In Gasiorowicz Quantum Mechanics, 3rd ed, pg.81, he finds the bound states for a delta function potential the following way: I have the following delta potential: $V(x)=-\frac{\hbar^2\lambda}{2ma}\delta(x)$ And now, to solve for the bound states $E<0$, i have: $$\frac{d^2u(x)}{dx^2}-k^2u(x)=-\frac{\lambda}{a}\delta(x)u(x),$$ where $k^2=\frac{2m|E|}{\hbar^2}$. Then:$$ u(x)= \begin{cases} e^{-kx},x>0\\ e^{kx},x<0\\ \end{cases} $$

And now, my doubt is the following: using this special condition for the derivative at the delta potential we get: $$(\frac{du}{dx})_{x=0^+}-(\frac{du}{dx})_{x=0^-}=-\frac{\lambda}{a}u(0) \equiv-k-k=-\frac{\lambda}{a},$$ . Although $u(0)$ is the same for $e^{-kx} $ and $e^{kx}$, in the case where my delta function is at $x\neq0$, which function would I choose? Or if the functions on the right and left of the delta potential were different?

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    $\begingroup$ Can you elaborate why you think this implies exp(-kx) was used? For x=0 -kx=0=kx $\endgroup$ – lalala Nov 7 '17 at 21:47
  • $\begingroup$ I'm so sorry! I wasn't clear! I edited my question further. $\endgroup$ – RicardoP Nov 7 '17 at 23:30
  • $\begingroup$ Thanks. Still a bit unclear. Delta function at x!=0 is zero. Are you asking why for positive x you use exp(-kx) and not exp(kx)? This is required so your wave function can be normalized (finite integeral of the square) $\endgroup$ – lalala Nov 8 '17 at 5:08

Since the wave function must be continuous everywhere, it can’t make a difference to choose one or the other as they must coincide at the discontinuity.

It the discontinuity is as $x=a$, the simply shifting $x \to x-a$ everywhere will do the trick.


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