### Thermodynamics and Wick Rotation

#### Posted by John Baez

My grad student Mike Stay is a busy guy with many interests. He has a full-time job at Google working on Caja, which is a ‘source-to-source translator for securing Javascript-based web content’. He’s working with me on monoidal bicategories and their applications to computer science. And on the side, he’s been pondering some questions about thermodynamics and the idea of ‘Wick rotation’.

I asked him to post those latter questions here. Can you help him out?

*guest post by Mike Stay*

In A Spring in Imaginary Time, John Baez showed how statics in $n$ dimensions turns into dynamics in $n-1$ dimensions. Consider a spring whose endpoints are fixed. The height of the spring is given by a function

$q(s):\mathbb{R} \to \mathbb{R}.$

The energy $E$ of the spring is given by

$E = \int \left[\frac{k}{2}\left(\frac{d q(s)}{d s}\right)^2 + V(s)\right]ds.$

where we interpret the data as follows:

$\array{s & unitless parametrization \\ q(s) & height of the spring at s, in meters \\ k & spring constant, in k g/s \\ V(s) & gravitational potential energy density at s \\ T(s) = \frac{k}{2}\left(\frac{d q(s)}{d s}\right)^2 & stretching potential energy density at s \\ S = \int (T + V)(s) d s & total energy in the spring }$

If we allow $s$ to be complex—that is, let $s = x + it$—and take $q$ and $V$ to be complex-analytic functions, then we can set $x = 0$ and reinterpret the equation as follows:

$\array{t & time, in seconds \\ q(it) & the height of the particle at time t, in meters \\ m & mass, in k g \\ V(i t) & potential energy at time t \\ T(i t) = \frac{m}{2}\left(\frac{d q(i t)}{i d t}\right)^2 & minus kinetic energy at time t \\ S = \int (T + V)(i t) i d t& \\ = -i \int \left[\frac{m}{2}\left(\frac{\partial q(i t)}{\partial t}\right)^2 -V(i t)\right] dt & -i times the total action of the path }$

If we let the potential energy be proportional to the height, then we find that the critical point $\delta S = 0$ is where $q(s) \propto s^2$; along the real and imaginary axes, it’s a parabola, opening upward for a hanging spring and downward for a thrown particle.

There’s another substitution we could do, however; Wick rotation relates quantum mechanics to statistical mechanics by replacing $i t / \hbar$ with $\beta/k_B$. If we do the same thing in this equation, the interpretation is less clear to me:

$\array{\beta & inverse temperature, in K^{-1} \\ q(\beta) & height of something at invtemp \beta, in meters \\ r & bits / m^2 K \\ T(\beta) = \frac{r}{2}\left(\frac{d q(s)}{d \beta}\right)^2 &"kinetic" energy \\ V(\beta) & potential energy \\ S = \int (T + V)(\beta) d \beta & total bits of entropy in the path }$

where one bit = $k_B ln(2) \; [J/K]$.

What kind of system is governed by this equation? Does the fact that
$r$ has units of information per square meter Kelvin have anything to
do with the holographic principle? What does it mean for there to be
entropy in a *single* path? When $\delta S = 0,$ does this
maximize the entropy?

Say we have a system with microstates indexed by $j$. The entropy of a distribution $p$ is

$S = - \sum_j p(j) ln p(j).$

We’re told that the average energy is $E$; the distribution that maximizes entropy subject to that constraint is the Gibbs distribution

$p(j) = \frac{exp(-\beta E)}{Z},$

where

$Z = \sum_j exp(-\beta E_j)$

and

$\beta = \frac{\partial S}{\partial E}.$

Wick rotation suggests there ought to be an analogous situation in which

$i t = \frac{\partial S}{\partial E};$

where $S$ is action. If so, then action should be a function of some distribution-like thing. Is it the wave function?

We can extend the analogy between quantum and statistical mechanics:

$\array{wave function modulo global phase & & distribution modulo multiplicative constant \\ \langle Q \rangle = \langle\phi| exp(-H i t/\hbar) Q|\phi\rangle & & \langle Q \rangle = \langle\phi| exp(-H \beta/k_B) Q|\phi\rangle}$

In the second line, $\phi$ is the uniform distribution; in the quantum case the equation only holds when $Q$ is an observable that commutes with $H$, while in the statistical case, it’s any observable.

What’s the complete picture? Do we get pairs of conjugate observables whose product has units of entropy? Can we talk about perturbation theory and Feynman diagrams for a statistical ensemble? How far does the analogy go? What are the important differences?

