The Finite Square Well

The finite square well potential is:

For bound states ($E < 0$), the wave function oscillates inside the well and decays exponentially outside. By symmetry, the solutions are either even or odd.

For even solutions: $\psi(x) = A \cos(l x)$ inside the well, and $\psi(x) = F e^{-\kappa |x|}$ outside, where $l = \sqrt{2m(V_0+E)}/\hbar$ and $\kappa = \sqrt{-2mE}/\hbar$.

Matching $\psi$ and $d\psi/dx$ at $x = a$ gives the transcendental equation:

This equation cannot be solved analytically for $E$. However, we can determine the number of bound states graphically. There is always at least one bound state, and the number increases as $V_0 a^2$ increases.

For scattering states ($E > 0$), a particle incident from the left has the form:

where $k = \sqrt{2mE}/\hbar$ and $l = \sqrt{2m(E+V_0)}/\hbar$. The reflection and transmission coefficients are found by matching $\psi$ and $d\psi/dx$ at $x = \pm a$.

The transmission coefficient is:

Quantum tunneling occurs when $E < V_0$ for a barrier (rather than a well). The wave function decays exponentially through the barrier, giving a non-zero probability of transmission even though classically the particle would be reflected.

For a rectangular barrier of height $V_0$ and width $2a$, the transmission coefficient is approximately $T \sim e^{-4\kappa a}$ when $\kappa a \gg 1$, where $\kappa = \sqrt{2m(V_0-E)}/\hbar$.

An remarkable feature of the finite square well is resonant transmission. When $2la = n\pi$, the sine term in the transmission formula vanishes and $T = 1$ — perfect transmission.

At these resonant energies, the reflected wave destructively interferes with itself, and the incident particle passes through the well as if it were not there. This is a purely quantum mechanical interference effect.

Resonant transmission is the one-dimensional analog of the Ramsauer-Townsend effect in three-dimensional scattering. It demonstrates that quantum particles can be transmitted through potentials with 100% probability at specific energies.

A finite square well has depth $V_0$ and width $2a$. Show that there is always at least one bound state, and determine the condition for a second bound state to appear.

Step 1 — Define dimensionless variables. Let $z = la$ and $z_0 = a\sqrt{2mV_0}/\hbar$. Then $\kappa a = \sqrt{z_0^2 - z^2}$.

Step 2 — Even state equation. $z \tan(z) = \sqrt{z_0^2 - z^2}$. The right side is a quarter-circle of radius $z_0$. The left side is a curve that starts at 0 and goes to infinity at $z = \pi/2$.

Step 3 — Existence of ground state. Since $z \tan(z)$ starts at 0 with slope 1, and $\sqrt{z_0^2 - z^2}$ starts at $z_0$ with slope 0, the curves always intersect for $z_0 > 0$. Thus there is always at least one bound state.

Step 4 — Second bound state. An odd solution appears when $z_0 > \pi/2$, because $-z\cot(z) = \sqrt{z_0^2 - z^2}$ has a solution only when the quarter-circle extends past $z = \pi/2$. Thus the condition for two bound states is $z_0 > \pi/2$, or $V_0 a^2 > \pi^2 \hbar^2 / (8m)$.