Perturbation theory with infinite potential 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?
 A: 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 $W[\psi_\lambda, \chi_\lambda]$. 
Now
$$
\phi(x)= \frac 1{2A} \left(e^{\beta L/2}\psi_\lambda(x)+e^{-\beta L/2}\chi_\lambda(x)\right)
$$
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.
A: 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.....
