Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 74
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For this we formulate again a homogeneous initial-value problem, but with boundary conditions dual to Gelfand and Yaglom’s in Eq. (3.918):˙ (τ ) = 0.Og (τ )D̄g (τ ) = 0; D̄g (τa ) = 1, D̄g a(3.937)In terms of the previous arbitrary set ηg (t) and ξg (t) of solutions of the homogeneousdifferential equation, the unique solution of (3.937) readsD̄g (τ ) =ξg (τ )η̇g (τa ) − ξ˙g (τa )ηg (τ ).Wg(3.938)This can be combined with the time derivative of (3.920) at τ = τb to yield¯ p,a (τ , τ )].Ḋg (τb ) + D̄g (τb ) = ±[2 − ∆ga b(3.939)By differentiating Eqs.
(3.937) with respect to g, we obtain the following inhomogeneous initial-value problem for D̄g0 (τ ) = ∂g D̄g (τ ):Og (τ )D̄g0 (τ ) = Ω2 (τ )D̄g0 (τ ); D̄g0 (τa ) = 1, D̄˙ 0 g (τa ) = 0,(3.940)whose general solution reads by analogy with (3.924)D̄g0 (τ ) = −Zττa˙ (τ , τ 0 ),dτ 0 Ω2 (τ 0 )∆g (τ, τ 0 )∆g a(3.941)˙ (τ , τ 0 ) acts on the first imaginary-time argument. With thewhere the dot on ∆g ahelp of identities (3.939) and (3.940), the combination Ḋ 0 (τ ) + D̄g0 (τ ) at τ = τb cannow be expressed in terms of the periodic and antiperiodic Green functions (3.166),by analogy with (3.925),Ḋg0 (τb )+D̄g0 (τb )¯ p,a (τ , τ )= ±∆ga bZτbτadτ Ω2 (τ )Gp,ag (τ, τ ).(3.942)H.
Kleinert, PATH INTEGRALS3473.27 Functional Determinants from Green FunctionsTogether with (3.939), this gives for the temporal integral on the right-hand side of(3.917) the simple expression analogous to (3.926)2Tr [Ωdet Λ̄p,ag= −∂g logWg0(τ )Gp,ag (τ, τ )]!hi= −∂g log 2 ∓ Ḋg (τb ) ∓ D̄g (τb ) ,(3.943)so that we obtain the ratio of functional determinants with periodic and antiperiodicboundary conditionshiDet (Õ−1 Og ) = C(tb , ta ) 2 ∓ Ḋg (τb ) ∓ D̄g (τb ) ,(3.944)where Õ = O0 − ω 2 = ∂τ2 − ω 2.
The constant of integration C(tb , ta ) is fixed inthe way described after Eq. (3.915). We go to g = 1 and set Ω2 (τ ) ≡ ω 2 . Forthe operator O1ω ≡ −∂τ2 − ω 2 , we can easily solve the Gelfand-Yaglom initial-valueproblem (3.918) as well as the dual one (3.937) byD1ω (τ ) =1sin ω(τ − τa ),ωD̄1ω (τ ) = cos ω(τ − τa ),(3.945)so that (3.944) determines C(tb , ta ) by1 = C(tb , ta )(4 sin2 [ω(τb − τa )/2]4cos2 [ω(τb − τa )/2]periodic case,antiperiodic case.(3.946)Hence we find the final results for periodic boundary conditionsdet Λ̄p1Det (Õ O1 ) =W1−1,DetΛ̄ω1 p2 − Ḋ1 (τb ) − D̄1 (τb ),=ωW14 sin2 [ω(τb − τa )/2](3.947)and for antiperiodic boundary conditionsdet Λ̄a1Det (Õ O1 ) =W1−1,DetΛ̄ω1 a2 + Ḋ1 (τb ) + D̄1 (τb ).=ωW14 cos2 [ω(τb − τa )/2](3.948)The intermediate expressions in (3.929), (3.947), and (3.948) show that the ratios of functional determinants are ordinary determinants of two arbitrary independent solutions ξ and η of the homogeneous differential equation O1 (t)y(t) = 0 orO1 (τ )y(τ ) = 0.
As such, the results are manifestly invariant under arbitrary lineartransformations of these functions (ξ, η) → (ξ 0 , η 0).It is useful to express the above formulas for the ratio of functional determinants(3.929), (3.947), and (3.948) in yet another form. We rewrite the two independentsolutions of the homogenous differential equation [−∂t2 − Ω2 (t)]y(t) = 0 as followsξ(t) = q(t) cos φ(t),η(t) = q(t) sin φ(t).(3.949)3483 External Sources, Correlations, and Perturbation TheoryThe two functions q(t) and φ(t) parametrizing ξ(t) and η(t) satisfy the constraintφ̇(t)q 2 (t) = W,(3.950)where W is the constant Wronski determinant. The function q(t) is a soliton of theErmankov-Pinney equation23q̈ + Ω2 (t)q − W 2 q −3 = 0.(3.951)For Dirichlet boundary conditions we insert (3.949) into (3.929), and obtain theratio of fluctuation determinants in the formDet (O0−1 O1 ) =1 q(ta )q(tb ) sin[φ(tb ) − φ(ta )].Wtb − ta(3.952)For periodic or antiperiodic boundary conditions with a corresponding frequencyΩ(t), the functions q(t) and φ(t) in Eq.
