Back to the on-screen lesson · Print

Average energies and why modes freeze

Averages from the Boltzmann distribution, $\bar{E} = -\partial\ln Z/\partial\beta$, the two-level system and the quantum oscillator, and equipartition as the high-temperature limit of quantum statistics.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to find the average energy of a system from its partition function, derive the average energy of a two-level system and a quantum oscillator, and decide whether a mode is frozen, partly active or classical at a given temperature.

2. What you already have

You can find the probability of any state from its Boltzmann factor and the partition function. You also know, from the first lesson, that equipartition gives each quadratic term $\tfrac{1}{2}kT$ but fails for vibrations at room temperature. This lesson computes averages from the Boltzmann distribution and shows exactly where equipartition holds and why it fails.

3. Words for this lesson

4. Averages from the partition function

Any average over a system in contact with a reservoir is a sum of values weighted by their Boltzmann probabilities: $\bar{X} = \sum_s X(s)e^{-\beta E(s)}/Z$, with $\beta = 1/kT$. For the energy there is a shortcut. Since $\partial Z/\partial\beta = -\sum_s E(s)e^{-\beta E(s)}$,

$$\bar{E} = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} = -\frac{\partial\ln Z}{\partial\beta}.$$

Find $Z$, take its logarithm, differentiate: no second sum needed.

The quantum harmonic oscillator. Its levels are $0, \epsilon, 2\epsilon, \ldots$ above the ground state, with $\epsilon = hf$. The partition function is a geometric series:

$$Z = 1 + e^{-\beta\epsilon} + e^{-2\beta\epsilon} + \cdots = \frac{1}{1 - e^{-\beta\epsilon}},$$

and differentiating $\ln Z = -\ln(1 - e^{-\beta\epsilon})$ gives

$$\bar{E} = \frac{\epsilon}{e^{\epsilon/kT} - 1}.$$

This one formula contains both behaviors the course has met. When $kT \gg \epsilon$, $e^{\epsilon/kT} \approx 1 + \epsilon/kT$ and $\bar{E} \approx kT$: the equipartition value for two quadratic terms. When $kT \ll \epsilon$, $\bar{E} \approx \epsilon e^{-\epsilon/kT}$, exponentially small: the mode is frozen out. Equipartition is not a separate law; it is the high-temperature limit of quantum statistics.

5. The method, step by step, and how to check it

1. Measure energies from the ground state. It keeps $Z$ starting at one and does not change any difference.
2. Write $Z$. Two levels: $1 + e^{-\beta\epsilon}$. An oscillator: $1/(1 - e^{-\beta\epsilon})$. A level with degeneracy $g$ contributes $ge^{-\beta E}$.
3. Differentiate $\ln Z$ with respect to $\beta$, not $T$, and put the minus sign in front.
4. Convert to numbers with $x = \epsilon/kT$, using $k = 8.617 \times 10^{-5}$ eV/K for energies in eV.

Checks. At high temperature every average must approach its classical value: $kT$ for an oscillator, $\tfrac{1}{2}kT$ per quadratic term. At low temperature it must approach zero above the ground state. A two-level system's average energy can never exceed $\epsilon/2$ at positive temperature, because the upper state is never more populated than the lower. And the oscillator's average is always below $kT$: quantization can only remove energy relative to the classical answer, never add it.

6. Why nitrogen's vibration is frozen and its rotation is not

Spectroscopy gives nitrogen's vibrational quantum as $\epsilon_v = 0.289$ eV and its rotational level spacing as about $\epsilon_r = 0.00025$ eV. At room temperature, $kT = 0.0259$ eV.

For the vibration, $x = 0.289/0.0259 = 11.2$, and $\bar{E}/kT = 11.2/(e^{11.2} - 1) = 1.6 \times 10^{-4}$. The vibration holds essentially no energy: frozen. For the rotation, $kT/\epsilon_r \approx 100$, so the rotor is deep in its classical regime and holds $kT$. That is exactly why the first lesson counted $f = 5$ for air: three translations, two rotations, and a vibration that the Boltzmann factor switches off.

Heat nitrogen and the vibration wakes. At $1000$ K, $x = 3.35$ and $\bar{E}/kT = 0.12$; at $3000$ K, $x = 1.12$ and $\bar{E}/kT = 0.54$. The molar heat capacity at constant volume rises smoothly from $\tfrac{5}{2}R$ toward $\tfrac{7}{2}R$ over this range, as measurements show. The characteristic vibrational temperature of nitrogen, $\epsilon_v/k = 3350$ K, is the scale on which the change happens.

