Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 85
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This integral will be calculated in Section 9.1. The resultisSD (z) = (2π)D/2 JD/2−1 (z)/z D/2−1 ,(4.264)where Jν (z) are Bessel functions. For small z, these behave like10Jν (z) ≈(z/2)ν,Γ(ν + 1)(4.265)thus ensuring that SD (kR) is indeed equal to SD at R = 0.Altogether, the classical limit of the bilocal density of states is1ρcl (E; xb , xa ) =2πh̄2D/2MJD/2−1 (p(E; x̄)R/h̄)(R/h̄)D/2−1.(4.266)At xb = xa , this reduces to the density (4.205).In three dimensions, the Bessel function becomesJ1/2 (z) =s2sin z,πz(4.267)and (4.266) yields1ρcl (E; xb , xa ) =2πh̄23/2M sin[p(E; x̄)R/h̄]1√.Γ(3/2) 2R/h̄(4.268)From the D-dimensional version of the short-time expansion (4.260) we obtain,after using once more the equivalence of t and ih̄d/dV ,3i d2h̄2 h 2h̄22 dρ(E; xb , xa ) = 1 −∇ V (x̄)[∇V (x̄)]−12MdV 2 24MdV 3)d12+ .
. . ρcl (E; xb , xa ).(4.269)+ [(xb −xa )∇] V (x̄)24dV(4.9.4Gradient Expansion of Tracelog of Hamiltonian OperatorStarting point is formula (1.586) for the tracelog of the Hamiltonian operator. By performing thetrace in the local basis |xi, we arrive at the useful formula involving the density of states (4.237)ZZ ∞DTr log Ĥ = d xdE ρ(E; x) log E.(4.270)−∞Inserting here the classical density of states (4.210), and integrating over the classical spectrumE ∈ (E0 , ∞), where E0 is the bottom of the potential V (x), we obtain the classical limit of thetracelog:D/2 ZZZ ∞M[Tr log Ĥ]cl = dD xdE ρcl (E; x) log E =dD x ID/2 (V (x)),(4.271)2πh̄2E010M.
Abramowitz and I. Stegun, op. cit., Formula 9.1.7.H. Kleinert, PATH INTEGRALS4.9 Quantum Corrections to Classical Density of States409where1Iα (V ) ≡Γ(α)Z∞VdE (E − V )α−1 log E.(4.272)The integrals ID/2 (V (x)) diverge, but can be calculated with the techniques explained in Section2.15 from the analytically regularized integrals11Z ∞Γ(−α + η)1dE (E − V )α−1 E −η = V α−η.(4.273)Iαη (V ) ≡Γ(α) VΓ(η)Since E −η = 1 − η log E + O(η 2 ), the coefficient of −η in the Taylor series of Iαη (V ) will yield thedesired integral. Since 1/Γ(η) ≈ η, we obtain directlyIα (V ) = −Γ(−α)V α ,(4.274)so that (4.271) becomes[Tr log Ĥ]cl = −M2πh̄2D/2Γ(−D/2)ZdD x [V (x)]D/2 .(4.275)The same result can be obtained with the help of formulas (4.237), (4.250), and (2.498) as[Tr log Ĥ]cl = −ZdD xZ∞dE∞Z−∞−∞dt eit[E−V (x)]/h̄2πh̄ (2πih̄t/M )D/2Z0∞dt0 −iEt0 /h̄e.t0(4.276)Integrating over the energy yields[Tr log Ĥ]cl = −ZDd xZ0∞1dte−itV (x)/h̄ .t (2πih̄t/M )D/2(4.277)Deforming the contour of integration by the substitution t = −iτ , we arrive at the integral representation of the Gamma function (2.490) which reproduces immediately the result (4.275).The quantum corrections are obtained by multiplying this with the prefactor in curly bracketsin the expansion (4.252):D/2MΓ(−D/2)Tr log Ĥ = −2πh̄2Zh̄2d2h̄2d3× dD x 1 −∇2 V (x) 2 −[∇V (x)]2+...[V (x)]D/2 .12MdV24MdV 3(4.278)The second term can be integrated by parts, which replaces ∇2 V (x) → −[∇V (x)]2 d/dV , so thatwe obtain the gradient expansionD/2MΓ(−D/2)Tr log Ĥ = −2πh̄2Zh̄2d3× dD x 1 +[∇V (x)]2+...[V (x)]D/2 .24MdV 3The curly brackets can obviously be replaced byh̄2 Γ(3 − D/2) [∇V (x)]21−+....24M Γ(−D/2) [V (x)]311I.S.
