While reading up on statistical physics, I am going through the calculation of the partition function of the harmonic oscillator in the microcanonical ensemble. The result for the partition function is:

$\Omega(E) = \frac{E_o}{\hbar \omega}$,

i.e. all one-dimensional harmonic oscillators of the same frequency $\omega$ have the same number of accessible microstates. Consequently, as $\omega$ grows, $\Omega(E)$ goes to zero. This is somewhat counter-intuitive to me. I would have expected that an oscillator with a higher frequency $\omega$ would have more accessible microstates than one with a lower frequency. What is wrong with my intuition?

Note that I am going through the classical calculation, the partition for the harmonic oscillator is then given as:

$\Omega(E) = \frac{E_0}{h}\int dp \int dx \delta \left( \frac{p^2}{2m} + \frac{1}{2}k x^2 - E\right) $


1 Answer 1


The classical calculation goes like this. Consider the Hamiltonian of the classical harmonic oscillator $$ H(q,p) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2 $$ The expression $\Omega(E)$ refers to the volume of the region in phase space corresponding to classical states $(q,p)$ with energies less than or equal to $E$, namely it is the volume of the interior of the ellipse specified by $$ \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2\leq E $$ The volume (area in this case) of the interior of this ellipse turns out to be $E/\hbar\omega$. One way to get intuition for this is that since $\omega^2$ multiplies $x^2$ in the hamiltonian, the ellipse constraint written above is satisfied only by points in phase space with smaller values of $x$ when you increase omega. This should become even more transparent if you actually perform the ellipse area calculation.

  • $\begingroup$ Why does the delta turn into an inequality? I thought we wanted to integrate along the (x, p) surface for which H(x, p) exactly equals E... $\endgroup$
    – Kris
    Commented Feb 13, 2020 at 17:01

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