A quantum mechanical particle of mass $M$ moves in one dimension in the presence of a negative delta function potential

$$V=-\frac{\hbar^{2}}{2 M \Delta} \delta(x),$$

where $\Delta$ is a parameter with dimensions of length.

(a) Write down the time-independent Schrödinger equation for energy eigenstates $\chi(x)$, with energy $E$. By integrating this equation across $x=0$, show that the gradient of the wavefunction jumps across $x=0$ according to

$$\lim _{\epsilon \rightarrow 0}\left(\frac{d \chi}{d x}(\epsilon)-\frac{d \chi}{d x}(-\epsilon)\right)=-\frac{1}{\Delta} \chi(0)$$

[You may assume that $\chi$ is continuous across $x=0 .$ ]

(b) Show that there exists a negative energy solution and calculate its energy.

(c) Consider a double delta function potential

$$V(x)=-\frac{\hbar^{2}}{2 M \Delta}[\delta(x+a)+\delta(x-a)] .$$

For sufficiently small $\Delta$, this potential yields a negative energy solution of odd parity, i.e. $\chi(-x)=-\chi(x)$. Show that its energy is given by

$$E=-\frac{\hbar^{2}}{2 M} \lambda^{2}, \quad \text { where } \quad \tanh \lambda a=\frac{\lambda \Delta}{1-\lambda \Delta}$$

[You may again assume $\chi$ is continuous across $x=\pm a$.]