Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 25
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Thus FG (T, µ, V ) must be directlyproportional to V . The proportionality constant defines the pressure p of the system:FG (T, µ, V ) ≡ −p(T, µ, V )V.(1.572)Under infinitesimal changes of the three variables, FG (T, µ, V ) changes as follows:dFG (T, µ, V ) = −SdT + µdN − pdV.(1.573)The first two terms on the right-hand side follow from varying Eq. (1.570) at a fixedvolume. When varying the volume, the definition (1.572) renders the last term.Inserting (1.572) into (1.570), we find Euler’s relation:E = T S + µN − pV.(1.574)The energy has S, N, V as natural variables.
Equivalently, we may writeF = −µN − pV,(1.575)where T, N, V are the natural variables.1.18Density of States and TracelogIn many thermodynamic calculations, a quantity of fundamental interest is the density of states To define it, we express the canonical partition functionZ(T ) = Tr e−Ĥ/kB T(1.576)as a sum over the Boltzmann factors of all eigenstates |ni of the Hamiltonian:, i.e.e−En /kB T .(1.577)dE ρ(E)e−E/kB T .(1.578)Z(T ) =XnThis can be rewritten as an integral:Z(T ) =ZThe quantityρ(E) =Xnδ(E − En )(1.579)H.
Kleinert, PATH INTEGRALS831.18 Density of States and Tracelogspecifies the density of states of the system in the energy interval (E, E + dE). Itmay also be written formally as a trace of the density of states operator ρ̂(E):ρ(E) = Tr ρ̂(E) ≡ Tr δ(E − Ĥ).(1.580)The density of states is obviously the Fourier transform of the canonical partitionfunction (1.576):ρ(E) =Z∞−i∞Z ∞dβ βEdβ βEe Tr e−β Ĥ =e Z(1/kB β).2πi−i∞ 2πiThe integralN(E) =EZdE 0 ρ(E 0 )(1.581)(1.582)is the number of states up to energy E. The integration may start anywhere belowthe ground state energy. The function N(E) is a sum of Heaviside step functions(1.310):XΘ(E − En ).(1.583)N(E) =nThis equation is correct only with the Heaviside function which is equal to 1/2 atthe origin, not with the one-sided version (1.303), as we shall see later.
Indeed, ifintegrated to the energy of a certain level En , the result isN(En ) = (n + 1/2).(1.584)This formula will serve to determine the energies of bound states from approximations to ω(E) in Section 4.7, for instance from the Bohr-Sommerfeld condition(4.184) via the relation (4.203). In order to apply this relation one must be surethat all levels have different energies. Otherwise N(E) jumps at En by half the degeneracy of this level. In Eq.
(4A.9) we shall exhibit an example for this situation.An important quantity related to ρ(E) which will appear frequently in this textis the trace of the logarithm, short tracelog, of the operator Ĥ − E.Tr log(Ĥ − E) =Xnlog(En − E).(1.585)It may be expressed in terms of the density of states (1.580) asTr log(Ĥ − E) = TrZ∞−∞dE 0 δ(E 0 − Ĥ) log(E 0 − E) =Z∞−∞dE 0 ρ(E 0 ) log(E 0 − E).(1.586)The tracelog of the Hamiltonian operator itself can be viewed as a limit of an operatorzeta function associated with Ĥ:ζ̂Ĥ (ν) = Tr Ĥ −ν ,(1.587)whose trace is the generalized zeta-functionhiζĤ (ν) ≡ Tr ζ̂Ĥ (ν) = Tr (Ĥ −ν ) =XnEn−ν .(1.588)841 FundamentalsFor a linearly spaced spectrum En = n with n = 1, 2, 3 .
. . , this reduces to Riemann’szeta function (2.514).From the generalized zeta function we can obtain the tracelog by forming thederivativeTr log Ĥ = −∂ν ζĤ (ν)|ν=0 .(1.589)By differentiating the tracelog (1.585) with respect to E, we find the trace of theresolvent (1.316):∂E Tr log(Ĥ − E) = TrX1 X111=Rn (E) = Tr R̂(E).=ih̄ nih̄E − Ĥn E − En(1.590)Recalling Eq. (1.326) we see that the imaginary part of this quantity slightly abovethe real E-axis yields the density of statesX1− Im ∂E Tr log(Ĥ − E − iη) =δ(E − En ) = ρ(E).πn(1.591)By integrating this over the energy we obtain the number of state function N(E) ofEq.
(1.582):X1Θ(E − En ) = N(E).(1.592)− Im Tr log(E − Ĥ) =πnAppendix 1ASimple Time Evolution OperatorConsider the simplest nontrivial time evolution operator of a spin-1/2 particle in a magnetic fieldB. The reduced Hamiltonian operator is Ĥ0 = −B · /2, so that the time evolution operator reads,in natural units with h̄ = 1,e−iĤ0 (tb −ta ) = ei(tb −ta )B· /2 .(1A.1)Expanding this as in (1.294) and using the fact that (B· )2n = B 2n and (B· )2n+1 = B 2n (B· ),we obtaine−iĤ0 (tb −ta ) = cos B(tb − ta )/2 + iB̂ · sin B(tb − ta )/2 ,(1A.2)where B̂ ≡ B/|B|. Suppose now the magnetic field is not constant but has a small time-dependentvariation δB(t). Then we obtain from (1.254) [or the lowest expansion term in (1.294)]Z tbδe−iĤ0 (tb −ta ) =(1A.3)dt e−iĤ0 (tb −t) δB(t) · e−iĤ0 (t−ta ) .taUsing (1A.2), the integrand on the right-hand side becomeshihcos B(tb −t)/2+iB̂ · sin B(tb −t)/2 δB(t) · cos B(t−ta )/2+iB̂ ·We simplify this with the help of the formula [recall (1.446)]σ i σ j = δij + iijk σ kisin B(t−ta )/2 .
