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The semi-classical approximation for using WKB in simple words says that in a region where the potential doesn't vary sharply compared to the wavelength of the wavefunction the momentum (or the wavelength) is simply

$p(x)=\sqrt{(2m(E-V(x)))}$

Using this one can then calculate the wavefunction for $E<V$ and $E>V$ regions and stitch them together with connecting formulas to give the wavefunction in the whole region.

It turns out though that the WKB can be used for the bound state of the Dirac delta well also. Now, certainly, the potential "suddenly" (very very fast) changes at a point in this case and it is not clear how one can use WKB in this situation.

Why and how can WKB aprroximation be used in this case?

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  • $\begingroup$ Reference to "It turns out"? Which page? $\endgroup$
    – Qmechanic
    Commented Mar 20, 2022 at 8:32

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A way to understanding is the following:

  1. First recall the gluing conditions are given by integrating around the point where the Dirac delta is centered. The gluing conditions at the singular point $x_0$ are obtained by integrating around it in the differential equation (e.g eigenvalue equation) $$\lim_{\epsilon\rightarrow 0}\int_{x_0-\epsilon}^{x_0+\epsilon} \mathcal{O}\,\psi(x) = \lambda\lim_{\epsilon\rightarrow 0}\int_{x_0-\epsilon}^{x_0+\epsilon}\psi(x) = 0$$ where ${\cal O}$ is your operator including the potential and $\psi(x)$ is the wavefunction. A second order differential operator will then lead to a discontinuity in the first derivative when traversing from the left to the right of the point $x_0$.
  2. WKB conditions really depend on neighborhoods of every point, not on single points, they thus apply for interior points of your domain, not the edge. In other words, to compute derivatives one needs that the point is in the domain and some arbitrarily small neighborhood as well, while WKB validity conditions depend only on derivatives. So so far no violation at any point of the domain.

All this means effectively that since the conditions for WKB are violated only at a single point at the edge of each solution region and the gluing does capture the information of the Dirac delta at the very edge, the WKB Ansatz is expected to work.

Observation: I am assuming you are dealing for example with a one-dimensional Schrödinger equation for example, the simplest Dirac Delta potential is probably \begin{align} V(x) = V_0\delta(x) \end{align} which is an infinite peak at the origin and 0 elsewhere. So the regions where you solve are not $E>V$ and $V>E$ but left and right of the singular point that is $x>0$ and $x<0$ in both cases $V=0$.

Observe how the point $x=0$ does not belong to either of the domains where you use your Ansatz.

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  • $\begingroup$ So, calculating at other points then is very trivial but how does one do the gluing in the dirac region. The connecting formulas I have studied talk about linearizing the potential at the E=V region and I am guessing you mean some other "gluing" ? $\endgroup$
    – Lost
    Commented Mar 9, 2022 at 13:26
  • $\begingroup$ Please add your connecting formulas to the question, then perhaps I can guide your further. Also, do you know how to solve the free particle time-independent Schrödinger equation with a Dirac potential? $\endgroup$
    – ohneVal
    Commented Mar 9, 2022 at 14:46
  • $\begingroup$ Yes. It hinges on finding the discontinuity in the wavefunction about the Dirac well. I do not know how to solve this using WKB. $\endgroup$
    – Lost
    Commented Mar 9, 2022 at 15:36
  • $\begingroup$ Also, I am not able to understand how my question about "why the sharp delta potential doesn't violate the WKB working requirements?" is answered by "WKB conditions really depend on neighborhoods of every point, not on single points, they thus apply for interior points of your domain, not the edge." Can you please throw more light on it specifically addressing why the discontinuity doesn't mess up the WKB approximation. I understand the part where you mention that for places other than where the Dirac well is situated the solution can be obtained without any difficulty. The solution at the.. $\endgroup$
    – Lost
    Commented Mar 9, 2022 at 15:38
  • $\begingroup$ ...point is then some stictching of the left and right solutions...but I don't understand how this stiching can be done when the rate of potential at the point totally breaks the approximation conditions of the WKB (..that V(x) should vary less rapidly compared to $\psi(x)$) $\endgroup$
    – Lost
    Commented Mar 9, 2022 at 15:42

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