Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 88
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Kleinert, PATH INTEGRALS4.10 Thomas-Fermi Model of Neutral Atoms425This expression can, incidentally, be obtained alternatively by analogy with thelocal expression (4.326) from a momentum integral over free wavefunctionsd3 p ip(xb −xa )/h̄e.(2πh̄)DZ(−)ρcl (xb , xa ) =|p|≤pF (x̄)(4.394)The simplest way to derive the exchange energy is to re-express the density ofstates ρ(−) (E; x) as the diagonal elements of the bilocal densityρ(−) (x) = ρ(−) (xb , xa )(4.395)and rewrite the electron-electron energy (4.356) as(−)Eeee2=4×2Zd3 xd3 x0 ρ(−) (x, x)1ρ(−) (x0 , x0 ).4π|x − x0 |(4.396)The factor 4 accounts for the four different spin pairs in the first and the secondbilocal density.↑ ↑; ↑ ↑;↑ ↑; ↓ ↓;↓ ↓; ↑ ↑;↓ ↓; ↓ ↓ .In the first and last case, there exists an exchange interaction which is obtained byinterchanging the second arguments of the bilocal densities and changing the sign.This yields(−)Eexch = −2 ×1e2 Z 3 3 0 (−)ρ(−) (x0 , x).d xd x ρ (x, x0 )24π|x − x0 |(4.397)The integral over x − x0 may be performed using the formulaZ0∞dzz 21 1(sin z − z cos z)z z321= ,4(4.398)and we obtain the exchange energy(−)Eexche2=− 34πZ"p (x̄)d x̄ Fh̄3#4.(4.399)InsertingpF (x) =the exchange energy becomes(−)Eexchvuut2Z f (ξ),aH aTF ξ24 aTF Z 25/3 e=− 2I≈ −0.3588 ZI,aH 2π aH 2 aHwhere I2 is the integralI2 ≡Z0∞dξ f 2 (ξ) ≈ 0.6154.(4.400)(4.401)(4.402)4264 Semiclassical Time Evolution AmplitudeHence we obtain(−)Eexch ≈ −0.2208Z 5/3e2,aH(4.403)giving rise to a correction factorCexch (Z) = 1 + 0.2872 Z −2/3(4.404)to the Thomas-Fermi energy (4.383).4.10.7Quantum Correction Near OriginThe Thomas-Fermi energy with exchange corrections calculated so far would be reliable for large-Z only if the potential was smooth so that the semiclassical approximation is applicable.
Near the origin, however, the Coulomb potential is singularand this condition is no longer satisfied. Some more calculational effort is necessaryto account for the quantum effects near the singularity, based on the following observation [2]. For levels with an energy smaller than some value ε < 0, which is largecompared to the ground state energy Z 2 e2 /aH , but much smaller than the averageThomas Fermi energy per particle Z 2 e2 /aZ ∼ Z 2 e2 /aH Z 2/3 , i.e., forZ 2 e2Z 2 e2 1−ε,aH Z 2/3aH(4.405)we have to recalculate the energy. Let us define a parameter ν byZ2−ε ≡,2aH ν 2(4.406)which satisfies1 ν 2 Z 2/3 .(4.407)The contribution of the levels with energyp2Ze2−< −ε2Mr(4.408)(−)to Ekin,TF is given by an integral like (4.331), where the momentum runs from 0qto p−ε (x) = 2M[−ε − V (x)]. In the kinetic energy (4.332), the potential V (x) issimply replaced by −ε − V (x), and the spatial integral covers the small sphere ofradius rmax , where −ε − V (x) > 0.
For the screening function f (ξ) = V (r)r/Ze2this implies the replacementf (ξ) → [−ε − V (r)]whereξm =r= f (ξ) − ξ/ξm ,Ze22ν 2 aHZ a(4.409)(4.410)H. Kleinert, PATH INTEGRALS4.10 Thomas-Fermi Model of Neutral Atoms427is small of the order Z −2/3 . Using relation (4.389) between total and kinetic energieswe find the additional total energy(−)∆Etot3 Z 2 e2=−5 aZξmax01dξ √ [f (ξ) − ξ/ξm]5/2 ,ξ(4.411)where ξmax ≡ rmax /a is the place at which the integrand vanishes, i.e., whereξmax = Ze2 εf (ξmax)/a,(4.412)this being the dimensionless version of−ε − V (rmax ) = 0.(4.413)Under the condition (4.407), the slope of f (ξ) may be ignored and we can use theapproximationξmax ≈ ξm(4.414)corresponding to rmax = Ze2 ε, with an error of relative order Z −2/3 .
