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In my Quantum Mechanics homework, I had to solve the following differential equation

$$ \left(\frac{d^2}{dx^2} + k\right) \psi = \lambda (\delta(x-a) + \delta(x+a)) $$

Which comes from the potential $V = -\lambda(\delta(x-a) + \delta(x+a))$, where $\lambda>0$. I attempted to solve the problem using Green's functions, so the operator $\mathcal{L} = \left(\frac{d^2}{dx^2} + k\right)$ and the boundary conditions I chose to be $\psi(x) = 0$ when $x \to \pm\infty$. I got the following Green's function for this -

$$ G(x, x^\prime) = -\frac{1}{2k} \exp(-ik |x-x^\prime|) $$

So with that as the Green's function my solution is

$$ \psi(x) = \frac{-\lambda}{2k}\int_{-\infty}^\infty \exp(-ik |x-x^\prime|) (\delta(x^\prime-a) + \delta(x^\prime+a)) \ dx^\prime = \frac{-\lambda}{2k} \left(\exp(-ik|x-a|) + \exp(-ik|x+a|)\right) $$

However I'm supposed to get two linearly independent wavefunctions, which I'm unsure of how to get. Could somebody help me understand how to get the two linearly independent solutions to this ODE using Green's functions?

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    $\begingroup$ This looks like a one-dimensional scattering problem (because rearranging the Schrodinger equation results in a positive-delta-function potential). Is it? Because in that case I'm not sure why (1) you are setting $\psi(x)=0$ and (2) why you expect only two solutions, when in scattering theory you will have a continuum of scattering states. If you are actually solving for bound state ($\lambda<0$), then there should still be some undetermined coefficients, because you should be integrating $G\psi V$ and not just $GV$. $\endgroup$
    – march
    Mar 29 at 20:28
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    $\begingroup$ In addition, if you are looking for bound states, then you should have real exponential, not complex ones. And if you are looking for scattering states, you are missing a source term (the incoming piece of the wave function). In either case, I think you meant for $k$ to be $k^2$ in the equation and the operator: your solutions imply that. If not, then in the solutions there should be $\sqrt{k}$'s. $\endgroup$
    – march
    Mar 29 at 20:30
  • $\begingroup$ I've updated the question to include my original potential as well and updated the $k$ to a $k^2$. It's a double delta function well, so the professor stated that I'm supposed to get two solutions if a certain condition is met (which he didn't give). And I also thought that a second order ODE would have two linearly independent solutions. I'm then supposed to combine the two solutions into a symmetric and an anti-symmetric solutions $\endgroup$
    – Cocoa
    Mar 30 at 10:54
  • $\begingroup$ Once the $k^2$ is corrected, Vladimirov: Equations of Mathematical Physics names two different fundamental solutions of the operator - one being the complex conjugate of the other. Maybe that's what your prof is after? :) $\endgroup$
    – kricheli
    Mar 30 at 13:00

1 Answer 1

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Note: It's likely I'm off by some factors of 2 in places (or other similar mistakes). That's something the OP can check on their own.

There are a couple of issues. In the comments, OP mentions that their instructor says that they are solving the double-well potential. In that case, the Schrodinger equation reads $$ -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} - g(\delta(x-a)+\delta(x+a))\psi(x) = E\psi\,, $$ which can be rearranged to get $$ \left(\frac{\partial^2}{\partial x^2} +k^2\right)\psi = -\lambda(\delta(x-a)+\delta(x+a))\psi(x) \,, $$ where $k^2 = 2mE/\hbar^2$ and $\lambda = {2m}g/{\hbar^2}$. Right away, we see a sign issue: there should be a negative sign on the right-hand side if this is meant to be a double-well potential, in which case we should get two bound states.

Second, without going through the details, the Green's function for the Helmholtz equation, satisfying $$ \left(\frac{\partial^2}{\partial x^2} +k^2\right)G_0(x) = -\delta(x)\,, $$ is given by $$ G_0(x) = i\frac{e^{ik|x|}}{2k}\,. $$ Then, we multiply both sides of the rearranged Schrodinger equation by $G_0$ and integrate. For the left-hand side, we get \begin{align} \int dx\,G_0(x-x')\left(\frac{\partial^2}{\partial x^2} +k^2\right)\psi =-\psi(x')\,, \end{align} after what amounts essentially to an integration by parts and applying the boundary conditions $\psi(x'\to\pm\infty)=0$ and $G_0(x'\to\pm\infty)=0$. On the right-hand side, we get \begin{align*} \int dx\,G_0(x-x')(-\lambda(\delta(x-a)+\delta(x+a))\psi(x)) &= -\lambda\int dx\,(\delta(x-a)+\delta(x+a))i\frac{e^{-ik|x-x'|}}{2k}\psi(x) \\ &= -\lambda\left(i\frac{e^{ik|x'-a|}}{2k}\psi(a)+i\frac{e^{ik|x'+a|}}{2k}\psi(-a)\right)\,. \end{align*} So, we have the following relationship: $$ \psi(x) = \lambda\left(i\frac{e^{ik|x-a|}}{2k}\psi(a)+i\frac{e^{ik|x+a|}}{2k}\psi(-a)\right)\,. $$

Finally, how do we extract two linearly independent solution bound states from this? First of all, note that for the state to be bound, we need to have $E<0$, which implies that $k = \sqrt{2mE/\hbar^2} = i\kappa$, where $\kappa >0$. This guarantees that the resulting real exponentials decay at infinity. The second thing has to do with what happens at the positions of the delta functions. Evaluating the equation at $x=\pm a$, we get \begin{align} \psi(a) &= \lambda\left(\frac{e^{-\kappa|a-a|}}{2\kappa}\psi(a)+\frac{e^{-\kappa|a+a|}}{2\kappa}\psi(-a)\right) =\frac{\lambda}{2\kappa}\left(\psi(a)+{e^{-2\kappa a}}\psi(-a)\right)\,, \\ \psi(-a) &= \lambda\left(\frac{e^{-\kappa|-a-a|}}{2\kappa}\psi(a)+\frac{e^{-\kappa|-a+a|}}{2\kappa}\psi(-a)\right) =\frac{\lambda}{2\kappa}\left({e^{-2\kappa}}\psi(a)+\psi(-a)\right)\,. \end{align} Solving the first for $\psi(-a)$, plugging it into the second, and rearranging yields $$ \psi(a) \left((\lambda -2 \kappa)e^{2 a \kappa }-\lambda \right) \left((\lambda -2 \kappa )e^{2 a \kappa }+\lambda \right)=0\,, $$ which is an implicit equation for $\kappa$. The first factor must be non-zero, and so we are left with the one of the second two factors being non-zero. One corresponds to the even bound state and the other corresponds the odd bound state.

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