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Simple question about the definitions, but I can't find it explicitly stated anywhere.

If you have a two level system described by $n_i = \frac{N}{Z}e^{-\beta E_i}$ and calculate the heat capacity

$$ c_v = \bigg( \frac{\partial U}{\partial T} \bigg)_{N,V} = \frac{\partial}{\partial T} NE₁\frac{e^{-\beta E_1}}{1 + e^{-\beta E_1}} = \frac{1}{Z²} Nk_B(\beta E_1)² e^{-\beta E_1}$$

Where:
$ T $ is temperature
$ \beta = \frac{1}{k_BT}$
$n_i$ is the number of particles in energy level $i$.
$ Z $ is the partition function

You get a peak at low temperature. In fact, I think this is true (I graphed it) for any system with a finite number of states of discrete energy. But I don't understand how a system with finite states can have a temperature. I understand temperature as the average translational kinetic energy per particle. When there are other degrees of freedom available, addition energy added to the system is partitioned among this translational energy and other modes like vibration and rotation. So it takes more total energy added to increase the average translational $E_K$ by the same amount.

If you have translational motion (around the centre of mass frame), then there are an infinite number of energy levels available, unless the velocities are bounded.

So by "two-level system" do we actually mean two states other than the translational energies? Like a cloud of electrons (non-interacting... neutrons?) with spin up and down in a magnetic field.

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    $\begingroup$ Is this the temperature def. where you can get wacky results, like going to negative infinity? Well, physics can make whatever def. is useful for description of real systems, even is results might be unorthodox. BTW I think in the future asking stuff like that on Physics.se could be a good idea. $\endgroup$
    – Mithoron
    Commented May 17, 2023 at 1:13
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    $\begingroup$ @FurrierTransform Hello. You are integrating the population fraction of a two-level system that will be found in the second level, i.e., $e^{-\beta \varepsilon}/(1 + e^{-\beta \varepsilon})$. However, that is not the way to relate $C_\mathrm{V}$ to the molecular partition function. Note even that the units don't make sense (you are integrating something unitless). Try to use perform $C_\mathrm{V} = 2kT \left(\dfrac{\partial \ln q}{\partial T}\right)_\mathrm{V} + kT^2 \left(\dfrac{\partial ^2 \ln q}{\partial T^2}\right)_\mathrm{V} $. $\endgroup$ Commented May 17, 2023 at 1:33
  • $\begingroup$ @FurrierTransform BTW, the anomaly that you refer to is another thing. It has to do with the decay to zero of $C_\mathrm{V}$ as $T\rightarrow \infty$, that will happen in that system. This is, you can heat something easier even if $T$ is really high (surprise!). If the math doesn't close to you, let me know in the comments and I can post something. (edit: in the last comment, I meant differentiate, not integrate, oops) $\endgroup$ Commented May 17, 2023 at 1:36
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    $\begingroup$ Related: Can somebody explain the low temperature Schottky anomaly for two level systems? $\endgroup$
    – ananta
    Commented May 17, 2023 at 6:15
  • $\begingroup$ I understand temperature as the average translational kinetic energy per particle. - rather T is classically proportional to average energy per a degree of freedom, later somewhat complicated by energy quantisation steps. There are no translational energy modes in solids. $\endgroup$
    – Poutnik
    Commented May 17, 2023 at 6:51

2 Answers 2

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To answer 'But I don't understand how a system with finite states can have a temperature', we can start with the thermodynamic definition of temperature as

$$\displaystyle T=\left(\frac{\partial U}{\partial S}\right)_V$$

where as usual $S$ is entropy and $U$ internal energy.

In a system such as a mole of an ideal gas or a crystal, for example, there are an infinite number of energy levels, and increasing the energy ($\Delta U $ is positive) increases the number of ways these energy levels can be populated and so the entropy increases. The slope $dU/dS$ is therefore positive and so is the temperature. From the viewpoint of the Boltzmann distribution the population of any two levels is, where $E_2>E_1$,

$$\displaystyle \frac{n_1}{n_2}=e^{-(E_2-E_1)/k_BT}$$

If there are an infinite number of energy levels then the temperature is always positive because the upper level is always populated less than any one below it. To make the populations equal would need an infinite temperature and so infinite energy.

What about a system with $n$ particles but with a finite number of energy levels? Suppose there are two levels, for example, with energy $0$ and $E$. In this case a plot of entropy vs. energy has the form of an inverted semicircle. At zero energy all the particles are in the lowest level and so the entropy is zero. Similarly, when the total energy is $nE$ all particles are in the highest levels and again the entropy is zero. At energy $nE/2$ the two levels are equally populated the entropy has its maximum value. In the energy range from $0\to nE/2$ the slope $dU/dS >0$ and the temperature is positive but in the range $nE/2\to nE$, the slope of $dU/dS$ is negative and so this corresponds to a negative temperature. The midpoint energy corresponds to infinite temperature and both levels are equally populated.

