The boundary condition for delta function

Question

Beginning with the Schrödinger equation for $N$ particles in one dimension interacting via a $delta$-function potential
$$left(-sum_{i=1}^{N}frac{partial^2}{partial x_i^2}+2csum_{<i,j>}delta(x_i-x_j)right)psi=Epsi$$
Why the boundary condition equivalent to the $delta$ function potential is
$$left(frac{partial}{partial x_j}-frac{partial}{partial x_k}right)psi |_{x_j=x_{k+}}-left(frac{partial}{partial x_j}-frac{partial}{partial x_k}right)psi |_{x_j=x_{k-}}=2cpsi |_{x_j=x_k}.$$

Wouter · Accepted Answer

Try integrating the original differential equation over an interval $[x_k-epsilon,x_k+epsilon]$, then take the limit for $epsilonrightarrow0$.

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).

The boundary condition for delta function

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