7. The heat capacity of a two-level system

For a two-level system, differentiating $\bar{E} = \epsilon/(e^{\epsilon/kT} + 1)$ with respect to $T$ gives a heat capacity

$$C = k\left(\frac{\epsilon}{kT}\right)^2\frac{e^{\epsilon/kT}}{(e^{\epsilon/kT} + 1)^2},$$

which is zero at both ends and peaks near $kT \approx 0.42\epsilon$ — the Schottky anomaly met with paramagnets. Unlike an oscillator's, a two-level system's heat capacity does not level off at high temperature: once both states are equally populated there is nowhere else for extra energy to go.

A measured bump of this shape in a material's heat capacity is a fingerprint of a pair of closely spaced levels, and its position gives their spacing directly. Physicists found the tunneling states of atoms in glasses, and the split levels of rare-earth ions in crystals, exactly this way.

8. The Einstein solid, revisited

An Einstein solid of $N$ oscillators, each in contact with the rest of the solid as its reservoir, has average energy $U = N\epsilon/(e^{\epsilon/kT} - 1)$ — the same result the counting of microstates gave, $q/N = 1/(e^{\epsilon/kT} - 1)$, now in two lines instead of a page. Differentiating gives Einstein's heat capacity,

$$C = Nk\left(\frac{\epsilon}{kT}\right)^2\frac{e^{\epsilon/kT}}{(e^{\epsilon/kT} - 1)^2},$$

which tends to $Nk$ per oscillator — $3R$ per mole of atoms — at high temperature and falls exponentially at low temperature. That fall is what Einstein used in 1907 to explain diamond's small heat capacity, one of the first successes of the quantum idea outside radiation.

9. Heat capacities that rise in steps

Put the pieces together for a real diatomic gas and the whole history of its heat capacity appears. Translation contributes $\tfrac{3}{2}R$ per mole at every temperature above a fraction of a kelvin, because the translational levels of a molecule in a box are spaced far below any practical $kT$. Rotation contributes $R$ once $kT$ is well above the rotational spacing, which for most molecules is a few kelvin; for hydrogen, the lightest, it is about $85$ K. Vibration contributes a further $R$ only once $kT$ approaches the vibrational quantum, which for most diatomic molecules means thousands of kelvin.

The measured molar heat capacity of hydrogen gas therefore climbs in two smooth steps: $\tfrac{3}{2}R$ below about $50$ K, $\tfrac{5}{2}R$ from about $250$ K to $750$ K, and rising toward $\tfrac{7}{2}R$ above a few thousand kelvin, just before the molecules break apart. Each step has the shape of the oscillator or rotor formula, centered near the mode's characteristic temperature.

This staircase was one of the great puzzles of late nineteenth-century physics. Equipartition predicted $\tfrac{7}{2}R$ at all temperatures, and nobody could explain why nature seemed to switch modes off. Maxwell called it the greatest difficulty the kinetic theory faced. The answer, the quantization of energy, is exactly the calculation of this lesson, and it came only after Planck and Einstein.

10. Zero-point energy

The oscillator's levels are really $(n + \tfrac{1}{2})\epsilon$, not $n\epsilon$: even in its ground state an oscillator keeps $\tfrac{1}{2}\epsilon$ of zero-point energy, which the uncertainty principle forbids it to lose. This lesson measured energies from the ground state, which shifts every level by the same amount and changes no probability, no heat capacity and no entropy. The zero-point energy is real — it is why liquid helium does not freeze at ordinary pressure, however cold — but thermal physics, which is about how energy changes with temperature, can almost always ignore it.

11. In the world: why chlorine and bromine have larger heat capacities than nitrogen

At room temperature the measured molar heat capacity at constant volume of nitrogen is $20.8$ J/(mol K), almost exactly $\tfrac{5}{2}R$. Chlorine's is $25.6$ and bromine's $27.7$ J/(mol K) — well above $\tfrac{5}{2}R$, even though all three are diatomic. The oscillator formula explains the difference.

Heavier atoms on weaker bonds vibrate more slowly, so their quanta are smaller. Chlorine's vibrational quantum is $0.0694$ eV, so at $300$ K, $x = 2.68$ and the vibration holds $x/(e^x - 1) = 0.20$ of $kT$. Bromine's quantum is $0.0402$ eV, $x = 1.56$, and it holds $0.42$ of $kT$. Nitrogen's, $0.289$ eV, gives $0.0002$. The vibrational heat capacity follows the same pattern, adding about $0.55R$ for chlorine and $0.8R$ for bromine — the extra measured beyond $\tfrac{5}{2}R$.

The same reasoning tells engineers when a gas's heat capacity will change with temperature. Combustion gases in a jet engine or rocket reach thousands of kelvin, where the vibrations of nitrogen, carbon dioxide and water all wake, and the heat capacities used in the design must follow the oscillator formula rather than any fixed value.

