6.2. Statistical Physics

6.2.1. Microcanonical Ensemble

The entropy is equal to:

S = k_{\mathrm{B}} \log W = {1\over\beta T} \log W

where W is the micronanonical partition function, the number of microstates within the range of energy.

6.2.2. Canonical Ensemble

The partition function is:

Z_{can} = \sum_n e^{-\beta E_n}

The Helmholtz free energy is equal to:

F = -{1\over\beta} \log Z_{can} = -{1\over\beta} \log \sum_n e^{-\beta E_n}

6.2.3. Grand Canonical Ensemble

The partition function for fermions is:

Z_{gr} = \sum_n e^{-\beta(E_n - \mu N_n)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^1 \right) e^{-\beta(E_n - \mu N_n)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^1 \right) e^{-\beta\left(\sum_\alpha n_\alpha \epsilon_\alpha - \mu \sum_\alpha n_\alpha\right)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^1 \right) \prod_\alpha e^{-\beta n_\alpha\left(\epsilon_\alpha - \mu\right)} = = \prod_\alpha \left(\sum_{n_\alpha=0}^1 e^{-\beta n_\alpha\left(\epsilon_\alpha - \mu\right)} \right) = = \prod_\alpha \left(1 + e^{-\beta (\epsilon_\alpha - \mu)} \right)

Similarly, for bosons we would get:

Z_{gr} = \sum_n e^{-\beta(E_n - \mu N_n)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^\infty \right) e^{-\beta(E_n - \mu N_n)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^\infty \right) e^{-\beta\left(\sum_\alpha n_\alpha \epsilon_\alpha - \mu \sum_\alpha n_\alpha\right)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^\infty \right) \prod_\alpha e^{-\beta n_\alpha\left(\epsilon_\alpha - \mu\right)} = = \prod_\alpha \left(\sum_{n_\alpha=0}^\infty e^{-\beta n_\alpha\left(\epsilon_\alpha - \mu\right)} \right) = = \prod_\alpha \left(1 - e^{-\beta (\epsilon_\alpha - \mu)} \right)^{-1}

The grand potential for fermions is then equal to:

\Omega = -{1\over\beta} \log Z_{gr} = = -{1\over\beta} \log\left( \prod_\alpha \left(1 + e^{-\beta (\epsilon_\alpha - \mu)} \right) \right) = = -{1\over\beta} \sum_\alpha \log\left( 1 + e^{-\beta (\epsilon_\alpha - \mu)} \right)

Similarly, the grand potential for bosons is equal to:

\Omega = -{1\over\beta} \log Z_{gr} = = -{1\over\beta} \log\left( \prod_\alpha \left(1 - e^{-\beta (\epsilon_\alpha - \mu)} \right)^{-1} \right) = = -{1\over\beta} \sum_\alpha \log\left( 1 - e^{-\beta (\epsilon_\alpha - \mu)} \right)^{-1} = = {1\over\beta} \sum_\alpha \log\left( 1 - e^{-\beta (\epsilon_\alpha - \mu)} \right)

6.2.4. Examples

Ideal Gas

Ideal gas is simply a system of classical particles, where for a given microstate specified by a set of coordinates \mathbf{x}_i and momenta \mathbf{p}_i , the total energy of the microstate is given by the following Hamiltonian:

H(\mathbf{x}_i, \mathbf{p}_i) = \sum_{i=1}^N {p_i^2 \over 2 m}\,,

that is, the particles are non-interacting, each has a mass m and a momentum \mathbf{p}_i. The canonical partition function is then equal to:

