Partition function (statistical mechanics)

Initially, let us assume that a thermodynamically large system is in thermal contact with the environment, with a temperature T, and both the volume of the system and the number of constituent particles are fixed. A collection of this kind of system comprises an ensemble called a canonical ensemble. The appropriate mathematical expression for the canonical partition function depends on the degrees of freedom of the system, whether the context is classical mechanics or quantum mechanics, and whether the spectrum of states is discrete or continuous.

For a canonical ensemble that is classical and discrete, the canonical partition function is defined as

$${\displaystyle Z=\sum _{i}e^{-\beta E_{i}},}$$

where

• ${\displaystyle i}$ is the index for the microstates of the system;
• ${\displaystyle e}$ is Euler's number;
• ${\displaystyle \beta }$ is the thermodynamic beta, defined as ${\displaystyle {\tfrac {1}{k_{\text{B}}T}}}$ where ${\displaystyle k_{\text{B}}}$ is Boltzmann's constant;
• ${\displaystyle E_{i}}$ is the total energy of the system in the respective microstate.

The exponential factor ${\displaystyle e^{-\beta E_{i}}}$ is otherwise known as the Boltzmann factor.

Derivation of canonical partition function (classical, discrete)

There are multiple approaches to deriving the partition function. The following derivation follows the more powerful and general information-theoretic Jaynesian maximum entropy approach.

According to the second law of thermodynamics, a system assumes a configuration of maximum entropy at thermodynamic equilibrium. We seek a probability distribution of states ${\displaystyle \rho _{i}}$ that maximizes the discrete Gibbs entropy

$${\displaystyle S=-k_{\text{B}}\sum _{i}\rho _{i}\ln \rho _{i}}$$

subject to two physical constraints:

1. The probabilities of all states add to unity (second axiom of probability):
   $${\displaystyle \sum _{i}\rho _{i}=1.}$$

   
2. In the canonical ensemble, the average energy is fixed (conservation of energy):
   $${\displaystyle \langle E\rangle =\sum _{i}\rho _{i}E_{i}\equiv U.}$$

   

Applying variational calculus with constraints (analogous in some sense to the method of Lagrange multipliers), we write the Lagrangian (or Lagrange function) ${\displaystyle {\mathcal {L}}}$ as

$${\displaystyle {\mathcal {L}}=\left(-k_{\text{B}}\sum _{i}\rho _{i}\ln \rho _{i}\right)+\lambda _{1}\left(1-\sum _{i}\rho _{i}\right)+\lambda _{2}\left(U-\sum _{i}\rho _{i}E_{i}\right).}$$

Varying and extremizing ${\displaystyle {\mathcal {L}}}$ with respect to ${\displaystyle \rho _{i}}$ leads to

$${\displaystyle {\begin{aligned}0&\equiv \delta {\mathcal {L}}\\&=\delta \left(-\sum _{i}k_{\text{B}}\rho _{i}\ln \rho _{i}\right)+\delta \left(\lambda _{1}-\sum _{i}\lambda _{1}\rho _{i}\right)+\delta \left(\lambda _{2}U-\sum _{i}\lambda _{2}\rho _{i}E_{i}\right)\\&=\sum _{i}{\bigg [}\delta {\Big (}-k_{\text{B}}\rho _{i}\ln \rho _{i}{\Big )}+\delta {\Big (}\lambda _{1}\rho _{i}{\Big )}+\delta {\Big (}\lambda _{2}E_{i}\rho _{i}{\Big )}{\bigg ]}\\&=\sum _{i}\left[{\frac {\partial }{\partial \rho _{i}}}{\Big (}-k_{\text{B}}\rho _{i}\ln \rho _{i}{\Big )}\,\delta (\rho _{i})+{\frac {\partial }{\partial \rho _{i}}}{\Big (}\lambda _{1}\rho _{i}{\Big )}\,\delta (\rho _{i})+{\frac {\partial }{\partial \rho _{i}}}{\Big (}\lambda _{2}E_{i}\rho _{i}{\Big )}\,\delta (\rho _{i})\right]\\&=\sum _{i}{\bigg [}-k_{\text{B}}\ln \rho _{i}-k_{\text{B}}+\lambda _{1}+\lambda _{2}E_{i}{\bigg ]}\,\delta (\rho _{i}).\end{aligned}}}$$

Since this equation should hold for any variation ${\displaystyle \delta (\rho _{i})}$, it implies that

$${\displaystyle 0\equiv -k_{\text{B}}\ln \rho _{i}-k_{\text{B}}+\lambda _{1}+\lambda _{2}E_{i}.}$$