(1A.4)(1A.5)so thatB̂ ·δB(t) ·= B̂ · δB(t) + i[B̂ × δB(t)] · , δB(t) ·B̂ ·= B̂ · δB(t) − i[B̂ × δB(t)] · , (1A.6)H. Kleinert, PATH INTEGRALS85Appendix 1B Convergence of Fresnel IntegralandδB(t) ·B̂ ·B̂ ·hi= B̂ · δB(t) B̂ ·+ i[B̂ × δB(t)] · B̂ ·no= i[B̂ × δB(t)] · B̂ + [B̂ · δB(t)]B̂ − [B̂ × δB(t)] × B̂ · . (1A.7)The first term on the right-hand side vanishes, the second term is equal to δB, since B̂2 = 1. Thuswe find for the integrand in (1A.4):cos B(tb − t)/2 cos B(t − ta )/2 δB(t) ·+i sin B(tb − t)/2 cos B(t − ta )/2{B̂ · δB(t) + i[B̂ × δB(t)] · }+i cos B(tb − t)/2 sin B(t − ta )/2{B̂ · δB(t) − i[B̂ × δB(t)] · }+ sin B(tb − t)/2 sin B(t − ta )/2 δB ·(1A.8)which can be combined tonocos B[(tb +ta )/2−t] δB(t)−sin B[(tb +ta )/2−t] [B̂ × δB(t)] · +i sin B(tb −ta )/2 B̂·δB(t).(1A.9)Integrating this from ta to tb we obtain the to variation (1A.3).Appendix 1BConvergence of Fresnel IntegralHere we prove the convergence of the Fresnel integrals (1.334) by relating it to the Gauss integral.According to Cauchy’s integral theorem, the sum of the integrals along the three pieces of theFigure 1.4 Triangular closed contour for Cauchy integral2closed contour shown in Fig.
1.4 vanishes since the integrand e−z is analytic in the triangulardomain:IZ AZ BZ O2222dze−z =dze−z +dze−z +dze−z = 0.(1B.1)0ABLet R be the radius of the arc. Then we substitute in the three integrals the variable z as follows:0 A:B 0:AB:z = p,z = peiπ/4 ,z = R eiϕ ,and obtain the equationZ RZ−p2iπ/4dp e+e00Rdp e−ip2dz = dp,z 2 = p2iπ/4dz = dp e, z 2 = ip2dz = i Rdp, z 2 = p2 ,+Z0π/42dϕ iR e−R(cos 2ϕ+i sin 2ϕ)+iϕ= 0.(1B.2)√The first integral converges rapidly to π/2 for R → ∞. The last term goes to zero in this limit.To see this we estimate its absolute value as follows:ZZ π/4 π/42−R2 (cos 2ϕ+i sin 2ϕ)+iϕ dϕ iR edϕ e−R cos 2ϕ .(1B.3)<R 00861 FundamentalsThe right-hand side goes to zero exponentially fast, except for angles ϕ close to π/4 where thecosine in the exponent vanishes.
In the dangerous regime α ∈ (π/4 − , π/4) with small > 0, onecertainly has sin 2ϕ > sin 2α, so thatRZπ/4dϕ e−R2 cos 2ϕα<RZπ/4αdϕsin 2ϕ −R2 cos 2ϕ.esin 2α(1B.4)The right-hand integral can be performed by parts and yields2αR e−Rcos 2α+hiϕ=π/421,e−R cos 2ϕR sin 2αϕ=α(1B.5)which goes to zero like 1/R for large R. Thus we find from (1B.2) the limiting formula√−e−iπ/4 π/2, orZ ∞√2dp e−ip = e−iπ/4 π,R0∞2dp e−ip =(1B.6)∞pwhich goes into Fresnel’s integral formula (1.334) by substituting p → p a/2.Appendix 1CThe Asymmetric TopThe Lagrangian of the asymmetric top with three different moments of inertia readsL=1[Iξ ωξ 2 + Iη ωη 2 + Iζ ωζ 2 ].2(1C.1)It has the Hessian metric [recall (1.12) and (1.385)]Iξ sin2 β + Iζ cos2 β − (Iξ − Iη ) sin2 β sin2 γ,−(Iξ − Iη ) sin β sin γ cos γ,g11g21==g31=Iζ cos β,g22g32==Iη + (Iξ − Iη ) sin2 γ,0,g33=Iζ ,(1C.2)rather than (1.460).
The determinant isg = Iξ Iη Iζ sin2 β,(1C.3)and the inverse metric has the componentsg 11=g 21=g 31=g 22=g 32=g 33=1{Iη + (Iξ − Iη ) sin2 γ}Iζ ,g1sin β sin γ cos γ(Iξ − Iη )Iζ ,g1{cos β[− sin2 γ(Iξ − Iη ) − Iη ]}Iζ ,g1{sin2 β[Iξ − sin2 γ(Iξ − Iη )]}Iζ ,g1{sin β cos β sin γ cos γ(Iη − Iξ )}Iζ ,g1{sin2 βIξ Iη + cos2 βIη Iζ + cos2 β sin2 γ (Iξ − Iη )Iζ }.g(1C.4)H.
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