Solve the time-independent Schr ̈odinger equation for a centered infinite square well with a delta-function barrier in the middle:

$\mathit{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{\alpha \delta }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{for}\mathbf{}\mathbf{-}\mathbf{a}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{+}\mathbf{a}\\ \mathbf{}\\ \mathbf{\infty }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{for}\mathbf{}\left|x\right|\mathbf{\ge }\mathbf{a}\end{array}$

Treat the even and odd wave functions separately. Don’t bother to normalize them. Find the allowed energies (graphically, if necessary). How do they compare with the corresponding energies in the absence of the delta function? Explain why the odd solutions are not affected by the delta function. Comment on the limiting cases α → 0 and α → ∞.

Since, the Schrodinger equation is given by,

$-\frac{ħ}{2m}\frac{d}{d{x}^2}\psi \left(x\right)+V\left(x\right)\psi \left(x\right)=E\psi \left(x\right)$

Where, the potential$V\left(x\right)$is given as:

${\int }_{-\epsilon }^{\epsilon }\left(\frac{ħ}{2m}\frac{d}{d{x}^2}\psi \left(x\right)+V\left(x\right)\psi \left(x\right)\right)dx={\int }_{-\epsilon }^{\epsilon }E\psi \left(x\right)dx$

Use the linearity of the integral to separate the left side of the previous equation into two integrals,

${\int }_{-\epsilon }^{\epsilon }\frac{ħ}{2m}\frac{d}{d{x}^2}\psi \left(x\right)dx+{\int }_{-\epsilon }^{\epsilon }\alpha \delta \left(x\right)\psi \left(x\right)dx={\int }_{-\epsilon }^{\epsilon }E\psi \left(x\right)dx$

The integral on the right side becomes zero. Hence the solution to the stationary Schrodinger equation gives continuous solutions, and therefore, the integral over an arbitrarily infinitesimal region is equal to zero. However, we can’t use the same logic for the term on the left side of the equation since the integral involves a delta function.

Therefore,

$$\begin{gathered} {\int }_{-\epsilon }^{\epsilon }\frac{{ħ}^2}{2m}\frac{d}{d{x}^2}\psi \left(x\right)dx=\alpha \psi \left(0\right) \\ \mathrm{Since}, \\ \frac{d}{dx}\psi {\left(0\right)}_{+}-\frac{d}{dx}\psi {\left(0\right)}_{-}=\frac{2m}{{ħ}^2}\alpha \psi \left(0\right) \end{gathered}$$

Now, we got an extra condition that needs to be satisfied – the spatial derivative of the wave function is discontinuous.

The wave function is equal to zero in the region where the potential is$\infty$since the probability of tunneling into those regions has to be zero. The solution must now consist of two sets – odd, and even wave functions as the potential are even. Denoting the region left of the delta function by l and right of the delta function by ll. Since the potential is equal to zero in these two regions (l and ll), the Schrodinger equation becomes,

$$\begin{gathered} -\frac{{ħ}^2}{2m}\frac{d}{dd{x}^2}\psi \left(x\right)=E\psi \left(x\right) \\ \mathrm{Since}\mathrm{energy}\mathrm{has}\mathrm{to}\mathrm{be}\mathrm{greater}\mathrm{than}\mathrm{zero}(\mathrm{as}\mathrm{the}\mathrm{potential}\mathrm{in}\mathrm{these}\mathrm{regions}\mathrm{is}\mathrm{zero}),\mathrm{Using}, \\ {\mathrm{k}}^2=\frac{2\mathrm{mE}}{{\mathrm{ħ}}^2} \\ \mathrm{\psi }\text{'}\text{'}\left(\mathrm{x}\right)+{\mathrm{k}}^2\mathrm{\psi }\left(\mathrm{x}\right)=0 \\ \mathrm{This}\mathrm{is}\mathrm{an}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{harmonic}\mathrm{oscillator}. \\ \mathrm{Therefore}, \\ \mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{Acos}\left(\mathrm{kx}\right)+\mathrm{B}\sin \left(\mathrm{kx}\right) \end{gathered}$$

The solutions in the regions l and ll are given by:

$$\begin{gathered} {\psi }_l\left(x\right)=A_1\cos \left(kx\right)+B_1\sin \left(kx\right) \\ {\psi }_{ll}\left(x\right)=A_2\cos \left(kx\right)+B_2\sin \left(kx\right) \end{gathered}$$

Since the solutions are odd, we must have ${\psi }_l\left(x\right)={\psi }_{ll}\left(x\right)$.

From that follows $B_1=-B_2=b\mathrm{and}{\mathrm{A}}_1={\mathrm{A}}_2=d$

Therefore,

$$\begin{gathered} {\psi }_l\left(x\right)=d\cos \left(kx\right)+b\sin \left(kx\right) \\ {\psi }_{ll}\left(x\right)=d\cos \left(kx\right)+b\sin \left(kx\right) \end{gathered}$$

As of now, we have three unknown variables, d, b, and k. Using the boundary condition that the wave functions must disappear at , together with the discontinuity condition and the normalization condition, we have 3 required equations.

From the first two conditions,

$$\begin{gathered} d\cos \left(ka\right)+b\sin \left(ka\right)=0 \\ \alpha \frac{2m}{ħ}a+2dk=0 \end{gathered}$$

In the second condition, the value of $\psi \left(0\right)$can be picked from either ${\psi }_l\mathrm{or}{\psi }_{ll}$or since both the wave functions are to be connected at the origin.

$$\begin{gathered} \left|\begin{array}{cc}\cos \left(KA\right)& \sin \left(KA\right)\\ \frac{2m}{{ħ}^2}\alpha & 2k\end{array}\right|=0 \\ \tan \left(ka\right)=-\frac{k{ħ}^2}{m\alpha } \end{gathered}$$

This is a transcendental equation. Hence it has no analytic solutions. The solutions can be determined either by graphical or numerical methods. Since this equation gives discrete values of k, this is the equation that determines the discrete energy spectrum of the system for even solutions.

Now, notice the limiting cases of the parameter$\alpha$.

As $\alpha \to 0$, the energy condition becomes$\tan \left(ka\right)\to \infty$, hence, producing the same results as the potential well but without the delta functions.

And for $\alpha \to \infty$, the energy condition becomes $\tan \left(ka\right)\to 0$, giving the energy spectrum of the potential well with the half-width, a.

This makes sense as $\alpha \to \infty$, the barrier becomes impenetrable.

Following the same logic, we obtain that $B_1=B_2=b\mathrm{and}{A}_1=A_2=0$, from the requirement that the wave function is to be odd and joined at the origin. Since the function disappears at the origin, the discontinuity requirement provides no new information. And hence, the problem reduces to the problem of the infinite square well without the delta function. The energies are determined by the condition that the wave function vanishes at $x=a\mathrm{or}x=-a$.

$$\begin{gathered} \sin \left(ka\right)=0 \\ k=\frac{n\pi }{a} \\ {E}_n=\frac{n^2{\pi }^2{ħ}^2}{2ma},n \end{gathered}$$, being an integer.

Therefore, we can conclude that the odd wave functions are unaffected by the delta function since they vanish at the origin.