# Quantum filter, potential barrier

I have been trying to teach myself quantum mechanics for quite a time now and I need help with a probably simple problem.

We are looking at the Schrödinger equation of particles of mass $m$ in one dimension with the potential $$V(x) = V_0\delta(x) + V_1\Theta(a-|x|),$$ where $V_0 < 0$ and $V_1 > 0$, $\delta(x)$ is the delta function and $\Theta(x)$ is the Heaviside function. Let a plain wave come in from the left ($x \to -\infty$) with energy $E$ ($0<E<V_1$).

Since this is new stuff to me, I'm wondering how to approach this problem. The goal is to determine the energy within the potential barrier and to derive the wave function $\psi(x)$ from this energy and from the boundary conditions.

My approach to this problem is the following general wave function: $$\psi(x) = \left\{ \begin{array}{ll} e^{ikx}+re^{-ikx}, & x <-a \\ Ce^{\kappa x}+De^{-\kappa x}, & -a < x <0 \\ Fe^{\kappa x}+Ge^{-\kappa x}, & 0<x<a\\te^{ikx},&x>a\end{array}\right. ,$$ with $C,D,F,G\in\mathbb{C}$, $r,t$ the reflection and transmission coefficients, $\kappa^2 = \frac{2m(V_1-E)}{\hbar^2}$ and $k^2 = \frac{2mE}{\hbar^2}$.

How do I determine the energy $E_0$ of the particle within the potential barrier? I think my problem is the delta potential and the fact that I have no idea whether my approach is correct. I hope somebody will show me how to deal with this kind of problems so that I can solve them myself in the future.

## 1 Answer

Be aware of continuity at $x = 0$. You have a wave function for negative and positive x (in the boundaries of a) and they must be the same for $x = 0$.

Knowing that, you need to look at the Schrödinger equation. Use $V(x)$ as it is given to you. Then you can use the definition of the delta distribution:

$$\int dx f(x)\delta(x-x_0) = f(x_0)$$

or

$$\int_a^b dx f(x)\delta(x-x_0) = \left\{\begin{array}{11}f(x_0), & a<x_0<b \\ 0, & else\end{array}\right.$$

Use $a=-\epsilon$ and $b=\epsilon$ and take $\lim_{\epsilon\to 0}$. You should end with an equation containing $\psi'(0^+)$ and $\psi'(0^-)$, which portray $\left. \frac{d\psi(x)}{dx} \right|_{x=0}$ depending on the direction you came from.

• Thanks a lot for the answer. When I integrate the Schrödinger equation, so $$-\frac{\hbar^2}{2m}\int_{-\epsilon}^\epsilon\frac{\partial^2\psi}{\partial x^2}dx+V_0\int_{-\epsilon}^\epsilon \delta(x)\psi(x) dx+V_1\int_{-\epsilon}^\epsilon \Theta(a-|x|)\psi(x) dx = E\int_{-\epsilon}^\epsilon\psi(x) dx,$$ I arrive at $$\psi'(0^-)-\psi'(0^+)=\frac{2m}{\hbar^2}V_0\psi(0),$$ which doesn't include $V_1$. Is this correct? Maybe I'm having trouble integrating the Heaviside function. – MeMeansMe May 22 '17 at 17:47
• Asume that $\epsilon \lt a$ So the Heaviside function adds the factor $1$: $$V_1 \int_{-\epsilon}^{\epsilon}... = V_1 \int_{-\epsilon}^{\epsilon} \psi(x)dx$$ so V_1 should remain. – Leeen gold May 22 '17 at 18:19
• But $$\lim_{\epsilon\to 0} \int_{-\epsilon}^\epsilon\psi(x) dx =0,$$ isn't it? – MeMeansMe May 22 '17 at 18:28
• No, you're right. The delta potential is relative to the potential $V_1$, so is the energy eigenvalue $E_0$ – Leeen gold May 22 '17 at 18:38
• So, it's correct that $V_1$ doesn't appear in the equation? – MeMeansMe May 22 '17 at 18:46