After this, theintegralZ01/c11 51dλ √ (1 − cλ)5/2 = √ B (1/2, 7/2) = √ π,cc8λ(4.415)yielding a Beta function B(x, y) ≡ Γ(x)Γ(y)/Γ(x + y), leads to an energy(−)∆Etot3 Z 2 e2 5 π=−5 a 8MrZ 2 e2aH ν=−ν,2a Z 1/3a(4.416)showing that the correction to the energy will be of relative order 1/Z 1/3 . Expressinga in terms of aH via (4.340), we find(−)∆Etot = −Z 2 e2ν.aH(4.417)The point is now that this energy can easily be calculated more precisely.
Sincethe slope of the screening function can be ignored in the small selected radius,the potential is Coulomb-like and we may simply sum all occupied exact quantummechanical energies En in a Coulomb potential −Ze2 /r which lie below the totalenergy −ε.
They depend on the principal quantum number n in the well-knownway:Z 2 e2 1En = −.(4.418)aH 2n2Each level occurs with angular momentum l = 0, . . . , n − 1, and with two spindirections so that the total degeneracy is 2n2 . By Eq. (4.406), the maximal energy4284 Semiclassical Time Evolution Amplitude−ε corresponds to a maximal quantum number nmax = ν. The sum of all energiesEn up to the energy ε is therefore given by(−)∆QM Etot= −2νZ 2 e2 1 X1aH 2 n=0Z 2 e2= −[ν] ,aH(4.419)where [ν] is the largest integer number smaller than ν.
The difference between thesemiclassical energy (4.416) and the true quantum-mechanical one (4.419) yields thedesired quantum correction(−)∆Ecorr=−Z 2 e2([ν] − ν).aH(4.420)For large ν, we must average over the step function [ν], and find1h[ν]i = ν − ,2(4.421)and thereforeZ 2 e2 1=.(4.422)aH 2This is the correction to the energy of the atom due to the failure of the quasiclassical expansion near the singularity of the Coulomb potential. With respect tothe Thomas-Fermi energy (4.383) which grows with increasing nuclear charge Z like−0.7687 Z 7/3 e2 /aH , this produces a correction factor(−)∆Ecorr7aTF≈ 1 − 0.6504 Z −1/36aH sCsing (Z) = 1 −(4.423)to the Thomas-Fermi energy (4.383).4.10.8Systematic Quantum Corrections to Thomas-FermiEnergiesJust as for the density of states in Section 4.9, we can derive the quantum corrections to the energiesin the Thomas-Fermi atom. The electrons fill up all negative-energy levels in the combined potentialV (x).
The density of states in these levels can be selected by a Heaviside function of the negativeHamiltonian operator as follows:ρ(−) (x) = hx|Θ(−Ĥ)|xi.(4.424)Using the Fourier representation (1.309) for the Heaviside function we writeZ ∞dtΘ(−Ĥ) =e−iĤt/h̄ ,−∞ 2πi(t − iη)(4.425)and obtain the integral representationρ(−)(x) =Z∞−∞dt(x t|x 0).2πi(t − iη)(4.426)H.
