Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 75
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(3.534), (3.535), and (3.536), but not the same. Extendingtheir method, we derive here a recursion relation for the perturbation coefficients of all energylevels of the anharmonic oscillator in any number of dimensions D, where the radial potential isl(l + D − 2)/2r2 + r2 /2 + (g/2)(a4 r4 + a6 r6 + . . . + a2q x2q ), where the first term is the centrifugalbarrier of angular momentum l in D dimensions.
We shall do this in several steps.3C.1One-Dimensional Interaction x4In natural physical units with h̄ = 1, ω = 1, M = 1, the Schrödinger equation to be solved reads1 21 d24+x+gxψ (n) (x) = E (n) ψ (n) (x).(3C.1)−2 dx22At g = 0, this is solved by the harmonic oscillator wave functionsψ (n) (x, g = 0) = N n e−x2/2Hn (x),(3C.2)with proper normalization constant N n , where Hn (x) are the Hermite polynomial of nth degreenXHn (x) =hpn xp .(3C.3)p=0Generalizing this to the anharmonic case, we solve the Schrödinger equation (3C.1) with the powerseries ansatzψ (n) (x)=e−x2/2∞X(n)(−g)k Φk (x),(3C.4)k=0E (n)=∞X(n)g k Ek .(3C.5)k=0(n)To make room for derivative symbols, the superscript of Φk (x) is now dropped.
Inserting (3C.4)and (3C.5) into (3C.1) and equating the coefficients of equal powers of g, we obtain the equationsxΦ0k (x) − nΦk (x) =kX01 00(n)Φk (x) − x4 Φk−1 (x) +(−1)k Ek0 Φk−k0 (x),20(3C.6)k =1where we have inserted the unperturbed energy(n)E0= n + 1/2,(3C.7)and defined Φk (x) ≡ 0 for k < 0. The functions Φk (x) are anharmonic versions of the Hermitepolynomials. They turn out to be polynomials of (4k + n)th degree:Φk (x) =4k+nXApk xp .(3C.8)p=024C.M. Bender and T.T. Wu, Phys.
Rev. 184 , 1231 (1969); Phys. Rev. D 7 , 1620 (1973).H. Kleinert, PATH INTEGRALS353Appendix 3C Recursion Relations for Perturbation CoefficientsIn a more explicit notation, the expansion coefficients Apk would of course carry the dropped(n)superscript of Φk . All higher coefficients vanish:Apk ≡ 0for p ≥ 4k + n + 1.(3C.9)From the harmonic wave functions (3C.2),Φ0 (x) = N n Hn (x) = N nnXhpn xp ,(3C.10)p=0we see that the recursion starts withAp0 = hpn N n .(3C.11)For levels with an even principal quantum number n, the functions Φk (x) are symmetric. It isconvenient to choose the normalization ψ (n) (0) = 1, such that N n = 1/h0n andA0k = δ0k .(3C.12)For odd values of n, the wave functions Φk (x) are antisymmetric.
Here we choose the normalizationψ (n)0 (0) = 3, so that N n = 3/h1n andA1k = 3δ0k .(3C.13)DefiningApk ≡ 0for p < 0 or k < 0,(3C.14)we find from (3C.6), by comparing coefficients of xp ,(p − n)Apk =kX01(n)p−4(p + 2)(p + 1)Ap+2+A+(−1)k Ek0 Apk−k0 .kk−120(3C.15)k =1The last term on the right-hand side arises after exchanging the order of summation as follows:kX4(k−k0 )+nk0(−1)(n)Ek0XApk−k0 xp =p=0k0 =14k+nXxpp=0kX0(n)(−1)k Ek0 Apk−k0 .(3C.16)k0 =1For even n, Eq. (3C.15) with p = 0 and k > 0 yields [using (3C.14) and (3C.12)] the desiredexpansion coefficients of the energies(n)Ek= −(−1)k A2k .(3C.17)For odd n, we take Eq.
