# How can temperature affect a single electron system?

When reading about the canonical ensemble once in a Statistical Physics book, the author stated that the important thing is that we have a system $A$ in contact with a much greater system $A'$. The much greater system is the one with temperature $T$ and one then derives the probability distribution for system $A$ as

$$p_i=\dfrac{1}{Z}e^{-\beta E_i},$$

being $p_i$ the probability that $A_i$ is in the $i$-th microstate with energy $E_i$ and $Z=\sum_{i}e^{-\beta E_i}$ is the partition function with $k_B T\beta = 1$.

Now the author states that the system $A$ could even be a single electron or single atom.

This is a little confusing. When talking about temperature and things like that, don't we always deal with macroscopic systems?

How can we talk about temperature in the context of a single electron? Furthermore, how can temperature affect a microscopic system if temperature is a macroscopic concept?

I'm having some trouble to bridge this gap between "temperature as something from a macroscopic world" interacting with microscopic systems.

• Yes, the purpose of statistical mechanics is to connect the microscopic quantities with the macroscopic quantities and the partition function acts like a bridge between the two. This is a very interesting result that appears magically. – UKH Sep 17 '16 at 2:18
• Regarding "don't we always deal with macroscopic systems?", that did use to be the case. But you should carefully consider the advances in e.g. A single-atom heat engine. (Not that "X is experimentally unaccessible" means that you should not contemplate what happens for that case, but "X is accessible" does de us to consider it seriously.) – Emilio Pisanty Jan 29 '19 at 7:08

How can we talk about temperature in the context of a single electron?

We can't. In this case, the author is talking about the temperature of the much greater, macroscopic system $A^{\prime}$ (a.k.a. reservoir).

Furthermore, how can temperature affect a microscopic system if temperature is a macroscopic concept?

Again, temperature is a property of the macroscipic system $A^{ \prime}$, and because $A$ and $A^\prime$ are in contact with each other, a property of $A^{\prime}$ will in general affect the behavior of $A$. Specifically, applying postulates of statistical physics on the isolated system $A + A^{\prime}$ leads to the probability distribution $p_i = e^{-\beta E_{i}}/Z$.

There is one important difference depending on whether $A$ itself is macroscopic or not.

If $A$ were macroscopic, various quantities of $A$ would have negligible fluctuations due to the central limit theorem. So, for example, the mean energy from the probability distribution $p_i$ would more or less mean "the" energy of the system. The detailed form of the distribution wouldn't have much meaning except as a calculational device.

On the other hand, if $A$ weren't macroscopic, the probability distribution itself would have significance. A prime example of this would be the Maxwell-Boltzmann speed distribution: $$f(v) = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-mv^2/2k_B T}.$$ It is simply a variant of the distribution $p_i = e^{-\beta E_{i}}/Z$ when applied to a single gas molecule of a classical ideal gas.

How thermodynamic variables (macroscopic quantities) are connected with the microscopic world?

An ensemble is a collection of identically prepared system. Consider a canonical ensemble of $M$ systems. If you take a system $A$ out of it, the remaining $(M-1)$ systems will be an exact replica of $A$. Since there is thermal contact between the systems, there is energy transfer taking place between the systems. So, the energy of $A$ fluctuates and is not a constant and the $(M-1)$ systems act like a heat bath. So, any system placed on a heat bath is a canonical system.

Each possible state of the system $A$ represents a microstate of the whole system. Let $i$ be such a quantum-mechanical state and $E_i$ be the energy eigen value of that state. Then the probability of observing the system in the microstate $|i\rangle$ is given by Boltzmann probability distribution:

$$p_1=\frac{exp({-E_i/k_BT})}{Z}\longrightarrow(1)$$

where

$$Z=\sum _j exp(E_j/k_BT)\longrightarrow(2)$$

which is just a some over states, and is called the Partition function, which primarily appeared as a normalization factor for probability conservation. But, this factor has interesting consequence other than just being a normalization constant. To derive that, we take the entropy of the system in a canonical ensemble (which originated as a consequence of Boltzmann's hypothesis which states that the probability of a quantum state is related to entropy, and is one of the essential basis of statistical mechanics)

$$S=-k_B \sum_{i=1}^W p_i ln(p_i)\longrightarrow(3)$$

where $W$ is the number of accessible quantum states possible for the system. Taking the logarithm $p_i$ given by equation $(1)$, we have

$$ln(p_i)=-\frac{E_i}{k_BT}-ln(Z)$$

Hence equation $(3)$ becomes

$$S=k_B\sum_{i=1}^W p_i\left(\frac{E_i}{k_BT}+ln(Z)\right)\longrightarrow(4)$$

Now, we have the average internal energy defined by

$$\bar{U}=\sum_i p_iE_i\longrightarrow(5)$$

Also, we have $\displaystyle{\sum_i p_i=1}$. Hence equation $(4)$ becomes

$$S=\frac{\bar{U}}{T}+k_B ln(Z)$$

which can be rewritten as

$$\bar{U}-TS=-kBT ln(Z)\longrightarrow(6)$$

The quantity $(\bar{U}-TS)$ is the average value of Helmholtz free energy, $F$. Hence equation $(6)$ can be written as, the Helmholtz free energy

$${\color{red}{F=-k_BT ln(Z)} }$$

which is a thermodynamic potential. This is very fundamental and great unification of statistical mechanics with thermodynamics. So the partition function $Z$ is not just a normalization factor, but it is something more. It bridges the microscopic world with measurable macroscopic thermodynamic properties of the system. It relates the quantum states of the system with free energy for a given temperature of the heat bath, and from free energy we can calculate all the thermodynamic quantities, as follows:

$$\textit{Pressure}, P=-\left(\frac{\partial{F}}{\partial{V}}\right)_T$$

$$\textit{Entropy}, S=-\left(\frac{\partial{F}}{\partial{T}}\right)_V$$

$$\textit{isothermal compressibility}, K=\left[-V\left(\frac{\partial{P}}{\partial{V}}\right)_T\right]^{-1}=\left[V\left(\frac{\partial^2{F}}{\partial{V^2}}\right)_T\right]^{-1}$$

$$\textit{heat capacity at constant volume}, C_V=T\left(\frac{\partial{S}}{\partial{T}}\right)_V=-T\left(\frac{\partial^2{F}}{\partial{T^2}}\right)_V$$

$$\textit{average internal energy}, \bar{U}=TS+F=k_BT^2\left(\frac{\partial{ln(Z)}}{\partial{T}}\right)_V$$

How can we talk about temperature in the context of a single electron?

The partition function just gives an astonishing connection between the microscopic and macroscopic world. The answer to this question is already made. Your system may be a single electron or an atom. To treat it with the techniques of canonical ensemble, you need to make sure that the system is in contact with a heat bath at a temperature $T$. At that temperature, the probability of the system to be found in a microstate is given by equation $(1)$ and once known the number of quantum states (which is the worst part), we can sum over the states which gives the partition function and from the partition function, we can find out the macroscopic variables associated with the system.

It is easier to understand the conundrum from the the statistical definition of temperature:

It is connected to the root mean square of the translational kinetic energy of the atom/molecule.

So it is dependent on a large ensemble of particles and can be identified with the thermodynamic temperature.

Nevertheless, one can use the expression for a small number of particles with the errors in the evaluation getting large from the small statistics. It has no meaning for one particle other than saying figuratively that "its kinetic energy could represent the temperature of an ensemble of particles ".