[Physics] Imaginary time in quantum and thermodynamics

partition functionpath-integralquantum mechanicsstatistical mechanicswick-rotation

The following question is about chapter 2 of Sakurai's Modern Quantum Mechanics. In the section about propagators and Feynman path integrals (p. 113 in my edition) he gives the following example:
$$
\begin{align}
G(t) \equiv& \int d^3 x' K(x', t; x',0)\\
=&\int d^3 x' \sum_{a'} |\langle x'|a'\rangle|^2 \exp\left(\frac{-iE_{a'}t}{\hbar}\right) \\
=&\sum_{a'}\exp\left( \frac{-iE_{a'}t}{\hbar} \right)
\end{align}
$$

He goes on to say that this is equivalent to taking the trace of the time evolution operator in the $\{|a'\rangle\}$ basis, or a "sum over states", reminiscent of the partition function in statistical mechanics. He then writes $\beta$ defined by
$$
\beta=\frac{it}{\hbar}
$$
real and positive, but with with $t$ purely imaginary, rewriting the last line of the previous example as the partition function itself:
$$
Z=\sum_{a'} \exp\left( -\beta E_{a'} \right).
$$
So my question is this: What is the physical significance (if any) of representing time as purely imaginary? What does this say about the connection between thermodynamics and quantum? The fact that you get the partition function exactly, save for the imaginary time, here seems too perfect to be just a trick. Can someone explain this to me?

Best Answer

I'm going to venture a way to think about this, hoping that I'm at a stage in my research to give a moderately coherent picture, however what follows is not a satisfactory detailed analysis. I'm not aware of an intuitive way to think about this being in textbooks or in the literature more generally; AFAIK thermal states are dealt with purely in terms of their algebraic structure (I note that I will be very happy to be corrected on this).

The Hamiltonian operator $\hat H$ describes the time-like evolution of the state (or of the operators if we take the Heisenberg PoV).

In relativistic models, a thermal equilibrium state is associated with a particular time-like direction. A thermal equilibrium state is invariant under space-like and time-like translations and under rotations orthogonal to a given time-like 4-vector. This definition applies as well to classical systems as to quantum systems.

The relativistic model makes it clear that a thermal state has inertia, because it requires energy and momentum to boost the equilibrium state to a different time-like direction.

In relativistic quantum theory, we have to introduce different Hamiltonian operators for different time-like directions, $\hat H(T^\mu)=\hat P_\mu T^\mu$. This operator commutes with generators of translations and of rotations orthogonal to $T^\mu$, but it does not commute with boosts, just as we require. From a symmetry group perspective, the only resources we have available are the six generators of the Lorentz group, and the definition of the thermal state is intimately connected with boosts (and with no other geometrical structure, such as, for example, conformal transformations). This is where I (very much) regret that I can't make the next step. We want to transform the vacuum from zero inertia to non-zero inertia (or perhaps transform infinite undirected vacuum zero-point energy into time-like directed inertia), which the Gibbs operator seems ideal for. It has felt as if it's on the tip of my tongue for well over five years, which is long enough that, of course, I've never really been close.

Like I say, this is too fragmentary. I'm sorry if it's not Useful to you in its current form. It's certainly frustrating to see for myself that this attempt to articulate the ideas looks like this. This way of thinking is partly grounded in a paper of mine that I have previously mentioned a few times on Physics SE, "A succinct presentation of the quantized Klein–Gordon field, and a similar quantum presentation of the classical Klein–Gordon random field", quant-ph/0411156, Phys. Lett. A 338, 8-12(2005), which discusses QFT free fields in sufficiently weird terms that it will twist your head a little, if you want that.