Finite Square Well

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 $$E = -\frac{\hbar^2\alpha^2}{2m} = -\frac{mV_0^2}{\hbar^2}$$