Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 41
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The next term is simply the Riemann zeta function ζ(ν) [recall (2.514)].The integral in the difference between sum and integral vanishes due to Veltman’rule (2.500). The same thing happens for sum minus integrals over nk−ν which yieldζ(ν − k) inside the sum, and we obtain the so-called Robinson expansion 29gν (βh̄ω) =Z∞Z ∞∞∞XX11dn ν e−nβh̄ω +−+nnν k=10n=1!gν (βh̄ω) = Γ(1 − ν)(βh̄ω)ν−1+ ζ(ν) +"∞XZ−∞∞X!k−νn#1(−βh̄ω)k ζ(ν − k).k=1 k!(2.564)This expansion will later play an important role in the discussion of Bose-Einsteincondensation [see Eq. (7.39)].
From this we can extract the desired sum (2.560)by going to the limit ν → 1. Close to the limit, the Gamma function has a poleΓ(1 − ν) = 1/(1 − ν) − γ + O(ν − 1). From the identity2z Γ(1 − z)ζ(1 − z) sin29J.E. Robinson, Phys. Rev. 83 , 678 (1951).πz= π 1−z ζ(z)2(2.565)1722 Path Integrals — Elementary Properties and Simple Solutionsand (2.515) we see that ζ(ν) behaves near ν = 1 like1+ γ + O(ν − 1) = −Γ(1 − ν) + O(ν − 1).ν −1(2.566)1(2p)!ζ(2p),ζ(1 − 2p) = (−1)pp(2π)2p(2.567)ζ(ν) =Hence the first two termsin (2.564) ican be combined to yield for ν → 1 the finitehresult limν→1 Γ(1 − ν) (βh̄ω)ν−1 − 1 =− log βh̄ω.
The remaining terms contain inthe limit the values ζ(0) = −1/2, ζ(−1), ζ(−2), etc. Here we use the property ofthe zeta function that it vanishes at even negative arguments, and that the functionat arbitrary negative argument is related to one at positive argument by the identity(2.565). This implies for the expansion coefficients in (2.564) with k = 1, 2, 3, .
. . inthe limit ν → 1:ζ(−2p) = 0,p = 1, 2, 3, . . . .Hence we obtain for the expansion (2.564) in the limit ν → 1:g(βh̄ω) = g1 (βh̄ω) = − log βh̄ω +∞βh̄ω X(−1)k+(βh̄ω)2k .ζ(2k)2k!k=1(2.568)This can now be inserted into Eq. (2.547) and we recover the previous expansion(2.549) for S(βh̄ω) which was derived there by a proper duality transformation.It is interesting to observe what goes wrong if one forgets the separation (2.561)of the sum into integral plus sum-minus-integral and its regularization (2.562) atn = 0, and re-expands (2.560) directly, and illegally, in powers of β. Then oneobtains the formal expansion∞X(−1)p(−1)p(βh̄ω)p = −ζ(1) +ζ(1 − p)(βh̄ω)p ,p!p!p=1p=0 n=1(2.569)which contains the infinite quantity ζ(1).
The correct result (2.568) is obtained fromthis by replacing the infinite quantity ζ(1) by − log βh̄ω, which may be viewed as aregularized ζreg (1):ζ(1) → ζreg (1) = − log βh̄ω.(2.570)g(βh̄ω) =∞X∞Xp−1n!It should be mentioned that the first two terms in the low-temperature expansion(2.547) can also be found from the sum (2.546) with the help of the Euler-Maclaurinformula30 for a sum over discrete points t = a + (k + κ)∆ of a function F (t) fromk = 0 to K ≡ (b − a)/∆:KXF (a + k∆) =k=0+301 Zb1dt F (t) + [F (a) + F (b)]∆ a2hi∆2p−1B2p F (2p−1) (b) − F (2p−1) (a) ,p=1 (2p)!∞X(2.571)M.