Isolating for ${\displaystyle \rho _{i}}$ yields

$${\displaystyle \rho _{i}=\exp \left({\frac {-k_{\text{B}}+\lambda _{1}+\lambda _{2}E_{i}}{k_{\text{B}}}}\right).}$$

To obtain ${\displaystyle \lambda _{1}}$, one substitutes the probability into the first constraint:

$${\displaystyle {\begin{aligned}1&=\sum _{i}\rho _{i}\\&=\exp \left({\frac {-k_{\text{B}}+\lambda _{1}}{k_{\text{B}}}}\right)Z,\end{aligned}}}$$

where ${\displaystyle Z}$ is a constant number defined as the canonical ensemble partition function:

$${\displaystyle Z\equiv \sum _{i}\exp \left({\frac {\lambda _{2}}{k_{\text{B}}}}E_{i}\right).}$$

Isolating for ${\displaystyle \lambda _{1}}$ yields ${\displaystyle \lambda _{1}=-k_{\text{B}}\ln(Z)+k_{\text{B}}}$.

Rewriting ${\displaystyle \rho _{i}}$ in terms of ${\displaystyle Z}$ gives

$${\displaystyle \rho _{i}={\frac {1}{Z}}\exp \left({\frac {\lambda _{2}}{k_{\text{B}}}}E_{i}\right).}$$

Rewriting ${\displaystyle S}$ in terms of ${\displaystyle Z}$ gives

$${\displaystyle {\begin{aligned}S&=-k_{\text{B}}\sum _{i}\rho _{i}\ln \rho _{i}\\&=-k_{\text{B}}\sum _{i}\rho _{i}\left({\frac {\lambda _{2}}{k_{\text{B}}}}E_{i}-\ln(Z)\right)\\&=-\lambda _{2}\sum _{i}\rho _{i}E_{i}+k_{\text{B}}\ln(Z)\sum _{i}\rho _{i}\\&=-\lambda _{2}U+k_{\text{B}}\ln(Z).\end{aligned}}}$$

To obtain ${\displaystyle \lambda _{2}}$, we differentiate ${\displaystyle S}$ with respect to the average energy ${\displaystyle U}$ and apply the first law of thermodynamics, ${\displaystyle dU=TdS-PdV}$:

$${\displaystyle {\frac {dS}{dU}}=-\lambda _{2}\equiv {\frac {1}{T}}.}$$

Thus the canonical partition function ${\displaystyle Z}$ becomes

$${\displaystyle Z\equiv \sum _{i}e^{-\beta E_{i}},}$$

where ${\displaystyle \beta \equiv 1/(k_{\text{B}}T)}$ is defined as the thermodynamic beta. Finally, the probability distribution ${\displaystyle \rho _{i}}$ and entropy ${\displaystyle S}$ are respectively

$${\displaystyle {\begin{aligned}\rho _{i}&={\frac {1}{Z}}e^{-\beta E_{i}},\\S&={\frac {U}{T}}+k_{\text{B}}\ln Z.\end{aligned}}}$$

In classical mechanics, the position and momentum variables of a particle can vary continuously, so the set of microstates is actually uncountable. In classical statistical mechanics, it is rather inaccurate to express the partition function as a sum of discrete terms. In this case we must describe the partition function using an integral rather than a sum. For a canonical ensemble that is classical and continuous, the canonical partition function is defined as

$${\displaystyle Z={\frac {1}{h^{3}}}\int e^{-\beta H(q,p)}\,\mathrm {d} ^{3}q\,\mathrm {d} ^{3}p,}$$

where

• ${\displaystyle h}$ is the Planck constant;
• ${\displaystyle \beta }$ is the thermodynamic beta, defined as ${\displaystyle {\tfrac {1}{k_{\text{B}}T}}}$;
• ${\displaystyle H(q,p)}$ is the Hamiltonian of the system;
• ${\displaystyle q}$ is the canonical position;
• ${\displaystyle p}$ is the canonical momentum.

To make it into a dimensionless quantity, we must divide it by h, which is some quantity with units of action (usually taken to be Planck's constant).