(3C.15) with p = 1 and odd k > 0 and find [using (3C.13) and (3C.14)]the expansion coefficients of the energies:(n)Ek= −(−1)k A3k .(3C.18)For even n, the recursion relations (3C.15) obviously relate only coefficients carrying evenindices with each other. It is therefore useful to set00pn = 2n0 , p = 2p0 , A2pk = Ck ,(3C.19)leading to000p −22(p0 − n0 )Ckp = (2p0 + 1)(p0 + 1)Ckp +1 + Ck−1−kXk0 =10pCk10 Ck−k0.(3C.20)3543 External Sources, Correlations, and Perturbation TheoryFor odd n, the substitution00+1n = 2n0 + 1 , p = 2p0 + 1 , A2p= Ckp ,k(3C.21)leads to000p −22(p0 − n0 )Ckp = (2p0 + 3)(p0 + 1)Ckp +1 + Ck−1−kX0pCk10 Ck−k0.(3C.22)k0 =1The rewritten recursion relations (3C.20) and (3C.22) are the same for even and odd n, except0for the prefactor of the coefficient Ckp +1 . The common initial values are 2p0 00hn /hn for 0 ≤ p0 ≤ n0 ,(3C.23)C0p =0otherwise.The energy expansion coefficients are given in either case by(n)Ek= −(−1)k Ck1 .(3C.24)The solution of the recursion relations proceeds in three steps as follows.
Suppose we havep0calculated for some value of k all coefficients Ck−1for an upper index in the range 1 ≤ p0 ≤02(k − 1) + n .0In a first step, we find Ckp for 1 ≤ p0 ≤ 2k + n0 by solving Eq. (3C.20) or (3C.22), startingwith p0 = 2k + n0 and lowering p0 down to p0 = n0 + 1. Note that the knowledge of the coefficientsCk1 (which determine the yet unknown energies and are contained in the last term of the recursion0relations) is not required for p0 > n0 , since they are accompanied by factors C0p which vanish dueto (3C.23).Next we use the recursion relation with p0 = n0 to find equations for the coefficients Ck1contained in the last term.
The result is, for even k,#"k−1X1n0 +1n0 −21001n0.(3C.25)Ck = (2n + 1)(n + 1)Ck+ Ck−1 −Ck0 Ck−k0n0C0k0 =1For odd k, the factor (2n0 + 1) is replaced by (2n0 + 3). These equations contain once more the0coefficients Ckn .0Finally, we take the recursion relations for p0 < n0 , and relate the coefficients Ckn −1 , . . . , Ck1 to0(n)Ckn . Combining the results we determine from Eq. (3C.24) all expansion coefficients Ek .The relations can easily be extended to interactions which are an arbitrary linear combinationV (x) =∞Xa2n n x2n .(3C.26)n=2A short Mathematica program solving the relations can be downloaded from the internet.25The expansion coefficients have the remarkable property of growing, for large order k, liker1 6 12(n)Ek −−−→ −(−3)k Γ(k + n + 1/2).(3C.27)π π n!This will be shown in Eq.
(17.323). Such a factorial growth implies the perturbation expansion tohave a zero radius of convergence. The reason for this will be explained in Section 17.10. At theexpansion point g = 0, the energies possess an essential singularity. In order to extract meaningfulnumbers from a Taylor series expansion around such a singularity, it will be necessary to find aconvergent resummation method. This will be provided by the variational perturbation theory tobe developed in Section 5.14.25See http://www.physik.fu-berlin/~kleinert/b3/programs.H. Kleinert, PATH INTEGRALSAppendix 3C Recursion Relations for Perturbation Coefficients3C.2355General One-Dimensional InteractionConsider now an arbirary interaction which is expandable in a power seriesv(x) =∞Xg k vk+2 xk+2 .(3C.28)k=1Note that the coupling constant corresponds now to the square root of the previous one, the lowestinteraction terms being gv3 x3 + g 2 v4 x4 + .
