Schrödinger equation of the double delta potential in momentum space

Question

This may be a silly question but it has gotten me stumped. I am currently trying to see if I can get anywhere by putting a simple symmetric double delta potential

\begin{align} V(x) = -\gamma (\delta(x+a) + \delta(x-a)) \end{align}

into the time independent Schrödinger equation and working in momentum space to find eigenvalues and eigenfunctions. The Schrödinger equation

\begin{align} \frac{\hbar}{2m}\Delta \Psi(x) + V(x) \Psi(x) = E \Psi(x) \end{align}

after some calculations becomes

\begin{align} \frac{\hbar p^2}{2m} \psi(p) - \frac{\gamma}{\pi \hbar} \int_{-\infty}^\infty \cos\left( \frac{a}{\hbar} (p'-p) \right) \psi(p') dp' = E \psi(p) \, . \end{align}

Now when I set the integral equal to $\bar{\psi}(p)$ I get

\begin{align} \psi(p) = -\frac{\gamma \bar{\psi}(p)}{\pi \hbar (p^2/2m - E)} \, . \end{align}

Putting this expression back into the integral is what puzzles me. The expression

\begin{align} \bar{\psi}(p) = \int_{-\infty}^\infty \cos\left( \frac{a}{\hbar} (p'-p) \right) \frac{-\gamma \bar{\psi}(p')}{\pi \hbar (p'^2/2m - E)} dp' \end{align}

seems to suggest that the function in the integral acts as a delta function $\delta(p' - p)$. But that can't be correct, since it has nonzero values for $p' \neq p$. So my question is if someone could give me any pointers on where I went wrong or if I made an incorrect assumption somewhere. I am aware that the answer usually gets calculated with an exponential Ansatz, but I was interested in knowing whether doing the calculations this way would also be possible.

Answer 1

One approach would be looking into the theory of integral equations (see Fredholm integral equation), in order to use an appropriate method for solving the equation in question. Indeed, plugging into the equations an expression that has a singularity is always problematic.

Another approach would be somewhat modifying problem and treating it as a scattering problem rather than as an eigenvalue problem (Which are not appropriately distinguished in introductory QM, but characterized by different boundary conditions and wave function normalization.) One could this start with a plane waves of a certain momentum in a free space, and consider adiabatic switching on of the potential $\propto e^{-\eta t}$. The denominators then would have regularizing imaginary unit $$\frac{1}{E-\frac{p^2}{2m}\pm i\eta}$$ and one could obtain interesting closed form solutions or a perturbative expansion.

Finally, a foolproof check is solving the problem in position representation and calculating Fourier transform of the solutions. It might be also a good idea to deal first with a single delta-function in momentum space.

Related:
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How to solve for the scattering solution of following Schrodinger equation?
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Answer 2

To answer your original question, no. Just because you rewrote the equation in this particular way that the factor in the integrand is a delta function. The reasoning would be true if $\bar\psi$ were arbitrary. However, in your case, $\bar\psi$ is defined to be a solution of this equation. You merely reformulated your eignevalue problem into a fixed point problem.

Btw, since you are interested in $\psi$, it is best to keep the penultimate equation. You can solve it perturbatively using a Born series which can be interpreted as Feynmann diagrams.

As a side not, the penultimate equation is essentially rederiving the Schwinger Lippmann equation. In the following, I’ll assume $E>0$ (diffusive states). The main difference is that you are not treating the singularity rigorously at $p=p_\pm:=\pm\sqrt{2mE}$. Technically, the inverse function: $$ \frac{1}{p^2/2m-E} $$ is undetermined as it is defined up to an additional terms: $$ A_+\delta(p-p_+)+A_-\delta(p-p_-) $$ with $A_\pm$ arbitrary complex numbers (exactly like forgetting the $+C$ when calculating antiderivatives). The ambiguity is lifted by prescribing an integration contour (equivalently shifting the poles) or using the Cauchy principal value for example. This correct treatment of the singularity is also necessary for your last equation (fixed point equation in $\bar\psi$).

Hope this helps.

Answer 3

Thanks to everyone who answered I eventually managed to figure out a solution. I will write it down here if anyone is interested: The trick was to split the cosine into exponentials using Eulers formula and then write the equation as

\begin{align} \frac{p^2}{2m}\psi(p) - \frac{\gamma}{2\pi\hbar}(\bar\psi_+ e^{-iap/\hbar} + \bar\psi_- e^{iap/\hbar})\psi(p) = E\psi(p) \end{align}

with

\begin{align} \bar\psi_\pm = \int_{-\infty}^\infty e^{\pm iap/\hbar} \psi(p) dp \, . \end{align}

Then we can write

\begin{align} \bar\psi_\pm = \frac{\gamma}{2\pi\hbar} (\bar\psi_\pm g(0) + \bar\psi\mp g(a)) \end{align}

with

\begin{align} g(a) = 2\int_{-\infty}^\infty \frac{\cos(2ap/\hbar)}{p^2/2m - E}dp \, . \end{align}

By requiring even or odd solutions for $\bar\psi_\pm$ ($\bar\psi_+ = \pm \bar\psi_-$) we can then find the eigenfunctions and eigenvalues.