The boundary condition for ##\delta## function

[Pring]

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?

 

[gonadas91]

I don't see clearly where the two terms you have written appear from. The integral you have to solve is:

\int_{-\varepsilon}^{\varepsilon}\partial^{2}_{x_{i}}\psi(x_{i})=\partial_{x_{i}}^{+\varepsilon}\psi(x_{i})-\partial_{x_{i}}^{-\varepsilon}\psi(x_{i})
so I don't clearly see where the substraction of the other coordinate in the derivative comes from. What do you mean by saying that x_{k} is considered asa derivative?

 

[gonadas91]

I missed to add a x_{k} summing the \varepsilon, 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).

 

[Pring]

I don't see clearly where the two terms you have written appear from. The integral you have to solve is:

\int_{-\varepsilon}^{\varepsilon}\partial^{2}_{x_{i}}\psi(x_{i})=\partial_{x_{i}}^{+\varepsilon}\psi(x_{i})-\partial_{x_{i}}^{-\varepsilon}\psi(x_{i})
so I don't clearly see where the substraction of the other coordinate in the derivative comes from. What do you mean by saying that x_{k} is considered asa derivative?

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?

 

[gonadas91]

Ok, since your function \psi depends on all variables (or coordinates), you boundary condition is evaluating the function at x_{k} (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 doesn't 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.

 

[Pring]

Ok, since your function \psi depends on all variables (or coordinates), you boundary condition is evaluating the function at x_{k} (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 doesn't 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.

I think $\frac{\partial}{\partial x_k}$ is inexistent, no reason for a integrable numerical value to be a variable.