# Partition function for a classical two-particle oscillator: Infinite limits?

Previous page: What is the partition function of a classical harmonic oscillator? Whenever I see the partition function of a classical two-particle oscillator,

$$Z(\beta) \, = \, \int dx \int dp ~ e^{-\beta H(x,p)} \, = \, \int dx ~ e^{-\beta k x^2} \int dp ~ e^{-\beta \frac{p^2}{2m}},$$

the $$x$$ and $$p$$ integration limits go from $$(-\infty,\infty)$$. I assume that these limits come from the approximation that the energy of the system can continuously range from $$[0,\infty)$$.

But I'm wondering if that's the whole story. Since

$$E \, = \, \frac{p^2}{2m} + kx^2 \, = \, \mathrm{constant}$$,

we find that $$x$$ and $$p$$ actually depend on each other. We can see this more easily when I rewrite the equation as

$$x \, \rightarrow \, x(p) \, = \, \pm \, \sqrt{\frac{1}{k}\left[E - \frac{p^2}{2m}\right]}$$ or $${p \, \rightarrow \, p(x) \, = \, \pm \, \sqrt{2m\left[E - k x^{2}\right]}}$$.

Note that when $$p = 0$$ we get the maximum $$x$$ limits $$x_{max} \, = \, \pm \, \sqrt{E/k}$$.

$$Z(\beta) \, = \, \int_{-x_{max}}^{+x_{max}} dx \int_{-p(x)}^{+p(x)} dp ~ e^{-\beta k x^2} ~ e^{-\beta \frac{p^2}{2m}}$$?

I used these limits when calculating the average energy. The result is more complicated, of course, compared to the average energy calculated using the infinite limits. As $$x_{max} \rightarrow \infty$$, the average energy using the dependent limits approaches the average energy using the infinite integrals.

Is there a reason why we don't use dependent limits?

• You're missing a factor of 1/2 in the potential energy term $k x^2$. Also, where are the two particles here? This looks like the partition function for a single harmonic oscillator. Commented Dec 25, 2023 at 3:15
• Yes, I didn't include the factor of 1/2, which I believe is standard. It doesn't change the results of this question. It is a simple harmonic oscillator, which can be viewed in a number of ways, including two particles on a spring or two atoms in a molecule. Commented Dec 25, 2023 at 5:05

The dependence of p on x (or the other way around) only comes from the condition of constant energy. This is a natural condition for a microcanonical ensemble, but it is wrong in the canonical ensemble. Remember that the canonical ensemble corresponds to the physical situation of a system (the harmonic oscillator) in contact with a thermostat at a fixed temperature $$T$$. Under such a condition, the oscillator's energy fluctuates, and every possible value of the energy is allowed (although with different probability). As a consequence, there is no relation between position and momentum, and integrations are over the unrestricted phase space.

• Excellent. Very clear and concise. Thanks. BTW, I wouldn't have thought about using dependent limits for a many-particle system because all of the interactions would make setting it up impossible. I guess the difference between a microcanonical and canonical ensemble still holds even for two particles. Commented Dec 24, 2023 at 19:21
• @NickElias I agree that the difference between a microcanonical and canonical ensemble still holds even for two particles. Commented Dec 25, 2023 at 0:48