A one-dimensional
harmonic oscillator has an infinite series of equally spaced energy
states, with \(\varepsilon_n = n\hbar\omega\), where \(n\) is an
integer \(\ge 0\), and \(\omega\) is the classical frequency of the
oscillator. We have chosen the zero of energy at the state \(n=0\)
which we can get away with here, but is not actually the zero of
energy! To find the true energy we would have to add a
\(\frac12\hbar\omega\) for each oscillator.
Show that for a harmonic oscillator the free energy is
\begin{equation}
F = k_BT\log\left(1 - e^{-\frac{\hbar\omega}{k_BT}}\right)
\end{equation} Note that at high temperatures such that
\(k_BT\gg\hbar\omega\) we may expand the argument of the logarithm
to obtain \(F\approx k_BT\log\left(\frac{\hbar\omega}{kT}\right)\).
From the free energy above, show that the entropy is
\begin{equation}
\frac{S}{k_B} =
\frac{\frac{\hbar\omega}{kT}}{e^{\frac{\hbar\omega}{kT}}-1}
- \log\left(1-e^{-\frac{\hbar\omega}{kT}}\right)
\end{equation}
Entropy of a simple harmonic oscillatorHeat capacity of a simple harmonic oscillator
This entropy is shown in the nearby figure, as well
as the heat capacity.
