# Delta function: Intuitive way for boundary conditions

Giving the Schrödinger equation

$$-\dfrac{\hbar^2}{2\,m}\,{\partial_x}^2\psi(x)+ V(x)\,\psi(x) = E\,\psi(x)$$

with potential $$V(x) = V_0\,\delta(x)$$.

Solving this equation using an ordinary Ansatz yields

$$\psi(x) = \begin{cases} A_1\,e^{\textstyle -i\,k\,x}+B_1\,e^{\textstyle +i\,k\,x} & x<0 \\\\B_2\,e^{\textstyle i\,k\,x} & x>0\end{cases}$$

Now my only question lays on: how to derive boundary conditions intuitively?

It is already assured the wave function has to be continuous at $$x = 0 \quad \Rightarrow \quad\psi(0^-) = \psi(0^+)$$

This is clear to me. But why for the derivative it has to hold: $$\psi'(0^-) -\psi'(0^+) = \dfrac{2\,m\,V_0}{\hbar^2}\,\psi(0)$$ ?

Under normal condition it just follows $$\psi'(0^-) = \psi'(0^+)$$, so what's the matter in this case?

Most textbooks explain it with integrating the Schrödinger equation with respect to $$x$$ and taking the limit as the delta function becomes narrower but it's quite a long calculation evaluating all this stuff. So I just wonder: is there an intuition?

• Related: Is continuity of the wavefunction "put in by hand" for the Dirac delta potentials? and links therein. May 22 at 10:52
• I do not see how one could look for intuition when discussing the boundary conditions of Schrödinger's equation. Already the continuity at the boundary between two regions where the potential has a jump has nothing to do with intuition. May 22 at 16:38

Let's integrate the Schrödinger equation from $$x=-\epsilon$$ to $$x=+\epsilon$$. We get:

$$\\$$

$$-\dfrac{\hbar^2}{2\,m}\,(\psi'(\epsilon)-\psi'(-\epsilon))+ V_0\,\psi(0) = E\,(\psi(\epsilon)-\psi(-\epsilon))$$

and after taking the limit $$\epsilon \rightarrow 0$$: $$\\$$

$$-\dfrac{\hbar^2}{2\,m}\,(\psi'_+-\psi'_-)+ V_0\,\psi(0) = 0$$

or:

$$\psi'(0^-) -\psi'(0^+) = \dfrac{2\,m\,V_0}{\hbar^2}\,\psi(0)$$

We can add some intuition: The Schrödinger equation tells us that the second derivative of the wavefunction behaves like a delta-function. This means that the first derivative must behave like the integral of a delta-function- which is a step-function. This means that there is a jump in the first derivative. A proper normalization of the constants gives the factor of $$\frac{2\,m\,V_0}{\hbar^2}$$