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I read in a book on quantum fluctuations and quantum noise that, at thermal equilibrium the classical canonical variables are uncorrelated, ie: $$\langle xp\rangle=\langle x\rangle\langle p\rangle$$ But I am not sure to understand the sense of at thermal equilibrium, for me it just means $$\langle x^2\rangle=\langle p^2\rangle=\frac{T}{2}$$ in the correct units with $T$ the temperature. On the other hand what I can derive easily in the canonical ensemble is: $$\langle xp \rangle = Z^{-1}\int {\cal{DxDp}}\;\text{e}^{-\beta H(x,p)}xp$$ and $$\langle x\rangle\langle p \rangle = Z^{-2}\int {\cal{DxDpDx'Dp'}}\;\text{e}^{-\beta H(x,p)}\text{e}^{-\beta H(x',p')}xp'$$ which are equal if we have the following equality $$H(x,p) = H_1(x) + H_2(p)$$ separability of $x$ and $p$ in the Hamiltonian, which seems unrelated to the previous hypothesis of thermal equilibrium (used here by taking the mean over the canonical distribution of course).

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I think you are right. You could study the case of a non separable Hamiltonian (particle in a constant magnetic field) and you (obviously) will find correlations. –  Fabian Dec 27 '12 at 21:50
@Fabian, Not in classical statistical mechanics for a charged particle in a magnetic field. See the Bohr-van Leeuwen theorem –  Vijay Murthy Dec 28 '12 at 9:56
@VijayMurthy: why should it follow from the Bohr-van Leeuwen theorem that the position and the canonical momentum are not correlated??? –  Fabian Dec 28 '12 at 16:56
My understanding was that the position-momentum correlation is related to the orbital angular momentum which is related the diamagnetic moment for classical charged particles. Bohr-van Leeuwen theorem says that there is no classical diamagnetism, and so I said that there cannot be non-zero $\langle xp\rangle$. But this is only for $(p-eA/c)^2$ kind of couplings for a charged particle in a magnetic field, and not a generic statement. –  Vijay Murthy Dec 28 '12 at 19:36

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