I'm trying to solve an excercise that involves first order perturbation theory and an infinite potential. To ease the problem, I tried to consider an easier one dimensional model. Consider an infinite square well potential $$V(x) = \begin{cases} 0\quad \text{if } 0<x<a \\ +\infty \quad \text{elsewhere} \end{cases} $$ which has ground state wavefunction $\psi_0$ and energy $E_0$. Now, the potential is perturbed and becomes $$V'(x) = \begin{cases} 0\quad \text{if } 0<x<a+\varepsilon \\ +\infty \quad \text{elsewhere} \end{cases} $$ where $\varepsilon \ll a$. What is the first order correction to the energy $E_0$? I know that for the infinite square well this problem can be solved analytically, but I'm trying to solve the same problem with a worse potential and it is required to use perturbation theory. My problem is that I'm not able to write the Hamiltonian as a sum of the potential plus a perturbation, since we are given directly the perturbated potential. What I tried is the following: I multiplied the unperturbated potential with a function $A(x)$ that is 0 between $a$ and $a+\varepsilon$, and 1 elsewhere. I wrote this function using the Heaviside $\theta(x)$: $$A(x) = \theta(-x+a)+\theta(x-a-\varepsilon) $$ such that I obtain $$A(x)V(x) = V(x)\theta(-x+a) + V(x)\theta(x-a-\varepsilon) $$ which is now a sum of almost the initial potential and another term but the problem is that the product $0\cdot +\infty$ is not well defined, so I don't think this is the correct approch. Do you have a better idea to approch this problem?

  • $\begingroup$ I'm not sure you can do this in a well-defined way with an infinite potential. You could do it for a finite potential well and then take the limit, though. $\endgroup$ – J. Murray Mar 18 '18 at 20:48
  • $\begingroup$ I agree with you, I tried using V'(x)- V(x) as suggested by the answer below, but I alway get infinite, no matter what. Unfortunately my excercise has an infinite potential link $\endgroup$ – user85231 Mar 18 '18 at 21:13
  • $\begingroup$ That exercise is a deformation in shape, not merely in radius - the suggestion given as a hint should be the way to go. $\endgroup$ – J. Murray Mar 18 '18 at 21:16
  • $\begingroup$ Yeah, I can guess that I should follow the hint, but I don't know how, if you write the Hamiltonian $H+V$, even if you are able to write the ellipsoid such that the sphere is a limit, I don't see how it is possible to get the form $H+V_0 + V'$ such that I can use perturbation theory $\endgroup$ – user85231 Mar 18 '18 at 21:22
  • $\begingroup$ I don't think that's the idea of the exercise. If you transform to coordinates $(x,y,u)$ (where $u=az/b$) then the problem is spherical again. You can then perturb the ground state wavefunction and apply the Hamiltonian operator to it. $\endgroup$ – J. Murray Mar 18 '18 at 21:31

It will help you to think in terms of V'(X) - V(X) .... Where is this function zero and where is it not? Where it is not, note that it takes a familiar form (up to a minus sign). It may look "bad," however your first order energy shift will involve the unperturbed wave function - for the infinite well you will find, then, that it is easy to work out the correction.....

  • $\begingroup$ By the way, there is nothing perturbative in your set up - what is the parameter in which you would make your expansion (epsilon cannot be used because it doesn't multiply the difference between potentials) $\endgroup$ – lux Mar 18 '18 at 17:19
  • $\begingroup$ I don't get how you can solve the problem in this way. First, if you calculate $V'(x) - V(x) $ you have substraction between infinites, which is not well defined, then how can this help writing the Hamiltonian in the form $H_0 + W$, where $W$ is the perturbation? What should I use in the perturbation formula $\langle\psi|W|\psi\rangle$? $\endgroup$ – user85231 Mar 18 '18 at 18:56
  • $\begingroup$ Since you want to compare like for like think of (both) infinite square wells in terms of a limit that their depth, H, goes to infinity (at the end of the calculation). This is justified since you are interested in the (non-adiabatic) change in the well's width, not it's height. You'll see that your W is not a perturbation, since it involves the height H which is a large parameter. However if you keep H large but finite and use your formula above you will see that there is actually nothing to do - don't forget the psi are the unperturbed wavefunctions..... think carefully about their form. $\endgroup$ – lux Mar 18 '18 at 19:10
  • $\begingroup$ Your comments don't make sense. I really don't understand how your comments should help solving the perturbation problem $\endgroup$ – user85231 Mar 18 '18 at 19:41
  • $\begingroup$ Nevermind my comment, maybe I understood, but when you calculate the integral, you still have an infinite quantity... $\endgroup$ – user85231 Mar 18 '18 at 19:47

