The Free Particle & Delta-Function Potential

For a free particle ($V = 0$), the time-independent Schrödinger equation is:

The general solution is a superposition of plane waves: $\phi(x) = A e^{ikx} + B e^{-ikx}$, where $k = \sqrt{2mE}/\hbar$. The energy is $E = \hbar^2 k^2 / 2m$.

Plane waves are momentum eigenstates: $\hat{p} e^{ikx} = \hbar k e^{ikx}$. However, they are not normalizable because $\int |e^{ikx}|^2 \, dx$ diverges. We handle this by using delta-function normalization or by constructing wave packets.

A wave packet is a superposition of plane waves with a spread of momenta:

If the momentum distribution $\tilde{\phi}(k)$ is peaked around $k_0$ with width $\Delta k$, the wave packet travels with group velocity $v_g = \hbar k_0 / m = p_0/m$, which is the classical particle velocity.

The delta-function potential $V(x) = -\alpha \delta(x)$ is the simplest model that exhibits both bound states ($E < 0$) and scattering states ($E > 0$). Despite its artificiality, it captures essential quantum features.

For the bound state ($E < 0$), the wave function decays exponentially away from the origin: $\psi(x) = A e^{-\kappa |x|}$, where $\kappa = \sqrt{-2mE}/\hbar$. Integrating the Schrödinger equation across $x = 0$ gives the discontinuity condition:

This yields $\kappa = m\alpha/\hbar^2$, so the bound state energy is $E = -m\alpha^2/(2\hbar^2)$. There is exactly one bound state, regardless of the strength of the potential.

For scattering states ($E > 0$), we consider a wave incident from the left: $\psi(x) = A e^{ikx} + B e^{-ikx}$ for $x < 0$, and $\psi(x) = F e^{ikx}$ for $x > 0$. The reflection and transmission coefficients are:

Remarkably, the delta-function potential becomes transparent ($T = 1$) at certain energies, a purely quantum effect.

Bound states are localized: the wave function decays to zero as $|x|$ goes to infinity. The particle is trapped in the potential. Bound state energies are discrete and negative (for attractive potentials).

Scattering states extend to infinity. The particle is free at large distances, but its wave function is modified by the potential in the interaction region. Scattering state energies are continuous and positive.

The delta-function well has exactly one bound state and a continuum of scattering states. More realistic potentials (like the finite square well) have multiple bound states when the potential is sufficiently deep or wide.

The distinction between bound and scattering states is crucial in quantum mechanics. Bound states determine the structure of atoms and molecules, while scattering states describe collisions and transport phenomena.

Find the bound state energy and normalized wave function for a particle of mass $m$ in the potential $V(x) = -\alpha \delta(x)$.

Step 1 — Ansatz. For $E < 0$, the wave function decays exponentially: $\psi(x) = A e^{-\kappa |x|}$, where $\kappa = \sqrt{-2mE}/\hbar > 0$.

Step 2 — Discontinuity condition. The derivative has a jump at $x = 0$: $d\psi/dx|_{0^+} - d\psi/dx|_{0^-} = -A\kappa - (A\kappa) = -2A\kappa$. This must equal $-(2m\alpha/\hbar^2) A \psi(0) = -(2m\alpha/\hbar^2) A$. Thus $2\kappa = 2m\alpha/\hbar^2$, giving $\kappa = m\alpha/\hbar^2$.

Step 3 — Energy. $E = -\hbar^2\kappa^2/(2m) = -\hbar^2(m\alpha/\hbar^2)^2/(2m) = -m\alpha^2/(2\hbar^2)$.

Step 4 — Normalization. $\int_{-\infty}^{+\infty} |\psi|^2 \, dx = 2|A|^2 \int_0^\infty e^{-2\kappa x} \, dx = 2|A|^2/(2\kappa) = |A|^2/\kappa = 1$. Thus $A = \sqrt{\kappa} = \sqrt{m\alpha/\hbar^2}$.

Step 5 — Final result. $\psi(x) = \sqrt{m\alpha/\hbar^2} \, e^{-m\alpha|x|/\hbar^2}$, with binding energy $|E| = m\alpha^2/(2\hbar^2)$.