Wilson-Sommerfeld Quantization of Dirac delta in Infinite Square Well (ISW)

Question

Inspired by this Phys.SE post I am curious to find the energies of Dirac delta potential inside the ISW (walls at $x=0,L$) $$ H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V_0\delta(x-L/2) $$

using Wilson-Sommerfeld Quantization (WSQ).

So I looked at this integral (Here, $p_1=\sqrt{2 m E}, p_2=\sqrt{2 m (E- V_0\delta(x-L/2))}~$) $$\oint p~ dx = 2 \int_0^{L/2-\epsilon} p_1~ dx + 2 \int_{L/2-\epsilon}^{L/2+\epsilon} p_2 ~dx+ 2 \int_{L/2+\epsilon}^{L} p_1~ dx $$ $$= 2* \sqrt{2 m E}*(L-2 \epsilon) + 2 \int_{L/2-\epsilon}^{L/2+\epsilon} p_2 (=\sqrt{2m (E-V_0 \delta(x-L/2))})~dx $$ The Integral term is zero E.g.- this integral (a particular case) , and WSQ gives the energies of ISW Only.

If you recall that momentum is discontinuous for a Dirac delta potential, so integral term should not be zero.

Now the Ques is: can we treat given Hamiltonian $H$ using WSQ? If yes, then how to tackle the integral (we have tackled this integral in Schrodinger's formalism)?

Answer 1

Somehow your expression doesn't make quite sense, see: Don't understand the integral over the square of the Dirac delta function , What is the square root of the Dirac Delta Function? ,and (https://www.wolframalpha.com/input/?i=Integrate+sqrt(f(x)*(DiracDelta%5Bt%5D))+dx ) Your input of Dirac function under square root doesn't make quite sense in standard mathematics, it automatically resolved into complex domain and then you almost automatically exited the WSQ.

Answer 2

At the physical (as opposed to rigorous mathematical) level, one can at least try to explore the weak and strong coupling limit of the Dirac delta potential.

1. On one hand, the quantum mechanically exact quantization condition is given by $$ \sin\frac{kL}{2}~=~0\qquad \qquad\vee\qquad \tan\frac{kL}{2}~=~-\frac{2k}{\kappa},\tag{1} $$ where $$ k~>~0, \qquad \kappa~:=~\frac{2m V_0}{\hbar^2}, \tag{2} $$ cf. e.g. Emilio Pisanty's answer here. The even-number modes $$k~\in\frac{\pi}{L}2\mathbb{N} \tag{3}$$ can not "feel" the Dirac delta potential (because their wave functions vanish at the Dirac delta singularity). The odd-number modes has the following expansions

   • Weak coupling limit ($V_0$ small): $$ k~=~\frac{2}{L}\left(\frac{\pi}{2}+\arctan\frac{\kappa}{2k}\right) ~\approx~\frac{\pi}{L}(2j+1)+\frac{\kappa}{kL}, \qquad j~\in~\mathbb{N}.\tag{4}$$

   • Strong coupling limit ($V_0$ large): $$ \frac{kL}{2}+\frac{2k}{\kappa}~\approx~\pi j \qquad\Leftrightarrow\qquad k~\approx~\frac{\pi j}{\frac{L}{2}+\frac{2}{\kappa}}~\approx~ \frac{\pi}{L} 2j\left(1- \frac{4}{\kappa L}\right) .\tag{5}$$

2. On the other hand, semiclassical WKB approximation yields the following.

   • Weak coupling limit ($V_0$ small): The Bohr-Sommerfeld quantization rule is $$ 2\pi n ~=~\oint \!\mathrm{d}x~ k(x) ~=~ \oint \!\mathrm{d}x~\sqrt{2m(E_n-V(x))} $$ $$ ~\approx~ k_n\oint \!\mathrm{d}x~ \left(1- \frac{V(x)}{2E_n}\right)~=~2k_n\left(L-\frac{V_0}{2E_n}\right) , \qquad E_n \equiv \frac{(\hbar k_n)^2}{2m}, \qquad n\in\mathbb{N},\tag{6}$$ which leads to $$ k^{(0)}_n~=~\frac{n\pi}{L}, \tag{7}$$ $$ k^{(1)}_n~=~\frac{n\pi}{L- \frac{V_0}{2E^{(0)}_n}} ~=~k^{(0)}_n\left(1- \frac{\kappa}{(k^{(0)}_n)^2L}\right)^{-1}~\approx~k^{(0)}_n+ \frac{\kappa}{k^{(0)}_n L},\tag{8}$$ perturbatively. In this limit products of Dirac delta distributions [from expanding the square root (6)] are effectively ignored.

3. Discussion. The zero-order approximation (7) correctly finds the even- & odd-number modes (3) & (4) when there is no Dirac delta potential. The first-order approximation (8) is correct for the odd-number modes, but this approach fails to take into account that the even-number modes are not affected by the Dirac delta potential.