. . . The powers of g count the number of loops of theassociated Feynman diagrams. Then Eqs. (3C.6) and (3C.15) becomexΦ0k (x) − nΦk (x) =kkXX0001 00(n)Φk (x) −(−1)k vk0 +2 xk +2 Φk−k0 +(−1)k Ek0 Φk−k0 (x),200k =1(3C.29)k =1and(p − n)Apk =kkXX01(n)p−j−2k00(p + 2)(p + 1)Ap+2−(−1)vA+(−1)k Ek0 Apk−k0 . (3C.30)k +2 k−k0k200k =1k =1The expansion coefficients of the energies are, as before, given by (3C.17) and (3C.18) for even andodd n, respectively, but the recursion relation (3C.30) has to be solved now in full.3C.3Cumulative Treatment of Interactions x4 and x3There exists a slightly different recursive treatment which we shall illustrate for the simplest mixedinteraction potentialV (x) =M 2 2ω x + gv3 x3 + g 2 v4 x4 .2(3C.31)Instead of the ansatz (3C.4) we shall now factorize the wave function of the ground state as follows:ψ(n)(x) =Mωπh̄1/4Mω 2(n)exp −x + φ (x) ,2h̄(3C.32)i.e., we allow for powers series expansion in the exponent:φ(n) (x) =∞X(n)g k φk (x) .(3C.33)k=1We shall find that this expansion contains fewer terms than in the Bender-Wu expansion of thecorrection factor in Eq.
(3C.4). For completeness, we keep here physical dimensions with explicitconstants h̄, ω, M .Inserting (3C.32) into the Schrödinger equationM 2 2h̄2 d2324(n)+ω x + gv3 x + g v4 x − E−ψ (n) (x) = 0,(3C.34)2M dx22we obtain, after dropping everywhere the superscript (n), the differential equation for φ(n) (x):−h̄2 00h̄2 02φ (x) + h̄ω x φ0 (x) −[φ (x)] + gv3 x3 + g 2 v4 x4 = nh̄ω + ,2M2Mwhere denotes the correction to the harmonic energy1+.E = h̄ω n +2(3C.35)(3C.36)3563 External Sources, Correlations, and Perturbation TheoryWe shall calculate as a power series in g:=∞Xg k k .(3C.37)k=1From now on we shall consuder only the ground state with n = 0.
Inserting expansion (3C.33) into(3C.35), and comparing coefficients, we obtain the infinite set of differential equations for φk (x):−k−1h̄2 00h̄2 X 0φk−l (x) φ0l (x) + δk,1 v3 x3 + δk,2 v4 x4 = k .φk (x) + h̄ω x φ0k (x) −2M2M(3C.38)l=1Assuming that φk (x) is a polynomial, we can show by induction that its degree cannot be greaterthan k + 2, i.e.,φk (x) =∞Xmc(k)m x ,m=1with c(k)m ≡0for m > k + 2 ,(3C.39)(k)The lowest terms c0 have been omitted since they will be determined at the end the normalizationof the wave function ψ(x). Inserting (3C.39) into (3C.38) for k = 1, we find(1)c1 = −v3,M ω2(1)c2 = 0,(1)c3 = −v3,3h̄ω1 = 0 .(3C.40)For k = 2, we obtain7v323v4v32v4(2)(2)−,c=0,c=−,348M 2 ω 4 4M ω 28M h̄ω 3 4h̄ω11v32 h̄23v4 h̄22 = −+.348M ω4M 2 ω 2(2)c1(2)= 0, c2 =(3C.41)(3C.42)For the higher-order terms we must solve the recursion relationsc(k)m =k−1X m+1X(m + 2)(m + 1)h̄ (k)h̄(k−l)cm+2 +n(m + 2 − n) c(l)n cm+2−n ,2mM ω2mM ωn=1(3C.43)l=1k = −k−1h̄2 X (l) (k−l)h̄2 (k)c2 −c1 c1.M2M(3C.44)l=1Evaluating this for k = 3 yields(3)c1 = −6v3 v4 h̄13v333v3 v45v33 h̄(3)(3)+,c=0,c=−+,23473536M ωM ω12M ω2M 2 ω 4v33v3 v4(3)(3)+, 3 = 0,c4 = 0, c5 = −10M 2 h̄ω 55M h̄ω 3(3C.45)and for k = 4:305v34 h̄123v32 v4 h̄21v42 h̄99v3447v32 v411v42(4)(4)−+, c3 = 0, c4 =−+,59473548632M ω8M ω8M ω64M ω16M ω16M 2 ω 4v32 v4v425v34(4)(4)−+,(3C.46)c5 = 0, c6 =48M 3 h̄ω 74M 2 h̄ω 512M h̄ω 3465v34 h̄3171v32 v4 h̄321v42 h̄34 = −+−.(3C.47)695732M ω8M ω8M 4 ω 5(4)c1(4)= 0, c2 =H.
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