Central delta barrier in a symmetric infinite well ($\tan (\lambda a)=-{\hslash }^2\lambda /(m\kappa )$)

Inside walls at $\pm a$, a central delta potential of strength $\kappa$ leaves odd parity operator eigenstates unchanged because their wavefunctions vanish at zero. For even states, continuity and the derivative jump ${\psi }^{\prime }(0+)-{\psi }^{\prime }(0-)=2m\kappa \psi (0)/{\hslash }^2$ give the displayed quantization equation. For $\kappa >0$, one root lies between $(n+1/2)\pi /a$ and $(n+1)\pi /a$ for every $n\ge 0$.

Put $c=m\kappa /{\hslash }^2$, $k_n=(n+1/2)\pi /a$, and ${\delta }_n=a({\lambda }_n-k_n)$. The quantization equation gives ${\delta }_n=\arctan (c/{\lambda }_n)$, so ${\lambda }_n{\delta }_n\to c$ and ${\delta }_n\to 0$. Therefore ${\lambda }_n^2-k_n^2\to 2c/a$, proving the displayed absolute energy shift. Its ratio to the unperturbed energy tends to zero. The normalized unperturbed even state has probability density $1/a$ at the origin, so its first-order delta-barrier expectation is also $\kappa /a$.