# The boundary condition for delta function

Beginning with the Schrödinger equation for $$N$$ particles in one dimension interacting via a $$\delta$$-function potential

$$\left(-\sum_{i=1}^{N}\frac{\partial^2}{\partial x_i^2}+2c\sum_{}\delta(x_i-x_j)\right)\psi=E\psi$$

Why the boundary condition equivalent to the $$\delta$$ function potential is

$$\left(\frac{\partial}{\partial x_j}-\frac{\partial}{\partial x_k}\right)\psi |_{x_j=x_{k+}}-\left(\frac{\partial}{\partial x_j}-\frac{\partial}{\partial x_k}\right)\psi |_{x_j=x_{k-}}=2c\psi |_{x_j=x_k}.$$

Try integrating the original differential equation over an interval $[x_k-\epsilon,x_k+\epsilon]$, then take the limit for $\epsilon\rightarrow0$.
The integral over the right-hand-side vanishes (if $\psi$ is continuous in $x_k$), the integral containing the delta function leads to the rhs of your resulting boundary condition and the integral over the second derivative leads to the first derivative terms approaching $x_k$ from above and from below (there is a discontinuity in the derivative of $\psi$ as a result of the delta function).
• Integrate $\int_{x_k-\varepsilon}^{x_k+\varepsilon}dx_j$, here, $x_k$ is a integrate limit. Why $x_k$ is considered as a derivative $\frac{\partial}{\partial x_k}$? It says that we can integrate the ordinate of j's particle with the boundary of k's particle? Oct 19, 2015 at 12:30