Negative temperatures are not uncommon, often called a 'population inversion' in a laser and occur also in many types of nmr experiments.

neg-temp

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  • $\begingroup$ Things get interesting if the disturbed energy distribution for some degree of freedom (laser electron excitation, IR laser/maser vibration/rotation one) are not in equilibrium with other degrees of freedom. In a way one could say the system have different TD temperatures for different DoF.. In other way, it has none. $\endgroup$
    – Poutnik
    Commented May 17, 2023 at 10:41
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    $\begingroup$ Yes, in jet cooled molecules, for example, there is a different vibrational and rotational temperature and v. low translational temperature because although the molecules are moving quickly their speed distribution can be tiny. It is also possible to produce molecules in the vapour phase in an excited state that is cooler than the ground state and which cools the gas, (but only by a bit) as it warms up by collisions. This takes a few tens of nanoseconds typically. $\endgroup$
    – porphyrin
    Commented May 17, 2023 at 11:52
  • $\begingroup$ Thanks for this clear and well-written answer, however I fear that I did not phrase my question as clearly. I understand why there is a Schottky anomaly, however I would like to clarify that by a "two-level system" we mean two states other than the translational kinetic energy levels. Like an ideal gas of particles with two spin states in a magnetic field. Because if there were only two level for each particle, that would be like all the particles having only two velocities: slow and fast. $\endgroup$ Commented May 22, 2023 at 12:57
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    $\begingroup$ yes states other than translation. $\endgroup$
    – porphyrin
    Commented May 22, 2023 at 16:44
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animation of raising temperature of a two-energy-state system

I understand temperature as the average translational kinetic energy per particle.

Temperature isn't just related to the kinetic energy of particles, it also changes the potential energy. The potential energies are incorporated into the energy levels $E_1$ and $E_2$. The kinetic energy require consideration other than the equations you have considered.

But I don't understand how a system with finite states can have a temperature.

All systems have a defined scaler temperature field (except for particles in high vacuum with long free flight where some particles absorb radiation (photons), for which it may not be defined, as pointed out in the comments). Now, to address the confusion, according to you, I think, a system with finite states shouldn't have a concept of temperature, because if we keep adding heat to the system, at some point, all particles in the lower-energy state will rise to the higher-energy state. What if we add more heat? This shouldn't be a problem because we can always add heat to the system, but the system has no more higher energy states to occupy. What now?

So by "two-level system" do we actually mean two states other than the translational energies?

Yes, when you say a two-level system, you mean something other than just kinetic energy. It is the potential energy that is represented by the energy levels, $E_1$ and $E_2$, which are quantized and finite here. For a purely kinetic energetic model, consider the case of ideal gases.


Remarks

The additional heat simply increases the kinetic energy of the system once maximum potential energy (all particles in $E_2$ state) is achieved, which, actually, never is because there is always some probability to detransition.

Related: Can somebody explain the low temperature Schottky anomaly for two level systems?.

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  • $\begingroup$ All systems have a temperature. Only those that are in thermal equilibrium. Particles in high vacuum may not be. $\endgroup$
    – Poutnik
    Commented May 17, 2023 at 6:46
  • $\begingroup$ @Poutnik very interesting! I am curious to read more about it. Would it be more accurate to say 'all systems have a defined temperature field?' $\endgroup$
    – ananta
    Commented May 17, 2023 at 6:50
  • $\begingroup$ If you mean a scalar T field or vector grad T field, sure, for macroscopic systems with dense enough matter. But for high enough vacuum with long enough mean free flight distance, eventually when some particles selectively absorb photons, there is no temperature. $\endgroup$
    – Poutnik
    Commented May 17, 2023 at 6:59
  • $\begingroup$ @Poutnik I have incorporated your suggestions in the answer. Could you please provide some reference, I searched for a while and couldn't find what I was looking for. $\endgroup$
    – ananta
    Commented May 17, 2023 at 7:04
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    $\begingroup$ @Poutnik I understood now. Yes, how did you know? Right now I am working on mathematical modelling of lithium-sulfur batteries, but I have worked on computational chemistry regarding chemical equilibria.. So I had to study quantum mechanics and statistical thermodynamics. Also some electron paramagnetic resonance in some compounds (also computationally). I'd like to know more about inorganic or organic chemistry, though. Maybe I am a frustrated chemist, but I just don't like the industry/chemical process design (yes theoretically). $\endgroup$ Commented May 17, 2023 at 13:33

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