# What is the solution to the Schrodinger equation for a dirac delta potential next to a hard wall?

Consider the Schrodinger equation for a particle confined in the following potential:

$$V(x) = - V_{0}\delta(x)$$ $$\space$$ $$x>-d$$

$$V(x) = \infty$$ $$\space$$ $$x< -d$$

When i solve the equation for the delta potential for the bound states I get that $$E =- \frac{mV_0^2}{2\hbar^2}$$.

I figured out that i need to fit the solutions to the boundary condition, i thus have that $$\psi(x)_+ = A e^{\sqrt{\frac{2m|E|}{\hbar^2}}x} + Be^{-\sqrt{\frac{2m|E|}{\hbar^2}}x}$$ and $$\psi(x)_- = C e^{\sqrt{\frac{2m|E|}{\hbar^2}}x} + De^{-\sqrt{\frac{2m|E|}{\hbar^2}}x}$$ for the postive and the negaive x-axis respectively. The solution has to be finite, so we need to put $$A = 0$$. Since we have an infinte wall, the wave function becomes zero at the boundry, thus we have :$$\psi(-d) = Ce^{-2kd} + De^{2kd}= 0$$.

The only way i can make this make sense is if $$C = e^{-2kd}$$ and $$D = -e^{2kd}$$. However, if i do this i get constants that are energy dependent. Is this allowed, or am i breaking some rule? , If i proceed, I have to use the continuity condition at the delta function. This give us: $$B = C+D$$

Then we also have the continuity condition for the derivative, $$\psi(0^+)_+^{'} - \psi(0^-)_-^{'} = \frac{2mV_0}{\hbar^2}$$.

If i do the math, i end up with $$k = -\frac{mV_0}{\hbar^2} + \frac{mV_0}{\hbar^2}e^{2dk}$$

This is a transcendental equation, which might arise sometimes in these type of problems, but it seems quite counter intuitive, since it suggest that the furher away the wall is the bigger the pertubation to the bound state is..which seems unphysical. Did i do my math wrong or am i missing something fundamental?

Boundary conditions at $$x = -d$$ and $$x=\infty$$ lead to the wave function $$\psi_-(x) = A\sinh(\kappa(x+d)), \quad \psi_+(x) = Be^{-\kappa x},\quad (1)$$ with $$\kappa = \sqrt{\frac{2m|E|}{\hbar^2}} > 0.$$ Conditions at $$x = 0$$ due to delta-function are $$\psi_-(0) = \psi_+(0),\quad -\frac{\hbar^2}{2m}(\psi_+\ \!\!'(0)-\psi_-\ \!\!'(0)) = V_0\psi_+(0).$$ These equations, together with (1), give $$B = A\sinh(\kappa d)$$ and $$\kappa = \frac{V_0m}{\hbar^2}(1-e^{-2\kappa d}).\quad (2)$$ The last equation gives the correct answer in the limit $$d\to \infty$$. Change $$\kappa = - k$$ leads to your equation. I believe you introduced in your correct calculations an unfortunate notation $$k = -\sqrt{\frac{2m|E|}{\hbar^2}} < 0.$$