Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 39

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Thus, at low temperature, we can replace the frequency sumin the exponent of (2.481) byXω0−−−→ h̄βT →0Z∞−∞dω 0.2π(2.485)This could have been expected on the basis of Planck’s rules for the phase spaceinvoked earlier on p. 98 to explain the measure of path integration. According totheseRrules, the volumeelement in the phase space of energy and time has the meaRsure dt dE/h = dt dω/2π. If the integrand is independent of time, the temporalintegral produces an overall factor , which for the imaginary-time interval (0, h̄β) ofstatistical mechanics is equal to h̄β = h̄/kB T , thus explaining the integral versionof the sum (2.485).The integral on the right-hand side of (2.484) diverges at large ω 0.

This is calledan ultraviolet divergence (UV-divergence), alluding to the fact that the ultravioletregime of light waves contains the high frequencies of the spectrum.The important observation is now that the divergent integral (2.484) can bemade finite by a mathematical technique called analytic regularization.20 This isbased on rewriting the logarithm log(ω 02 + ω 2 ) in the derivative form:d.log(ω + ω ) = − (ω 02 + ω 2 )− d=0022(2.486)Equivalently, we may obtain the logarithm from an → 0 -limit of the function11lMS () = − (ω 02 + ω 2)− + .(2.487)20G.

’t Hooft and M. Veltman, Nucl. Phys. B 44 , 189 (1972). Analytic regularization is atpresent the only method that allows to renormalize nonabelian gauge theories without destroyinggauge invariance. See also the review by G. Leibbrandt, Rev. Mod. Phys. 74 , 843 (1975).H. Kleinert, PATH INTEGRALS2.15 Field-Theoretic Definition of Harmonic Path Integral by Analytic Regularization 161The subtraction of the pole term 1/ is commonly referred to a minimal subtraction.Indicating this process by a subscript MS, we may write1lMS () = − (ω 02 + ω 2 )− MS,.(2.488)→0Using the derivative formula (2.486), the trace of the logarithm in the free energy(2.484) takes the formd1Tr log(−∂τ2 + ω 2) = −h̄βdZ∞−∞dω 0 02(ω + ω 2)− .2π=0(2.489)We now set up a useful integral representation, due to Schwinger, for a power a−generalizing (2.460).

Using the defining integral representation for the Gamma functionZ ∞dτ µ −τ ω2τ e= ω −µ/2 Γ(µ),(2.490)τ0the desired generalization is−a1 Z ∞ dτ −τ aτ e .=Γ() 0 τ(2.491)This allows us to re-express (2.489) as1d 1 Z ∞ dω 0 Z ∞ dτ −τ (ω0 2 +ω2 ) 22Tr log(−∂τ + ω ) = −τ e.h̄βd Γ() −∞ 2π 0 τ=0(2.492)As long as is larger than zero, the τ -integral converges absolutely, so that we caninterchange τ - and ω 0 -integrations, and obtain1d 1 Z ∞ dτ Z ∞ dω 0 −τ (ω0 2 +ω2 ) 22Tr log(−∂τ + ω ) = −τe.h̄βd Γ() 0 τ−∞ 2π=0(2.493)At this point we can perform the Gaussian integral over ω 0 using formula (1.335),and findd 11Tr log(−∂τ2 + ω 2) = −h̄βd Γ()Z0∞dτ 12τ √ e−τ ω .τ 2 τπ=0(2.494)For small , the τ -integral is divergent at the origin.

It can, however, be defined byan analytic continuation of the integral starting from the regime > 1/2, where itconverges absolutely, to = 0. The continuation must avoid the pole at = 1/2.Fortunately, this continuation is trivial since the integral can be expressed in terms ofthe Gamma function, whose analytic properties are well-known. Using the integralformula (2.490), we obtain1d 11Tr log(−∂τ2 + ω 2) = − √ ω 1−2Γ( − 1/2) .h̄β2 πd Γ()=0(2.495)1622 Path Integrals — Elementary Properties and Simple SolutionsThe right-hand side has to be continued analytically from > 1/2 to = 0. This iseasily done using the defining property of the Gammafunction Γ(x) = Γ(1 + x)/x,√from which we find Γ(−1/2) = −2Γ(1/2) = −2 π, and 1/Γ() ≈ /Γ(1 + ) ≈ .The derivative with respect to leads to the free energy of the harmonic oscillatorat low temperature via analytic regularization:1Tr log(−∂τ2 + ω 2 ) =h̄βZdω 0log(ω 02 + ω 2 ) = ω,2π∞−∞(2.496)so that the free energy of the oscillator at zero-temperature becomesh̄ω.2Fω =(2.497)This agrees precisely with the result obtained from the lattice definition of the pathintegral in Eq.

