Poles Arising in a Scattered Particle in a Delta Potential

[dreinh]

I am working on problem a professor gave me to get an idea for the research he does, and have hit a point where I'm having a difficult time seeing where I need to go from where I'm at. I would also like to go ahead and apologize for not knowing how to format correctly.

I was given that a particle is scattered with the given Hamiltonian:
$$
H = P^2 - g\delta(x)
$$
Where $\delta(x)$ is the [Dirac delta function](http://en.wikipedia.org/wiki/Dirac_delta_function). I was able to find the states in momentum representation, and was supposed to Fourier transform them to get the position representation states. Doing this leads to an integral with simple poles on the real axis, and can be solved by moving the poles above or below the real axis by some constant, applying the residue theorem, and taking the constant to zero. This leads to three different solutions, depending on if I move one pole up and one down (two possibilities), or both poles into a contour.

While I was talking to my professor, he mentioned that the solutions only work for certain values of $x$, and that the range of $x$ is given by what makes the arc in the contour go to zero. I'm actually having trouble seeing this, since that introduces an ambiguity in the solutions. If I shift the left pole up and the right pole down and close the contour in the upper plane, this means that x has to be positive, in order to get a decaying exponential in the integral. However there is nothing stopping me from moving the poles in the opposite way, and closing above to obtain a different solution for $x>0$.

Is there something wrong in the math, or is the way I move the poles governed by the physical situation I'm interested in? (Which would be no plane waves moving the left for $x>0$ d.)

I should mention that I am assuming:
$$
|\psi> = |p> + |\psi_{sc}>
$$

Where p is the incoming momentum and $|\psi_{sc}>$ is the scattered portion of the wave function.


Working in momentum representation, I obtain:
$$
\psi_{sc}(k) = \frac{g - \frac{g^2 i}{2p+ g i}}{2\pi(k^2-p^2)}
$$

Where k is the momentum variable, and p is the fixed momentum of the incoming particle. The transform I obtain is:
$$
\psi_{sc}(x) = \eta \int_{-\infty}^{\infty} \frac{e^{i k x}}{k^2-p^2} dk
$$

This is where I run into the problem with the poles. I have decided that the left pole should move up in order to have a plane wave moving to the right for x>0 (closing the contour upwards) and a plane wave moving to the left for x<0 (closing the contour downwards). I also know that if I move both poles down and close down, I obtain a superposition of waves moving left and right. I'm not sure how these pieces fit together for my solution and could use some ideas.

 

[azntoon]

The way I see it, the choice of how to move the poles is dictated by the physical situation. The idea is that the particle is initially moving in with momentum $p$, and then is scattered by the delta function potential. This means that for $x>0$, the wave should only have a component going to the right, because the particle was initially moving to the right. For $x<0$, the wave should only have a component going to the left, because the particle was initially moving to the right. So if I close the contour above, I should get a wave that only has a component going to the right for $x>0$ and only a component going to the left for $x<0$. If I close the contour below, I should get a superposition of both components for all $x$. In addition, the range of x is also determined by the pole movement. If I move one pole up and one down, then the exponential decay in the integral should be such that the arc at infinity goes to zero. This means that x should have a maximum value, beyond which the solution does not hold. This maximum value is determined by the pole movement. I hope this helps clarify the situation. Ultimately, the choice of how to move the poles should be dictated by the physical situation you are interested in. In this case, it seems that the physics dictates that the poles be moved in such a way that the resulting wave function has a single component going to the right for $x>0$ and a single component going to the left for $x<0$. The range of x is then determined by the exponential decay of the integral, which in turn is determined by the pole movement.