0
$\begingroup$

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

$$(-\sum_{1}^{N}\frac{\partial^2}{\partial x_i^2}+2c\sum_{<i,j>}\delta(x_i-x_j))\psi=E\psi$$

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

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

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – pring Oct 19 '15 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.