# How do you solve the Schrödinger equation with a position space delta function potential in momentum space? [closed]

I am solving the Schrodinger equation in position space with an attractive delta function potential energy, $$-\frac{h^2}{2m} \frac{d^2}{dx^2} \psi(x)-\lambda \delta(x) \psi(x)=E \psi(x),$$ for a bound state. I need to solve for E, using fourier transform. I took the fourier transform of the whole equation and ended up with, $$-\frac{h^2}{2m} (ik)^2\phi(k)-\lambda \delta(k) \phi(k)=E\phi(k).$$ After using, $$\delta(k) =1 \text{ or } 2\pi$$ I don't get the required energy for this attractive potential, I would be glad if anyone could help? As I remember the energy expression should be $$E=-\frac{m\lambda^2}{2h^2}.$$

## closed as off-topic by stafusa, Jon Custer, John Rennie, sammy gerbil, Kyle KanosDec 26 '17 at 15:55

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Start from the position space SE with potential $-\lambda \delta(x)$. Define the Fourier transform as: $$\tilde \psi(k)=\frac{1}{\sqrt{2\pi\hbar}} \int dx \,\,e^{-ikx/\hbar}\psi(x)$$ The Fourier transform of the product is given by: $$\mathcal{F}[\delta(x)\psi(x)](k)=\frac{1}{\sqrt{2\pi\hbar}} \int dx \,\,e^{-ikx/\hbar}\delta(x)\psi(x)=\frac{\psi(0)}{\sqrt{2\pi\hbar}}$$ The momentum space SE therefore gives: $$\tilde \psi(k)= \frac{\lambda\psi(0)}{\sqrt{2\pi\hbar}}\frac{1}{(k^2/2m-E)}$$ Inverting the transform: $$\psi(x)= \frac{\lambda\psi(0)}{2\pi\hbar}\int dk \,\frac{e^{ikx/\hbar}}{k^2/2m-E}=\frac{m\lambda\psi(0)}{\hbar\sqrt{-2mE}}\exp{\left(-\frac{\sqrt{-2mE}}{\hbar}|x|\right)}$$ where the result of the integral in the second step is standard and can be obtained from tables of Fourier transforms. Setting $x=0$, we obtain $E$ in terms of $\lambda$: $$E=-\frac{m\lambda^2}{2\hbar^2}$$