In one dimension a particle of mass $m$ and momentum $\hbar k, k>0$, is scattered by a potential $V(x)$ where $V(x) \rightarrow 0$ as $|x| \rightarrow \infty$. Incoming and outgoing plane waves of positive $(+)$ and negative $(-)$ parity are given, respectively, by

$$\begin{array}{ll} I_{+}(k, x)=e^{-i k|x|}, & I_{-}(k, x)=\operatorname{sgn}(x) e^{-i k|x|} \\ O_{+}(k, x)=e^{i k|x|}, & O_{-}(k, x)=-\operatorname{sgn}(x) e^{i k|x|} \end{array}$$

The scattering solutions to the time-independent Schrödinger equation with positive and negative parity incoming waves are $\psi_{+}(x)$ and $\psi_{-}(x)$, respectively. State how the asymptotic behaviour of $\psi_{+}$and $\psi_{-}$can be expressed in terms of $I_{+}, I_{-}, O_{+}, O_{-}$and the S-matrix denoted by

$$\boldsymbol{S}=\left(\begin{array}{cc} S_{++} & S_{+-} \\ S_{-+} & S_{--} \end{array}\right)$$

In the case where $V(x)=V(-x)$ explain briefly why you expect $S_{+-}=S_{-+}=0$.

The potential $V(x)$ is given by

$$V(x)=V_{0}[\delta(x-a)+\delta(x+a)]$$

where $V_{0}$ is a constant. In this case, show that

$$S_{--}(k)=e^{-2 i k a}\left[\frac{\left(2 k-i U_{0}\right) e^{i k a}+i U_{0} e^{-i k a}}{\left(2 k+i U_{0}\right) e^{-i k a}-i U_{0} e^{i k a}}\right]$$

where $U_{0}=2 m V_{0} / \hbar^{2}$. Verify that $\left|S_{--}\right|^{2}=1$ and explain briefly the physical meaning of this result.

For $V_{0}<0$, by considering the poles or zeros of $S_{--}(k)$ show that there exists one bound state of negative parity in this potential if $U_{0} a<-1$.

For $V_{0}>0$ and $U_{0} a \gg 1$, show that $S_{--}(k)$ has a pole at

$$k a=\pi+\alpha-i \gamma$$

where, to leading order in $1 /\left(U_{0} a\right)$,

$$\alpha=-\frac{\pi}{U_{0} a}, \quad \gamma=\left(\frac{\pi}{U_{0} a}\right)^{2}$$

Explain briefy the physical meaning of this result, and why you expect that $\gamma>0$.