Abramowitz and I. Stegun, op. cit., Formulas 23.1.30 and 23.1.32.H. Kleinert, PATH INTEGRALS2.16 Finite-N Behavior of Thermodynamic Quantities173or, more generally for t = a + (k + κ)∆,K−1XF (a + (k + κ)∆) =k=01∆Zbadt F (t) +hi∆p−1B2p (κ) F (2p−1) (b) − F (2p−1) (a) ,p=1 (2p)!∞X(2.572)where Bn (κ) are the Bernoulli numbers defined by the expansion∞Xteκttn=B (κ) ,et − 1 n=0 n n!(2.573)2and Bn = Bn (0). Using the first formula with a = ω12, b = ωM, and ∆ = ω1 , we findM hX2log(ωm2+ ω )−2log(ωm)m=0i)m=M (()m=Mi ω ω h2− ω=0= π + m log(ωm+ ω 2 )−2 ω1 ω1m=1o no1n2+log(ω12 + ω 2 ) + log(ωM+ ω2) − ω = 0 .2m=1(2.574)For small T , the leading two terms on the right-hand side are1ω2ωπ − log 2 ,ω1 2ω1(2.575)in agreement with the first two terms in the low-temperature series (2.547).
Notethat the Euler-Maclaurin formula is unable to recover the exponentially small termsin (2.547), since they are not expandable in powers of T .The transformation of high- into low-temperature expansions is an importanttool for analyzing phase transitions in models of statistical mechanics.312.16Finite-N Behavior of Thermodynamic QuantitiesThermodynamic fluctuations in Euclidean path integrals are often imitated in computer simulations.
These are forced to employ a sliced time axis. It is then important to know in which waythe time-sliced thermodynamic quantities converge to their continuum limit. Let us calculate theinternal energy E and the specific heat at constant volume C for finite N from (2.476). Using(2.476) we have∂(β ω̃e )∂β∂(ω̃e )∂β==ω,cosh(ω̃e /2)2tanh(ω̃e /2),β(2.576)and find the internal energyE==31∂h̄∂(β ω̃e )(βF ) = coth(βh̄ω̃e /2)∂β2∂βh̄ω coth(βh̄ω̃e /2).2 cosh(ω̃e /2)(2.577)See H. Kleinert, Gauge Fields in Condensed Matter , Vol. I Superflow and Vortex Lines, WorldScientific, Singapore, 1989, pp. 1–742 (http://www.physik.fu-berlin.de/~kleinert/b1).1742 Path Integrals — Elementary Properties and Simple SolutionsFigure 2.4 Finite-lattice effects in internal energy E and specific heat C at constant volume,as a function of the temperature for various numbers N + 1 of time slices.
Note the nonuniformway in which the exponential small-T behavior of C ∝ e−ω/T is approached in the limit N → ∞.The specific heat at constant volume is given by∂∂21EC = −β 2 2 (βF ) = −β 2kB∂β∂β111= β 2 h̄2 ω 2+coth(βh̄ω̃/2)tanh(ω̃/2).ee4h̄β cosh2 (ω̃e /2)sinh2 (βh̄ω̃e /2)(2.578)Plots are shown in Fig. 2.5 for various N using natural units with h̄ = 1, kB = 1. At hightemperatures, F, E, and C are independent of N :FEC1log β,β1= T,→β→ 1.→(2.579)(2.580)(2.581)These limits are a manifestation of the Dulong-Petit law : An oscillator has one kinetic and onepotential degree of freedom, each carrying an internal energy T /2 and a specific heat 1/2.
At lowtemperatures, on the other hand, E and C are strongly N -dependent (note that since F and Eare different at T = 0, the entropy of the lattice approximation does not vanish as it must in thecontinuum limit). Thus, the convergence N → ∞ is highly nonuniform. After reaching the limit,the specific heat goes to zero for T → 0 exponentially fast, like e−ω/T . The quantity ω is calledactivation energy.32 It is the energy difference between ground state and first excited state of theharmonic oscillator. For large but finite N , on the other hand, the specific heat has the large value32Note that in a D-dimensional solid the lattice vibrations can be considered as an ensemble ofharmonic oscillators with energies ω ranging from zero to the Debye frequency.Integrating overRthe corresponding specific heats with the appropriate density of states, dωω D−1 e−ω/kB T , givesthe well-known power law at low temperatures C ∝ T D .H. Kleinert, PATH INTEGRALS1752.17 Time Evolution Amplitude of Freely Falling ParticleN + 1 at T = 0.
