This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2021-01-13 15:33 +0800.

We try solving the even function solutions to the time-independent Schrödinger equation for the potential

V=−α∑j=−nnδ ⁣(x−ja)V=-\alpha\sum_{j=-n}^n\delta\!\left(x-ja\right)

such that E<0E<0 (bound states).

Obviously, the solutions have the form

ψ={A⌊∣x∣a⌋eκ∣x∣+B⌊∣x∣a⌋e−κ∣x∣,∣x∣<na,Aneκ∣x∣+Bne−κ∣x∣,∣x∣>na,\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,\\ A_n\mathrm e^{\kappa\left|x\right|}+B_n\mathrm e^{-\kappa\left|x\right|}, &\left|x\right|>na, \end{cases}

where Aj,BjA_j,B_j (j=0,1,…,nj=0,1,\ldots,n) are constants of integration, and κ≔−2mEℏ\kappa\coloneqq\frac{\sqrt{-2mE}}{\hbar}.

Noting that we are finding bound states, we should have lim⁡x→∞ψ=0\lim_{x\to\infty}\psi=0. Therefore,

An=0.A_n=0. (1)(1)

Function ψ\psi is naturally continuous at x=0x=0. Considering the continuity of ψ\psi at ∣x∣=ja\left|x\right|=ja (j=1,2,…,nj=1,2,\ldots,n), we have

Aj−1eκja+Bj−1e−κja=Ajeκja+Bje−κja.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}. (2)(2)

For j=−n,…,nj=-n,\ldots,n, integrate both sides of the time-independent Schrödinger equation over interval [ja−ε,ja+ε]\left[ja-\varepsilon,ja+\varepsilon\right] and let ε→0\varepsilon\to0, and we have

dψdx∣ja−ja+=−βψ∣ja,\left.\frac{\mathrm d\psi}{\mathrm dx}\right|_{ja^-}^{ja^+}=-\beta\left.\psi\right|_{ja},

where β≔2mαℏ2\beta\coloneqq\frac{2m\alpha}{\hbar^2}.

For j=0j=0, the formula above gives

(A0κ−B0κ)−(−A0κ+B0κ)=−β(A0+B0).\left(A_0\kappa-B_0\kappa\right)-\left(-A_0\kappa+B_0\kappa\right)=-\beta\left(A_0+B_0\right). (3)(3)

For j=1,2,…,nj=1,2,\ldots,n, on the other hand,

(Ajκeκja−Bjκe−κja)−(Aj−1κeκja−Bj−1κe−κja)=−β(Ajeκja+Bje−κja).\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). (4)(4)

Equations 1, 2, 3, and 4 together form a homogeneous linear equation w.r.t. Aj,BjA_j,B_j (j=0,1,…,nj=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 κ\kappa by this property. However, the equation is transcendental for n>0n>0.

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