# Taylor expansion for a double well/perturbed infinite square well

I'm trying to estimate the ground state energy for a perturbed infinite square well directly. The potential is piecewise constant $$V(x)=\begin{cases}&\infty, \qquad x<-a/2\\ &0 \qquad -a/2a/2\end{cases}.$$ 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?

• 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. – K_inverse Dec 6 '18 at 9:04