Z_{can}(T, V, N) = \sum_n e^{-\beta E_n} = = \int {\d^{3N} x\, \d^{3N} p \over N! (2\pi\hbar)^{3N}} e^{-\beta H(\mathbf{x}_i, \mathbf{p}_i)} = = \int {\d^{3N} x\, \d^{3N} p \over N! (2\pi\hbar)^{3N}} e^{-\beta \sum_{i=1}^N {p_i^2 \over 2 m}} = = {1\over N!}\left( \int {\d^3 x \d^3 p \over (2\pi\hbar)^3} e^{-\beta {p^2 \over 2 m}}\right)^N = = {1\over N!}\left(V \int_0^\infty {4\pi p^2 \d p \over (2\pi\hbar)^3} e^{-\beta {p^2 \over 2 m}}\right)^N = = {1\over N!}\left(V {4\pi \over (2\pi\hbar)^3} {\sqrt\pi (2m)^{3\over 2}\over 4 \beta^{3\over2}} \right)^N = = {1\over N!}\left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)^N\,,

where we used the following integral:

\int_0^\infty p^2 e^{-\alpha p^2} \d p = {\sqrt\pi\over4\alpha^{3\over2}}\,.

The Helmholtz free energy is then equal to:

F(T, V, N) = -{1\over\beta} \log Z_{can}(T, V, N) = = -{1\over\beta} \log\left( {1\over N!} \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)^N\right) = = {N\over\beta} \left({\log N!\over N} - \log \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)\right) = = {N\over\beta} \left(\log N - 1 + O\left(\log N\over N\right) - \log \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)\right) = = {N\over\beta} \left( - \log \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} {V e \over N}\right) + O\left(\log N\over N\right) \right) = = {N\over\beta} \left({3\over2} - \log \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} {V e^{5\over2} \over N}\right) + O\left(\log N\over N\right) \right) = = N k_\mathrm{B} T \left({3\over2} - \log \left( {V T^{3\over2} \over N \left(2\pi\hbar^2\over m k_\mathrm{B}e^{5\over3}\right)^{3\over2} }\right) + O\left(\log N\over N\right) \right)\,,

where we used the Stirling’s approximation for N!. For large N this is equal to the Helmholtz free energy of the ideal gas (see Ideal Gas):

F(T, V, N) = N k_\mathrm{B} T \left(c_V - \log \left({VT^{c_V}\over N\Phi}\right) \right)\,,

with c_V={3\over2} and \Phi=\left(2\pi\hbar^2\over m k_\mathrm{B}e^{5\over3}\right)^{3\over2}. See that section where all other thermodynamic properties are derived from it.

We can also start from the grand canonical partition function:

Z_{gr}(T, V, \mu) = \sum_{N=0}^\infty e^{\beta \mu N} Z_{can}(T, V, N) = = \sum_{N=0}^\infty e^{\beta \mu N} {1\over N!}\left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)^N = = \sum_{N=0}^\infty {1\over N!}\left(e^{\beta \mu} \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)^N = = e^{e^{\beta \mu} \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V}

And the grand potential is:

\Omega(T, V, \mu) = -{1\over\beta} \log Z_{gr}(T, V, \mu) = = -{1\over\beta} {e^{\beta \mu} \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V} = = -{k_\mathrm{B} T} {e^{\mu\over k_\mathrm{B} T} \left(m k_\mathrm{B} T\over2\pi\hbar^2\right)^{3\over2} V} = = - {k_\mathrm{B} V T^{5\over2} \over \left(2\pi\hbar^2\over m k_\mathrm{B}e^{5\over3}\right)^{3\over2} e^{{5\over2}-{\mu\over k_\mathrm{B} T}}} \,.

This is equal to the grand potential of an ideal gas:

\Omega(T, V, \mu) = - {k_\mathrm{B} V T^{c_p} \over \Phi e^{c_p-{\mu\over k_\mathrm{B} T}}} \,,

with c_p={5\over2} and \Phi=\left(2\pi\hbar^2\over m k_\mathrm{B}e^{5\over3}\right)^{3\over2}. The thermodynamics section then shows that the corresponding Helmholtz free energy is the same as we obtained above from the canonical ensemble. Note that we also obtained the same \Phi as before.