What is the boltzmann equilibrium distribution of two diffusing particles connected by a 'spring'?

Question

Imagine you are tyring to draw the Boltzmann equilibrium distribution for two diffusing particles connected by a spring. In this case, how do you calculate the partition function?

Here is where I am at:

If we have two particles separated by distance $r$ and spring potential $\phi(r) =\frac{K}{2}(r-l)^2$, with constant $l$, we get the Boltzmann Distribution given by $\rho(r) =\frac{1}{Z}\exp[-\phi(r)/k_BT]$ with the partition function given by $Z =\sum\limits_{i}g_i\exp[-\phi(r)/k_BT]$

Heres where I am lost:

To calculate the partition function we need to know the number of energy states and the number of instances of each state. I imagine that the energy states are distinguished by different values of $r$, but not sure.

My questions are what are the energy states?What is $g_i$ in this case?

I appreciate any help. Thanks!

Answer 1

First of all, you have to realize that you are considering a classical continuous system. Thus, the partition function is an integral instead of a sum. For a single particle, this is: $$ Z = \frac{1}{h^3}\int e^{-\beta H(\mathbf{p},\mathbf{q})}\mathrm{d}^3p\mathrm{d}^3q \quad, $$ where $H(\mathbf{p},\mathbf{q})$ is the Hamiltonian with $\mathbf{p}$ and $\mathbf{q}$ being the canonical momentum and position, respectively. The prefactor $1/h^3$ corresponds to the $g_i$ in your question. As a common convention, $h$ is chosen to be Planck's constant.

In the case of two particles connected by a spring, the Hamiltonian is: $$ H(\mathbf{p}_1,\mathbf{p}_2,\mathbf{q}_1,\mathbf{q}_2) = \frac{\mathbf{p}_1^2}{2m_1} + \frac{\mathbf{p}_2^2}{2m_2} + \frac{K}{2}\left(\lvert\mathbf{q}_1 - \mathbf{q}_2\rvert - l \right)^2 $$ This Hamiltonian nicely separates, when we calculate the partition function: $$ Z = \frac{1}{h^6}\int e^{-\beta\frac{\mathbf{p}_1^2}{2m_1}}\mathrm{d}^3p_1\int e^{-\beta\frac{\mathbf{p}_2^2}{2m_2}}\mathrm{d}^3p_2\iint e^{-\beta\frac{K}{2}(\lvert\mathbf{q}_1 -\mathbf{q}_2\rvert - l)^2}\mathrm{d}^3q_1\mathrm{d}^3q_2 $$

The first two terms correspond to the kinetic energy. They contribute a factor of $(2\pi m_ik_BT)^{3/2}$ each. Thus, the partition sum is

$$ Z = \frac{1}{h^6} \left(2\pi k_BT\sqrt{m_1m_2}\right)^3\iint e^{-\beta\frac{K}{2}(\lvert\mathbf{q}_1 - \mathbf{q}_2\rvert -l)^2}\mathrm{d}^3q_1\mathrm{d}^3q_2 $$

To calculate the contribution of the last term in the Hamiltonian, we can first use the translational invariance, considering position $\mathbf{q}_1$ and the relative position $\mathbf{r} = \mathbf{q}_1 - \mathbf{q}_2$. The energy term depends only on $\mathbf{r}$. $$ \frac{K}{2}\left(\lvert\mathbf{q}_1-\mathbf{q}_2\rvert -l\right)^2 = \frac{K}{2}\left(\lvert\mathbf{r}\rvert - l\right)^2 $$

and the translational invariance of $\mathbf{q}_1$ adds factor $V$ which is the available volume:

$$ \iint e^{-\beta\frac{K}{2}(\lvert\mathbf{q}_1-\mathbf{q}_2\rvert - l)^2}\mathrm{d}^3q_1\mathrm{d}^3q_2 = \int\mathrm{d}^3q_1\int e^{-\beta\frac{K}{2}(\lvert\mathbf{r}\rvert -l)^2}\mathrm{d}^3r = V\int e^{-\beta\frac{K}{2}(\lvert\mathbf{r}\rvert -l)^2}\mathrm{d}^3r $$

In the next step, we can use the rotational symmetry of the spring potential and convert to spherical coordinates:

$$ \int e^{-\beta\frac{K}{2}(\lvert\mathbf{r}\rvert -l)^2}\mathrm{d}^3\mathbf{r} = \int\limits_0^\infty\int\limits_0^\pi\int\limits_0^{2\pi} e^{-\beta\frac{K}{2}(r - l)^2} r^2\sin{\theta}\,\mathrm{d}r\mathrm{d}\phi\mathrm{d}\theta = 4\pi \int\limits_0^\infty r^2 e^{-\beta\frac{K}{2}(r - l)^2}\mathrm{d}r = 4\pi I $$

where the integral $I$ can be solved with standard techniques (even though it takes some effort). Thus, the total partition function is $$ Z = \frac{4\pi}{h^6} V \left(2\pi k_BT\sqrt{m_1m_2}\right)^3\cdot I $$