Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 86
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Replacing x by the Euclideantime τ , b by h̄β, and V (x) by the time-dependent square frequency ω 2 (τ ), we obtain the gradientexpansionZ h̄β 2h̄βω(τ )[∂τ ω 2 (τ )]2 322.(4.316)∂ω2 +. . .log 2 sinhTr log[−∂τ + ω (τ )] =dτ 1 +12h̄β204.9.6Quantum Corrections to Bohr-SommerfeldApproximationThe expansion (4.302) can be used to obtain a higher-order expansion of the density of states ρ(E),thereby extending Eq. (4.249).
For this we recall Eq. (1.591) according to which we can calculatethe exact density of states from the formula1ρ(E) = − ∂E Im Tr log −∂x2 + [V (x) − E] .π(4.317)Integrating this over the energy yields, according to Eq. (1.592), the number of states times π,and thus the simple exact quantization condition for a nondegenerate one-dimensional system:−Im Tr log −∂x2 + [V (x) − E] = π(n + 1/2).(4.318)pThepleft-hand side is simply obtained from (4.302) by replacing V (x) → V (x)−E, so that V (x) →−i E − V (x), and we find("#Z2p 21V 005V 02Re i Tr log −∂x + [V (x) − E] =dx E − V (x) 1 + h̄3 +2h̄32(E −V )8(E −V )"#)224221V 0 V 0019V 007V 0 V (3)V (4)1105V 0++.
(4.319)−h̄4654 +4 +3 + ...2048(E −V )256(E −V )128(E −V )32(E −V )32(E −V )The first term in the expansion corresponds to the Bohr-Sommerfeld approximation, the remainingones yield the quantum corrections. The integrand agrees, of course, with the WKB expansion ofthe eikonal (4.14) with the expansion terms (4.16), (4.18).Let us calculate the corrections for a purely quartic potential V (x) = gx4 /4, where the integralover the first term in (4.319) between the turning points ±xE = ±(4E/g)1/4 gave the BohrSommerfeld approximation (4.29). The integrals of the higher terms in (4.319) are divergent, butcan be calculated in analytically regularized form from Formula (4.30).
This extends the BohrSommerfeld equation toN (y) = y −−14697π390065π 411π 2+−+12πy 10368Γ8( 34 )y 31866240Γ8( 34 )y 5501645312Γ16( 43 )y 753352893π 3+ . . . = n + 1/2.7739670528Γ16( 34 )y 9(4.320)The left-hand side is the number of states in Eqs. (1.584) and (4.203) as a function of y ratherthan E. The function is plotted in Fig. 4.320 for increasing orders in y. Given a solution y (n) ofthis equation, we obtain the energy E (n) from Eq. (4.31) with√(4.321)κ(n) = [y (n) 3Γ(3/2)2 /2 π]4/3 .H. Kleinert, PATH INTEGRALS4.10 Thomas-Fermi Model of Neutral Atoms24152N (y)n + 1/21.52146810-20.5-4(n)0.511.5y-0.522.53-6log[E (n) /EBS − 1]-8-1Figure 4.1 Determination of energy eigenvalues E (n) for purely quartic potential gx4 /4in semiclassical expansion.
The intersections of N (y) with the horizontal lines yield y (n) ,from which E (n) is obtained via Eqs. (4.321) and (4.31). The increasing dash lengthsshow the expansions of N (y) to increasing orders in y. For the ground state with n = 0,the expansion is too divergent to give improvements to the lowest approximation withoutresumming the series; Right: Comparison between exact and semiclassical energies. The(n)plot is for log[E (n) /EBS − 1].For large n, where y (n) → n, we recover the Bohr-Sommerfeld result (4.32). We can invert theseries (4.320) and obtain0.026525823 0.002762954 0.001299177−−y (n) = (n + 21 ) 1 +(n + 21 )2(n + 21 )4(n + 12 )60.003140091 0.007594497+++....(4.322)(n + 21 )8(n + 21 )10The results are compared with the exact ones in Table 4.1, which approach rapidly the Bohr-Table 4.1 Particle energies in purely anharmonic potential gx4 /4 for n = 0, 2, 4, 6, 8, 10.n0246810E (n) /(gh̄/4M 2 )1/30.667 986 259 155 777 108 34.696 795 386 863 646 196 210.244 308 455 438 771 076 016.711 890 073 897 950 947 123.889 993 634 572 505 935 531.659 456 477 221 552 442 8κ(n) /2(n + 1/2)4/30.841 609 948 950 895 5260.692 125 685 914 981 3140.689 449 772 359 340 7650.688 828 486 600 234 4660.688 590 146 947 993 6760.688 474 290 179 981 433Sommerfeld limit 0.688 253 702 .
. . The approach is illustrated in the right-hand part of Fig. 4.1(n)where log[(y (n) /(n + 21 ))4/3 − 1] = log[E (n) /EBS − 1] is plotted once for the exact values and oncefor the semiclassical expansion in Fig. 4.1. The second excited states is very well represented bythe series. For a detailed study of the convergence of the semiclassical expansion see Ref. [4].4.10Thomas-Fermi Model of Neutral AtomsThe density of states calculated in the last section forms the basis for the ThomasFermi model of neutral atoms.
If an atom has a large nuclear charge Z, most of4164 Semiclassical Time Evolution Amplitudethe electrons move in orbits with large quantum numbers. For Z → ∞, we expectthem to be described by semiclassical limiting formulas, which for decreasing valuesof Z require quantum corrections. The largest quantum correction is expected forelectrons near the nucleus which must be calculated separately.4.10.1Semiclassical LimitFilling up all negative energy states with electrons of both spin directions producessome local particle density n(x), which is easily calculated from the classical localdensity (4.210) over all negative energies, yielding the Thomas-Fermi density ofstates(−)ρcl (x)M=dE ρcl (E; x) =V (x)2πh̄2Z0D/21[−V (x)]D/2 .Γ(D/2 + 1)(4.323)This expression can also be obtained directly from the phase space integral over theaccessible free-particle energies.
At each point x, the electrons occupy all levels upto a Fermi energyp (x)2EF = F+ V (x).(4.324)2MThe associated local Fermi momentum is equal to the local momentum function(4.208) at E = EF :pF (x) = p(EF ; x) =q2M[EF − V (x)].(4.325)The electrons fill up the entire Fermi sphere |p| ≤ pF (x):Z pF (x)dD p112π D/2 pDD−1F (x).=Sdpp=DD DD|p|≤pF (x) (2πh̄)0(2πh̄)(2πh̄) Γ(D/2) D(4.326)For neutral atoms, the Fermi energy is zero and we recover the density (4.323).By occupying each state of negative energy twice, we find the classical electrondensity(−)n(x) = 2ρcl (x).(4.327)(−)ρcl (x) =ZThe potential energy density associated with the levels of negative energy is obviously(−)Epot TF (x)(−)= V (x)ρM(x) = −2πh̄2D/21[−V (x)]D/2+1 .
(4.328)Γ(D/2 + 1)To find the kinetic energy density we integrate(−)Ekin TF (x)0=Z=D/2MD/2 + 1 2πh̄2V (x)dE [E − V (x)]ρcl (E; x)D/21[−V (x)]D/2+1 .Γ(D/2 + 1)(4.329)H. Kleinert, PATH INTEGRALS4.10 Thomas-Fermi Model of Neutral Atoms417As in the case of the density of states (4.326), this expression can be obtaineddirectly from the phase space integral over the free-particle energies(−)Ekin TF (x) =Z|p|≤pF (x)d D p p2.(2πh̄)D 2M(4.330)Performing the momentum integral on the right-hand side yields the energy density(−)Ekin TF (x)11=D SD2M(2πh̄)ZpF (x)0dp pD+1 =SD pD+21(x)F, (4.331)D(2πh̄) D + 2 2Min agreement with (4.329).
The sum of the two is the Thomas-Fermi energy density(−)ETF (x) =Z0V (x)= −dE E ρcl (E; x)M1D/2 + 1 2πh̄2D/21[−V (x)]D/2+1 .Γ(D/2 + 1)(4.332)The three energies are related by(−)ETF (x) = −11(−)(−)Ekin TF (x) =Epot TF (x).D/2D/2 + 1(4.333)Note that the Thomas-Fermi model can also be applied to ions. Then the energylevels are filled up to a nonzero Fermi energy EF , so that the density of states(4.323) and the kinetic energy (4.329) have −V replaced by EF − V .
This followsimmediately from the representations(4.326) and (4.330) where the right-hand sidesqdepend only on pF (x) = 2M[EF − V (x)]. In the potential energy (4.328), theexpression (−V )D/2+1 is replaced by (−V )(EF − V )D/2 , whereas in the ThomasFermi energy density (4.332) it becomes (1 − EF ∂/EF )(EF − V )D/2+1 .4.10.2Self-Consistent Field EquationThe total electrostatic potential energy V (x) caused by the combined charges of thenucleus and the electron cloud is found by solving the Poisson equation∇2 V (x) = 4πe2 [Zδ (3) (x) − n(x)] ≡ 4πe2 [nC (x) − n(x)].(4.334)The nucleus is treated as a point charge which by itself gives rise to the CoulombpotentialZe2VC (x) = −.(4.335)rRecall that in these units e2 = αh̄c, where α is the dimensionless fine-structureconstant (1.503).
A single electron near the ground state of this potential has orbitswith diameters of the order naH /Z, where n is the principal quantum number andaH the Bohr radius of the hydrogen atom, which will be discussed in detail in4184 Semiclassical Time Evolution AmplitudeChapter 13. The latter is expressed in terms of the electron charge e and mass Mas1 Ch̄2λ .(4.336)aH =2 =α MMeThis equation implies that aH is about 137 times larger than the Compton wavelengthof the electron−13λCcm.(4.337)M ≡ h̄/Mc ≈ 3.861 593 23 × 10It is convenient to describe the screening effect of the electron cloud upon theCoulomb potential (4.335) by a multiplicative dimensionless function f (x). Restricting our attention to the ground state, which is rotationally symmetric, we shall writethe solution of the Poisson equation (4.334) asV (x) = −Ze2f (r).r(4.338)At the origin the function f (r) is normalized to unity,f (0) = 1,(4.339)to ensure that the nuclear charge is not changed by the electrons.It is useful to introduce a length scale of the electron cloudaTF1 2πh̄2 Γ(5/2)= 2 1/3M2 · 4πeZ"#2/3=1 3π2 42/3aHaH≈ 0.8853 1/3,1/3ZZ(4.340)which is larger than the smallest orbit aH /Z by roughly a factor Z 2/3 .
All lengthscales will now be specified in units of aTF , i.e., we setr = aTF ξ.(4.341)In these units, the electron density (4.327) becomes simplyn(x) = −2Ze2 M3/2 "3π 2h̄3f (ξ)aTF ξ#3/2"f (ξ)Z3=3 aTFξ4πaTF#3/2.(4.342)The left-hand side of the Poisson equation (4.334) reads1 dZe2 1 00∇ V (x) =f (ξ),rV (x) = − 3r dr 2aTF ξ2(4.343)so that we obtain the self-consistent Thomas-Fermi equation1f 00 (ξ) = √ f 3/2 (ξ),ξξ > 0.(4.344)The condition ξ > 0 excludes the nuclear charge from the equation, whose correctsize is incorporated by the initial condition (4.339).H. Kleinert, PATH INTEGRALS4.10 Thomas-Fermi Model of Neutral Atoms419f (ξ)ξFigure 4.2 Solution for screening function f (ξ) in Thomas-Fermi model.Equation (4.344) is solved by the function shown in Fig. 4.2.
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