# Example: The Boltzmann distribution

## Understanding the situation

In our previous readings (Fluctuations and The Boltzmann distribution), we have read how a degree of freedom interacting in a thermal bath (imbedded in a fluid at some temperature, $T$) will be continually exchanging energy with the elements of the fluid and, as a result, its energy will fluctuate.

At thermal equilibrium, the average energy in each degree of freedom will be the same and the fluctuations will be governed by the Boltzmann factor, $e^{-E/k_BT}$. To see how this works, let's do an example with one degree of freedom.

Let's consider a gas of diatomic molecules where the interaction of the atoms in the molecule looks like a Lennard-Jones potential. Potential energies like the L-J potential look a lot like the potential energy of a spring. Give them a little energy and the atoms will vibrate back and forth. The coordinate in the graph of the atom-atom PE at the right represents the separation of the two atoms ($r$) and the vertical coordinate represents the interaction potential energy of the two atoms, ($U$). If the total energy of the atoms were given by the red line, $E_1$, the atoms would oscillate: that their separation would go back and forth along the red line to the walls of the PE well.

But atoms and molecules are quantum systems. They behave like an oscillator (as if connected by a spring), but their energy can not be equal to any value; they can only take on specific (quantized) values — for example the values $E_1$ and $E_2$ shown. (See Quantum oscillators - discrete states.)

If one of these molecules is in a thermal bath, we would expect that it most likely would start out in the state $E_1$, its ground state (the lowest energy possible). But through collisions with other molecules, it would sometimes get bumped up into its excited state,$E_2$. Let's see by how much.

## Presenting a sample problem

Suppose that a two-atom molecule interacts by an attractive potential that has a number of excited states. The ground state has an energy $E_1$. If the molecule gets excited into a state $E_2$, there is an enhanced likelihood that it will undergo a chemical reaction. If the system is in a thermal bath at temperature $T$, what is the probability that the molecule will be found in the excited state $E_2$ compared to the probability that it will be found in its ground state? If the system is at room temperature (so $k_BT$ ~ 250 meV) and $E_2 - E_1 = 1.65 \mathrm{eV}$, what is the relative probability of finding the excited state compared to the ground state?

## Solving this problem

Since the probability of finding the molecule is an energy state $E$ is proportional to $e^{-E/k_BT}$, the ratio of the probabilities of finding the molecule in any state of energy $E_n$  is

$$P_n = P_0 e^{-E_n/k_BT}$$

where $P_0$ is the normalizing factor. Then the ratio of the probabilities that the system is in the state $E_2$ compared to in the state $E_1$ is

$$\frac{P_2}{P_1} = \frac{P_0 e^{-E_2/k_BT}}{P_0 e^{-E_1/k_BT}}$$

The $P_0$ cancels out and since $e^a/e^b = e^{a-b}$ our ratio is

$$\frac{P_2}{P_1} = e^{-(E_2 - E_1)/k_BT} = e^{-\Delta E/k_BT}$$

For the numbers we are given, $ΔE$ = 1650 meV (milli-eV) and $ΔE/k_BT = (1650/250) = 6.6$. Our ratio of probabilities is therefore

$$\frac{P_2}{P_1} = e^{-6.6} = 1.4 x 10^{-3}$$.

or about 0.1% of the time.

Joe Redish 2/15/16

Article 612