(2.399), or from the path integral (3.805) with the Fourier measure(2.439).With the above procedure in mind, we shall often use the sloppy formula expressing the derivative of Eq. (2.491) at = 0:Zlog a = −∞0dτ −τ ae .τ(2.498)This formula differs from the correct one by a minimal subtraction and can beused in all calculations with analytic regularization. Its applicability is based onthe possibility of dropping the frequency integral over 1/ in the alternative correctexpression11Tr log(−∂τ2 + ω 2) = −h̄βZ∞−∞dω 02π1 021(ω + ω 2)− −.(2.499)→0In fact, within analytic regularization one may set all integrals over arbitrary purepowers of the frequency equal to zero:Z0∞dω 0 (ω 0 )α = 0 for all α.(2.500)This is known as Veltman’s rule.21 It is a special limit of a frequency integral whichis a generalization of the integral in (2.489):Z∞−∞dω 0 (ω 02 )γΓ(γ + 1/2) 2 γ+1/2−(ω ).022 =2π (ω + ω )2πΓ()(2.501)This equation may be derived by rewriting the left-hand side as1Γ()Z∞−∞dω 0 02 γ(ω )2πZ0∞dτ −τ (ω0 2 +ω2 )τ e.τ(2.502)21See the textbook H.

Kleinert and V. Schulte-Frohlinde, Critical Properties of φ4 -Theories,World Scientific, Singapore, 2001 (http://www.physik.fu-berlin.de/~kleinert/b8).H. Kleinert, PATH INTEGRALS2.15 Field-Theoretic Definition of Harmonic Path Integral by Analytic Regularization 163The integral over ω 0 is performed as follows:dω 0 02 γ −τ (ω0 2 +ω2 )1 Z ∞ dω 0 2 02 γ+1/2 −τ (ω0 2 +ω2 ) τ −γ−1/2(ω )e=(ω ) e=Γ(γ + 1/2),2π 0 ω 0 22π−∞ 2π(2.503)leading to a τ -integral in (2.502)Z∞Z∞0dτ −γ−1/2 −τ ω2τe= (ω 2)γ+1/2+ ,τ(2.504)and thus to the formula (2.501). The Veltman rule (2.500) follows from this directlyin the limit → 0, since 1/Γ() → 0 on the right-hand side.

This implies that thesubtracted 1/ term in (2.499) gives no contribution.The vanishing of all integrals over pure powers by Veltman’s rule (2.500) wasinitially postulated in the process of developing a finite quantum field theory ofweak and electromagnetic interactions. It has turned out to be extremely usefulfor the calculation of critical exponents of second-order phase transitions from fieldtheories.21An important consequence of Veltman’s rule is to make the logarithms of dimensionful arguments in the partition functions (2.481) and the free energy (2.483)Rmeaningful quantities. First, since d(ω 0/2π) log ω 2 = 0, we can divide the argumentof the logarithm in (2.484) by ω 2 without harm, and make them dimensionless.

Atfinite temperatures, we use the equality of sum and integral over an ωm -independentquantity ckB T∞Xc=m=−∞Z∞−∞dωmc2π(2.505)to show that alsokB T∞Xlog ω 2 = 0,(2.506)m=−∞so that we have, as a consequence of Veltman’s rule, that the Matsubara frequencysum over the constant log ω 2 vanishes,Xlog ω 2 = 0,(2.507)ω0for all temperatures. For this reason, also the argument of the logarithm in the freeenergy (2.483) can be divided by ω 2 without change, thus becoming dimensionless.2.15.2Finite-Temperature Evaluation of Frequency SumAt finite temperature, the free energy contains an additional term consisting of thedifference between the Matsubara sum and the frequency integralk T∆Fω = B22ωmh̄log2 +1 −2ωm=−∞∞X!Z∞−∞2ωmdωmlog+1 ,2πω2!(2.508)1642 Path Integrals — Elementary Properties and Simple Solutionswhere we have used dimensionless logarithms as discussed at the end of the lastsubsection.

