# The Partition Function

Consider two systems (1 and 2) in thermal contact such that

 $$N_2$$ $$\gg$$ $$N_1$$ $$E_2$$ $$\gg E_1$$ $$N$$ $$N_1 + N_2; E = E_1 + E_2$$ $$\text {dim} (x_1)$$ $$\gg$$ $$\text {dim} (x_2)$$

and the total Hamiltonian is just $$H (x) = H_1 (x_1) + H_2 (x_2)$$

Since system 2 is infinitely large compared to system 1, it acts as an infinite heat reservoir that keeps system 1 at a constant temperature $$T$$ without gaining or losing an appreciable amount of heat, itself. Thus, system 1 is maintained at canonical conditions, $$N, V, T$$.

The full partition function $$\Omega (N, V, E )$$ for the combined system is the microcanonical partition function

$\Omega(N,V,E) = \int dx \delta(H(x)-E) = \int dx_1 dx_2 \delta (H_1(x_1) + H_2(x_2)-E)$

Now, we define the distribution function, $$f (x_1)$$ of the phase space variables of system 1 as

$f(x_1) = \int dx_2 \delta (H_1(x_1)+ H_2(x_2)-E)$

Taking the natural log of both sides, we have

$\ln f(x_1) = \ln \int dx_2 \delta (H_1(x_1) + H_2(x_2) - E)$

Since $$E_2 \gg E_1$$, it follows that $$H_2 (x_2) \gg H_1 (x_1)$$, and we may expand the above expression about $$H_1 = 0$$. To linear order, the expression becomes
 $$\ln f (x_1)$$ $$\ln \int dx_2 \delta (H_2(x_2)-E) + H_1(x_1) \frac {\partial }{ \partial H_1 (x_1)} \ln \int dx_2 \delta (H_1(x_1) + H_2(x_2) - E) \vert _{H_1(x_1)=0}$$ $$\ln \int dx_2 \delta (H_2(x_2)-E) -H_1(x_1) \frac {\partial}{\partial E} \ln \int dx_2 \delta (H_2(x_2)-E)$$

where, in the last line, the differentiation with respect to $$H_1$$ is replaced by differentiation with respect to $$E$$. Note that

 $$\ln \int dx_2 \delta (H_2( _2)-E)$$ $$\frac {S_2 (E)}{k}$$ $$\frac {\partial}{\partial E} \ln \int dx_2 \delta (H_2(x_2)-E)$$ $$\frac {\partial}{\partial E} \frac {S_2(E)}{k} = \frac {1}{kT}$$

where $$T$$ is the common temperature of the two systems. Using these two facts, we obtain

 $$\ln f (x_1)$$ $$\frac {S_2 (E)}{k} - \frac {H_1 (x_1)}{kT}$$ $$f (x_1)$$ $$e^{\frac {S_2(E)}{k}}e^{\frac {-H_1(x_1)}{kT}}$$

Thus, the distribution function of the canonical ensemble is

$f(x) \propto e^{\frac {-H(x)}{kT}}$

The prefactor $$exp (\frac {S_2 (E) }{k} )$$ is an irrelevant constant that can be disregarded as it will not affect any physical properties.

The normalization of the distribution function is the integral:

$\int dxe^{\frac {-H(x)}{kT}} \equiv Q(N,V,T)$

where $$Q (N, V, T )$$ is the canonical partition function. It is convenient to define an inverse temperature $$\beta = \frac {1}{kT}$$.  $$Q (N, V, T )$$ is the canonical partition function. As in the microcanonical case, we add in the ad hoc quantum corrections to the classical result to give

$Q(N,V,T) = \frac {1}{N!h^{3N}} \int dx e^{-\beta H(x)}$

The thermodynamic relations are thus,

### Hemlholtz free energy

$A (N, V, T ) = - \frac {1}{\beta} \ln Q (N, V, T )$

$P = -\left ( \frac {\partial A}{\partial V} \right )_{N,T} = kT \left( \frac {\partial \ln Q(N,V,T)}{\partial V} \right )_{N,T}$

To see that this must be the definition of $$A (N, V, T )$$ , recall the definition of $$A$$:

$A = E - TS = \langle H (x) \rangle - TS$

But we saw that

$S = - \left ( \frac {\partial A}{\partial T } \right ) _{N,V}$

Substituting this in gives

$\frac {\partial A}{\partial T} = \frac {\partial A}{\partial \beta} \frac {\partial \beta }{\partial T} = - \frac {1}{kT^2} \frac {\partial A}{\partial \beta}$
it follows that

$A = \langle H(x) \rangle + \beta \frac {\partial A}{\partial \beta}$

This is a simple differential equation that can be solved for $$A$$. We will show that the solution is

$A = - \frac {1}{\beta} \ln Q (\beta)$

Note that

$\beta \frac {\partial A}{\partial \beta} = \frac {1}{\beta} \ln Q (\beta) - \frac {1}{Q} \frac {\partial Q}{\partial \beta} = A - \langle H(x)\rangle$

Substituting in gives, therefore

$A = \langle H(x)\rangle + A - \langle H(x)\rangle = A$
so this form of $$A$$ satisfies the differential equation.Other thermodynamics follow:

$A = \langle H(x) \rangle - T \frac {\partial A}{\partial T}$

or, noting that

Average energy:

 $$E$$ $$\langle H(x)\rangle = \frac {1}{Q} C_N \int dx H(x) e^{-\beta H(x)}$$ $$- \frac {\partial}{\partial \beta} \ln Q(N,V,T)$$

#### Pressure

$P = -\left ( \frac {\partial A}{\partial V} \right )_{N,T} = kT \left ( \frac {\partial \ln Q (N,V,T)}{\partial V} \right )_{N,T}$

#### Entropy

 $$S$$ $$- \frac {\partial A}{\partial T} = - \frac {\partial A}{\partial \beta} \frac {\partial \beta}{\partial T} = \frac {1}{kT^2} \frac {\partial A}{\partial \beta}$$ $$k \beta^2 \frac {\partial}{\partial \beta} \left( -\frac {1}{\beta} \ln Q(N,V,T)\right ) = -k \beta \frac {\partial \ln Q}{\partial \beta} + k \ln Q$$ $$k \beta E + k \ln Q = k \ln Q + \frac {E}{T}$$

#### Heat capacity at constant volume

$C_V = \left ( \frac {\partial E}{\partial T} \right )_{N,V} = \frac {\partial E}{\partial \beta} \frac {\partial \beta}{\partial T} = k \beta^2 \frac {\partial}{\partial \beta^2} \ln Q (N,V,T)$