There is a related problem that sugests how to to proceed. I learned it from Alan MacKane and I think he told me that he found it in a paper by Sidney Coleman:

Let $\psi_\lambda(x)$ be a normalized bound-state solution to the Schr{\"o}dinger equation on the entire real line $$ [-\partial_x^2 +q(x)]\psi_\lambda=\lambda \psi_\lambda, \quad \psi_\lambda\in {\rm L}^2({\mathbb R}), $$ with $ \psi_\lambda(x) \sim Ae^{-\beta|x|}$ at large $|x|$. Then the small shift in $\lambda$ that arises from confining the system in a box of length $L$, so that $\psi_{\lambda+\delta\lambda}(L/2)=\psi_{\lambda+\delta\lambda}(-L/2)=0$, is $$ \delta \lambda \sim 4A^2 \beta e^{-\beta L}. $$ This is another case where the potential perturbation is infinite even though $1/L$ is a small parameter.

To obtain this result, we proceed as follows: We need to change $\lambda$ so that a solution that is zero at $x=-L/2$ evolves to zero at $x=+L/2$. To do this we require a Green function for the initial value problem $$ [-\partial_x^2 +q(x)-\lambda]\psi=f(x),\qquad \psi(-L/2)=0. $$ To construct the Green function we make use of the second solution
$$ \chi_\lambda(x) \propto \psi_\lambda(x) \int^x_0 [\psi_\lambda(\xi)]^{-2} d\xi, $$ which we scale so that $$ \chi_\lambda(x)\sim {\rm sgn}(x) A e^{\beta |x|}, \quad |x| \gg 0. $$ The initial-value Green function is then $$ G(x,\xi) = \frac{1}{2A^2\beta}[ \psi_\lambda(x) \chi_\lambda(\xi)-\chi_\lambda(x) \psi_\lambda(\xi)] \theta(x-\xi), $$ where $2A^2\beta$ is the Wronskian $W[\psi_\lambda, \chi_\lambda]$.

$$ \phi(x)= \frac 1{2A} \left(e^{\beta L/2}\psi_\lambda(x)+e^{-\beta L/2}\chi_\lambda(x)\right) $$ is zero at $x=-L/2$ and unity at $x=+L/2$.
We use $G(x,\xi)$ with $f(x) \to \delta\lambda\, \phi(x)$ to solve, to first order in $\delta\lambda$, the equation with $\lambda \to \lambda+\delta \lambda$ and $ \psi_{\lambda+\delta\lambda}(-L/2)=0$. We obtain
$$ \psi_{\lambda+\delta\lambda}(x) = \phi(x) +\delta \lambda \int_{-L/2}^x \frac{1}{2A^2\beta}[ \psi_\lambda(x) \chi_\lambda(\xi)-\chi_\lambda(x) \psi_\lambda(\xi)] \phi(\xi)\,d\xi. $$

By construction this $\psi_{\lambda+\delta\lambda}(x) $ satisfies the boundary condition at $x=-L/2$, while at $x=+L/2$ we have $$ \psi_{\lambda+\delta\lambda}(L/2) = \phi(L/2)+\delta\lambda \int_{-L/2}^{L/2}\frac{1}{2A^2\beta} \left[ \textstyle{\frac 12} e^{-\beta L}\chi_\lambda^2(\xi)- \textstyle{\frac 12} e^{\beta L} \psi^2_\lambda(\xi)\right]\,d\xi. $$ Now $\phi(L/2)=1$, $$ e^{-\beta L} \int_{-L/2}^{L/2} \chi_\lambda ^2(\xi)\,d\xi $$ is $O(1)$ and so is negligible compared to the $e^{\beta L}$ term, and $$ \int_{-L/2}^{L/2} \psi^2_\lambda(\xi)\,d\xi\sim 1 $$ by the normalization assumption. Thus $$ \psi_{\lambda+\delta\lambda}(L/2)= 1 - \frac{\delta\lambda}{4A^2\beta} e^{\beta L}+O(\delta\lambda^2). $$ Requiring this to be zero gives $$ \delta \lambda = 4A^2 \beta e^{-\beta L}, $$ as claimed.

So you can do essentially the same thing for your $\epsilon$ boundary shift.


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