This is due to ω̃e and cosh2 (ω̃e /2) behaving, for a finite N and T → 0 (where becomes large) like1log(2 ω 2 ),cosh(ω̃e /2) → ω/2.ω̃e→(2.582)HenceT →01coth[(N + 1) log(ω)] −−−→ 0,βT →0E−−−→C−−−→ N + 1.T →0(2.583)(2.584)The reason for the nonuniform approach of the N → ∞ limit is obvious: If we expand (2.476) inpowers of , we find1ω̃e = ω 1 − 2 ω 2 + . . . .(2.585)24When going to low T at finite N the corrections are quite large, as can be seen by writing (2.585),with = h̄β/(N + 1), as1h̄2 ω 2+ ...
.ω̃e = ω 1 −24 kb2 T 2 (N + 1)2(2.586)Note that (2.585) contains no corrections of the order . This implies that the convergence ofall thermodynamic quantities in the limit N → ∞, → 0 at fixed T is quite fast — one order in1/N faster than we might at first expect [the Trotter formula (2.26) also shows the 1/N 2 -behavior].2.17Time Evolution Amplitude of Freely Falling ParticleThe gravitational potential of a particle on the surface of the earth isV (x) = V0 + M g · x,(2.587)where −g is the earth’s acceleration vector pointing towards the ground, and V0some constant. The equation of motion readswhich is solved byẍ = −g,(2.588)gx = xa + va (t − ta ) + (t − ta )2 ,2(2.589)with the initial velocityva =Inserting this into the actionA=Ztbtaxb − xa g− (tb − ta ).tb − ta2(2.590)M 2dtẋ − V0 − g · x ,2(2.591)1762 Path Integrals — Elementary Properties and Simple Solutionswe obtain the classical action1M (xb − xa )2 1Acl = −V0 (tb − ta ) +− (tb − ta )g · (xb + xa ) − (tb − ta )3 g2 .
(2.592)2 tb − ta224Since the quadratic part of (2.591) is the same as for a free particle, also the fluctuation factor is the same [see (2.120)], and we find the time evolution amplitude1− h̄i V0 (tb −ta )(xb tb |xa ta ) = q3e2πih̄ω(tb − ta )/MiM (xb − xa )21× exp− (tb − ta )g · (xb + xa ) − (tb − ta )3 g22h̄tb − ta12("#). (2.593)The potential (2.587) can be considered as a limit of a harmonic potentialV (x) = V0 +M 2ω (x − x0 )22(2.594)forω → 0,x0 = −g/ω 2 → −∞ ≈ ĝ,V0 = −Mx20 /2 = −Mg 2 /2ω 4 → −∞, (2.595)keepingg = −Mω 2 x0 ,andv0 = V0 +(2.596)M 2 2ω x02(2.597)fixed. If we perform this limit in the amplitude (2.170), we find of course (2.593).The wave functions can be obtained most easily by performing this limitingprocedure on the wave functions of the harmonic oscillator.
In one dimension, weset n = E/ω and find that the spectral representation (2.287) goes over into(xb tb |xa ta ) =ZdEAE (xb )A∗E (xa )e−iE(tb −ta )/h̄ ,(2.598)with the wave functionsx E1−.AE (x) = √ Ailεlε(2.599)Here ε ≡ (h̄2 g 2 M/2)1/3 and l ≡ (h̄2 /2M 2 g)1/3 = ε/Mg are the natural units ofenergy and length, respectively, and Ai(z) is the Airy function solving the differentialequationAi00 (z) = zAi(z),(2.600)H.
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