# Why is Euclidean Time Periodic?

I've been reading a bit about finite temperature quantum field theory, and I keep coming across the claim that when one Euclideanizes time $$it\to\tau,$$ the time dimension becomes periodic, with period related to the inverse temperature $$\beta$$. Can someone please explain where the periodicity comes from and how we know to identify the period with $$\beta$$?

I don't think that Wick rotated time $$\tau$$ is periodic by itself. But it turns out that thermal averages of operators are periodic with respect to the variable $$\tau$$. Consider a generic time dependent operator $$\hat{A}(\tau)$$ with the standard time evolution expansion $$\hat{A}(\tau) = e^{\hat{H}\tau} \hat{A}(0) e^{-\hat{H}\tau}$$ and consider its thermal average $$A(\tau) \equiv \hat{\left\langle A (\tau) \right\rangle } = Z^{-1} \mathrm{Tr}[e^{-\beta \hat{H} }\hat{A}(\tau)]$$, where $$Z$$ is the parition function. You can prove rather simply that $$A(\tau + \beta) = A(\tau)$$ by exploiting firstly the fact that $$e^{-\beta\hat{H}} e^{\beta\hat{H}} = 1$$ and secondly the cyclic property of the trace (I'll leave this as an exercise).

However, not all the objects that we are interested in are necessarily periodic. A remarkable example is the Green function at positive time $$\tau \geq 0$$ $$G_{kp}(\tau) = - \left\langle \hat{\psi}_k(\tau) \hat{\psi}_p^{\dagger}(0) \right\rangle$$ which is written in terms of time dependent field operators. In fact you can prove that $$G_{kp}(\tau+\beta) = \zeta G_{kp}(\tau)$$, where $$\zeta = +1$$ if $$\hat{\psi}$$ is a bosonic operator, and $$\zeta = -1$$ if it is fermionic, so that the function is either periodic or antiperiodic.

In conclusion, the (anti)periodicity of functions with respect to euclidean time relies on how you compute thermal averages.

• So then if fermionic Greens functions have antiperiodic boundary conditions, why can we not identify the Euclidean time as having a period of $2\beta$? Or is there something I'm missing? Jun 11, 2020 at 21:19
• You are right. However it's remarkable that if you take intervals of length $\beta$ in imaginary time, the fermionic Green function still has some symmetry property. Also taking intervals of length $\beta$ is more standard in books and papers as far as I know Jun 11, 2020 at 21:28
• Is the antiperiodic nature of fermionic Greens functions because fermionic fields anticommute? Jun 12, 2020 at 15:28
• Yes, exactly! If you do the calculation to prove it, at some point you need to swap the positions of the field operators, and a minus sign comes up from the fact that they anticommute Jun 12, 2020 at 15:30
• Why isn't $A(\tau)$ entirely independent of $\tau$? After all $A(\tau) = Z^{-1}\mathrm{Tr}(e^{-\beta H} e^{H \tau}\hat{A}(0) e^{-H \tau}) = Z^{-1}\mathrm{Tr}(e^{-\beta H} \hat{A}(0) )$. This is because $e^{\tau H}$ and $e^{-\beta H}$ commute. Jan 31, 2021 at 21:20

If you make the substitution: $$\frac{\mathrm{i}t}{\hbar}\rightarrow\frac{1}{k_{\mathrm{B}}T} = \beta,$$ the quantum evolution operator becomes: $$\mathrm{e}^{-\mathrm{i}\frac{\hat{H}t}{\hbar}} \rightarrow \mathrm{e}^{-\beta \hat{H}},$$ which looks familiar in the context of thermal statistical field theory since the partition function $$Z$$ is given by: $$Z = \mathrm{Tr}\left [ \mathrm{e}^{-\beta \hat{H}} \right ].$$

$$\tau$$ is related to $$\mathrm{i}t$$ with maybe a minus sign and some conventions on units.

If you think of $$\mathrm{e}^{-\beta \hat{H}}$$ as an evolution operator $$\mathrm{e}^{-\tau \hat H}$$, taking a state and evolving it from $$\tau = 0$$ to $$\tau = \beta$$, then it's clear that the boundary conditions are periodic.

Because if you take $$\tau = 2\beta$$, then: $$\mathrm{e}^{-2\beta \hat{H}} = \mathrm{e}^{-\beta \hat{H}}\cdot \mathrm{e}^{-\beta \hat{H}},$$ i.e. the first "evolution operator" evolves from $$\tau = 0$$ to $$\tau = \beta$$, and the second one also. I.e. the end-time of the first evolution operator becomes the new $$\tau = 0$$ for the second evolution operator. So the evolution is periodic. In the context of thermal statistical field theory. Don't know about generic Euclidean time.

Suppose you're doing Euclidean single-particle quantum mechanics and you want to compute the partition function $$Z = \mathrm{Tr}(e^{-\beta H})$$. First start by writing $$Z = \int dq $$ The integrand can in turn be written, following the famous Feynman derivation of the path integral $$ = \int_{x(0) = x(\beta) = q} \mathcal{D}[x(\tau)] e^{-S_E[x]}$$ where $$S_E[x]$$ is the Euclidean action of the trajectory $$x(\tau)$$. So we have here a sum over trajectories that start and finish at $$q$$, with the time elapsed in the trajectory being $$\beta$$. When we integrate over $$q$$ we get a "sum" over all loops (paths that start and finish at the same point) which take time $$\beta$$. $$Z = \int_{x(0) = x(\beta)} \mathcal{D}[x(\tau)] e^{-S_E[x]}$$ Now because a starting point of the loop is arbitrary and $$S_E$$ does not depend on which point we label as the starting point, we could always reparametrise our Euclidean trajectory so that the "starting position" was another point along the loop. The only requirement is that the overall time of the loop is $$\beta$$. In this sense the Euclidean time parameter $$\tau$$ of each loop is periodic with period $$\beta$$. This argument generalises to path integrals of fields and also applies to quantities other than the partition function.