12. In the world: reading molecular vibrations in the infrared

The vibrational quanta used in this lesson are measured, not guessed. A molecule absorbs infrared light whose photon energy matches its vibrational quantum, and an infrared spectrometer records the wavelengths absorbed. Carbon monoxide absorbs strongly at $4.67$ μm, a photon energy of $hc/\lambda = 1240/4670 = 0.266$ eV — its vibrational quantum.

At room temperature, $x = 0.266/0.0259 = 10.3$, so the fraction of carbon monoxide molecules already vibrating is about $e^{-10.3} = 3.4 \times 10^{-5}$. Nearly every molecule starts in its ground state, which is why room-temperature absorption spectra are clean: each molecule absorbs from the same starting level. The carbon monoxide detectors required in American homes do not use infrared, but industrial and automotive emissions analyzers do, measuring exactly this absorption band to find the concentration of the gas.

In a hot flame, where the excited vibrational levels are populated, extra "hot bands" appear at slightly different wavelengths, and their strength relative to the main band is a Boltzmann factor — a thermometer that works inside a flame where no solid probe would survive.

13. Equipartition is a limit, not a law

Equipartition is so often correct that it is easy to treat it as exact: every quadratic term, $\tfrac{1}{2}kT$, always. It is the high-temperature limit of the quantum result, and it holds only for modes whose quantum is small beside $kT$. For a mode whose quantum is large, the true average is exponentially smaller, and treating equipartition as a law predicts heat capacities that experiments flatly contradict — the historical puzzle that helped launch quantum theory.

The opposite mistake is to think a quantum mode is simply off below some temperature and on above it. The transition is smooth: the average energy $\epsilon/(e^{\epsilon/kT} - 1)$ rises continuously from zero toward $kT$, and at the characteristic temperature $\epsilon/k$ it is already $0.58kT$. Heat capacity curves show gradual steps, never sudden jumps.

14. A two-level system when kT equals the gap

1. A two-level system has $\epsilon = kT$. Write its partition function.

   $Z = 1 + e^{-1} = 1 + 0.368 = 1.368$

   One factor for each state.

2. Find the probability of each state.

   $P_0 = \dfrac{1}{1.368} = 0.731, \qquad P_1 = \dfrac{0.368}{1.368} = 0.269$

   Factor over $Z$.

3. Find the average energy.

   $\bar{E} = 0 \times 0.731 + \epsilon \times 0.269 = 0.269\epsilon$

   Each energy weighted by its probability.

4. Check with the closed formula.

   $\dfrac{\epsilon}{e + 1} = \dfrac{\epsilon}{3.718} = 0.269\epsilon$

   The same result.

5. Compare with the upper limit.

   $0.269\epsilon < 0.5\epsilon$

   Even at infinite temperature only half the systems would be excited.

15. The vibration of nitrogen, cold and hot

1. Nitrogen's vibrational quantum is $0.289$ eV. Find $x$ at $300$ K.

   $x = \dfrac{0.289}{8.617 \times 10^{-5} \times 300} = \dfrac{0.289}{0.02585} = 11.18$

   The quantum in units of $kT$.

2. Find the average vibrational energy.

   $\bar{E} = \dfrac{0.289}{e^{11.18} - 1} = \dfrac{0.289}{7.2 \times 10^{4}} = 4.0 \times 10^{-6}\ \text{eV}$

   Almost nothing: the vibration is frozen.

3. Compare with equipartition.

   $\dfrac{\bar{E}}{kT} = \dfrac{4.0 \times 10^{-6}}{0.02585} = 1.6 \times 10^{-4}$

   Equipartition would give $kT$; the quantum oscillator gives a ten-thousandth of it.

4. Find $x$ at $3000$ K.

   $x = \dfrac{0.289}{0.2585} = 1.118$

   Ten times hotter, ten times smaller.

5. Find the average energy at $3000$ K.

   $\bar{E} = \dfrac{0.289}{e^{1.118} - 1} = \dfrac{0.289}{2.059} = 0.140\ \text{eV}$

   Now the vibration holds a real share.

6. Compare with equipartition again.

   $\dfrac{0.140}{0.2585} = 0.54$

   Over half of $kT$: the mode is waking up, and the heat capacity is rising toward $\tfrac{7}{2}R$.

16. The oscillator's average energy from its partition function

1. Write the oscillator's partition function as a series.

   $Z = \sum_{n = 0}^{\infty} e^{-n\beta\epsilon} = 1 + e^{-\beta\epsilon} + e^{-2\beta\epsilon} + \cdots$

   Levels $n\epsilon$ above the ground state.

2. Sum the geometric series.

   $Z = \dfrac{1}{1 - e^{-\beta\epsilon}}$

   Ratio $e^{-\beta\epsilon} < 1$, so the series converges.

3. Take the logarithm.

   $\ln Z = -\ln\left(1 - e^{-\beta\epsilon}\right)$

   The logarithm of a reciprocal is minus the logarithm.

