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In physics a free particle is a particle that is never under the influence of an
external force
Classical Free Particle
The classical free particle is characterized simply by a fixed velocity. The momentum is
given by
- <math>\mathbf{p}=m\mathbf{v}<math>
and the energy by
- <math>E=\frac{1}{2}mv^2<math>
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:
- <math>
- \frac{\hbar^2}{2m} \nabla^2 \ \psi(\mathbf{r}, t) =
i\hbar\frac{\partial}{\partial t} \psi (\mathbf{r}, t)
<math>
The solution for a particular momentum is given by a plane wave:
- <math>
\psi(\mathbf{r}, t) = e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}
<math>
with the constraint
- <math>
\frac{\hbar^2 k^2}{2m}=\hbar \omega
<math>
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
- <math>
\langle\mathbf{p}\rangle=\langle \psi |-i\hbar\nabla|\psi\rangle = \hbar\mathbf{k}
<math>
The expectation value of the energy E is
- <math>
\langle E\rangle=\langle \psi |i\hbar\frac{\partial}{\partial t}|\psi\rangle = \hbar\omega
<math>
Solving for k and ω and substituting into the constraint equation yields the familiar relationship between energy and momentum for non-relativistic massive particles
- <math>
\langle E \rangle =\frac{\langle p \rangle^2}{2m}
<math>
where p=|p|. The group velocity of the wave is defined as
- <math>\left.\right.
v_g= d\omega/dk = dE/dp = v
<math>
where v is the classical velocity of the particle.
The phase velocity of the wave is defined as
- <math>\left.\right.
v_p=\omega/k = E/p = p/2m = v/2
<math>
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:
- <math>\left.\right.
\psi(\mathbf{r}, t) = \int
A(\mathbf{k})e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}
d\mathbf{k}
<math>
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
- <math>
\nabla^2\psi-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\psi
= \frac{m^2c^2}{\hbar^2}\psi
<math>
with the same solution as in the non-relativistic case:
- <math>
\psi(\mathbf{r}, t) = e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}
<math>
except with the constraint
- <math>
-k^2+\frac{\omega^2}{c^2}=\frac{m^2c^2}{\hbar^2}
<math>
Just as with the non-relativistic particle, we have for energy and momentum:
- <math>
\langle\mathbf{p}\rangle=\langle \psi |-i\hbar\nabla|\psi\rangle = \hbar\mathbf{k}
<math>
- <math>
\langle E\rangle=\langle \psi |i\hbar\frac{\partial}{\partial t}|\psi\rangle = \hbar\omega
<math>
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:
- <math>\left.\right.
\langle E \rangle^2=m^2c^4+\langle p \rangle^2c^2
<math>
For massless particles, we may set m=0 in the above equations. We then recover the relationship between energy and momentum for massless particles:
- <math>\left.\right.
\langle E \rangle=\langle p \rangle c
<math>
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