Analytic approximation of bound state energies for finite square well

Question

Considering a fairly deep finite potential well given by: $$U(x)= \begin{cases} U_0 \ \ , \ \ |x|>\frac{a}{2}\\ 0 \ \ , \ \ |x|\le \frac{a}{2} \end{cases}$$

We know that the energies of the bound states can be calculated numerically with equations:

$$ \sqrt{K^2-k^2}=k\ \text{tan}(k\frac{a}{2})\ \ \ , \ \ \ \text{For even bound eigenstates}\\ \sqrt{K^2-k^2}=-k\cot(k\frac{a}{2}) \ \ \ , \ \ \ \text{For odd bound eigenstates}\\ \text{Where} \ \ \ K=\sqrt{\frac{2mU_0}{\hbar^2}} \ \ \ \text{and} \ \ \ k=\sqrt{\frac{2mE}{\hbar^2}} $$

Is it possible to write an approximation in the form: $$E_n=E_0n^2+\text{correction term}$$ Where $E_0=\frac{\hbar^2\pi^2}{2ma^2}$ is the ground state energy of the infinite potential well.

Here is my attempt. If we look at the points of intersection of the plot that give us the allowed energies:

Here we can see that the black points of intersection represent a discrete value of $k$, say $k_n$. By the definition of $k$ we can see that there is a certain energy associated with each $k_n$: $$E_n=\frac{\hbar^2 k_n^2}{2m}$$

Moreover we can see that for each point of intesection the value of $k_n$ obeys $\frac{(n-1)\pi}{a}\leq k_n <\frac{n\pi}{a}$. So there is some number $0<c_n\leq 1$ such that:

$$E_n=\frac{\hbar^2 k_n^2}{2m}=\frac{\hbar^2 \pi^2(n-c_n)^2}{2ma^2}=\frac{\hbar^2 \pi^2 n^2}{2ma^2}-\frac{\hbar^2 \pi^2 2nc_n}{2ma^2}+\frac{\hbar^2 \pi^2 c_n^2}{2ma^2}$$ Now since the well is assumed to be deep we see that for the first couple states $c_n<<1$ so the last term can be ignored. I think I've got to somehow find a taylor series expansion which will give me an approximation for $c_n$ but I'm not sure how to do that. Do you have any suggestions or an entirely different approach?

Answer 1

I will set $a=2$ with no loss of generality. Your equations are: $$ \sqrt{K^2-k^2} = k\tan k \quad \sqrt{K^2-k^2} = -k\cot k $$ and you want the roots in the limit $K\to\infty$ i.e. the infinite well limit. You just need to apply perturbation theory about eigenvalue you singled out.

I will treat the even case but the same discussion applies for the odd case. Take the $n$-th solution, which exists for sufficiently large $K$ ($K>\pi n$) so that: $$ k_n\to \frac\pi2(2n+1) := k_n^\infty $$ with $n\in\mathbb N$ (eigenvalue of the infinite well). You just need to estimate at leading order the eigenvalue equation: $$ \sqrt{K^2-k_n^2}\sim K \quad k_n\tan k_n \sim \frac{k_n^\infty}{k_n^\infty-k_n} $$ so the next leading order term is: $$ k_n = k_n^\infty+\frac{k_n^\infty}K+o(K) $$ which you can check numerically. By systematically doing the asymptotic expansion of the RHS and LHS, you can iteratively get the next leading order terms. You will need for this the general expansion: $$ \sqrt{K^2-k_n^2} = \sum_{m=0}^M\binom{1/2}mK^{m+1}k_n^{-m}+o(K^{M+1}k_n^{-M}) \\ k_n\tan k_n = [k_n^\infty-(k_n^\infty-k_n)]\sum_{m=0}^\infty\frac{(-1)^m4^mB_{2m}}{(2m)!}(k_n^\infty-k_n)^{2m-1} $$ A word of warning, the perturbative correction is not uniform in $n$, since after all the solutions do not appear uniformly in $n$. You will need to take this into account when exploiting such formulas.