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What is a simple intuitive way to see the relation between imaginary time (periodic) and temperature relation?

+ 13 like - 0 dislike

I guess I never had a proper physical intuition on, for example, the "KMS condition". I have an undergraduate student who studies calculation of Hawking temperature using the Euclidean path integral technique, and shamefully his teacher is not able to give him a simple, intuitive argument for it. What would it be?

Added on October 21:

First of all, thanks Moshe and S Huntman for answers. My question was, however, looking for more "intuitive" answer. As Moshe pointed out it may not be possible, since after all time is "imaginary" in this case. So, let me be more specific, risking my reputation.

I should first say I understand there are formal relation between QFT and statistical mechanics as in well-known review like "Fulling & Ruijsenaars". But, when you try to explain this to students with less formal knowledge, it sometimes helps if we have an explicit examples. My motivation originally comes from "Srinivasan & Padmanabhan". In there, they says tunneling probability calculation using complex path (which is essentially a calculation of semi-classical kernel of propagator) can give a temperature interpretation because "In a system with a temperature $\beta^{-1}$ the absorption and emission probabilities are related by

P[emission] = $\exp(-\beta E)$P[absorption]. (2.22) "

So, I was wondering whether there is a nice simple example that shows semi-classical Kernel really represents temperature. I think I probably envisioned something like two state atom in photon field in thermal equilibrium, then somehow calculate in-in Kernel from |1> to |1> with real time and somehow tie this into distribution of photon that depends on temperature. I guess what I look for was a simple example to a question by Feynman $ Hibbs (p 296) pondering about the possibility of deriving partition function of statistical mechanics from real time path integral formalism.

This post has been migrated from (A51.SE)
asked Oct 18, 2011 in Theoretical Physics by Demian Cho (265 points) [ no revision ]

2 Answers

+ 11 like - 0 dislike

Lots of different ways to answer, but none of them can be too intuitive since imaginary time is, well, imaginary. But here is one attempt to make the result more or less self-evident.

The basic object to calculate in quantum statistical mechanics (in thermal equilibrium, in the canonical ensemble) is the partition function (with potential insertions if you want to calculate correlation functions):

$Z= Tr\,(e^{-\beta H})= \sum_\psi \langle \psi(0)|e^{-\beta H}|\psi(0) \rangle$

where $H$ is the Hamiltonian and we have a sum over any complete set of states $\psi$, written in the Schrödinger picture at some fixed time which we take to be $t=0$. In that picture the time evolution of a state is

$\psi(t) = e^{-i t H}|\psi(0)\rangle$

The basic observation now is that the Boltzmann factor $e^{-\beta H}$ can be regarded as an evolution of the state $\psi$ over imaginary time period $-i \beta$. Therefore we can write:

$Z= \sum_\psi \langle \psi(0)|\psi(-i\beta) \rangle$

This is now the vacuum amplitude (with possible insertions) which is the sum over all states $\psi$ in some arbitrary complete basis. Except that you propagate any final states with time $ i \beta$ with respect to the initial state. In other words however you choose to calculate your vacuum amplitude (or correlation function) — a popular method is a path integral — you have to impose the condition that the initial and final states are the same up to that imaginary time shift. This is the origin of the imaginary time periodicity.

This post has been migrated from (A51.SE)
answered Oct 18, 2011 by Moshe (2,375 points) [ no revision ]
Sorry for the nit-picking, but I think Moshe meant to say the _Schrödinger picture._ (The _Schrödinger representation,_ on the other hand, is an important representation of the Heisenberg algebra, cf. Stone-von Neumann Theorem.)

This post has been migrated from (A51.SE)
Sure, I’ll fix that just in case it does generate confusion for someone.

This post has been migrated from (A51.SE)
Nice, Moshe! Qmechanic: fun terminology. I guess the opposite is not true: the Heisenberg representation isn't an important representation of the Schrödinger algebra. However, what is true is that Wigner's friend is an important representation of Schrödinger's cat and Wigner's friend could even be Heisenberg himself, see e.g. this picture http://en.wikipedia.org/wiki/File:Heisenberg,W._Wigner,E._1928.jpg :-D

This post has been migrated from (A51.SE)
+ 5 like - 0 dislike

The time evolution of an observable $\hat A$ in the Heisenberg picture is given as usual by $\tau_t(\hat A) := e^{i\mathcal{\hat H}t/\hbar} \hat A e^{-i\mathcal{\hat H}t/\hbar}$. The quantum Gibbs rule $\langle \hat A \rangle = \mbox{Tr}(\hat \rho \hat A)$, with $\hat \rho := Z^{-1}e^{-\beta \mathcal{\hat H}}$ and $Z := \mbox{Tr}(e^{-\beta \mathcal{\hat H}})$, is generalized by the KMS condition

$\left \langle \tau_t(\hat A) \hat B \right \rangle = \left \langle \hat B\tau_{t+i\hbar\beta}(\hat A) \right \rangle$.

For convenience, we recall a formal derivation of this from the Gibbs rule and the cyclic property of the trace:

$\left \langle \tau_t(\hat A) \hat B \right \rangle$ $ = Z^{-1}\mbox{Tr}(e^{-\beta \mathcal{\hat H}} e^{i\mathcal{\hat H}t/\hbar} \hat A e^{-i\mathcal{\hat H}t/\hbar} \hat B)$ $ = Z^{-1}\mbox{Tr}(\hat B e^{i\mathcal{\hat H}z/\hbar} \hat A e^{-i\mathcal{\hat H}t/\hbar})$ $ = Z^{-1}\mbox{Tr}(\hat B e^{i\mathcal{\hat H}z/\hbar} \hat A e^{-i\mathcal{\hat H}z/\hbar} e^{-\beta \mathcal{\hat H}})$ $ = \left \langle \hat B\tau_z(\hat A) \right \rangle$

where here we have written $z := t+i\hbar\beta$.

The analyticity of $\tau_z$ in a strip forms the actual substance of the KMS condition.

This post has been migrated from (A51.SE)
answered Oct 18, 2011 by S Huntsman (405 points) [ no revision ]

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