2
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ Linked? $\endgroup$ – Cosmas Zachos Mar 8 at 17:31
  • $\begingroup$ @CosmasZachos Yes, I had seen that post before posting the question and I have solved the Sch. Equ. for $H$ as they did, but my ques is about retrieving the result semi-classically. $\endgroup$ – lambda Mar 8 at 18:22
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @J C, is there any way to resolve this problem? $\endgroup$ – Epsilon Mar 12 at 3:49
  • $\begingroup$ @SachinKumar Probably using Cauchy en.wikipedia.org/wiki/Cauchy%27s_integral_formula and calculate it out. My suspension was not about the root of $\delta$, but the negative sign in front of it. I doubt weather it will be a satiable answer. Notice you will need to use the definition of $\delta$ and take limit though, then again, headache. $\endgroup$ – J C Mar 12 at 6:01
  • $\begingroup$ @J C, Thanks, I will try $\endgroup$ – Epsilon Mar 13 at 5:54
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.