4. Differentiate with respect to $\beta$.

   $\dfrac{\partial\ln Z}{\partial\beta} = -\dfrac{\epsilon e^{-\beta\epsilon}}{1 - e^{-\beta\epsilon}}$

   The chain rule on $\ln(1 - e^{-\beta\epsilon})$.

5. Change the sign for the average energy.

   $\bar{E} = -\dfrac{\partial\ln Z}{\partial\beta} = \dfrac{\epsilon e^{-\beta\epsilon}}{1 - e^{-\beta\epsilon}}$

   The shortcut $\bar{E} = -\partial\ln Z/\partial\beta$.

6. Multiply top and bottom by $e^{\beta\epsilon}$.

   $\bar{E} = \dfrac{\epsilon}{e^{\beta\epsilon} - 1}$

   The standard form.

7. Take the high-temperature limit.

   $\beta\epsilon \ll 1: \quad e^{\beta\epsilon} - 1 \approx \beta\epsilon \quad\Rightarrow\quad \bar{E} \approx \dfrac{1}{\beta} = kT$

   Equipartition, for two quadratic terms.

8. Take the low-temperature limit.

   $\beta\epsilon \gg 1: \quad \bar{E} \approx \epsilon e^{-\beta\epsilon}$

   Exponentially small: the mode is frozen out.

17. Your turn: a two-level system is at a temperature where $e^{\epsilon/kT} = 3$. What is its average energy?

1. Write the average energy of a two-level system.

   $\bar{E} = \dfrac{\epsilon}{e^{\epsilon/kT} + 1}$

   From $Z = 1 + e^{-\epsilon/kT}$.

2. Substitute the exponential.

   $\bar{E} = \dfrac{\epsilon}{3 + 1}$

   Given directly.

3. Your turn: work this step out. Its working is at the end of the packet.

   Simplify the fraction.

18. Guided practice

A quantum oscillator with quantum $\epsilon$ is in contact with a reservoir at a temperature so high that $kT = 600\epsilon$. What is its average energy above the ground state, very nearly?

• $kT$
• $\tfrac{1}{2}kT$
• $\epsilon$, one quantum
• $0$, since the oscillator is quantum

19. Guided practice

Complete the worked solution: a two-level system has states at $0$ and $18$ meV, and at its temperature $e^{-\epsilon/kT} = 0.28$. Find its partition function, the probability of the upper state and the average energy.

1. Add the two Boltzmann factors.

   $Z = 1 + 0.28 =$ _z

   The partition function sums every state's factor.

2. Divide the upper state's factor by $Z$.

   $P_1 = \dfrac{0.28}{Z} =$ _p

   A probability is a factor over the partition function.

3. Multiply the probability by the upper state's energy.

   $\bar{E} = P_1 \times 18 =$ _u meV

   The lower state has zero energy and contributes nothing.

20. Guided practice

Each system is in contact with a reservoir at $515$ K. Match each to its average energy.

21. Practice

For a quantum oscillator with $x = \epsilon/kT$, fill in $\bar{E}/\epsilon$ and $\bar{E}/kT$ to four decimal places. The exponentials are $e^{0.5} = 1.6487$, $e^1 = 2.7183$, $e^2 = 7.3891$ and $e^4 = 54.598$. The row to start with is $x = 1$.

22. Practice

A defect in a crystal has two states, $8$ meV apart. At $60$ K, what is its average energy above the lower state, in meV? Take $e^{1.5473} = 4.6989$.

Answer: meV

23. Practice

The vibration of a chlorine molecule has quantum $\epsilon = 69.4$ meV. At $300$ K, what is its average vibrational energy above the ground state, in meV? Use $k = 8.617 \times 10^{-5}$ eV/K.

Answer: meV

24. Somewhere new

The measured heat capacities of the halogen gases rise from fluorine to iodine. The vibration of a iodine molecule has quantum $\epsilon = 0.0265$ eV. At $300$ K, what fraction of its equipartition value $kT$ does its vibrational energy reach?

Answer:

25. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

26. Test question

For a quantum oscillator with $x = \epsilon/kT$, fill in $\bar{E}/\epsilon$ and $\bar{E}/kT$ to four decimal places. The exponentials are $e^{0.5} = 1.6487$, $e^1 = 2.7183$, $e^2 = 7.3891$ and $e^4 = 54.598$. The row to start with is $x = 1$.

27. What you can do now

You can find the average energy of simple quantum systems at any temperature. Explain to someone why nitrogen's vibration stores almost no energy at room temperature while its rotation stores the full equipartition share.

Working for the steps left to you

17. Your turn: a two-level system is at a temperature where $e^{\epsilon/kT} = 3$. What is its average energy?, step 3