Free particle

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In physics, a free particle is a particle that, in some sense, is not bound. In classical physics, this means the particle is not subject to any external force.

1 Classical Free Particle
2 Non-Relativistic Quantum Free Particle
3 Relativistic free particle
4 See also

[edit] Classical Free Particle

The classical free particle is characterized simply by a fixed velocity. The momentum is given by

$\mathbf{p}=m\mathbf{v}$

and the energy by

$E=\frac{1}{2}mv^2$

where m is the mass of the particle and v is the vector velocity of the particle.

[edit] Non-Relativistic Quantum Free Particle

The Schrödinger equation for a free particle is:

$- \frac{\hbar^2}{2m} \nabla^2 \ \psi(\mathbf{r}, t) = i\hbar\frac{\partial}{\partial t} \psi (\mathbf{r}, t)$

The solution for a particular momentum is given by a plane wave:

$\psi(\mathbf{r}, t) = e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$

with the constraint

$\frac{\hbar^2 k^2}{2m}=\hbar \omega$

where r is the position vector, t is time, k is the wave vector, and ω is the angular frequency. Since the integral of ψψ^* over all space must be unity, the wave function must first be normalized. This is not a problem for general free particles somewhat localized in momentum and position. (See particle in a box for a further discussion.)

The expectation value of the momentum p is

$\langle\mathbf{p}\rangle=\langle \psi |-i\hbar\nabla|\psi\rangle = \hbar\mathbf{k}$

The expectation value of the energy E is

$\langle E\rangle=\langle \psi |i\hbar\frac{\partial}{\partial t}|\psi\rangle = \hbar\omega$

Solving for k and ω and substituting into the constraint equation yields the familiar relationship between energy and momentum for non-relativistic massive particles

$\langle E \rangle =\frac{\langle p \rangle^2}{2m}$

where p=|p|. The group velocity of the wave is defined as

$\left.\right. v_g= \frac{d\omega}{dk} = \frac{dE}{dp} = v$

where v is the classical velocity of the particle. The phase velocity of the wave is defined as

$\left.\right. v_p=\frac{\omega}{k} = \frac{E}{p} = \frac{p}{2m} = \frac{v}{2}$

A general free particle need not have a specific momentum or energy. In this case, the free particle wavefunction may be represented by a superposition of free particle momentum eigenfunctions:

$\left.\right. \psi(\mathbf{r}, t) = \int A(\mathbf{k})e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)} d\mathbf{k}$