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A Similarity in form of time-evolution and Gibbs weight?

  1. Jul 13, 2017 #1
    Why do the time-evolution operator in quantum mechanics ##\exp{iHt}## and the Gibbs-weight operator in statistical physics ##\exp{-H/T}## have the same functional form? – i.e. both exponentials of the Hamiltonian operator.
    The Matsubara trick/method just takes this as a fact in thermal QFT; but one might wonder whether there’s a deeper reason why these two operators (with apparently different physical meanings) look so similar. My feeling is that it is related to generating functions in some way, but not sure how.
    Thanks
     
  2. jcsd
  3. Jul 14, 2017 #2

    vanhees71

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    The time evolution operator for states in the Schrödinger picture is
    $$\hat{U}=\exp(-\mathrm{i} \hat{H} t).$$
    This is due to the equations of motion of QT.

    The statistical operator in the canonical ensemble is
    $$\hat{\rho}(T)=\frac{1}{Z} \exp(-\beta \hat{H}),$$
    where ##\beta=1/T## is the inverse temperature due to the maximum-entropy principle for the case that the average energy of the system is given. ##\beta## is the Lagrange parameter for this constraint.

    It's just luck that both operators are so similar, and you can do perturbation theory along the Matsubara temperature to calculate finite-temperature QFT almost as in vacuum QFT. However in thermal field theory you have (anti-)periodic boundary conditions for the field operators, so that instead of an energy integral in the vacuum QFT you have the sum over the Matsubara frequency in thermal QFT. For more details, see

    http://th.physik.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf
     
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