Free particle

In physics, a free particle is a particle that in some sense, is not bound. In the classical case, this is represented with the particle not being influenced by any external force.

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.

Non-Relativistic Quantum Free Particle

The Schroedinger 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, there will be a problem normalizing this momentum eigenfunction. This will not be a problem for a general free particle which is 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

$\left.\right.v_{g}=d\omega /dk=dE/dp=v$

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

$\left.\right.v_{p}=\omega /k=E/p=p/2m=v/2$

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:

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

where the integral is over all k-space.

Relativistic free particle (Klein-Gordon equation)

If the particle is charge-neutral and spinless, and relativistic effects cannot be ignored, we may use the Klein-Gordon equation to describe the wave function. The Klein-Gordon equation for a free particle is written

$\nabla ^{2}\psi -{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\psi ={\frac {m^{2}c^{2}}{\hbar ^{2}}}\psi$

with the same solution as in the non-relativistic case:

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

except with the constraint

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

Just as with the non-relativistic particle, we have for energy and momentum:

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

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

Except that now when we solve for k and ω and substitute into the constraint equation, we recover the relationship between energy and momentum for relativistic massive particles:

$\left.\right.\langle E\rangle ^{2}=m^{2}c^{4}+\langle p\rangle ^{2}c^{2}$

For massless particles, we may set m=0 in the above equations. We then recover the relationship between energy and momentum for massless particles:

$\left.\right.\langle E\rangle =\langle p\rangle c$