# What's the size to talk about thermal equilibrium?

I am confused about what's the time we can talk about the thermal equilibrium? I mean how many particles the system will contain?

This link is a lecture note to teach thermal physics following primarily the Kittle's textbook. He shows an example to demonstrate how to calculate the multiplicity function. Here you can see this line: $\textbf{A hydrogen atom in thermal equilibrium.}$

How is this possible? Can we talk about the thermal equilibrium for just one atom?If we cannot, then what's the suitable size and how many particles the system will contain?

The another confused point is related the Gibb's ensemble theory, by which one can derive the whole thermodynamics based on some model.

But what's the reason we can apply the ensemble theory to deal with the following single-particle problem? I mean how can we talk about the thermal equilibrium for just one harmonic oscillator like just one atom below?

A single classical hamrmonic oscillator in $3$D(internal energy and specific heat?):

1.Hamiltonian: $$H = \dfrac{\textbf{p}\cdot\textbf{p}}{2m} + \dfrac{m\omega^2}{2}\textbf{r}\cdot\textbf{r}$$

2.Partition function: $$Z = \dfrac{1}{\hbar^3} \int d\textbf{r} d\textbf{p} e^{-\beta H} = \dfrac{1}{\hbar^3} \int d\textbf{r} d\textbf{p} e^{-\beta \left( \dfrac{\textbf{p}\cdot\textbf{p}}{2m} + \dfrac{m\omega^2}{2}\textbf{r}\cdot\textbf{r} \right)} = (\dfrac{2\pi}{\beta \hbar \omega})^3$$

3.Internal energy and specific heat: $$U = -\dfrac{\partial \ln Z}{\partial \beta} = 3 k_B T \quad ; \quad C = \dfrac{\partial U}{\partial T} = 3k_B$$

PS: A pedestrian thought of mine: There is some relation between the thermal equilibrium (many degrees of freedom) and mechanical equilibrium.(one or just a few degrees of freedom)

• I can think of the zeroth law of thermodynamics and the Equipartition Theorem which could be adopted as the definition of thermal equilibrium. In neither of them a single atom would be enough to verify a thermal equilibrium. For the first you need thermometric variables which are averages over many microscopic states. For the second you need average over many particles. – Diracology Feb 11 '17 at 13:31
• Seemingly the hydrogen can exist in many different states depend on the energy level of the electron. – Jack Feb 11 '17 at 13:46

I think the confusion here is that, in the statement by Kittel, it is not emphasized that there exists a large temperature bath at temperature $T$. Namely, the single hydrogen atom is in thermal equilibrium with the temperature bath. It is precisely this large, macroscopic temperature bath that allows us to have a well-defined temperature and continue to do statistical mechanics. In other words, in the canonical ensemble where heat is allowed to flow, you can talk about "microscopic" things. Perhaps a clearer example of this is a single atom of H$_2$ in a gas of O$_2$ held at temperature $T$. The speed of the H$_2$ particle is still given by the equipartition theorem, even though we only have a "microscopic" number of H$_2$, namely $1$.

Of course, one can be perverse and apply statistical mechanics to exceedingly small physical systems, such as a single particle in the microcanonical ensemble. This a perfectly valid thing to do, and there is nothing inherently wrong with it, but in many ways it is not useful. For example, the temperature of a single particle is always $0$, cause it only ever has one microstate at each energy. In addition, many key results require the number of particles $N \rightarrow \infty$. But speaking purely in principle, there isn't anything wrong with looking at small systems.

• +1 There is nothing perverse about applying statistical mechanics to small systems, even those containing one particle. Only the term "probability" has to be appropriately interpreted. You either think in terms of an ensemble of (macroscopically) identical systems, or think of the whole procedure as one of statistical inference given limited information about the system, such as temperature of its bath (see Jaynes' work). – Deep Feb 15 '17 at 6:21
• @Deep I was thinking primarily of it being perverse in the microcanonical ensemble. If you have external couplings to a bath (eg canonical or grand canonical), I agree there is nothing wrong with thinking of single particle systems, as the physical intuition is that the bath has some macroscopic quantity associated with it. I can't think of physical systems where small systems in the microcanonical ensemble are practically useful. However, as you note in other contexts (information) it could be useful to think of small systems. – Aaron Feb 15 '17 at 6:33

To understand statistical mechanics and many-body physics, one often starts with examples which "decouple" in nice ways to yield a simple solution. In this case, this "hiding" of details is a bit confusing if you actually think about it. I will, therefore, give a longer mathematical walkthrough and some intuition in the end:

Let's ignore quantum mechanics! Consider any system of non-interacting, identical sub-systems. It can be described by a Hamiltonian function:

$$H([p, q]) = \sum_j^N H_0(\vec{p}_j, \vec{q}_j)$$

where $N$ is the number of subsystems (related to the thermodynamic limit), $\vec{p}_j, \vec{q}_j$ are the variables describing subsystem $j$ and I've written $[p,q]$ symbolically to indicate that the Hamiltonian, of course, depends on all coordinates and momenta! Note that $H_0$ is the same function for all particles and that there are no "cross-terms" since the subsystems do not interact.

