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?


1 Answer 1


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 $\epsilon \rightarrow 0$: $\\$

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


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

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 $\frac{2\,m\,V_0}{\hbar^2}$


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