Our strategy is to solve the TISE for the regions $x < 0$ and $x > 0$ and apply the boundary conditions at $x=0$ where the delta function resides. Since $E$ is negative, we define the positive constant $k = \sqrt{-2m\frac{E}{\hbar^2}}$ and the general solution to the TISE will then be

\[\psi(x) = \begin{cases} \psi_-(x) = Ae^{kx} + Be^{-kx}, & x < 0, \\ \psi_+(x) = C e^{kx} + D e^{-kx}, & x>0. \end{cases}\]

Next, we look for divergences in the solution. The term $Be^{-kx}$ diverges as $x\rightarrow -\infty$ and the term $C e^{kx}$ diverges as $x\rightarrow +\infty$, so we can throw both of them out, leaving us with

\[\psi(x) = \begin{cases} \psi_-(x) = Ae^{kx} , & x < 0, \\ \psi_+(x) = D e^{-kx}, & x>0. \end{cases}\]

Next, we apply the boundary conditions. Continuity of $\psi(x)$ at $x=0$ gives us $\psi_-(0) = \psi_+(0)$, which easily gives us $A = D$, so we now have

\[\psi(x) = \begin{cases} \psi_-(x) = Ae^{kx} , & x < 0, \\ \psi_+(x) = A e^{-kx}, & x>0. \end{cases}\]

We can now find $A$ by normalization $\int_{-\infty}^{+\infty} | \psi(x)|^2 = 1$. It is left as an in class activity to show that $A = \sqrt{k}$, so we have

\[\psi(x) = \begin{cases} \psi_-(x) = \sqrt{k}e^{kx} , & x < 0, \\ \psi_+(x) = \sqrt{k} e^{-kx}, & x>0. \end{cases}\]

We still have to deal with the boundary condition on the derivative, which is a little more tricky. We cannot apply continuity of the derivative because the potential is infinite at $x=0$. However, recall that in section 4.iii we derived that, if the wavefunction is continuous at $x = x'$ then

\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x'-\epsilon}\right ]=\frac{2m}{\hbar^2} \lim_{\epsilon \rightarrow 0_+} \left [ \int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x)\,\mathrm{d}x \right ]\]

Applying this at $x=0$ gives

\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{-\epsilon}\right ] = \left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} =-\frac{2m}{\hbar^2}\lim_{\epsilon\rightarrow 0_+} \left [\int_{-\epsilon}^{+\epsilon} \alpha \delta(x)\psi(x)\,\mathrm{d}x \right ] = -\frac{2m\alpha}{\hbar^2}\psi(0) = -\frac{2m\alpha}{\hbar^2}\sqrt{k}.\]

Evidently, we must have

\[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} = -2k^{\frac{3}{2}} = -\frac{2m\alpha}{\hbar^2}\sqrt{k},\]

which yields

\[k = \frac{m\alpha}{\hbar^2}.\]

Using $k = \sqrt{-2m\frac{E}{\hbar^2}}$ we can convert this back into an energy, which yields

\[E = -\frac{m\alpha^2}{2\hbar^2}.\]

Since $E$ is determined uniquely, there is only one bound state that has this exact energy.

Unlike in the classical case, a particle with this energy is not confined to $x=0$, since its wavefunction extends into the regions $x < 0$ and $x>0$, although it does decay exponentially in these regions.

For the scattering case we define the positive constant $k = \sqrt{\frac{2mE}{\hbar^2}}$ and then the general solution to the TISE is

\[\psi(x) = \begin{cases} \psi_-(x) = Ae^{ikx} + Be^{-ikx}, & x < 0, \\ \psi_+(x) = C e^{ikx} + D e^{-ikx}, & x>0. \end{cases}\]

These solutions are oscillatory and unnormalizable, but non-divergent so we cannot throw any of them out on that basis. However, if we consider a particle scattering from the left then there will be no incident wave on the right, so we can remove the term $De^{-ikx}$ leaving

\[\psi(x) = \begin{cases} \psi_-(x) = Ae^{ikx} + Be^{-ikx}, & x < 0, \\ \psi_+(x) = C e^{ikx}, & x>0. \end{cases}\]

Continuity of $\psi(x)$ at $x=0$ gives $\psi_+(0) = \psi_-(0)$, or $A + B = C$, so we have

\[\psi(x) = \begin{cases} \psi_-(x) = Ae^{ikx} + Be^{-ikx}, & x < 0, \\ \psi_+(x) = (A+B) e^{ikx}, & x>0. \end{cases}\]

We now have to deal with the boundary condition on the derivative. We must have

\[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} = -\frac{2m\alpha}{\hbar^2}\psi(0) = -\frac{2m\alpha}{\hbar^2}(A+B)\]

It is left as an in class activity to show that this implies that

\[B = \frac{-\beta}{\beta + ik}A, \qquad C = (A+B) = \frac{ik}{\beta + ik},\]

where $\beta = \frac{m\alpha}{\hbar^2}$.

With this, we can calculate the transmission and reflection coefficients:

\[T = \frac{|C|^2}{|A|^2} = \frac{k^2}{\beta^2 + k^2}, \qquad R = \frac{\beta^2}{\beta^2 + k^2}.\]

Substituting $k = \sqrt{2m\frac{E}{\hbar^2}}$ and $\beta = \frac{m\alpha}{\hbar^2}$ gives

\[T = \frac{2E\hbar^2}{m\alpha^2 + 2E\hbar^2}, \qquad R = \frac{m\alpha^2}{m\alpha + 2E\hbar^2}.\]

Classically, there would be no reflection from the delta function potential, so this is an interference effect. However, for large $E$, when $2E\hbar^2$ is large compared to $m\alpha^2$, we have $T \approx 1$ and $R \approx 0$, as expected.

Since we have the equations to hand, it is no more difficult to study the delta function barrier

\[V(x) = +\alpha \delta(x),\] where $\alpha$ is a positive constant.

In this case, there are no bound states, and the only difference this will make to the scattering states is that, when we evaluate the boundary condition on the derivative we will get

\[\lim_{\epsilon \rightarrow 0_+} \left [ \int_{-\epsilon}^{+\epsilon}V(x)\psi(x)\,\mathrm{d}x \right ] = +\frac{2m\alpha}{\hbar^2}(A+B),\]

instead of $-\frac{2m\alpha}{\hbar^2}(A+B)$, which has the effect of changing $B$ and $C$ to