Kleinert, PATH INTEGRALS4.10 Thomas-Fermi Model of Neutral Atoms429Inserting the short-time expansion (4.250), and the correspondence t → ih̄d/dV in the timeintegral, we find3h̄2d2h̄222 dρ(x) = 1 −∇ V (x) 2 −[∇V (x)]+ . . . ρTF (x).12MdV24MdV 3(4.427)The potential energy is simply given by(−)Epot (x)= hx|V (x)Θ(−Ĥ)|xi =Zd3 xV (x)ρ(−) (x).(4.428)With (4.427), this becomesd2h̄2d3h̄2(−)∇2 V (x) 2 −[∇V (x)]2+...ρTF (x).Epot (x) = V (x) 1 −12MdV24MdV 3(4.429)For the energy of all negative-energy states we may introduce a density functionE (−) (x) = hx|ĤΘ(−Ĥ)|xi.(4.430)The derivative of this with respect to V (x) is equal to the density of states (4.424):∂E (−) (x) = ρ(−) (x).∂V (x)(4.431)This follows right-away from ∂ Ĥ/∂V (x) = 1 and ∂[xΘ(x)]/δx = Θ(x).Inserting the representation (4.425), the factor Ĥ can be obtained by applying the differentialoperator ih̄∂t to the exponential function. After a partial integration, we arrive at the integralrepresentationZ ∞dtE (−) (x) = ih̄(x t|x 0).(4.432)2−∞ 2πi(t − iη)Inserting on the right-hand side the expansion (4.250), the leading term produces the local ThomasFermi energy density (4.332):(−)ETF (x) = −M2πh̄D/21[−V (x)]D/2+1 .Γ(D/2 + 2)(4.433)The short-time expansion terms yield, with the correspondence t → ih̄d/dV , the energy includingthe quantum correctionsE(−)3h̄2d2h̄2(−)22 d(x) = 1 −∇ V (x) 2 −[∇V (x)]+ .
. . ETF (x).12MdV24MdV 3(4.434)One may also calculate selectively the kinetic energy density from the expressionE (−) (x) =1hx|p̂2 Θ(−Ĥ)|xi.2M(4.435)This can obviously be extracted from (4.432) by a differentiation with respect to the mass:Z ∞∂dt(−)Ekin (x) = −Mih̄hx|e−iĤt/h̄ |xi∂M2πi(t− iη)2−∞∂E (−) (x).(4.436)= −M∂M4304 Semiclassical Time Evolution Amplitude(−)According to Eq. (4.434), the first quantum correction to the energy Ee isZ3h̄2d2h̄22(−)32 d∆Ee= − d x −∇ V−(∇V )12MdV 224MdV 33/2M12(−V )5/2× 25 2πh̄Γ(5/2)√Z2M1−3/2231/2 2=(∇V ) .d x (−V ) ∇ V − (−V )12h̄π 24(4.437)It is useful to bring the second term to a more convenient form. For this we note that by the chainrule of differentiationii 3h3 hV −1/2 (∇V )2 + 2V 1/2 ∇2 V .(4.438)∇2 V 3/2 = ∇ V 1/2 ∇V =24As a consequence we find∆Ee(−)=√Z2 22M31/2 23/2d x (−V ) ∇ V − ∇ (−V ).24h̄π 23(4.439)This energy evaluated with the potential V determined above describes directly the lowestcorrection to the total energy.
To prove this, consider the new total energy [recall (4.358)](−)Etot = Ee(−) + ∆Ee(−) − Eϕϕ [ϕ].(4.440)Extremizing this in the field ϕ(x) and denoting the new extremal field by ϕ(x) + ∆ϕ(x), we obtainfor ∆ϕ(x) the field equation:δ1[Ee(−) + ∆Ee(−) ] + 2 ∇2 [ϕ(x) + ∆ϕ(x)] = 0.δV (x)e(4.441)Taking advantage of the initial extremality condition (4.359), we derive the field equation(−)δ∆Ee1 2∇ ∆ϕ(x) = −.e2δV (x)(4.442)(−)If we now expand the corrected energy up to first order in ∆Ee(−)Etot = Ee(−) + ∆Ee(−) +Z(−)d3 xδEe1∆ϕ(x) − Eϕϕ [ϕ] −δV (x)4πe2Zand ∆ϕ(x), we obtaind3 x∇ϕ(x)∇∆ϕ(x).(4.443)Due to the extremality property (4.359) of the uncorrected energy at the original field ϕ(x) , thesecond and fourth terms cancel each other, and the correction to the total energy is indeed givenby (4.439).Actually, the statement that the energy (4.439) is the next quantum correction is not quite true.When calculating the first quantum correction in Subsection 4.10.7, we subtracted the contributionof all orbits with total energiesE < −ε.(4.444)After that we calculated in (4.419) the exact quantum corrections coming from the neighborhoodr < rmax = Ze2 ε(4.445)of the origin.
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