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If a degree of freedom xn appears only as a quadratic term anxn2 in the Hamiltonian H, then the first of these formulae implies that : k_{B} T = \Bigl\langle x_{n} \frac{\partial H}{\partial x_{n}}\Bigr\rangle = 2\langle a_n x_n^2 \rangle, which is twice the contribution that this degree of freedom makes to the average energy \langle H\rangle.
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