Finite Potential Well and Sketching Wave Functions

The general solutions to the finite potential well problem include terms with $c2 \cos(kx)$ , $c3 e^{-\alpha x}$ , and $c_4 e^{\alpha x}$ .
The first solution resembles a sinusoidal function within the well (up to $-l/2$ ) and decaying exponential functions outside the well.
This is similar to the infinite box problem but with non-zero probability outside the box.
Classically forbidden regions: areas where classical mechanics predicts a particle cannot exist because its energy would be less than the potential.
For the $n=1$ solution:
- Inside the box: the wave function looks like $c_2 \cos(kx)$ .
- Outside the box (right): $c_3 e^{-\alpha x}$ .
- Outside the box (left): $e^{\alpha x}$ .
Energy levels in the finite well are discrete, similar to the infinite well, but only up to a certain point. Above the potential, the energy levels become a continuum, resembling a free particle.

To solve the finite well, boundary conditions must be matched:
- The wave function must be continuous at $x = \pm l/2$ .
- The derivative of the wave function must also be continuous at $x = \pm l/2$ .
Continuity of the wave function:
- $c2 \cos(kl/2) = c3 e^{-\alpha l/2}$
$\alpha$ describes how far the wave function extends into the classically forbidden region.
- Imagine $e^{-x/(1/\alpha)}$ where $\frac{1}{\alpha}$ represents a characteristic length. The wave function declines by 68% over this length, similar to a capacitor losing charge over one time constant.

Differentiability of the wave function:
- $\frac{d\psi}{dx}$ evaluated at $x = +l/2$ (inside the box) must equal $\frac{d\psi}{dx}$ evaluated at $x = +l/2$ (outside the box).
Taking the derivative of the wave function inside the box:
- $-c_2 k \sin(kl/2)$
Taking the derivative of the wave function outside the box (on the right):
- $- \alpha c_3 e^{-\alpha l/2}$
By dividing the differentiability equation by the continuity equation, we eliminate $c2$ and $c3$ .
- $\frac{-k \sin(kl/2)}{\cos(kl/2)} = \frac{-\alpha c3 e^{-\alpha l/2}}{c3 e^{-\alpha l/2}}$
- $\alpha = k \tan(kl/2)$

$k$ is a function of $E$ , while $\alpha$ depends on both the potential $V_0$ and the energy level $E$ .
Graphing $\alpha$ as a function of $k$ :
- The tangent function approaches infinity when its argument approaches $\pi/2$ radians.

$k^2 + \alpha^2 = \frac{2mV_0}{\hbar^2}$
This equation represents a circle.
Let $k0 = \sqrt{\frac{2mV0}{\hbar^2}}$ , where $k_0$ is the wave number associated with the potential.
The equation becomes: $k^2 + \alpha^2 = k_0^2$

The solutions for $k$ (and thus, $E$ ) are found where the circle intersects with the tangent function.
Increasing the potential increases the radius of the circle, leading to more intersections and thus more discrete energy levels.
Solutions for E < V_0 cannot be found analytically, but they can be found graphically.

For a given potential, there will be a finite number of bound state solutions.
As the potential approaches infinity, the finite well problem approaches the infinite well problem.
In the limit as $V_0 \rightarrow \infty$ , the solutions correspond to $\sin(n \pi x / L)$ .
There will always be at least one bound state, even for a very small potential.
Bound states imply that the probability of finding the particle at $\pm \infty$ is zero.

$\alpha$ is considered positive because E < V_0 for bound state solutions.
The first solution has a cosine function, the second has a sine function, the third has a cosine function, and so on.
Only solutions with cosine in the middle region were initially found (k1, k3, k5, etc.).

For even solutions (E2, E4, E6), use $c_1 \sin(kx)$ .
The same mathematical process is followed, but the boundary condition equations change slightly due to switching from cosine to sine.
$\alpha = -k \cot(kl/2)$
The equation $k^2 + \alpha^2 = k_0^2$ still holds (the circle function remains the same).

To find even solutions, graph $-\cot(kl/2)$ .
Intersections between the circle and the cotangent curves give the allowed energy levels.
The number of bound states depends on the potential energy. A larger potential leads to more bound states.
As potential goes to infinity, the number of bound states approaches infinity, and the finite well merges with the infinite well problem.

For any potential with bound states:
- The first solution has no nodes (zeros) between the boundaries.
- The next bound state has one node.
- The one after that has two nodes, and so on.

For the eigenvalue problem: $-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = (E - V) \psi$
Rearranging, we get: $\frac{d^2 \psi}{dx^2} = \frac{2m}{\hbar^2} (V - E) \psi$
The second derivative is related to the curvature of the wave function.
- If the second derivative is positive, the function is concave up.
- If the second derivative is negative, the function is concave down.
When E > V, the solution exhibits sinusoidal behavior.
When E < V, the behavior is exponential.
The magnitude of the second derivative (and thus the tightness of the curvature) is proportional to $|E - V|$ .

Regions where E is much bigger than V gives large negative number, The function has a larger curvature or tighter curvature
In classically allowed regions (E > V), the wave function oscillates.
In classically forbidden regions (E < V), the wave function decays exponentially and is concave up.

Estimate the first energy level by pretending that the function goes to zero at a characteristic length beyond the well.
$E_1 = \frac{\hbar^2 \pi^2}{2m L^2}$
Replace L with an effective length: $L + 2(1/\alpha)$
$E_1 \approx \frac{\hbar^2 \pi^2}{2m (L + 2/\alpha)^2}$
This approximation treats the finite well like an infinite box problem with extended boundaries. The smaller the potential, the longer the extended box.