For a gas of ${\displaystyle N}$ identical classical particles in three dimensions, the partition function is

$${\displaystyle Z={\frac {1}{N!h^{3N}}}\int \,\exp \left(-\beta \sum _{i=1}^{N}H({\textbf {q}}_{i},{\textbf {p}}_{i})\right)\;\mathrm {d} ^{3}q_{1}\cdots \mathrm {d} ^{3}q_{N}\,\mathrm {d} ^{3}p_{1}\cdots \mathrm {d} ^{3}p_{N}}$$

where

• ${\displaystyle h}$ is the Planck constant;
• ${\displaystyle \beta }$ is the thermodynamic beta, defined as ${\displaystyle {\tfrac {1}{k_{\text{B}}T}}}$;
• ${\displaystyle i}$ is the index for the particles of the system;
• ${\displaystyle H}$ is the Hamiltonian of a respective particle;
• ${\displaystyle q_{i}}$ is the canonical position of the respective particle;
• ${\displaystyle p_{i}}$ is the canonical momentum of the respective particle;
• ${\displaystyle \mathrm {d} ^{3}}$ is shorthand notation to indicate that ${\displaystyle q_{i}}$ and ${\displaystyle p_{i}}$ are vectors in three-dimensional space.

The reason for the factorial factor N! is discussed below. The extra constant factor introduced in the denominator was introduced because, unlike the discrete form, the continuous form shown above is not dimensionless. As stated in the previous section, to make it into a dimensionless quantity, we must divide it by h^3N (where h is usually taken to be Planck's constant).

For a canonical ensemble that is quantum mechanical and discrete, the canonical partition function is defined as the trace of the Boltzmann factor:

$${\displaystyle Z=\operatorname {tr} (e^{-\beta {\hat {H}}}),}$$

where:

• ${\displaystyle \operatorname {tr} (\circ )}$ is the trace of a matrix;
• ${\displaystyle \beta }$ is the thermodynamic beta, defined as ${\displaystyle {\tfrac {1}{k_{\text{B}}T}}}$;
• ${\displaystyle {\hat {H}}}$ is the Hamiltonian operator.

The dimension of ${\displaystyle e^{-\beta {\hat {H}}}}$ is the number of energy eigenstates of the system.

For a canonical ensemble that is quantum mechanical and continuous, the canonical partition function is defined as

$${\displaystyle Z={\frac {1}{h}}\int \langle q,p|e^{-\beta {\hat {H}}}|q,p\rangle \,\mathrm {d} q\,\mathrm {d} p,}$$

where:

• ${\displaystyle h}$ is the Planck constant;
• ${\displaystyle \beta }$ is the thermodynamic beta, defined as ${\displaystyle {\tfrac {1}{k_{\text{B}}T}}}$;
• ${\displaystyle {\hat {H}}}$ is the Hamiltonian operator;
• ${\displaystyle q}$ is the canonical position;
• ${\displaystyle p}$ is the canonical momentum.

In systems with multiple quantum states s sharing the same energy E_s, it is said that the energy levels of the system are degenerate. In the case of degenerate energy levels, we can write the partition function in terms of the contribution from energy levels (indexed by j) as follows:

$${\displaystyle Z=\sum _{j}g_{j}\cdot e^{-\beta E_{j}},}$$

where g_j is the degeneracy factor, or number of quantum states s that have the same energy level defined by E_j = E_s.

The above treatment applies to quantum statistical mechanics, where a physical system inside a finite-sized box will typically have a discrete set of energy eigenstates, which we can use as the states s above. In quantum mechanics, the partition function can be more formally written as a trace over the state space (which is independent of the choice of basis):

$${\displaystyle Z=\operatorname {tr} (e^{-\beta {\hat {H}}}),}$$

where Ĥ is the quantum Hamiltonian operator. The exponential of an operator can be defined using the exponential power series.

The classical form of Z is recovered when the trace is expressed in terms of coherent states[1] and when quantum-mechanical uncertainties in the position and momentum of a particle are regarded as negligible. Formally, using bra–ket notation, one inserts under the trace for each degree of freedom the identity:

$${\displaystyle {\boldsymbol {1}}=\int |x,p\rangle \langle x,p|{\frac {dx\,dp}{h}},}$$

where |x, p⟩ is a normalised Gaussian wavepacket centered at position x and momentum p. Thus

$${\displaystyle Z=\int \operatorname {tr} \left(e^{-\beta {\hat {H}}}|x,p\rangle \langle x,p|\right){\frac {dx\,dp}{h}}=\int \langle x,p|e^{-\beta {\hat {H}}}|x,p\rangle {\frac {dx\,dp}{h}}.}$$

A coherent state is an approximate eigenstate of both operators ${\displaystyle {\hat {x}}}$ and ${\displaystyle {\hat {p}}}$, hence also of the Hamiltonian Ĥ, with errors of the size of the uncertainties. If Δx and Δp can be regarded as zero, the action of Ĥ reduces to multiplication by the classical Hamiltonian, and Z reduces to the classical configuration integral.

