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In statistical mechanics, the thermodynamic is inverse of the temperature: $\beta \propto T^{-1}$.

In Euclidean QFT, I have often run into the expression like $\beta \propto g^{-2}$ where $g$ is the coupling constant.

Therefore, the coupling constant is proportional to temperature so that we have "weak coupling = low tempertaure" / "strong coupling = high temperature".

However, I do not immediately see why coupling constant can be given the meaning of temperature. That is,

What is physical justification for the relation $\beta \propto g^{-2}$ in the context of Euclidean QFT regarded as statistical mechanics?

Could anyone clarify for me?

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I'm not sure in what context you have been seeing $\beta \sim g^{-2}$, but to me (lattice-QCD person) it seems like you're getting two different $\beta$'s confused. In Lattice-QCD, which is naturally formulated in Euclidean spacetime, the continuum action is discretised onto a lattice by introducing $SU(3)$-valued wilson lines on each of the edges of the lattice. The action is written as: $$S[U] = \frac{\beta}{3} \sum_{n \in \Lambda} \sum_{\mu < \nu} \mathrm{Re}\mathrm{Tr}(1 - U_{\mu \nu}(n)) = \frac{a^4}{2 g^2} \sum_{n \in \Lambda} \sum_{\mu,\nu} \mathrm{Tr}(F_{\mu \nu}(n)^2) + O(a^2) $$ where $\Lambda$ indicates the set of lattice points, $\mu,\nu$ are euclidean spacetime directions, and $U_{\mu\nu}(n)$ is the plaquette in the $\mu,\nu$-th direction, $\beta$ is defined to be $6/g^2$, and $a$ is the lattice spacing (see (2.54,3.4) of Gattringer and Lang). The path integral is then expressed as: $$Z[U] = \int dU e^{-S[U]}$$ where $dU$ is the product haar measure over all the links. Because it is formulated in Euclidean spacetime, this looks just like a statistical mechanics system, where $\beta \to \infty$ is the "$T \to 0$ limit" (where the path integral limits to sampling over only configurations with very small actions), this is also the continuum limit. There are two different perturbative expansions available, the strong-coupling expansions (large $g$, small $\beta$) by expanding observables in powers of $\beta$, or weak-coupling expansion (also known as lattice perturbation theory, by expanding in small $g$, or large $\beta$).

Note however that this $\beta$ has nothing to do with the actual temperature of the Euclidean path integral. By placing the lattice in a box with finite extent in the Euclidean time direction (and periodic boundary conditions), then you are actually simulating at a temperature $T$ corresponding to $1/L$, where $L$ is the length of the box in the Euclidean time direction. Sometimes people refer to this length $L$ by the variable $\beta$, but this is not the same $\beta$ as the one that appears in the lattice action.

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  • $\begingroup$ Thank you for your reply. I didn't mean to have anything to do with $1/L$ in my question. QFT people (not jus gauge theory, I believe) do seem to use the terminolgy "high / low temperature expansion" for "strong / weak coupling expansion" in your post. So, you are saying that this is a mere analogy rising from notational overlap without direct physical meaning? $\endgroup$
    – Keith
    Commented Jun 27 at 15:02
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    $\begingroup$ @Keith Sorry I misunderstood the question slightly then, but yes - high/low temperature expansions in this context are just analogies to statistical mechanics systems, and don't relate to the actual physical temperature of the system. In the lattice-QCD action I wrote above, as you turn the beta appearing in the action smaller, the bare gauge coupling g gets larger, and you introduce more fluctuations to your path integral (it looks like the path integral is sampling at "higher temperature") $\endgroup$ Commented Jun 27 at 15:05
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    $\begingroup$ (and yes this is a general thing in QFT, I'm just answering from what I know about lattice-QCD. maybe if other people know about similar expansions in other QFTs they can comment more) $\endgroup$ Commented Jun 27 at 15:07
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    $\begingroup$ If you forget about taking the continuum (scaling) limit (beta -> infinity) of lattice-QCD, and just think about it as a statistical mechanics problem, then yes beta plays exactly the role of inverse temperature. Interestingly enough, for the Wilson lattice action I wrote above, people believe there is a phase transition at finite beta called the "roughening transition" (see e.g. inspirehep.net/literature/9920) which is expected to be an infinite-order phase transition. I don't think this has actually been formally proved yet though! $\endgroup$ Commented Jun 27 at 15:11
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    $\begingroup$ In lattice-QCD it is both, the bare lattice action I wrote above $S[U] = (\beta/3) \sum \mathrm{ReTr}(1-\mathrm{plaq.})$ is written in terms of a single variable, $\beta$. The value of $\beta$ that you choose defines your lattice spacing $a$ (you don't have separate control over the lattice spacing $a$). Taking $\beta \to \infty$ forces the gauge fields to become more smooth, and the physical lattice spacing $a \to 0$. (see e.g. Section 3.5 in Gattringer and Lang) $\endgroup$ Commented Jun 27 at 15:26

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