# 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

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

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

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

\(\ln f (x_1)\) | \( \ln \int dx_2 \delta (H_2(x_2)-E) + H_1(x_1) \frac {\partial }{ \partial H_1 (x_1)} \ln \int dx_2 \delta (H_1(x_1) + H_2(x_2) - E) \vert _{H_1(x_1)=0}\) | ||

\( \ln \int dx_2 \delta (H_2(x_2)-E) -H_1(x_1) \frac {\partial}{\partial E} \ln \int dx_2 \delta (H_2(x_2)-E)\) |

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 normalization of the distribution function is the integral:

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

*ad hoc*quantum corrections to the classical result to give

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

**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

\(S\) | \(- \frac {\partial A}{\partial T} = - \frac {\partial A}{\partial \beta} \frac {\partial \beta}{\partial T} = \frac {1}{kT^2} \frac {\partial A}{\partial \beta} \) | ||

\(k \beta^2 \frac {\partial}{\partial \beta} \left( -\frac {1}{\beta} \ln Q(N,V,T)\right ) = -k \beta \frac {\partial \ln Q}{\partial \beta} + k \ln Q\) | |||

\( k \beta E + k \ln Q = k \ln Q + \frac {E}{T}\) |