# Calculating the mean separation using a canonical ensemble [closed]

I am having trouble with the following physics question.

Two atoms, each of mass $$m$$, interact with each other by a force that can be derived by the mutual potential energy equation given by

$$U = U_{0}\left[\left(\frac{a}{x}\right)^{12} - 2\left(\frac{a}{x}\right)^{6}\right],$$

where $$a$$ is a constant, and $$x$$ is the separation between two particles. The particles are in contact with a heat bath at temperature $$T$$, low enough so that $$kT << U_{0}$$, but the temperature is high enough for statistical mechanical techniques to be applied. I need help calculating the mean separation $$x(T)$$ of the particles.

I was able to recognize the above equation as a Lennard-Jones Potential equation. Furthermore, since the system is in contact with a heat bath, I am aware that we should use a canonical ensemble approach. I am having some trouble making progress on this question. Help would be very appreciated :)

• Have you attempted to determine the partition function? Commented Sep 30, 2018 at 4:58
• You only have two particles, but you have a continuum of possible energy levels so your expression for the partition function should be an integral over $x$. What you have written is too simple Commented Sep 30, 2018 at 5:12
• Also, you'd square it, not multiply by 2 to go from the single to the two particle partition function Commented Sep 30, 2018 at 5:13
• Something like $Z=\int\int dr_1 dr_2 e^{-\beta U(|r_1-r_2|)}$ Commented Sep 30, 2018 at 5:15
• Sorry for all the comments! It seems like they specify the mass $m$ of each particle which suggests kinetic energy must also be accounted for. Commented Sep 30, 2018 at 5:16

This is a very difficult problem in general, but the prompt says to take $$kT\ll U_0$$ which suggests we treat the potential like a quadratic well. This means that the system asks like a quantum harmonic oscillator which we know has $$\langle x \rangle=0$$. This means the atoms will spend equal times closer than equilibrium and farther than equilibrium so the average will just be the equilibrium separation.
We can find the equilibrium position $$x_0$$ by requiring $$\left.\frac{\partial U}{\partial x}\right|_{x_0}=0\to U_0 \left(\frac{12 a^6}{x_0^7}-\frac{12 a^{12}}{x_0^{13}}\right)=0\to x_0=a$$ This says $$\langle x \rangle=a$$. However this didn't really make use of any statistical mechanics. From here you could get the partition function if you like by expanding $$U(x)$$ about $$x=a$$ gives (skipping lots of algebra) $$U(x)\approx U_0\left[\frac{36 x^2}{a^2}-1\right]\to k=\frac{72U_0}{a^2}$$ So we have $$H\approx \frac{p^2}{2m}+\frac{1}{2}\frac{72U_0}{a^2} x^2-U_0$$ The extra $$-U_0$$ can just be absorbed into the eigenvalues and then we have a quantum harmonic oscillator as promised. This gives energy eigenvalues $$\varepsilon_n=\hbar\omega(n+1/2)-U_0$$ with $$\omega=\sqrt{k/m}=\sqrt{72U_0/ma^2}$$.
From here we can use the standard techniques to derive the partition function $$Z = \sum_{n=0}^\infty e^{-\beta \varepsilon_n}=\sum_{n=0}^\infty e^{-\beta \left(\hbar\omega(n+1/2)-U_0\right)}=\frac{e^{-\beta\left(\hbar\omega/2-U_0\right)}}{1-e^{-\beta\hbar\omega}}$$ From which any quantity of interest can be derived.
If your professor really wants to see $$x(T)$$ where $$x(T)\neq a$$ i.e. having a real functional dependence on $$T$$, then you might need to do your Taylor series to higher order than 2 in which case you'd get anharmonic terms and be forced (I believe) to do some sort of perturbation to get the energy levels and from there the partition function. I won't go into that here but you can find resources online about how to potentially go about that.