Equipartition of energy

Let $x_i=q_i,p_i$, the following

$$⟨x_i\frac{\partial H}{\partial x_k}⟩={\delta }_{ij}{k}_{B}T$$

Proof (For the canonical ensemble)

$$⟨x_i\frac{\partial H}{\partial x_k}⟩=\frac{1}{Z}\int x_i\frac{\partial H}{\partial x_k}{e}^{-\beta H}d\Gamma$$

using that

$$\frac{\partial }{\partial x_i}\left(x_{i}H\right)={\delta }_{ij}H+x_i\frac{\partial H}{\partial x_k}$$ $$=\frac{1}{Z}\int \frac{\partial }{\partial x_i}\left(x_{i}H\right)e^{-\beta H}d\Gamma -{\delta }_{ij}\frac{1}{Z}\int H{e}^{-\beta H}d\Gamma$$

the first term is zero, the second is just the average energy $U$

$$={\delta }_{ij}U$$

but in the canonical ensemble this is the same as

$$={\delta }_{ij}\frac{1}{\beta }={\delta }_{ij}{k}_{B}T$$

A famouse use of this theorem is that any quadratic term in the hamiltonian

H = \sum_i \alpha_i x_i^2 $$result in a $\frac{1}{2} k_B T$ term in the total average energy, since

\sum_i\left\langle x_i \frac{\partial H}{\partial x_i}\right\rangle = \sum_i\langle x_i 2\alpha x_i \rangle = 2 \langle H \rangle

$$sothat$$

\langle H \rangle = \sum_i \frac 12 k_B T