Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The boundary condition for ##\delta## function

Tags:
  1. Oct 20, 2015 #1
    Beginning with the Schrodinger equation for N particles in one dimension interacting via a δ-function potential

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

    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}.##

    Integrate ##\int_{x_k-\varepsilon}^{x_k+\varepsilon}##, 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?
     
  2. jcsd
  3. Oct 20, 2015 #2
    I don't see clearly where the two terms you have written appear from. The integral you have to solve is:

    [tex] \int_{-\varepsilon}^{\varepsilon}\partial^{2}_{x_{i}}\psi(x_{i})=\partial_{x_{i}}^{+\varepsilon}\psi(x_{i})-\partial_{x_{i}}^{-\varepsilon}\psi(x_{i})[/tex]
    so I dont clearly see where the substraction of the other coordinate in the derivative comes from. What do you mean by saying that [tex] x_{k}[/tex] is considered asa derivative?
     
  4. Oct 20, 2015 #3
    I missed to add a [tex]x_{k}[/tex] summing the [tex]\varepsilon[/tex], but what this condition is saying is that there is a finite discontinuity in the derivative for the ith particle (and for every of them).
     
  5. Oct 20, 2015 #4
    We integrate the Schrodinger equation as ##\int_{x_k-\varepsilon}^{x_k+\varepsilon}dx_j##, here ##x_j## is a integrable variable, ##x_k## is a integrable number. The boundary contains one term ##\frac{\partial}{\partial x_k}##, it means that ##x_k## also is a integrable variable or else? How to deduce to this term?
     
  6. Oct 21, 2015 #5
    Ok, since your function [tex]\psi[/tex] depends on all variables (or coordinates), you boundary condition is evaluating the function at [tex]x_{k}[/tex] (since epsilong tends to zero, but you have to take the limits carefully, in fact this condition is talking about a discontinuity in the derivative, so the answer is different if you approach xk by the left or the right.) The boundary doesnt contain the derivative respect to xk, since you are integrating the second derivative respect to xi, so the answer (the integral) is the first derivative respect to xi evaluated at xk in two different limits. I hop this can help a bit.
     
  7. Oct 21, 2015 #6
    I think ##\frac{\partial}{\partial x_k}## is inexistent, no reason for a integrable numerical value to be a variable.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook