10.3 Most probable distributions
Method. Maximise \(\ln W\) subject to fixed \(N\) and \(U\), using Lagrange multipliers \(\alpha\) and \(\beta\):
\[\delta\Big[\ln W-\alpha\sum_in_i-\beta\sum_in_iE_i\Big]=0 .\]
FD. With Stirling, \(\ln W_{\rm FD}=\sum_i\big[g_i\ln g_i-n_i\ln n_i-(g_i-n_i)\ln(g_i-n_i)\big]\). Then
\[\frac{\partial\ln W}{\partial n_i}=\ln\frac{g_i-n_i}{n_i}=\alpha+\beta E_i\quad\Rightarrow\quad n_i=\frac{g_i}{e^{\alpha+\beta E_i}+1}.\]
BE. Similarly, \(\partial\ln W_{\rm BE}/\partial n_i=\ln\dfrac{n_i+g_i}{n_i}\) (for \(g_i\gg1\)), giving \(n_i=\dfrac{g_i}{e^{\alpha+\beta E_i}-1}\).
MB. \(\partial\ln(W_{\rm MB}/N!)/\partial n_i=\ln(g_i/n_i)\), giving \(n_i=g_ie^{-\alpha-\beta E_i}\).
Identifying the multipliers. Comparison with thermodynamics gives \(\beta=1/k_BT\) and \(\alpha=-\mu/k_BT\). Writing \(A=e^\alpha\) and the fugacity \(z=1/A=e^{\beta\mu}\):
\[\boxed{\bar n_i=\frac{g_i}{e^{\beta(E_i-\mu)}+a}},\qquad a=+1\ (\text{FD}),\quad 0\ (\text{MB}),\quad -1\ (\text{BE}).\]
The same result follows exactly, without Stirling, from the grand canonical ensemble (Ch. 9).
Classical limit. All three distributions agree when \(e^{\beta(E-\mu)}\gg1\) for all \(E\), i.e. when \((E-\mu)\gg k_BT\). Then \(\bar n\ll1\): states are sparsely occupied, so the question of double occupancy never arises. This is equivalent to \(z=n\lambda^3/g\ll1\).
Restrictions on \(\mu\). For bosons, the requirement \(\bar n\ge0\) demands \(\mu<E_{\min}\) (or \(\mu\le0\) if \(E_{\min}=0\)). For fermions, \(\mu\) can have any sign; at \(T=0\), \(\mu=E_F>0\). For photons and phonons the number is not conserved, so there is no constraint on \(N\) and no multiplier \(\alpha\); hence \(\mu=0\).
Occupation of a single state (\(g=1\)), with \(q=e^{-\beta(\epsilon-\mu)}\):
| Allowed \(n\) |
0, 1 |
0, 1, 2, … |
| \(P(n)\) |
\(P(1)=\bar n\), \(P(0)=1-\bar n\) |
\(P(n)=(1-q)q^n\) (geometric) |
| Mean |
\(\bar n=\dfrac{1}{q^{-1}+1}\) |
\(\bar n=\dfrac{q}{1-q}\) |
| Variance |
\(\bar n(1-\bar n)\) |
\(\bar n(1+\bar n)\) |
For bosons, \(P(0)=1-q\) directly gives \(\bar n=\dfrac{1-P(0)}{P(0)}\).