From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Path integrals and Feynman diagrams Date: 25 May 2000 23:52:58 GMT Newsgroups: sci.physics.research In article <8gb2fd$kj7$1@crib.corepower.com>, Nathan Urban wrote: >I'm not terribly comfortable >with why and how you can derive everything of thermodynamic interest >from the partition function, but I've done a lot of it at least. Okay. You've seen how it works in zillions of examples. Maybe a high-level view will help fit all those zillions of examples into the proper perspective. In classical stat mech, we have a space of states X and a function H: X -> R called the Hamiltonian, and for each temperature T we get a probability distribution exp(-H(x)/kT) on the space X which says how likely it is for the system to be in a given state x. Actually, I've forgotten to *normalize* the probability distribution! We really need to use (1/Z) exp(-H(x)/kT) where Z = integral exp(-H(x)/kT) dx Anyway, our job is compute expectation values of observables. An observable is a function f: X -> R and its expectation value is an integral like this: = (1/Z) integral f(x) exp(-H(x)/kT) dx Now, how can we do this starting from the partition function? Well, here's how. First, we cleverly consider a perturbed system where the Hamiltonian is not H but H + cf for some "coupling constant" c. The partition function of this modified system depends on c, and it's Z(c) = integral exp(-(H(x) + cf(x)/kT) dx Suppose we take the derivative of this with respect to c and set c = 0. Then a little scribbling shows that Z'(0) = - integral f(x) exp(-(H(x)/kT) dx Lo and behold! This is almost the expectation value of f! In fact, = -Z'(0)/Z(0) where Z(0) is just what I was previously calling Z - the partition function of the original unperturbed system. See? The expectation value of any observable is closely related to the partition function of the perturbed system where we use that observable to perturb the Hamiltonian! So once you are good at perturbatively computing partition functions, you are good at computing expectation values of observables. And this applies not only to stat mech but also quantum mechanics and quantum field theory. Just stick an "i" in the formulas. Schwinger called this idea "source theory". Feynman mechanized it and sold it to the masses. (Or so Schwinger grumpily remarked.) ============================================================================== From: baez@galaxy.ucr.edu Subject: Re: Path integrals and Feynman diagrams Date: 26 May 2000 04:57:45 GMT Newsgroups: sci.physics.research In article <8gkecq$414$1@pravda.ucr.edu>, John Baez wrote: >Z(c) = integral exp(-(H(x) + cf(x)/kT) dx > >Suppose we take the derivative of this with respect to c and >set c = 0. Then a little scribbling shows that > >Z'(0) = - integral f(x) exp(-(H(x)/kT) dx ... and a little more scribbling shows I left out a factor of 1/kT here. Grr. Ugly little constants. Shoulda set the damn things to 1. ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Path integrals and Feynman diagrams Date: 28 May 2000 06:28:14 GMT Newsgroups: sci.physics.research In article <8gosa1$901$1@bob.news.rcn.net>, Michael Weiss wrote: >John Baez wrote, after secretly setting kT = 1: >| = -Z'(0)/Z(0) >Curious. Z'(0)/Z(0) is the logarithmic derivative of Z, i.e., (d/dc) log Z(c) >at c=0. Usually it simplifies things to compute it out that way (at least >formally), but since the logarithm of the integral is not the integral of the >logarithm, that doesn't work here. Excellent! The wise mathematician seeks patterns, and knows that much of what seems like "coincidence" at first glance is but the beckoning hand of the muse. You're noticing that some things simplify if we work not with the partition function Z but its logarithm. But you're puzzled because Z is not the exponential of anything you've already met - instead, it's the *integral* of the exponential exp(-H/kT). Now, you could shrug, mutter "coincidence", and stop here - but you decided to post an article about this puzzlement instead. Wise! In statistical mechanics, the logarithm of the partition function has a name: it's called the Helmholtz free energy! Well, actually I'm leaving out some constants again. Let's get it right: let's define the Helmholtz free energy, F, by: Z = exp(-F/kT) so F = -kT ln(Z) The first formula says that the Helmholtz free energy is what we'd have to replace the Hamiltonian by if we replaced the phase space by a SINGLE POINT and we wanted to get the same partition function. Replacing the phase space by a single point makes the integral go away: Z = integral exp(-H(x)/kT) dx becomes Z = exp(-F/kT) Anyway, it turns out there's a really nice story about Helmholtz free energy and the concept of "effective action" in quantum field theory. I would tell this story, but it's already been told very nicely in chapter 11.3 of Peskin and Schroeder's "An Introduction to Quantum Field Theory", so I won't.... ... I just wanted to say you're onto something here.