We try solving the even function solutions to the time-independent Schrödinger equation for the potential $V=-\alpha\sum_{j=-n}^n\delta\!\left(x-ja\right)$ such that $E<0$ (bound states).

Obviously, the solutions have the form $\psi= \begin{cases} A_{\left\lfloor\frac{\left|x\right|}{a}\right\rfloor}\mathrm e^{\kappa\left|x\right|} +B_{\left\lfloor\frac{\left|x\right|}{a}\right\rfloor}\mathrm e^{-\kappa\left|x\right|}, &\left|x\right|na, \end{cases}$ where $A_j,B_j$ ($j=0,1,\ldots,n$) are constants of integration, and $\kappa\coloneqq\frac{\sqrt{-2mE}}{\hbar}$.

Noting that we are finding bound states, we should have $\lim_{x\to\infty}\psi=0$. Therefore, $A_n=0.$$\p{1}$ Function $\psi$ is naturally continuous at $x=0$. Considering the continuity of $\psi$ at $\left|x\right|=ja$ ($j=1,2,\ldots,n$), we have $A_{j-1}\mathrm e^{\kappa ja}+B_{j-1}\mathrm e^{-\kappa ja}=A_j\mathrm e^{\kappa ja}+B_j\mathrm e^{-\kappa ja}.$$\p{2}$

For $j=-n,\ldots,n$, integrate both sides of the time-independent Schrödinger equation over interval $\left[ja-\varepsilon,ja+\varepsilon\right]$ and let $\varepsilon\to0$, and we have $\left.\frac{\mathrm d\psi}{\mathrm dx}\right|_{ja^-}^{ja^+}=-\beta\left.\psi\right|_{ja},$ where $\beta\coloneqq\frac{2m\alpha}{\hbar^2}$.

For $j=0$, the formula above gives $\left(A_0\kappa-B_0\kappa\right)-\left(-A_0\kappa+B_0\kappa\right)=-\beta\left(A_0+B_0\right).$$\p{3}$

For $j=1,2,\ldots,n$, on the other hand, $\left(A_j\kappa\mathrm e^{\kappa ja}-B_j\kappa\mathrm e^{-\kappa ja}\right) -\left(A_{j-1}\kappa\mathrm e^{\kappa ja}-B_{j-1}\kappa\mathrm e^{-\kappa ja}\right) =-\beta\left(A_j\mathrm e^{\kappa ja}+B_j\mathrm e^{-\kappa ja}\right).$$\p{4}$

Equations 1, 2, 3, and 4 together form a homogeneous linear equation w.r.t. $A_j,B_j$ ($j=0,1,\ldots,n$). To require that the equation has non-zero solutions, the determinant of the coefficient matrix should be zero, and we can find $\kappa$ by this property. However, the equation is transcendental for $n>0$.

If we found the value of $\kappa$, the solution space for the homogeneous linear equation should be one-dimensional, and then we can determine all the constants by normalizing $\psi$.