The thermal partition function (in the canonical ensemble) can then be written as

$$Z(T) = Tr e^{-H/{k_B T}} = \sum_{\text{all config.}} e^{-H[p,q]/{k_B T}}$$

You may not have seen this expression in the general form of a trace, but to be concrete: the sum over configurations could for example be multiple integrals over all $\vec{p}_i, \vec{q}_i$ for an ideal gas:

$$Z(T) \propto \dfrac{1}{\hbar^{3N}N!} \int \prod_j d\vec{p}_j d\vec{q}_j \exp\left(-H[p,q]/{k_B T}\right)$$

Then we do a trick! We rewrite the sum over all possible configurations as a product, using the factorization of $H$:

$$\dfrac{1}{\hbar^{3N}N!} \int \prod_j d\vec{p}_j d\vec{q}_j e^{-H[p,q]/{k_B T}} = \dfrac{1}{\hbar^{3N}N!} \prod_j^N \left[ \int d\vec{p}_j d\vec{q}_j \exp\left(-H_0(p_j, q_j)\right)\right]$$

Here I just split up the exponential of total Hamiltonian into a product of exponentials of subsystem Hamiltonian:

$$\exp(-H/{k_B T}) = \prod_j \exp(-H_0(p_j, q_j)/{k_B T})$$

and shuffled around the order in the multiple integrals a bit. But then we see that this is nothing else than the product of $N$ identical integrals!

$$Z(T) = Z_0(T)^N$$

where the single-subsystem partition function is:

$$Z_0(T) = \hbar^{-3} \int dp_j dq_j \exp\left(-H_0(p_j, q_j)\right)$$

Compare the form of this to the partition function for a single harmonic oscillator which is written down in your included image!

With just one more step of calculation, one sees that the density matrix $\rho$, from which all expectation values are calculated, also factorizes into a product. This means that any averages for one subsystem are statistically independent of the others. We might as well ignore the other subsystems in the description, which is somewhat intuitive since the whole system is just $N$ copies of the same subsystem!

Thermodynamic equilibrium is the result of a competition between mechanical energy, pressure and sheer combinatorics (which state is most common statistically). Often this is summarized in the free (Gibbs/Helmholtz) potential. It can be written as (using $Z = Z_0^N$):

$$F = - k_B T \log Z = - k_B T N \log Z_0$$

Thermal equilibrium means that the free energy is minimized with respect to variations. From the above expression, it is clear that it is enough to minimize $\log Z_0$. In this case, the "combinatorics" of entropy is independent of $N$, so the limit does not change anything! This is of course highly unnatural: any physical system has at least some communication between its components. Other people are probably better at explaining when it is meaningful to talk about thermal equilibrium and when it is not.

So there it is, this is the reason why we sometimes talk about statistical physics and ensembles for single hydrogen atoms or harmonic oscillators.

Sources:

• You give me an answer for the second part, but seemingly you have ignored something in your partition function (absolutely no Plank's constant). – Jack Feb 15 '17 at 5:25
• You are absolutely right, each double integral should be over $\hbar^{-1} dp_j dq_j$. I changed it in the derivation. – plan Feb 15 '17 at 10:36
• For classical cases, there is an N factorial in the denominator. – Jack Feb 17 '17 at 11:41

Luckily, I have received the answer delivered by Prof. Hsiu-Hau Lin in his wikispace (in Chinese, sorry for this).He listed three conclusions (translated by me):

1.Considering a system large enough can be described by Hamiltonian mechanics, then the entropy of the system is constant, the system will never arrive at thermal equilibrium.

2.The thermal equilibrium is the characteristic of open quantum system, so only the open quantum system can arrive at the thermal equilibrium.

3.Even a single atom or spin also can arrive at thermal equilibrium. (Take Stern–Gerlach experiment as an example.)

He also told me by reading this book: The Theory of Open Quantum Systems (written by Breuer and Petruccione), I can understand fully these three conclusions above.

• The conclusion about thermalisation is flawed, in my opinion. It just postpones the problem of how thermalisation actually occurs in quantum systems. I have read Breuers & Petrucciones book and don't agree with this conclusion. There is at least 10 years of research about how thermalisation happens in quantum systems. The basic idea is the eigenstate thermalisation hypothesis which is based on the studies of the statistics of random Hamiltonians and their spectra. – plan Feb 17 '17 at 11:21
• You can post your thoughts in the answer here, many thanks for that. – Jack Feb 17 '17 at 11:40

Thermal Equilibrium exists for systems. Now, those systems can be as big as small as you want to consider. If you consider a system that consists of only 1 particle, it's always at Equilibrium. If you consider a body along with some medium in which it's placed then it's at Equilibrium if there is no energy variation throughout and the entropy is maximum.

• So can we define the temperature for just one atom? – Jack Feb 11 '17 at 13:39
• You sure can.. It wouldn't serve much of a purpose but, you can. – Rippr Feb 12 '17 at 11:22