Poles for a particle scattered in a delta potential

Question

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. 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 know there shouldn't be any waves traveling to the left for $x>0$. My final goal is to check my states by checking the reflection and transmission coefficients and confirming they add up to 1.

Answer 1

So one detail I omitted from the question was that: $$ \psi_{sc}(k)=\frac{g+g I}{2\pi(k^2-p^2)} $$ Where: $$ I=\int^{\infty}_{-\infty}\psi(q)dq \space\space\space\space (1) $$ (I had used in arbitrary prescription in the original description of the problem, this is what I obtain before solving for $I$)$$\\$$ Using equation (1) we can solve for I, obtaining: $$ I=\frac{\frac{g}{2\pi}\int^{\infty}_{\infty}\frac{dq}{q^2-p^2}}{1-\frac{g}{2\pi}\int^{\infty}_{-\infty}\frac{dq}{q^2-p^2}} $$

Now, the integrals in $I$ can be solved by shifting the poles on the real axis. We determine the shifts by Fourier Transforming $\psi(k)$ to obtain $\psi(x)$, and move the poles to meet our boundary conditions: a plane wave moving to the right for $x>0$ and a plane wave moving to the left for $x<0$.

The Fourier Transform we obtain is proportional to: $$ \int^{\infty}_{-\infty}dk\space \frac{e^{ikx}}{(k+p)(k-p)} $$ If we close the contour up, this gives our wave function for $x>0$, since the arc of the contour must go to zero when the radius is taken to infinity. Closing down gives the wave function for $x<0$.

Through some reasoning, we obtain that the correct prescription is given by: $$ \lim_{\epsilon \to 0}\int^{\infty}_{-\infty}\frac{e^{ikx}}{(k+p+i\epsilon)(k-p-i\epsilon)} $$

Now, this is the prescription we have to use in $I$, and when we do the Fourier Transform. The first time I did the problem, I originally thought we had 3 possible prescriptions for the Fourier Transform and for $I$. However, we have to pick the one prescription that fits our boundary conditions and stick with it throughout the whole problem, since the prescription constrains our wave function to the conditions. Working the rest of the problem now yields the correct reflection and transmission coefficients.