Question 5.24: (a) Using Equations 5.66 and 5.70, show that the wave functi...

(a) Equations 5.66 and 5.70 \Rightarrow \psi=A \sin k x+B \cos k x ; \quad A \sin k a=\left[e^{i q a}-\cos k a\right] B .

\psi(x)=A \sin (k x)+B \cos (k x), \quad(0<x<a)                  (5.66).

A \sin (k a)=\left[e^{i q a}-\cos (k a)\right] B                  (5.70).

So

\psi=A \sin k x+\frac{A \sin k a}{\left(e^{i q a}-\cos k a\right)} \cos k x=\frac{A}{\left(e^{i q a}-\cos k a\right)}\left[e^{i q a} \sin k x-\sin k x \cos k a+\cos k x \sin k a\right] .

=C\left\{\sin k x+e^{-i q a} \sin [k(a-x)]\right\}, \text { where } C \equiv \frac{A e^{i q a}}{e^{i q a}-\cos k a} .

(b) If z = ka = jπ, then sin ka = 0, Eq. 5.71 \Rightarrow \cos q a=\cos k a=(-1)^{j} \Rightarrow \sin q a=0, \text { so } e^{i q a}=\cos q a+i \sin q a=(-1)^{j} , and the constant C involves division by zero. In this case we must go back to Eq. 5.70, which is a tautology (0=0) yielding no constraint on A or B, Eq. 5.68 holds automatically, and Eq. 5.69 gives

\cos (q a)=\cos (k a)+\frac{m \alpha}{\hbar^{2} k} \sin (k a)                  (5.71).

B=e^{-i q a}[A \sin (k a)+B \cos (k a)]                (5.68).

k A-e^{-i q a} k[A \cos (k a)-B \sin (k a)]=\frac{2 m \alpha}{\hbar^{2}} B                (5.69).

A \sin (k a)=\left[e^{i q a}-\cos (k a)\right] B                  (5.70).

k A-(-1)^{j} k\left[A(-1)^{j}-0\right]=\frac{2 m \alpha}{\hbar^{2}} B \Rightarrow B=0 . \quad \text { So } \psi=A \sin k x .

Here ψ is zero at each delta spike, so the wave function never \feels” the potential at all.