Gradshteyn and I.M. Ryzhik, op. cit., Formula 3.196.2.(4.279)(4.280)4104 Semiclassical Time Evolution AmplitudeIn one dimension and with M = 1/2, this amounts to the formulaZph̄2 [V 0 (x)]212 2+ ... .dx V (x) 1 +Tr log[−h̄ ∂x + V (x)] =h̄32 V 3 (x)(4.281)It is a useful exercise to rederive this with the help of the Gelfand-Yaglom method in Section 2.4.There exists another method for deriving the gradient expansion (4.278).
We split V (x) intoa constant term V and a small x-dependent term δV (x), and rewriteh̄2 2h̄2 2∇ + V (x) = Tr log −∇ + V + δV (x)Tr log −2M2Mh̄2 2= Tr log −∇ + V +Tr log (1 + ∆V δV ) ,(4.282)2Mwhere ∆V denotes the functional matrix−1 Z0h̄2 2dD p eip(x−x )/h̄∆V (x, x0 ) = −∇ +V≡ ∆V (x − x0 ).=2M(2πh̄)D p2 /2M + V(4.283)This coincides with the fixed-energy amplitude (i/h̄)(x|x0 )E at E = −V [recall Eq. (1.345)].The first term in (4.282) is equal to (4.275) if we replace V (x) in that expression by theconstant V , so that we may write+ Tr log (1 + ∆V δV ) .(4.284)Tr log Ĥ = [Tr log Ĥ]cl V (x)→VWe now expand the remainder1Tr log (1 + ∆V δV ) = Tr ∆V δV − Tr (∆V δV )2 + . .
. ,2and evaluate the expansion terms. The first term is simplyZZDTr ∆V δV = d x ∆V (x, x)δV (x) = ∆V (0) dD x δV (x).(4.285)(4.286)where∆V (0) =Z1dD p= ∂V [Tr log Ĥ]cl V (x)→V .(2πh̄)D p2 /2M + VThe result of the integration was given in Eq. (1.349).The second term in the remainder (4.285) reads explicitlyZZ112− Tr (∆V δV ) = −dD x dD x0 ∆V (x, x0 )δV (x0 )∆V (x0 , x)δV (x).22(4.287)(4.288)We now make use of the operator relation1[f (Â), B̂] = f 0 (Â)[Â, B̂] − f 00 (Â)[Â, [Â, B̂]] + . . . ,2(4.289)to expandhihiδV ∆V = ∆V δV + ∆2V T̂ , δV + ∆3V T̂ , [T̂ , δV ] + .
. . ,(4.290)where T̂ is the operator of the kinetic energy p̂2 /2M . It commutes with any function f (x) asfollows:h̄2 2 ∇ f + 2 (∇f ) · ∇ ,(4.291)[T̂ , f ] = −2Mh̄4 2 2 [T̂ , [T̂ , f ]] =(∇ ) f + 4[∇∇2 f ] · ∇ + 4[∇i ∇i f ]∇i ∇j ,(4.292)4M 2....H. Kleinert, PATH INTEGRALS4.9 Quantum Corrections to Classical Density of States411Inserting this into (4.288), we obtain a first contributionZZ1dD x dD x0 ∆V (x, x0 )∆V (x0 , x)[δV (x)]2 .−2(4.293)The spatial integrals are performed by going to momentum space, where we derive the generalformulaZZZDDd x d x1 · · · dD xn ∆V (x, x1 )∆V (x1 , x2 ) · · · ∆V (xn−1 , xn )∆V (xn , x)ZdD p1(−1)n n=∂V ∆V (0).(4.294)=(2πh̄)D (p2 /2M +V )n+1n!This simplifies (4.293) to1∂V ∆V (0)2ZdD x [δV (x)]2 .(4.295)We may now combine the non-gradient terms of δV (x) consisting of the first term in (4.284),of (4.286), and of (4.295), and replace in the latter ∆V (0) according to (4.287), to obtain the firstthree expansion terms of [Tr log Ĥ]cl with the full x-dependent V (x) in Eq.