## Re: Thermodynamics and Wick Rotation

A very partial answer to your question “Can we talk about perturbation theory and Feynman diagrams for a statistical ensemble?” is “yes, as for any integral”. More precisely, recall that in Dyson’s original work on Feynman diagrams, he rejected Feynman’s idea that the diagrams were pictures of physical events in favor of a purely formal mathematical interpretation: the diagrams for Dyson do nothing more nor less than keep track of the terms of an asymptotic expansion of an integral. Let $f$ be any real-valued smooth function on $\mathbb{R}^n$, $C \subseteq \mathbb{R}^n$ compact and the closure of its interior, and suppose that $f$ has a unique critical point in $C$. Then as $\epsilon \to 0$, the asymptotics of $\int_C \exp( i\epsilon^{-1} f)$ depend only on the values of $f$ in a formal neighborhood of its critical point, i.e. only on the Taylor coefficients. When the critical point is a minimum, the same statement holds upon replacing $i$ by $-1$, and the dependence on the Taylor coefficients is the same. In particular, when the critical point is nondegenerate, and now letting $\epsilon$ be pure imaginary or negative as the case may be, we have:

$\epsilon \log \int \exp( \epsilon^{-1} f) = \sigma + f^{(0)} - \frac{\epsilon}{2} \log \det f^{(2)} + \sum_{\text{connected diagrams }\Gamma} \frac{\operatorname{ev}(\Gamma) \, \epsilon^{\beta(\Gamma)}}{|\operatorname{Aut}\Gamma|} + O(\epsilon^\infty)$

Here $f^{(k)}$ is the $n$th Taylor coefficient of $f$ at the critical point (a symmetric tensor in $(\mathbb{R}^n)^{\otimes k}$; $\sigma$ depends only on $\epsilon, n$ and the signature of $f^{(2)}$; a

diagramis a graph with all vertices trivalent or higher; theevaluation$\operatorname{ev}(\Gamma)$ of a diagram $\Gamma$ places a weight on each vertex, a balloon on each edge, and reads it from top to bottom as a “string” diagram in Vect where the vertical edges as $\mathbb{R}^n$, the caps are the $(-f^{(2)})^{-1}$, and an $k$-valent vertex is $f^{(k)}$; and thefirst Betti number$\beta(\Gamma)$ is the dimension of the first homology of $\Gamma$ (one plus the number of edges minus the number of vertices). The sum is of course a Baez-and-Dolan “groupoid integral” over the groupoid of all connected diagrams. The LHS is not analytic in $\epsilon$; the RHS makes it clear that the failure of analyticity is in $\sigma$ alone. Exponentiating, we have $\exp\sigma = (2\pi\epsilon)^{n/2} i^{-\nu(f^{(2)})}$, where $\nu(f^{(2)})$ is the Morse index of $f$ at the critical point, i.e. the number of negative eigenvalues of $f^{(2)}$, i.e. the maximal dimension of any subspace of $\mathbb{R}^n$ on which $f^{(2)}$ is negative-definite, if I’m remembering the signs correctly, which I probably am not. It’s worth emphasizing that upon exponentiating the sum of connected diagrams, you get the sum of all diagrams, still weighted in the groupoid-integral sense.(Incidentally, it’s tempting to interpret $f^{(0)}$ as the value of the connected Feynman diagram with a unique 0-valent vertex. In fact, remembering that $f^{(1)} = 0$, you can try to consider the sum of

allconnected diagrams, without the condition that all vertices be trivalent. When you do this, you should evaluate a $k$-valent vertex as $f^{(k)}$, and evaluate an edge not as $-(f^{(2)})^{-1}$ but as “$0^{-1}$” — more precisely, add and subtract from $f^{(2)}$ some metric on $\mathbb{R}^n$, and use the inverse metric for a cap and the rest for a bivalent vertex. Then there are infinitely many diagrams per Betti number, and the sums do not converge, but formally they equal what is above. Actually, if all functions were analytic, and if all reasonable series converged, then you could pickanypoint in $C$, and do the same thing, now allowing 1-valent vertices to represent $f^{(1)}$.)Thus, since one formalism for statistical mechanics is in a “path integral” sense, it is perfectly reasonable to employ Feynman diagrams for computing asymptotics. But I don’t believe that the diagrams depict pictures of physical events in statistical mechanics. Similarly, from the path-integral formalism for quantum mechanics, you can compute asymptotics via Feynman diagrams — this is implicit in the work by e.g. DeWitt-Morette and also Kleinert, and I have some papers on the arXiv that make the diagrammatic approach very explicit. But again the diagrams do not depict physical events. It is only for formal QFT on a Riemannian manifold that perturbs a free theory that it is reasonable to interpret the pictures as events.