76031 |
3.67303 |
3.14159 |
z |
(e)
P_{r}=\frac{I_{r}}{I_{r}+I_{l}}=\frac{1}{1+\left(I_{l} / I_{r}\right)} , where
I_{l}=\int_{0}^{a / 2+\epsilon} A^{2} \sin ^{2} k x d x=\left.A^{2}\left[\frac{1}{2} x-\frac{1}{4 k} \sin (2 k x)\right]\right|_{0} ^{a / 2+\epsilon} .
=A^{2}\left\{\frac{1}{2}\left(\frac{a}{2}+\epsilon\right)-\frac{1}{4 k} \sin \left[2 k\left(\frac{a}{2}+\epsilon\right)\right]\right\}=\frac{a}{4} A^{2}\left[1+\frac{2 \epsilon}{a}-\frac{1}{k a} \sin \left(k a+\frac{2 \epsilon}{a} k a\right)\right] .
=\frac{a}{4} A^{2}\left[1+\delta-\frac{1}{z} \sin (z+z \delta)\right] .
I_{r}=\int_{a / 2+\epsilon}^{a} F^{2} \sin ^{2} k(a-x) d x . \quad \text { Let } \quad u \equiv a-x, d u=-d x .
=-F^{2} \int_{a / 2-\epsilon}^{0} \sin ^{2} k u d u=F^{2} \int_{0}^{a / 2-\epsilon} \sin ^{2} k u d u=\frac{a}{4} F^{2}\left[1-\delta-\frac{1}{z} \sin (z-z \delta)\right] .
\frac{I_{l}}{I_{r}}=\frac{A^{2}[1+\delta-(1 / z) \sin (z+z \delta)]}{F^{2}[1-\delta-(1 / z) \sin (z-z \delta)]} . \quad \text { But (from (b)) } \frac{A^{2}}{F^{2}}=\frac{\sin ^{2} k(a / 2-\epsilon)}{\sin ^{2} k(a / 2+\epsilon)}=\frac{\sin ^{2}[z(1-\delta) / 2]}{\sin ^{2}[z(1+\delta) / 2]} .
=\frac{I_{+}}{I_{-}} , where I_{\pm} \equiv\left[1 \pm \delta-\frac{1}{z} \sin z(1 \pm \delta)\right] \sin ^{2}[z(1 \mp \delta) / 2] , P_{r}=\frac{1}{1+\left(I_{+} / I_{-}\right)} .
Using δ = 0.01 and the z’s from (d), Mathematica gives
| 1000 |
100 |
20 |
5 |
1 |
0 |
T |
| 0.00248443 |
0.146529 |
0.401313 |
0.471116 |
0.486822 |
0.490001 |
P_r |
\text { As } t: 0 \rightarrow \infty(\text { so } T: 0 \rightarrow \infty) , the probability of being in the right half drops from almost 1/2 to zero-the particle gets sucked out of the slightly smaller side, as it heads for the ground state in (a).
(f)
