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If I know trivial (anti)commutation relation for some operators (let's say Fermi operators), I can only use it if they are in the same moment of time. If I have their time dependence and they don't have to be observed in the same moment of time, then those (anti)commutation relations are not that trivial and so I can't use them that way.

But what about imaginary time dependence (via the exponential factor)? For example, let's consider Fermi operators in Heisenberg picture, or Dirac (interaction) picture. Do they have some trivial (anti)commutation relations that can be used or not?

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  • $\begingroup$ You can use them in a time-ordered operator $T_\tau$. $\endgroup$
    – Jack
    Aug 7, 2021 at 23:57

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I assume that you are talking about something related to many-body Green's function method. Usually, we learn the (anti)commutation from basic quantum mechanics, which are formulated in the Schrodinger picture, and hence the operator is time independence. However, when you want to perform many-body perturbation expansion, you need Dirac and Heisenberg pictures, in which all the operators are time-dependent. Now, what are the corresponding commutation relations between creation and annihilation operators? This is what you want to know? Simply, it will be written as:

$$\{c_{k_1}(t_1),c^\dagger_{k_2}(t_2)\}=\delta_{k_1k_2}\delta(t_1-t_2)\tag{1}$$

or

$$\{c_{k_1}(\tau_1),c^\dagger_{k_2}(\tau_2)\}=\delta_{k_1k_2}\delta(\tau_1-\tau_2)\tag{2}$$

The imaginary time technique doesn't change the commutation relation and is just used to unify the relation between the time-evolution operator and the Boltzmann factor. You can find more from the famous book: Quantum Statistical Mechanics (Leo P. Kadanoff and Gordon Baym).

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