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

muscaria

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

Related Quantum Physics News on Phys.org

vanhees71

Gold Member
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

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

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving