Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 87
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Near the origin, itstarts out likef (ξ) = 1 − sξ + . . . ,(4.345)with a slopes ≈ 1.58807.For large ξ, it goes to zero likef (ξ) ≈(4.346)144.ξ3(4.347)This power falloff is a weakness of the model since the true screened potential shouldfall off exponentially fast. The right-hand side by itself happens to be an exactsolution of (4.344), but does not satisfy the desired boundary condition f (0) = 1.4.10.3Energy Functional of Thomas-Fermi AtomLet us derive an energy functional whose functional extremization yields theThomas-Fermi equation (4.344). First, there is the kinetic energy of the spin-upand spin-down electrons in a potential V (x).
It is given by the volume integral overtwice the Thomas-Fermi expression (4.329):(−)Ekin = 23 M5 2πh̄23/21Γ(5/2)Zd3 x [−V (x)]5/2 .(4.348)This can be expressed in terms of the electron density (4.327) as(−)Ekin3 Z 3 5/3= κ d x n (x),5(4.349)h̄ 2 2/33π.2M(4.350)Z(4.351)whereκ≡The potential energy(−)Epot =d3 x V (x)n(x)4204 Semiclassical Time Evolution Amplitude(−)is related to Ekin via relation (4.333) as5 (−)(−)Epot = − Ekin ,3(4.352)and the total electron energy in the potential V (x) is(−)(−)Ee(−) = Ekin + Epot =2 (−)E .5 pot(4.353)We now observe that if we consider the energy as a functional of an arbitrary densityn(x),ZZ3(−)35/3Ee = Ee [n] ≡ κ d x n (x) + d3 x V (x)n(x),(4.354)5the physical particle density (4.327) constitutes a minimum of the functional, whichsatisfies κn2/3 (x) = −V (x).
In the Thomas-Fermi atom, V (x) on the right-handside of (4.354) is, of course, the nuclear Coulomb potential, i.e.,(−)(−)Epot = EC ≡Zd3 x VC (x) n(x).(4.355)The energy Ee(−) has to be supplemented by the energy due to the Coulombrepulsion between the electrons(−)Eee= Eee [n] =e22Zd3 xd3 x0 n(x)100 n(x ).|x − x |(4.356)The physical energy density should now be obtained from the minimum of thecombined energy functionalZe2 Z 3 3 03 Z 3 5/3130Etot [n] = κ d x n (x) + d x VC (x) n(x) +d xd x n(x)0 n(x ).52|x − x |(4.357)Since we are not very familiar with extremizing nonlocal functionals, it will beconvenient to turn this into a local functional.
This is done as follows. We introducean auxiliary local field ϕ(x) and rewrite the interaction term asEee [n, ϕ] = Eϕ [n, ϕ] − Eϕϕ [ϕ] ≡Z1d x ϕ(x)n(x) −8πe23Zd3 x ∇ϕ(x)∇ϕ(x). (4.358)Extremizing this in ϕ(x), under the assumption of a vanishing ϕ(x) at spatial infinity, yields the electric potential of the electron cloud∇2 ϕ(x) = −4πe2 n(x),(4.359)which is the same as (4.334), but without the nuclear point charge at the origin.Inserting this into (4.358) we reobtain precisely to the repulsive electron-electroninteraction energy (4.356).H.
Kleinert, PATH INTEGRALS4.10 Thomas-Fermi Model of Neutral Atoms421Replacing the last term in (4.357) by the functional (4.358), we obtain a totalenergy functional Etot [n, ϕ], for which it is easy to find the extremum with respectto n(x). This lies atκ n2/3 (x) = −V (x),(4.360)whereV (x) = VC (x) + ϕ(x)(4.361)is the combined Thomas-Fermi potential of the nucleus and the electron cloud solvingthe Poisson equation (4.334).If the extremal density (4.360) is inserted into the total energy functionalEtot [n, ϕ], we may use the relation (4.352) to derive the following functional of ϕ(x):Etot [ϕ] = −25Zd3 x V (x)n(x) +18πe2Zd3 x ϕ(x)∇2 ϕ(x).(4.362)When extremizing this expression with respect to ϕ(x) we must remember thatV (x) = VC (x) + ϕ(x) is also present in n(x) with a power 3/2. The extremum liestherefore again at a field satisfying the Poisson equation (4.359).4.10.4Calculation of EnergiesWe now proceed to calculate explicitly the energies occuring in Eq.
(4.362). Theyturn out to depend only on the slope of the screening function f (ξ) at the origin.(−)Consider first Epot . The common prefactor appearing in all energy expressions canbe expressed in terms of the Thomas-Fermi length scale aTF of Eq. (4.340) as2M 3/24π 3 Z −1/2e = 3/2 .(2πh̄)3/2 Γ(5/2)aTF(4.363)We therefore obtain the simple energy integral involving the screening function f (ξ):(−)EpotZ 2 e2 Z ∞1dξ √ f 5/2 (ξ).=−a 0ξ(4.364)The interaction energy between the electrons at the extremal ϕ(x) satisfying (4.359)becomes simplyZ1(−)Eee=d3 x n(x)ϕ(x),(4.365)2which can be rewritten as(−)Eee1Z 31 (−) 1 (−)=d x n(x)[V (x) − VC (x)] = Epot − EC .222(4.366)Inserting this into (4.362) we find the alternative expression for the total energy2 (−)1 (−) 1 (−)(−)(−)Etot = Epot − Eee= − Epot + EC .5102(4.367)4224 Semiclassical Time Evolution Amplitude(−)The energy EC of the electrons in the Coulomb potential is evaluated as follows.Replacing n(x) by −∇2 ϕ(x)/4πe2 , we have, after two partial integrations withvanishing boundary terms and recalling (4.334),(−)EC = −14πe2Zd3 x ϕ(x)∇2 VC (x) = −Zd3 x ϕ(x)nC (x).(4.368)Now, sinceZe2ϕ(x) = V (x) − VC (x) = −[f (ξ) − 1],rnC = Zδ (3) (x),(4.369)(−)we see that the Coulomb energy EC depends only on ϕ(0), which can be expressedin terms of the negative slope (4.346) of the function f (ξ) as:ϕ(0) =Ze2s.a(4.370)Thus we obtainZ 2 e2s.(4.371)aWe now turn to the integral associated with the potential energy in Eq.
(4.364):(−)EC = −ZI[f ] =∞01dξ √ f 5/2 (ξ).ξ(4.372)By a trick it can again be expressed in terms of the slope parameter s. We expressthe energy functional (4.362) in terms of the screening function f (ξ) asEtot [ϕ] = −Z 2 e2ε[f ],aTF(4.373)with the dimensionless functional21ε[f ] ≡ I[f ] + J[f ] =52Z0∞)(12 1 5/2√ f (ξ) − [f (ξ) − 1]f 00 (ξ) .dξ5 ξ2(4.374)The second integral can also be rewritten asJ[f ] =Z∞0dξ [f 0 (ξ)]2 .(4.375)This follows from a partial integrationJ[f ] = −Z0∞dξ [f (ξ) − 1]f 00 (ξ) =Z0∞dξ [f 0 (ξ)]2 − [f (ξ) − 1]f 0 (ξ)|∞0 ,(4.376)inserting the boundary condition f (ξ) − 1 = 0 at ξ = 0 and f 0 (ξ) = 0 at ξ = ∞.We easily verify that the Euler-Lagrange equation following from ε[f ] is theThomas-Fermi differential equation (4.344).H.
Kleinert, PATH INTEGRALS4.10 Thomas-Fermi Model of Neutral Atoms423As a next step in calculating the integrals I and J, we make use of the fact thatunder a scaling transformation¯ = f (λξ),f (ξ) → f(ξ)(4.377)the functional ε[f ] goes over into12ελ[f ] = λ−1/2 I[f ] + λJ[f ].52(4.378)This must be extremal at λ = 1, from which we deduce that for f (ξ) satisfying thedifferential equation (4.344):5I[f ] = J[f ].(4.379)2This relation permits us to express the integral J in terms of the slope of f (ξ) atthe origin. For this we separate the two terms in (4.376), and replace f 00 (ξ) via theThomas-Fermi differential equation (4.344) to obtainJ =−Z∞0dξ f (ξ)f 00(ξ) +Z0∞Zdξ f 00 (ξ) = −0∞1dξ √ f 5/2 (ξ) −f 0 (0) = −I + s.
(4.380)ξTogether with (4.379), this implies25s,J = s.77Thus we obtain for the various energies:(4.381)I=(−)Ekin(−)EC3 Z 2 e2 5=s,5 aTF 7Z 2 e2=−s,aTF(−)Epot(−)EeeZ 2 e2 5=−s,aTF 7Z 2 e2 1=s,aTF 7Ee(−)2 Z 2 e2 5=−s,5 aTF 7(4.382)and the total energy is(−)Etot =2 (−)1 (−) 1 (−)3 Z 2 e2e2(−)Epot −Eee= − Epot + EC = −s ≈ −0.7687 Z 7/3 . (4.383)51027 aTFaHThe energy increases with the nuclear charge Z like Z 2 /aTF ∝ Z 7/3 .At the extremum, we may express the energy functional ε[f ] with the help of(4.380) as1 Z∞11dξ √ f 5/2 (ξ) − f (0)f 0 (0),ε̄[f ] = −(4.384)10 02ξor in a form corresponding to (4.367):ε̄¯[f ] ≡ −111I[f ] + JC [f ] = −10210Z0∞11dξ √ f 5/2 (ξ) +2ξZ0∞1dξ √ f 3/2 (ξ).
(4.385)ξUsing the Thomas-Fermi equation (4.344), the second integral corresponding to theCoulomb energy can be reduced to a surface term yielding JC [f ] = s, so that (4.385)gives the same total energy as (4.374).4244.10.54 Semiclassical Time Evolution AmplitudeVirial TheoremNote that the total energy is equal in magnitude and opposite in sign to the kineticenergy. This is a general consequence of the so-called viral theorem for Coulombsystems. The kinetic energy of the many-electron Schrödinger equation contains theLaplace differential operator proportional to ∇2 , whereas the Coulomb potentialsare proportional to 1/r. For this reason, a rescaling x → λx changes the sum ofkinetic and total potential energiesEtot = Ekin + Epot(4.386)λ2 Ekin + λEpot .(4.387)intoSince this must be extremal at λ = 1, one has the relation2Ekin + Epot = 0,(4.388)Etot = −Ekin .(4.389)which proves the virial theoremIn the Thomas-Fermi model, the role of total potential energy is played by the(−)(−), and Eq.
(4.382) shows that the theorem is satisfied.combination Epot − Eee4.10.6Exchange EnergyIn many-body theory it is shown that due to the Fermi statistics of the electronicwave functions, there exists an additional electron-electron exchange interactionwhich we shall now take into account. For this purpose we introduce the bilocaldensity of all states of negative energy by analogy with (4.323):(−)ρcl (xb , xa ) =Z0V (x̄)(−)dEρcl (E; xb , xa ).(4.390)In three dimensions we insert (4.268) and rewrite the energy integral as1 Z pF (x̄)dE =dp p ,M 0V (x̄)Z0(4.391)with the Fermi momentum pF (x̄) of the neutral atom at the point x̄ [see (4.325)].In this way we find(−)ρcl (xb , xa ) =p3F (x̄) 1(sin z − z cos z),2π 2h̄3 z 3(4.392)z ≡ pF (x̄)R/h̄.(4.393)whereH.
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