For simplicity, we will use the discrete form of the partition function in this section. Our results will apply equally well to the continuous form.

Consider a system S embedded into a heat bath B. Let the total energy of both systems be E. Let p_i denote the probability that the system S is in a particular microstate, i, with energy E_i. According to the fundamental postulate of statistical mechanics (which states that all attainable microstates of a system are equally probable), the probability p_i will be inversely proportional to the number of microstates of the total closed system (S, B) in which S is in microstate i with energy E_i. Equivalently, p_i will be proportional to the number of microstates of the heat bath B with energy E − E_i:

$${\displaystyle p_{i}={\frac {\Omega _{B}(E-E_{i})}{\Omega _{(S,B)}(E)}}.}$$

Assuming that the heat bath's internal energy is much larger than the energy of S (E ≫ E_i), we can Taylor-expand ${\displaystyle \Omega _{B}}$ to first order in E_i and use the thermodynamic relation ${\displaystyle \partial S_{B}/\partial E=1/T}$, where here ${\displaystyle S_{B}}$, ${\displaystyle T}$ are the entropy and temperature of the bath respectively:

$${\displaystyle {\begin{aligned}k\ln p_{i}&=k\ln \Omega _{B}(E-E_{i})-k\ln \Omega _{(S,B)}(E)\\[5pt]&\approx -{\frac {\partial {\big (}k\ln \Omega _{B}(E){\big )}}{\partial E}}E_{i}+k\ln \Omega _{B}(E)-k\ln \Omega _{(S,B)}(E)\\[5pt]&\approx -{\frac {\partial S_{B}}{\partial E}}E_{i}+k\ln {\frac {\Omega _{B}(E)}{\Omega _{(S,B)}(E)}}\\[5pt]&\approx -{\frac {E_{i}}{T}}+k\ln {\frac {\Omega _{B}(E)}{\Omega _{(S,B)}(E)}}\end{aligned}}}$$

Thus

$${\displaystyle p_{i}\propto e^{-E_{i}/(kT)}=e^{-\beta E_{i}}.}$$

Since the total probability to find the system in some microstate (the sum of all p_i) must be equal to 1, we know that the constant of proportionality must be the normalization constant, and so, we can define the partition function to be this constant:

$${\displaystyle Z=\sum _{i}e^{-\beta E_{i}}={\frac {\Omega _{(S,B)}(E)}{\Omega _{B}(E)}}.}$$

In order to demonstrate the usefulness of the partition function, let us calculate the thermodynamic value of the total energy. This is simply the expected value, or ensemble average for the energy, which is the sum of the microstate energies weighted by their probabilities:

$${\displaystyle \langle E\rangle =\sum _{s}E_{s}P_{s}={\frac {1}{Z}}\sum _{s}E_{s}e^{-\beta E_{s}}=-{\frac {1}{Z}}{\frac {\partial }{\partial \beta }}Z(\beta ,E_{1},E_{2},\cdots )=-{\frac {\partial \ln Z}{\partial \beta }}}$$

or, equivalently,

$${\displaystyle \langle E\rangle =k_{\text{B}}T^{2}{\frac {\partial \ln Z}{\partial T}}.}$$

Incidentally, one should note that if the microstate energies depend on a parameter λ in the manner

$${\displaystyle E_{s}=E_{s}^{(0)}+\lambda A_{s}\qquad {\text{for all}}\;s}$$

then the expected value of A is

$${\displaystyle \langle A\rangle =\sum _{s}A_{s}P_{s}=-{\frac {1}{\beta }}{\frac {\partial }{\partial \lambda }}\ln Z(\beta ,\lambda ).}$$

This provides us with a method for calculating the expected values of many microscopic quantities. We add the quantity artificially to the microstate energies (or, in the language of quantum mechanics, to the Hamiltonian), calculate the new partition function and expected value, and then set λ to zero in the final expression. This is analogous to the source field method used in the path integral formulation of quantum field theory.

In this section, we will state the relationships between the partition function and the various thermodynamic parameters of the system. These results can be derived using the method of the previous section and the various thermodynamic relations.

