Potential
The potential is
$$V(x) =
\begin{cases}
-V_0, & \text{if $-L < x < L$} \\
0, & \text{if $|x|>L$}
\end{cases}$$
The Schrodinger's equation inside the well is
$$\phi_E''(x) = \frac{2m}{\hbar^2}(-V_0 - E)\phi_E(x)$$
So the general solution is
$$\phi_E(x) = A \cos(kx) + B\sin(dx)$$
where
$$k = \sqrt{\frac{2m}{\hbar^2}(V_0 + E)}$$
Outside the square well is
$$\phi_E''(x) = -\frac{2m}{\hbar^2}E\phi_E(x)$$
So the general solution is
$$\phi_E(x) = Ce^{\alpha x} + De^{-\alpha x}$$
where
$$\alpha = \sqrt{\frac{2m}{\hbar^2}|E|}$$
So, the general solution is
$$\phi_E(x) =
\begin{cases}
Ee^{\alpha x} + Fe^{-\alpha x}, & \text{if $x > L$} \\
A \cos (kx) + B \sin (kx), & \text{if $-L < x < L$} \\
Ce^{\alpha x} + De^{-\alpha x}, & \text{if $x < -L$} \\
\end{cases}
$$
Boundary Condition
The boundary conditions are:
- Since the wave function must be normalizable to be able to represent probability, \(\phi_E\) must tend to zero on \(\pm\infty\)
- Since the second derivative of the wave function is a well-defined function, \(\phi_E\) and its first derivative must be continuous
By the first boundary condition, \(D\) and \(E\) must be zero.
Parity
There are two possible solutions: even or odd. Let's consider the even solution first. This means \(B = 0\) and \(C = F\).
$$\phi_E(x) =
\begin{cases}
Ce^{-\alpha x}, & \text{if $x > L$} \\
A \cos (kx), & \text{if $-L < x < L$} \\
Ce^{\alpha x}, & \text{if $x < -L$} \\
\end{cases}
$$
By the second boundary condition,
$$A \cos kL = C e^{-\alpha L}$$
and
$$kA \sin kL = \alpha C e^{-\alpha L}$$
So, we have
$$k \tan kL = \alpha$$
Also
$$\alpha^2 + k^2 = \frac{2m}{\hbar^2}V_0$$
With two equations, one may solve for the solutions of the two variables.
Delta Function
If potential is
$$V(x) = -V_0\delta(x)$$
The Schrodinger's equation becomes
$$\phi_E''(x) = \frac{2m}{\hbar^2}(-V_0\delta(x) - E)\phi_E(x)$$
The previous second boundary condition no longer holds as the second derivative of the wave function is a delta function.
The boundary conditions are:
- Since the wave function must be normalizable to be able to represent probability, \(\phi_E\) must tend to zero on \(\pm\infty\)
- Since the second derivative of the wave function is a delta function, \(\phi_E'\) is a step function and \(\phi_E\) is a continuous function
When the solution is even
$$\phi_E(x) =
\begin{cases}
Ce^{-\alpha x}, & \text{if $x > 0$} \\
Ce^{\alpha x}, & \text{if $x < 0$} \\
\end{cases}
$$
By the second boundary condition, \(\phi_E\) is continuous at zero, so \(\phi_E(0) = Ce^{\alpha (0)} = C\)
So,
$$\phi_E(x) =
\begin{cases}
Ce^{-\alpha x}, & \text{if $x > 0$} \\
C, & \text{if $x = 0$} \\
Ce^{\alpha x}, & \text{if $x < 0$} \\
\end{cases}
$$
\begin{align}
\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial^2x}-V_0\delta(x)\right)\phi_E &= E \phi_E \\
-\frac{2m}{\hbar^2}\left(E+V_0\delta(x)\right)\phi_E &= \phi_E''\\
\int^a_{-a}\phi_E'' &= -\frac{2m}{\hbar^2}\int^a_{-a}\left(V_0\delta(x)+E\right)\phi_E \\
\phi_E'(a) - \phi_E'(-a) &= -\frac{2m}{\hbar^2}V_0\phi_E(0) + -\frac{2m}{\hbar^2}\int^a_{-a}E\phi_E \\
\lim_{a \to 0}(\phi_E'(a) - \phi_E'(-a)) &= \lim_{a \to 0} \left(-\frac{2m}{\hbar^2}V_0C + -\frac{2m}{\hbar^2}\int^a_{-a}E\phi_E \right) \\
&= -\frac{2m}{\hbar^2}V_0C \\
\phi'_{E+}(0) - \phi'_{E-}(0) &= -\frac{2m}{\hbar^2}V_0C\\
-\alpha C e^{-\alpha (0)} - \alpha C e^{\alpha (0)} &= -\frac{2m}{\hbar^2}V_0C\\
2\alpha &= \frac{2m}{\hbar^2}V_0\\
\alpha &= \frac{mV_0}{\hbar^2}
\end{align}
Since
$$\alpha = \frac{\sqrt{2m|E|}}{\hbar}$$
we have
$$E = -\frac{\hbar^2\alpha^2}{2m} = -\frac{mV_0^2}{\hbar^2}$$
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