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Suppose you have non-relativistic fermions scattering off a delta function potential.

It is an easy job to solve $H=-\partial_x^2+\epsilon \delta(x)$ by starting with an eigenfunction of the form $\psi(x)=(A e^{-ikx}+B e^{ikx})\theta(-x)+(C e^{-ikx}+D e^{ikx})\theta(x)$ and looking at the continuity of the function at $x=0$ and discontinuity of the slope of the wavefunction at $x=0$. Then one can compute the S-matrix.

It is also easy to solve Hamiltonian describing two fermions with attraction/repulsion at the point of contact i.e. $H=-\partial_{x_1}^2-\partial_{x_2}^2+g\delta(x_1-x_2)$.

For example to solve above you start with the eigenstate of the form \begin{aligned} &\psi(x_1,x_2)=\theta\left(x_{2}-x_{1}\right)\left\{A \mathrm{e}^{\mathrm{i}\left(k_{1} x_{1}+k_{2} x_{2}\right)}+B \mathrm{e}^{\mathrm{i}\left(k_{2} x_{1}+k_{1} x_{2}\right)}\right\} \\ &+\theta\left(x_{1}-x_{2}\right)\left\{C \mathrm{e}^{\mathrm{i}\left(k_{1} x_{2}+k_{2} x_{1}\right)}+D\mathrm{e}^{\mathrm{i}\left(k_{2} x_{2}+k_{1} x_{1}\right)}\right\} \end{aligned}

and then demand the continuity of the eigenfunction at $x_1=x^2$ and the condition that the above function is eigenstate with eigenvalue $k_1^2+k_2^2$, we can compute the S-matrix relative the incoming and outgoing amplitudes.

Now, consider a problem mixing both of the cases. You have fermions scattering off the delta function potentials and also attracting/repelling each other at the point of contact i.e. take the Hamiltonian of the form \begin{equation} H=-\partial_{x_1}^2-\partial_{x_2}^2+\epsilon(\delta(x_1)+\delta(x_2))+ g \delta(x_1-x_2). \end{equation}

Now, how would one solve this problem?

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1 Answer 1

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The general way of solving such problems is by defining the boundary conditions at the points singled out by delta functions: the wave function should be continuous at these points, while its derivative experiences a jump (the corresponding boundary condition is easily found by integrating the SE in the infinitesimal interval around delta function). One then solves the SE outside the special points (plane wave solutions) and connects the solutions using the boundary conditions. In case of two fermions one might have also to antisymmetrize the solution.

Details
Let's take SE $$ \left[-\partial_x^2+\epsilon\delta(x)\right]\psi(x)=E\psi(x) $$ The first boundary condition is $$ \psi(+\eta)=\psi(-\eta) =\psi(0), $$where $\eta$ is infinitesimal. I.e., the wave function is continuous at $x=0$. The second boundary condition is obtained by integrating the SE: $$ \int_{-\eta}^\eta dx\left[-\partial_x^2+\epsilon\delta(x)\right]\psi(x)=\int_{-\eta}^\eta dx E\psi(x)\\\Rightarrow -\partial_x\psi(x)|_{x=\eta}+\partial_x\psi(x)|_{x=-\eta}+\epsilon\psi(0)\approx2\eta\psi(0)\rightarrow 0\\ \Rightarrow \partial_x\psi(x)|_{x=\eta}-\partial_x\psi(x)|_{x=-\eta}=\epsilon\psi(0) $$ This also can be used to find the bound states in a delta-potential well. See, e.g., Intuition for the number of bound states to the double Dirac potential well

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