The sum is conveniently split into a subtracted, manifestly convergentexpression∆1 Fω = kB T∞Xm=122∞Xωmω2ωmloglog1+, (2.509)+1−log=kTB2ω2ω2ωmm=1"!!#and a divergent sum∞X∆2 Fω = kB Tlogm=12ωm.ω2(2.510)The convergent part is most easily evaluated. Taking the logarithm of the productin Eq. (2.400) and recalling (2.401), we findω21+ 2ωm∞Ym=1and therefore∆F1 =!=sinh(βh̄ω/2),βh̄ω/2(2.511)sinh(βh̄ω/2)1log.ββh̄ω/2(2.512)The divergent sum (2.510) is calculated by analytic regularization as follows: Werewrite2∞ωmd Xωmlog 2 = − 2d m=1 ωωm=1∞X"−and express the sum over m− #→0d= − 2d2πβh̄ω!−∞Xm=1− m, (2.513)→0in terms of Riemann’s zeta function∞Xζ(z) =m−z .(2.514)m=1This sum is well defined for z > 1, and can be continued analytically into the entirecomplex z-plane.

The only singularity of this function lies at z = 1, where in theneighborhood ζ(z) ≈ 1/z. At the origin, ζ(z) is regular, and satisfies221ζ 0(0) = − log 2π,2ζ(0) = −1/2,(2.515)such that we may approximate1ζ(z) ≈ − (2π)z ,2z ≈ 0.(2.516)Hence we find222d 2πωmlog 2 = − 2d βh̄ωωm=1∞X!−ζ()→0d= (βh̄ω) = log h̄ωβ.d→0(2.517)I.S. Gradshteyn and I.M. Ryzhik, op. cit., Formula 9.541.4.H. Kleinert, PATH INTEGRALS2.15 Field-Theoretic Definition of Harmonic Path Integral by Analytic Regularization 165thus determining ∆2 Fω in Eq. (2.510).By combining this with (2.512) and the contribution −h̄ω/2 from the integral(2.508), the finite-temperature part (2.483) of the free energy becomes∆Fω =1log(1 − e−h̄βω ).β(2.518)Together with the zero-temperature free energy (2.497), this yields the dimensionallyregularized sum formulaFω =∞1h̄ω 11 X2log(ωm+ ω2) =Tr log(−∂τ2 + ω 2 ) =+ log(1 − e−h̄ω/kB T )2β2β m=−∞2β!h̄ωβ1log 2 sinh,=β2(2.519)in agreement with the properly normalized free energy (2.477) at all temperatures.2.15.3Tracelog of First-Order Differential OperatorThe trace of the logarithm in the free energy (2.484) can obviously be split into twotermsTr log(−∂τ2 + ω 2) = Tr log(∂τ + ω) + Tr log(−∂τ + ω).(2.520)Since the left-hand side is equal to βh̄ω by (2.496), and the two integrals must bethe same, we obtain the low-temperature resultdωTr log(∂τ + ω) = Tr log(−∂τ + ω) = h̄βlog(−iω 0 + ω) = h̄β−∞ 2πh̄βω.=2Z∞dωlog(iω 0 + ω)−∞ 2πZ∞(2.521)The same result could be obtained from analytic continuation of the integrals over∂ (±iω 0 + ω) to = 0.For a finite temperature, we may use Eq.

(2.519) to find!βh̄ω1Tr log(∂τ + ω) = Tr log(−∂τ + ω) = Tr log(−∂τ2 + ω 2 ) = log 2 sinh, (2.522)22which reduces to (2.521) for T → 0.In Subsection 3.3.2, this formula will be generalized to arbitrary positive timedependent frequencies Ω(τ ), where it reads [see (3.133)]("#)1 h̄β 00Tr log [∂τ + Ω(τ )] = log 2 sinhdτ Ω(τ 00 )2 0ZR h̄β1 h̄β 0000− 0 dτ 00 Ω(τ 00 )dτ Ω(τ ) + log 1 − e=. (2.523)2 0Z1662 Path Integrals — Elementary Properties and Simple SolutionsThe result is also the same if there is an extra factor i in the argument ofthe tracelog. To see this we consider the case of time-independent frequency whereVeltman’s rule (2.500) tells us that it does not matter whether one evaluates integralsover log(iω 0 ∓ ω) or over log(ω 0 ± iω).Let us also replace ω 0 by iω 0 in the zero-temperature tracelog (2.496) of thesecond-order differential operator (−∂τ2 + ω 2 ). Then we rotate the contour of integration clockwise in the complex plane to find∞Z−∞dω 0log(−ω 02 + ω 2 − iη) = ω,2πω ≥ 0,(2.524)where an infinitesimal positive η prescribes how to bypass the singularities at ω 0 =±ω ∓ iη along the rotated contour of integration.

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