Does statistical mechanics equal one-loop quantum field theory\?

Details Does statistical mechanics equal one-loop quantum field theory\? Does statistical mechanics equal one-loop quantum field theory\?

Jun 1986

adsabs.harvard.edu/cgi-bin/nph-data_query?bibcode=1986phrvd..33.3640a&link_type=abstract

Physical Review D (Particles and Fields), Volume 33, Issue 12, 15 June 1986, pp.3640-3644

Physics

Condensed Matter

Statistical Mechanics

34

Quantum Statistical Mechanics, Theory Of Quantized Fields, Lagrangian And Hamiltonian Approach

Scientific paper

The ``thermodynamic'' partition function ZT(β)=Jnexp(-βEn) is compared to the Euclidean ``quantum'' path integral ZQ(β)=Fd[φ]exp(-S) over (anti)periodic fields φ(τ+β)=+/-φ(τ). We assume (1) free spin-0 or spin-(1/2) fields and (2) an ultrastatic spacetime. Our main result is that ZT(β) does not equal ZQ(β). Nevertheless, they are simply related: we prove that lnZQ(β)=lnZT(β)+(A+B lnμ2)β. Thus, the logarithms of the two partition functions differ only by a term proportional to β. The constant A arises from vacuum energy and the constant B from the renormalization-scale (μ) dependence of ZQ. We derive a simple formula for A and B in terms of the ``energy'' ζ function ζE(z)=JkEk -z. In particular we show that A and B are determined by the behavior of the energy ζ function near z=-1: for small ɛ, +/-ζE(-1+ɛ)=(1/4)Bɛ-1+( 1/2)A+O(ɛ) (where the upper sign applies to bosons and the lower sign applies to fermions). We also give a high-temperature expansion of Z(β) in terms of ζE(z). Finally we argue that ZT and ZQ are interchangeable in any situation where gravitational effects are unimportant. This is because adding a term linear in β to lnZ is equivalent to shifting all energies by a constant; but if gravity is neglected, then the physics only depends upon the difference between energies, which is unchanged.

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