(4.275).The next contribution to (4.288) coming from (4.291) ish̄24MZDd x×ZDd x1ZdD x2 ∆V (x, x1 )∆V (x1 , x2 )∆V (x2 , x)on∇2 δV (x) δV (x) + 2 [∇δV (x)]2 + 2[∇δV (x)]δV (x)∇ ,(4.296)where the last ∇ acts on the first x in ∆V (x, x1 ), due to the trace.
It does not contribute to theintegral since it is odd in x − x1 .We now perform the integrals over x1 and x2 using formula (4.294) and findh̄28MZdD xno2∇2 δV (x) [δV (x)]+2 [∇δV (x)] ∂V2 ∆V (0).(4.297)The first term can be integrated by parts, after which it removes half of the second term.A third contribution to (4.288) which contains only the lowest gradients of δV (x) comes fromthe third term in (4.292):−h̄48M 2ZdD xZdD x1ZdD x2ZdD x3 ∆V (x, x1 )∆V (x1 , x2 )∆V (x2 , x3 )∆V (x3 , x)× 4 [∇i ∇j δV (x)] δV (x)∇i ∇j ,(4.298)where the last ∇i ∇j acts again on the first x in ∆V (x, x1 ), as a consequence of the trace.
Inmomentum space, we encounter the integral#"ZZdD pdD p8M δijV−4pi pj /h̄21= − 2−(2πh̄)D (p2 /2M + V )4(2πh̄)D (p2 /2M +V )3 (p2 /2M +V )4h̄ D8M δij 1 213= − 2∂V + V ∂V ∆V (0).(4.299)6h̄ D 2so that the third contribution to (4.288) reads, after an integration by parts,h̄2−MZδijd x [∇δV (x)]DD21 21∂V + V ∂V326∆V (0).(4.300)4124 Semiclassical Time Evolution AmplitudeCombining all gradient terms in [∇V (x)]2 and replacing ∆V (0) according to (4.287), we recoverthe previous result (4.279) with the curly brackets (4.280).For the one-dimensional tracelog, this leads to the formulaZph̄2 [V 0 (x)]22 2Tr log[−h̄ ∂x + V (x)] = dx V (x) 1 ++ ... .(4.301)32 V 3 (x)It is a useful exercise to rederive this with the help of the Gelfand-Yaglom method in Section 2.4.This expansion can actually be deduced, and carried to much higher order, with the help ofthe gradient expansion of the trace of the logarithm of the operator −h̄2 ∂τ2 + w2 (τ ) derived inSubsection 2.15.3.
If we replace τ by x, v(τ ) by V (x), h̄ by 1, we obtain from (2.535):!!(Z32pV 009V 0 V 00V (3)15V 0V05V 0132−− h̄−+dx V (x) 1 − h̄ 3/2 − h̄h̄32V 38V 24V64V 9/232V 7/216V 5/2!)4221105V 0221V 0 V 0019V 007V 0 V (3)V (4)− h̄4.(4.302)−++−2048V 6256V 5128V 432V 432V 3The h̄-term in the curly brackets vanishes if V (x) is the same at the boundaries, and the h̄2 termgoes over into the h̄2 -term in (4.301).For completeness let us mention that by performing the operation (4.317) on the full expansion(4.302) we obtain the density of states!(Z42112155 V 025 V 03 V 00132ρ(E) =+h̄1 − h̄+dx √2πh̄E−V32 (E − V )38 (E − V )22048 (E − V )6!)221989 V 0 V 00133 V 0049 V 0 V (3)5 V (4)+.(4.303)5 +4 +4 +3256 (E − V )128 (E − V )32 (E − V )32 (E − V )4.9.5Local Density of States on CircleFor future use, let us also calculate this determinant for x on a circle x = (0, b), so that, as aside result, we obtain also the gradient expansion of the tracelog of the operator (−∂τ2 + ω 2 (τ )) ata finite temperature.
For this we recall that for a τ -independent frequency, the starting point isEq. (2.541), according to which the tracelog of the operator (−∂τ2 + ω 2 ) with periodic boundaryconditions in τ ∈ (0, h̄β) is given by"#Z ∞∞X2h̄dτ −1/2−(nh̄β)2 /4τFω = − √τ1+2ee−τ ω .(4.304)π 0 τn=1The first term is the zero-temperature expression, the second comes from the Poisson summationformula and givesp the finite-temperature effects. In the first (classical) term of the density (4.245),the factor 1/ 2πh̄τ /M came from the integral over the Boltzmann factor involving the kineticR∞2energy −∞ (dk/2π)e−τ h̄k /2M . For periodic boundary conditions in x ∈ (0, b), this is changed toP −τ h̄k2 /2Mm(1/b) m e, where km = 2πm/b.