As we have already seen, the thermodynamic energy is

$${\displaystyle \langle E\rangle =-{\frac {\partial \ln Z}{\partial \beta }}.}$$

The variance in the energy (or "energy fluctuation") is

$${\displaystyle \langle (\Delta E)^{2}\rangle \equiv \langle (E-\langle E\rangle )^{2}\rangle ={\frac {\partial ^{2}\ln Z}{\partial \beta ^{2}}}.}$$

The heat capacity is

$${\displaystyle C_{v}={\frac {\partial \langle E\rangle }{\partial T}}={\frac {1}{k_{\text{B}}T^{2}}}\langle (\Delta E)^{2}\rangle .}$$

In general, consider the extensive variable X and intensive variable Y where X and Y form a pair of conjugate variables. In ensembles where Y is fixed (and X is allowed to fluctuate), then the average value of X will be:

$${\displaystyle \langle X\rangle =\pm {\frac {\partial \ln Z}{\partial \beta Y}}.}$$

The sign will depend on the specific definitions of the variables X and Y. An example would be X = volume and Y = pressure. Additionally, the variance in X will be

$${\displaystyle \langle (\Delta X)^{2}\rangle \equiv \langle (X-\langle X\rangle )^{2}\rangle ={\frac {\partial \langle X\rangle }{\partial \beta Y}}={\frac {\partial ^{2}\ln Z}{\partial (\beta Y)^{2}}}.}$$

In the special case of entropy, entropy is given by

$${\displaystyle S\equiv -k_{\text{B}}\sum _{s}P_{s}\ln P_{s}=k_{\text{B}}(\ln Z+\beta \langle E\rangle )={\frac {\partial }{\partial T}}(k_{\text{B}}T\ln Z)=-{\frac {\partial A}{\partial T}}}$$

where A is the Helmholtz free energy defined as A = U − TS, where U = ⟨E⟩ is the total energy and S is the entropy, so that

$${\displaystyle A=\langle E\rangle -TS=-k_{\text{B}}T\ln Z.}$$

Furthermore, the heat capacity can be expressed as

$${\displaystyle C_{v}=T{\frac {\partial S}{\partial T}}=-T{\frac {\partial ^{2}A}{\partial T^{2}}}.}$$

Suppose a system is subdivided into N sub-systems with negligible interaction energy, that is, we can assume the particles are essentially non-interacting. If the partition functions of the sub-systems are ζ₁, ζ₂, ..., ζ_N, then the partition function of the entire system is the product of the individual partition functions:

$${\displaystyle Z=\prod _{j=1}^{N}\zeta _{j}.}$$

If the sub-systems have the same physical properties, then their partition functions are equal, ζ₁ = ζ₂ = ... = ζ, in which case

$${\displaystyle Z=\zeta ^{N}.}$$

However, there is a well-known exception to this rule. If the sub-systems are actually identical particles, in the quantum mechanical sense that they are impossible to distinguish even in principle, the total partition function must be divided by a N! (N factorial):

$${\displaystyle Z={\frac {\zeta ^{N}}{N!}}.}$$

This is to ensure that we do not "over-count" the number of microstates. While this may seem like a strange requirement, it is actually necessary to preserve the existence of a thermodynamic limit for such systems. This is known as the Gibbs paradox.

It may not be obvious why the partition function, as we have defined it above, is an important quantity. First, consider what goes into it. The partition function is a function of the temperature T and the microstate energies E₁, E₂, E₃, etc. The microstate energies are determined by other thermodynamic variables, such as the number of particles and the volume, as well as microscopic quantities like the mass of the constituent particles. This dependence on microscopic variables is the central point of statistical mechanics. With a model of the microscopic constituents of a system, one can calculate the microstate energies, and thus the partition function, which will then allow us to calculate all the other thermodynamic properties of the system.

The partition function can be related to thermodynamic properties because it has a very important statistical meaning. The probability P_s that the system occupies microstate s is

$${\displaystyle P_{s}={\frac {1}{Z}}e^{-\beta E_{s}}.}$$

Thus, as shown above, the partition function plays the role of a normalizing constant (note that it does not depend on s), ensuring that the probabilities sum up to one:

$${\displaystyle \sum _{s}P_{s}={\frac {1}{Z}}\sum _{s}e^{-\beta E_{s}}={\frac {1}{Z}}Z=1.}$$

This is the reason for calling Z the "partition function": it encodes how the probabilities are partitioned among the different microstates, based on their individual energies. The letter Z stands for the German word Zustandssumme, "sum over states". The usefulness of the partition function stems from the fact that it can be used to relate macroscopic thermodynamic quantities to the microscopic details of a system through the derivatives of its partition function. Finding the partition function is also equivalent to performing a Laplace transform of the density of states function from the energy domain to the β domain, and the inverse Laplace transform of the partition function reclaims the state density function of energies.