**The one-dimensional harmonic
oscillator in thermodynamic equilibrium**

An important case of a statistical mixture is that of a system in thermodynamic
equilibrium with a heat reservoir at temperature T.
The various possible dynamical
states are the eigenstates of the Hamiltonian H.
The statistical weight of a given eigenstate depends upon the corresponding eigenvalue of H.
It is proportional to
the Boltzmann factor exp(-E/(kT)), in which E
is the eigenvalue of H and k is the Boltzmann constant. (k = 1.38*10^{-23}J/K)

A system in thermodynamic equilibrium can be represented by the density operator

ρ = ∑_{n}p_{n}ρ^{n} = ∑_{n}p_{n}|n><n|,

where {|n>} are the eigenstates of H,

H|n> = E_{n}|n>.

Then p^{n} = Nexp(-E_{n}/(kT)), where N is a
normalization constant to make the total probability equal to one.

ρ = ∑_{n}Nexp(-E_{n}/(kT))|n><n| = ∑_{n}Nexp(-H/(kT))|n><n|
= Nexp(-H/(kT)),

since ∑_{n}|n><n| = I.

We need Tr(ρ) = 1, Tr(Nexp(-H/(kT)) = 1, Tr(exp(-H/(kT))
= N^{-1},

N^{-1} is called the **partition function Z**,
and we write ρ = Z^{-1}exp(-H/(kT)).

Let us calculate the partition function for the harmonic oscillator.

Z = ∑_{n}<n|exp(-E_{n}/(kT)|)n> = ∑_{n}exp(-(n+½)ħω/(kT))
= exp(-ħω/(2kT)) ∑_{n}exp(-(nħω/kT).

(1 - x)^{-1} = 1 + x + x^{2} + x^{3} + ... =
∑_{n}x^{n}.

Therefore ∑_{n}exp(-(nħω/(kT) ) = ∑_{n}exp(-(ħω/(kT))^{n} =
(1 - exp(-(ħω/(kT))^{-1},

and

Z = exp(-ħω/(2kT))/[1 - exp(-ħω/(kT))].

Now let us calculate the average energy.

<H> = Tr(ρH) = Z^{-1}Tr(exp(-H/(kT)) H) = Z^{-1}∑_{n}(n+½)ħω exp(-(n+½)ħω/(kT))
= kT^{2}(1/Z)dZ/dT,

since

dZ/dT = d/dT ∑_{n}exp(-(n+½)ħω/(kT)) = (1/(kT^{2}))∑_{n}(n+½)ħω exp(-(n+½)ħω/(kT)).

Using Z = exp(-ħω/(2kT))/[1 - exp(-ħω/(kT))]
we
find a simpler expression for dZ/dT.

dZ/dT = (ħω/(2kT^{2})) exp(-ħω/(2kT))/[1 - exp(-ħω/(kT))]
+

(ħω/(kT^{2}))exp(-ħω/(kT))exp(-ħω/(2kT))/[1 - exp(-ħω/(kT))]^{2
}= Z(ħω/(2kT^{2})) + Z(ħω/(kT^{2})))exp(-ħω/(kT))/[1 - exp(-ħω/(kT))].

We now have <H> = ½ħω + ħω/[exp(ħω/(kT)) - 1].

This is Planck’s formula (to within a constant ½ħω) for the average energy of a quantized
oscillator.

The energy of a classical one dimensional oscillator is E(x,p) = ½p^{2}/m
+ ½mω^{2}x^{2}.
The mean energy of such an oscillator in
thermodynamic equilibrium at temperature T is

<E> = ∫_{∞}^{∞}∫_{∞}∞E(x,p)exp(-E(x,p)/(kT))dxdp/∫_{∞}^{∞}∫_{∞}∞exp(-E(x,p)/(kT))dxdp
= kT.

Temperature | QM oscillator <H> | Classical oscillator <E> |

T --> 0 | ½ħω | 0 |

kT << ħω | ½ħω + ħω exp(-ħω/(kT)) | kT |

kT >> ħω | ½ħω + ħω/(ħω/(kT)) ≈ kT | kT |

Note: For the three-dimensional harmonic oscillator in thermodynamic equilibrium the mean energy <H> is three times that of a one-dimensional oscillator with the same frequency.