0
$\begingroup$

I'm trying to estimate the ground state energy for a perturbed infinite square well directly. The potential is piecewise constant $$V(x)=\begin{cases}&\infty, \qquad x<-a/2\\ &0 \qquad -a/2<x<-b/2 \\&V_0 \qquad-b/2<x<b/2 \\&0 \qquad b/2<x<a/2 \\&\infty \qquad x>a/2\end{cases}.$$ Matching boundary each of the boundary conditions results in the transcendental equation $$2 k_1 k_2 \sin(k_1(1-b))+\tanh(k_2 b)[k_1^2+k_2^2+(k_1^2-k_2^2)\cos(k_1(1-b))]=0$$ where $$k_1=\sqrt{2ma^2E/\hbar^2}$$ $$k_2=\sqrt{2ma^2(V_0-E)/\hbar^2}.$$ We see that as $b$ tends to zero, we get back the original infinite square well energies, given by the equation $\sin(k_1)=0$. I want to see how the ground state will change for very small $b\ll 1$ to compare to the perturbation theory result. How would one go about Taylor expanding the above expression to approximate the lowest $E$, I assume $b\ll 1$ but can I also assume $k_1 \approx 0$ as well?

$\endgroup$
  • $\begingroup$ Since the perturbation involves two parameters, $V_{0}$ and $b$. I would try $k_{2} b \ll 1$. I also notice that when $V_{0} \to \infty$, it results in two separated infinite wells. $\endgroup$ – K_inverse Dec 6 '18 at 9:04

Your Answer

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

Browse other questions tagged or ask your own question.