# Solution of Schrödinger equation for Dirac delta potential $V(x) = \sum_{i=1}^P \sigma_i\delta(x-x_i)$

So, I am trying to solve Schrödinger Equation for Dirac delta potential.

The Schrödinger equation: $$-\frac{\hbar^2}{2m}\frac{d^2\Psi(x)}{dx^2} + V(x)\Psi(x) = E\Psi(x)$$ And, the potential looks like: $$V(x) = \sum_{i=1}^P \sigma_i\delta(x-x_i)$$

• If the number of delta functions goes to infinity, this is a slightly generalized version of the Dirac Comb - that might help in your search. – J. Murray Feb 26 '18 at 16:12
• This is essentially a boundary matching problem – By Symmetry Feb 26 '18 at 16:34
• The double delta potential is treated in plenty of online resources and textbooks, and the tools to expand from two to $n$ deltas are identical to those used in an $n$-piece piecewise-constant potential. So: what exactly do you want to know? What is it about the procedure that is causing you trouble? – Emilio Pisanty Feb 26 '18 at 16:41
• – Cosmas Zachos Feb 26 '18 at 16:44
• thank you for your responses. I am taking Quantum Mechanics for the first time and in book of Stephen Gasiorowicz, I couldn't find anything similar to this or even in wikipedia. My problem is that I don't know how to manipulate this V(x) potential to solve the differential equation. – Zulfidin Khodzhaev Feb 26 '18 at 17:09

This is the Kronig-Penney model. Solve the Schrödinger equation between two Dirac peaks (easy: the particle is free) then write the continuity of the wavefunction. Integrate the Schrödinger equation around a Dirac peak (i.e. over $[x_i-\epsilon;x_i+\epsilon]$ to show that $\psi'$ is not continuous and that the discontinuity is proportional to $\sigma_i$. With the continuity of $\psi$, you now have enough equations to determine all coefficients.