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?

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    $\begingroup$ I'll add that the Wick rotation isn't just interesting in QM - in classical mechanics, a problem in statics becomes a problem in dynamics. math.ucr.edu/home/baez/classical/spring.pdf $\endgroup$
    – dbrane
    Commented May 23, 2011 at 16:11
  • $\begingroup$ i usually see this in a different context; imaginary components of energy eigenvalue appear as a half life variable. the mapping $\beta = \frac{it}{ \hbar}$ looks like a fixed temperature dependence inversely proportional with time (real time, that is, which is where energy wavefunctions oscillate rather than decay), and nothing else beyond that. I don't see how an ad hoc dependence can be in general relevant unless the physical system under study can be approximated with that relationship because some well-defined mechanism $\endgroup$
    – lurscher
    Commented May 23, 2011 at 16:21
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    $\begingroup$ It does seem too perfect, and it's such a fantastically useful trick with far reaching consequences in actually performing quantum calculations -- but it is just a trick; a happy coincidence of the equations we use to describe physical phenomena. $\endgroup$
    – wsc
    Commented May 23, 2011 at 19:21
  • $\begingroup$ @wsc Much to my disappointment, this is the precise answer I've gotten from my professors before. I guess really wanting it to be meaningful in a deeper physical way is not enough to make it so ;-) $\endgroup$
    – C Earnest
    Commented May 23, 2011 at 19:35

2 Answers 2


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.

  • $\begingroup$ thanks for sharing your research, i was wondering about these issues recently when looking at the answers to this question physics.stackexchange.com/questions/8049/… i think this is precisely a case where there is an unavoidable ambiguity in our descriptions of physical systems which can be adjusted to vary from a thermal description to a quantum description (not continuously it seems, although the $\xi$ from your paper seems to suggest a parameter describing such freedom, although my skimmed reading might have completely misunderstood that) $\endgroup$
    – lurscher
    Commented May 23, 2011 at 18:55
  • $\begingroup$ Dunno, lurscher. I think the relationship between thermal and quantum fluctuations is hyperbolic, or, that is, non-compact. Just as one can boost a time-like 4-vector as much as you like it will never become a space-like 4-vector, I suppose there'll always be a vacuum component. The math doesn't look like a circle. $\endgroup$ Commented May 23, 2011 at 19:40
  • $\begingroup$ for what systems can you establish that hyperbolic relation? $\endgroup$
    – lurscher
    Commented May 23, 2011 at 21:59

The Green's functions in QM and in the heat conduction equations describe the relaxation of a $\delta$-like perturbation at some point at t=0. In QM it is a superposition of waves, in the heat conduction (or diffusion) equation it is spreading out the perturbation over the system. In the latter case the "regular regime" of the relaxation is described "only" with the slowest exponential $\exp(-t/\tau_0)$.

The statistical sum $Z$, however makes sense only at finite T. Nobody considers large variations of T. Instead, it is interesting to study $Z(T)$ as a function of the system parameters involved in $E_n$.

  • $\begingroup$ I still don't understand from your answer what the physical significance of representing time as a purely imaginary number could mean. Could you maybe expand on that a little? $\endgroup$
    – C Earnest
    Commented May 23, 2011 at 17:02
  • $\begingroup$ In QM equation and in the heat conduction equation the time is real. It is the equations that are different. Formally the Green's function of one's is obtained from the other's with a simple time variable change. Of course, physically the corresponding solutions are very different - an oscillating (wavy) and a decaying (dying out) with time. $\endgroup$ Commented May 23, 2011 at 17:06

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