Ensemble average

The ensemble average of a dynamical variable O(q,p)O(\mathbf{q},\mathbf{p}) over an ensemble is the weighted average over all possible microstates of the system in the thermodynamic limit:

⟨O⟩=∫ρ(q,p,t)O(q,p) dp dq∫ρ(q,p,t) dp dq=1Z∫Oρ dpdq\langle O \rangle =\frac{\int\rho(\mathbf{q},\mathbf{p},t)O(\mathbf{q},\mathbf{p})\,dp\,dq}{\int \rho(\mathbf{q},\mathbf{p},t)\,dp\,dq}=\frac{1}{Z}\int O\rho\ dpdq

where ρ\rho is the density function of the ensemble and ZZ is the partition function. For discrete states, this becomes

⟨O⟩=∑ρ(t)O∑ρ(t)=1Z∑Oρ\langle O \rangle =\frac{\sum \rho(t)O }{\sum \rho(t)}=\frac{1}{Z}\sum O\rho

where the sums go over some label to index the states.

In the quantum case, the average in a system state ∣ψ⟩\ket{\psi} for an observable O^\hat{O}, expressed in a basis of eigenstates {∣n⟩}n∈N\{ \ket{n} \}_{n\in \mathbb{N}}, is

⟨O^⟩=Tr(O^ρ^)Tr(ρ^)=∑np(n)⟨n∣O^∣n⟩\langle \hat{O} \rangle=\frac{\text{Tr}(\hat{O}\hat{\rho})}{\text{Tr}(\hat{\rho})} =\sum_{n}p(n)\braket{ n | \hat{O}| n }

where ρ^\hat{\rho} is the density matrix of the system and p(n)p(n) is the Probability of being in the microstate ∣n⟩\ket{n}1 . Note that this is just the weighted sum of state averages ⟨n∣O^∣n⟩\braket{ n | \hat{O} | n } where the likeliness of each state is the weight. The probabilities can be computed from the coefficients of the Fourier series expansion of ∣ψ⟩\ket{\psi} in the ∣n⟩\ket{n} basis:

∣ψ⟩=∑ncn∣n⟩→p(n)=∣cn∣2\ket{\psi} =\sum_{n}c_{n}\ket{n} \quad\to \quad p(n)=\lvert c_{n} \rvert ^{2}

Time independence

Note how ⟨O⟩\langle O \rangle is in general time dependent, as it depends on ρ(q,p,t)\rho(\mathbf{q},\mathbf{p},t). When ⟨O⟩\langle O \rangle does not depend on time, the system is in thermal equilibrium. Knowing when ⟨O⟩\langle O \rangle isn't time dependent can be useful. For instance, if ρ\rho is itself time independent (ρ(q,p)\rho(\mathbf{q},\mathbf{p}) instead of ρ(q,p,t)\rho(\mathbf{q},\mathbf{p},t)), then it follows that ⟨O⟩\langle O \rangle is itself independent. But then the question is, when does ρ\rho not depend on time? One such case is when

ρ(q,p)=ρ′(H(q,p))\rho(\mathbf{q},\mathbf{p})=\rho'(H(\mathbf{q},\mathbf{p}))

where HH is the Hamiltonian of the system. In this case the prime means differentiation. In fact, if this is true then

dρdt=−∑i=13N(∂ρ′∂H∂H∂qi∂H∂pi−∂ρ′∂H∂H∂pi∂H∂qi)=0\frac{ d \rho }{ d t } =-\sum_{i=1}^{3N} \left( \frac{ \partial \rho' }{ \partial H } \frac{ \partial H }{ \partial q_{i} } \frac{ \partial H }{ \partial p_{i} } -\frac{ \partial \rho' }{ \partial H } \frac{ \partial H }{ \partial p_{i} } \frac{ \partial H }{ \partial q_{i} } \right)=0

This verifies Liouville's theorem.

Footnotes

  1. Note that this probability has nothing to do with quantum indeterminacy. ↩