The partition function, degeneracy and characteristic temperature

q counts the states a molecule can actually reach

Intuition

A cinema has seats at many prices. On a cold, poor night only the cheapest seats fill. On a rich night nearly every seat fills. The partition function is like “how many seats are in play tonight”. It starts at 1 when only the front row is used and climbs towards the total number of seats.

The sum in the denominator of the Boltzmann distribution is the partition function:

$q = \sum_i e^{-\beta\varepsilon_i}$

Each state adds a score equal to its population relative to the ground state:

$e^{-\beta\varepsilon_i} = \frac{n_i}{n_0}$

So $q$ grows when excited states fill, which means it grows with $T$. If $T \to 0$, then $q \to 1$, because all molecules are in the ground state. If $T \to \infty$, then $q \to$ the number of molecular states. The lecture’s summary is that $q$ is, roughly, the number of thermally accessible states per molecule.

Try it: worked example

A harmonic oscillator has hν equal to 1/β. Find q and the probabilities of the first three levels. Then bring the levels closer together (Case 2).

Show answer and steps

AnswerCase 1: q = 1.58, p0 = 0.633, p1 = 0.233, p2 = 0.086. Case 2: q = 2.54, p0 = 0.394, p1 = 0.239, p2 = 0.145.

1. Level $i$ has energy $\varepsilon_i = ih\nu$. Since $h\nu = \beta^{-1}$, this is $\varepsilon_i = i\beta^{-1}$.
2. So $\beta\varepsilon_i = i$ and each score is $e^{-\beta\varepsilon_i} = e^{-i}$.
3. Add the scores: $q = \sum_{i=0}^{\infty} e^{-i} = 1 + e^{-1} + e^{-2} + \cdots$. This is a geometric series with ratio $e^{-1}$.
4. Its sum is $q = \dfrac{1}{1 - e^{-1}} = 1.58$.
5. Level 0: $p_0 = \dfrac{e^{-0}}{1.58} = \dfrac{1}{1.58} = 0.633$.
6. Level 1: $p_1 = \dfrac{e^{-1}}{1.58} = 0.233$.
7. Level 2: $p_2 = \dfrac{e^{-2}}{1.58} = 0.086$. Roughly 1.58 states are appreciably occupied.
8. Case 2, with levels closer together, has $q = 2.54$: $p_0 = \dfrac{e^{-0}}{2.54} = 0.394$, $p_1 = \dfrac{e^{-1/2}}{2.54} = 0.239$, $p_2 = \dfrac{e^{-1}}{2.54} = 0.145$.
9. Closer levels mean more states are reachable, so $q$ is bigger and each probability is smaller.

Bulk properties come from derivatives of q

Why bother with $q$? Because it holds all the bulk thermodynamics in terms of single-molecule properties. Take the total energy. The plain idea: the energy is just “how many molecules at each level, times that level’s energy”, added up. The slide shows how to get it from the slope of $q$.

Start with $E = \sum_i \varepsilon_i n_i$ and put in $n_i = N e^{-\beta\varepsilon_i}/q$:

$E = N\sum_i \varepsilon_i\frac{e^{-\beta\varepsilon_i}}{q}$

Differentiating $q$ with respect to $\beta$ brings down a factor $-\varepsilon_i$ from each exponent:

$\frac{dq}{d\beta} = -\sum_i \varepsilon_i e^{-\beta\varepsilon_i}$

So the sum in $E$ is $-\dfrac{1}{q}\dfrac{dq}{d\beta}$, which equals $-\dfrac{d\ln q}{d\beta}$. Then

$E = -\frac{N}{q}\frac{dq}{d\beta} = -N\frac{d\ln q}{d\beta} = NkT^2\frac{d\ln q}{dT}$

The last step swaps $\beta$ for $T$ using $\beta = 1/kT$, which gives $d\beta/dT = -1/kT^2$. Usually the important thermodynamic functions are derivatives of $q$.

Degeneracy: several states can share one level

Intuition

Two seats in the same row cost the same. In the partition function you must count both seats, not one, because molecules can sit in either.

Spectroscopy sees energy levels, and several states may share one level. The number of states with energy $\varepsilon$ is the degeneracy $g(\varepsilon)$. The slide’s ladder has levels 0, 1, 1’, 2, 3, where the first excited level is doubly degenerate. The partition function can be summed over states or over levels:

$q = \sum_i e^{-\beta\varepsilon_i} \ \text{(states)}, \qquad q = \sum_{\varepsilon_i} g(\varepsilon_i)\,e^{-\beta\varepsilon_i} \ \text{(levels)}$

The population ratio of two levels picks up a degeneracy ratio:

$\frac{n(\varepsilon_j)}{n(\varepsilon_i)} = \frac{g(\varepsilon_j)}{g(\varepsilon_i)}\,e^{-\beta(\varepsilon_j-\varepsilon_i)}$

Pitfall

When you sum over levels you must include $g$. Forget it and a doubly degenerate level is counted as if it had half its real population. With $g = 2$ for the upper level, $g = 1$ for the lower, and a gap of $kT$, the ratio is $2e^{-1} = 0.74$, not $e^{-1} = 0.37$. (The numbers here come from the lecture formula, not from a slide.)

The characteristic temperature says what hot and cold mean

Is 300 K hot? For a system with a big energy gap, no. Whether it is hot depends on the system’s own energy spacing. The characteristic temperature of a system with typical spacing $\Delta E$ is

$\Theta = \frac{\Delta E}{k}$

and the population of an upper level relative to a lower one is

$\frac{n_j}{n_i} \sim e^{-\Delta E/kT} = e^{-\Theta/T}$

$\Theta$ is roughly the lowest temperature at which excited states begin to fill. If $T \gg \Theta$ many states are occupied and the system is “hot”. If $T \ll \Theta$ only the ground state is significantly occupied and the system is “cold”.

Try it: worked example

A system has a characteristic temperature of 300 K. Find the first-excited to ground-state population ratio at 300 K, 3000 K and 30 K.

Show answer and steps

Answer0.37 at 300 K, 0.90 at 3000 K, 4.5 x 10^-5 at 30 K.

1. Use $n_j/n_i \sim e^{-\Theta/T}$ with $\Theta = 300$ K.
2. At 300 K: $\Theta/T = 1$, so the ratio is $e^{-1} = 0.37$.
3. At 3000 K: $\Theta/T = 0.1$, so the ratio is $e^{-0.1} = 0.90$. The first excited state and many others are significantly occupied.
4. At 30 K: $\Theta/T = 10$, so the ratio is $e^{-10} = 4.5 \times 10^{-5}$. Only the ground state is appreciably occupied.

Exam detail

Lecture 13 summary slide: temperature is a property of the most probable configuration; zeroth law; the Kelvin scale and absolute zero; the partition function is roughly the number of thermally accessible states; degeneracy should be taken into account; the characteristic temperature tells you whether something is at a high or a low temperature. The numbers 1.58, 0.633, 0.233, 0.086 and 2.54, 0.394, 0.239, 0.145 are from the oscillator example.

Recall

What is the partition function in one phrase, and what are its limits at T = 0 and T = infinity?

Roughly the number of thermally accessible states per molecule. It tends to 1 at $T = 0$ and to the total number of molecular states as $T \to \infty$.