By Poisson’s formula (1.205), this can be replaced bythe integral and an auxiliary sum∞ Z ∞∞XX22211 X −τ h̄kmdk −τ h̄k2 /2M+ibkn/2Mpe=e−n Mb /2h̄τ. (4.305)e=b m2π2πh̄τ /M n=−∞n=−∞ −∞If the sum is inserted into the integral (4.245), we obtain the density ρ(E; x) on a circle of circumference b, with the classical contributionZ ∞22∞dτ X e−n Mb /2h̄τ −τ [E−V (x)]/h̄ρcl (b, E; x) = 2.(4.306)2πh̄ n=−∞(2πh̄τ /M )1/20H.
Kleinert, PATH INTEGRALS4.9 Quantum Corrections to Classical Density of States413The n = 0 -term in the sum leads back to the original expression (4.245) on an infinite x-axis. Theτ -integrals are now done with the help of formula (2.542) which yields, due to Kν (z) = K−ν (z),ZνnM bdτ ν −n2 Mb2 /2h̄τ −[E−V (x)]τ /h̄τ e=2Kν (np(E; x)b/h̄),τp(E; x)∞0(4.307)and we obtain, instead of (4.246),1/2∞ X11nM bpρcl (b, E; x) =2K1/2 (np(E; x)b/h̄).πh̄ 2πh̄/M n=0 p(E; x)pπ/2z e−z [recall (2.544)], this becomesInserting K1/2 (z) =1Mρcl (b, E; x) =πh̄ p(E; x)The sumP∞n=11+2∞Xe−np(E;x)b/h̄n=1!.(4.308)(4.309)αn is equal to α/(1 − α), so that we obtainpM coth 2M [E − V (x)] b/2h̄M coth[p(E; x)b/2h̄]p=.ρcl (b, E; x) =πh̄p(E; x)πh̄2M [E − V (x)](4.310)For b → ∞, this reduces to the previous density (4.246).If we include the higher powers of τ in (4.245), we obtain the generalization of expression(4.247):ρ(b, E; x) =h̄2d2h̄2d31−V 00 (x) 2 −[V 0 (x)]2+...ρcl (b, E; x).12MdV24MdV 3(4.311)The tracelog is obtained by integrating this over dE log E from V (x) to infinity.
The integraldiverges, and we must employ analytic regularization. We proceedby using theR ∞ as in (4.276),0real-time version of (4.306) and rewriting log E as an integral − 0 (dt0 /t0 )e−iEt /h̄ , so that theleading term in (4.311) is given by[Tr log Ĥ]cl = −Zb0Zdx∞dE−∞Z∞−∞Z22∞dt X e−in Mb /2h̄t+it[E−V (x)]/h̄ ∞ dt0 −iEt0 /h̄e. (4.312)2πh̄ n=−∞t0(2πih̄t/M )1/20The integral over E leads now to[Tr log Ĥ]cl = −∞ ZXn=−∞bdx∞Z0022dt1e−in Mb /2h̄t−itV (x)/h̄ .t (2πih̄t/M )1/2(4.313)Deforming again the contour of integration by the substitution t = −iτ , creating the τ −1 in thedenominator by an integration over V (x)/h̄, we see that[Tr log Ĥ]cl = −π= πZZ0bdx0bdxZZ∞V (x)∞V (x)dV ρcl (b, E; x)E−V (x)→VpdV 1 coth 12 V /Ebp,1Eb 4πbV /Eb2(4.314)where Eb ≡ h̄/2M b2 is the energy associated with the length b.
The integration over V producesa factor h̄/τ in the integrand.4144 Semiclassical Time Evolution AmplitudeThus we obtainTr log Ĥ =Z0b s3V(x)2d1h̄2 .[V 0 (x)]2+...log2sinhdx 1 +24MdV 3b2Eb (4.315)For M = h̄2 /2, this give us the finite-b correction to formula (4.301).
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