The results of the quantum harmonic oscillator can be used to look at the equilibrium situation for a quantum ideal gas in a harmonic trap, which is a harmonic potential containing a large number of particles that do not interact with each other except for instantaneous thermalizing collisions. This situation is of great practical importance since many experimental studies of Bose gases are conducted in such harmonic traps.
Using the results from either Maxwell–Boltzmann statistics, Bose–Einstein statistics or Fermi–Dirac statistics we use the Thomas–Fermi approximation (gas in a box) and go to the limit of a very large trap, and express the degeneracy of the energy states (
gi
For massive particles in a harmonic well, the states of the particle are enumerated by a set of quantum numbers
[nx,ny,nz]
E=\hbar\omega\left(nx+ny+nz+3/2\right)~~~~~~~~ni=0,1,2,\ldots
Suppose each set of quantum numbers specify
f
f
f=2
n
E
g=f | n3 | =f |
6 |
(E/\hbar\omega)3 | |
6 |
f
E
E+dE
dg= | 1 |
2 |
| |||||
fn | ~ |
1 | |
2 |
~\beta3E2dE
Notice that in using this continuum approximation, we have lost the ability to characterize the low-energy states, including the ground state where
ni=0
Without using the continuum approximation, the number of particles with energy
\epsiloni
Ni=
gi | |
\Phi |
| for particles obeying Maxwell–Boltzmann statistics | |||||
-1 | for particles obeying Bose–Einstein statistics | |||||
+1 | for particles obeying Fermi–Dirac statistics |
with
\beta=1/kT
k
T
\mu
dN
E
E+dE
dN=
dg | |
\Phi |
We are now in a position to determine some distribution functions for the "gas in a harmonic trap." The distribution function for any variable
A
PAdA
A
A
A+dA
PA~dA=
dN | |
N |
=
dg | |
N\Phi |
It follows that:
\intAPA~dA=1
Using these relationships we obtain the energy distribution function:
PE~dE=
1 | \left( | |
N |
f | \right)~ | |
(\hbar\omega\beta)3 |
1 | |
2 |
\beta3E2 | |
\Phi |
dE
The following sections give an example of results for some specific cases.
For this case:
\Phi=e\beta(E-\mu)
Integrating the energy distribution function and solving for
N
N=
f | |
(\hbar\omega\beta)3 |
~e\beta\mu
Substituting into the original energy distribution function gives:
PE~dE=
\beta3E2e-\beta | |
2 |
dE
For this case:
\Phi=e\beta/z-1
z
z=e\beta\mu
Integrating the energy distribution function and solving for
N
N=
f | |
(\hbar\omega\beta)3 |
~Li3(z),
Lis(z)
\zeta(3)
z
\beta
\beta
\betac
z=1
N=
f | ||||||
|
~\zeta(3).
The temperature at which
\beta=\betac
N= | g0z | + |
1-z |
f | |
(\hbar\omega\beta)3 |
~Li3(z)
For this case:
\Phi=e\beta(E-\mu)+1
Integrating the energy distribution function gives:
1= | f |
(\hbar\omega\beta)3 |
~\left[-Li3(-z)\right]
Lis(z)