(3.949) have the same periodicity. The initialvalue φ(ta ) may always be assumed to vanish, since otherwise ξ(t) and η(t) couldbe combined linearly to that effect. Substituting (3.949) into (3.947) and (3.948),the function q(t) drops out, and we obtain the ratios of functional determinants forperiodic boundary conditionsφ(tb )ω(tb − ta )Det (Õ O1 ) = 4 sin4 sin2,22−12(3.953)and for antiperiodic boundary conditionsDet (Õ−1 O1 ) = 4 cos2ω(tb − ta )φ(tb ).4 cos222(3.954)For a harmonic oscillator with Ω(t) ≡ ω, Eq.
(3.951) is solved byq(t) ≡sW,ω(3.955)and Eq. (3.950) yieldsφ(t) = ω(t − ta ).(3.956)Inserted into (3.952), (3.953), and (3.954) we reproduce the known results:Det (O0−1 O1 ) =23sin ω(tb − ta ),ω(tb − ta )Det (Õ−1 O1 ) = 1.For more details see J. Rezende, J. Math. Phys. 25, 3264 (1984).H.
Kleinert, PATH INTEGRALS349Appendix 3A Matrix Elements for General PotentialAppendix 3AMatrix Elements for General PotentialThe matrix elements hn|V̂ |mi can be calculated for an arbitrary potential V̂ = V (x̂) as follows:We represent V (x̂) by a Fourier integral as a superposition of exponentialsV (x̂) =i∞Z−i∞dkV (k) exp(kx̂),2πi(3A.1)√†and express√ exp(kx̂) in terms of creation and annihilation operators as exp(kx̂) = exp[k(â+â )/ 2],set k ≡ 2, and write down the obvious equation√∂n ∂m1 2x̂αâ (â+↠) βâ†.(3A.2)hn|e|mi = √h0|eee|0in!m! ∂αn ∂β mα=β=0We now make use of the Baker-Campbell-Hausdorff Formula (2A.1) with (2A.6), and rewrite11e eB̂ = eÂ+B̂+ 2 [Â,B̂]+ 12 ([Â,[Â,B̂]]+[B̂,[B̂,Â]])+....
.(3A.3)Identifying  and B̂ with â and ↠, the property [â, ↠] = 1 makes this relation very simple:†e(â+â)†= eâ eâ e−2/2,(3A.4)and the matrix elements (3A.2) become†††h0|eαâ e(â+â ) eβâ |0i = h0|e(α+)â e(β+)â |0ie−2/2.(3A.5)The bra and ket states on the right-hand side are now eigenstates of the annihilation operator âwith eigenvalues α + and β + , respectively. Such states are known as coherent states. Usingonce more (3A.3), we obtain†h0|e(α+)â e(β+)â |0i = e(+α)(+β),(3A.6)and (3A.2) becomes simplyhn|e√ 2x̂1∂ n ∂ m (α+)(β+) −2 /2 .|mi = √een!m! ∂αn ∂β mα=β=0(3A.7)We now calculate the derivatives∂ n ∂ m (+α)(+β) e∂αn ∂β mα=β=0∂nm (+α) =(+α)e∂αn.(3A.8)α=0Using the chain rule of differentiation for products f (x) = g(x) h(x):f(n)n Xn (l)(x) =g (x)h(n−l) (x),l(3A.9)l=0the right-hand side becomes∂nm (+α) (+α)e∂αn=α=0=n ln−lXn ∂∂m(+α) (+α)el ∂αl∂αn−ll=0α=0nX2nm(m − 1) · · · (m − l + 1)n+m−2l e .ll=0(3A.10)3503 External Sources, Correlations, and Perturbation TheoryHence we findhn|e√2x̂n 21 X n m|mi = √l!n+m−2l e /2 .ln!m! l=0 l(3A.11)From this we obtain the matrix elements of single powers x̂p by forming, with the help of (3A.9)2and (∂ q /∂q )e /2 |=0 = q!!, the derivativesp!∂ p n+m−2l 2 /2 pe[2l − (n + m − p)]!! = l−(n+m−p)/2.
=∂pn+m−2l2[l − p − (n+m−p)/2]!=0(3A.12)The result ismin(n,m) Xp!n m1l! l+p−(n+m)/2.hn|x̂p |mi = √ll2[l−(n + m − p)/2]!n!m! l=(n+m−p)/2For the special case of a pure fourth-order interaction, this becomes√√√√hn|x̂4 |n − 4i =n − 3 n − 2 n − 1 n,√√hn|x̂4 |n − 2i = (4n − 2) n − 1 n,hn|x̂4 |ni =hn|x̂4 |n + 2i =hn|x̂4 |n + 4i =6n2 + 6n + 3,√√(4n + 6) n + 1 n + 2,√√√√n + 1 n + 2 n + 3 n + 4.(3A.13)(3A.14)For a general potential (3A.1) we findZ i∞n 21dk1 X n m√l! l−(n+m)/2hn|V (x̂)|mi =V (k)k n+m−2l ek /4 .ll2πi2n!m! l=0−i∞Appendix 3B(3A.15)Energy Shifts for gx4 /4 -InteractionFor the specific polynomial interaction V (x) = gx4 /4, the shift of the energy E (n) to any desired order is calculated most simply as follows. Consider the expectations of powers x̂4 (z1 )x̂4 (z2 ) · · · x̂4 (zn )of the operator x̂(z) = (↠z + âz −1 ) between the excited oscillator states hn| and |ni.
Hereâ and ↠are the√ usual creation and annihilation operators of the harmonic oscillator, and|ni = (a† )n |0i/ n! . To evaluate these expectations, we make repeated use of the commutationrules [â, ↠] = 1 and of the ground state property â|0i = 0. For n = 0 this giveshx4 (z)iω = 3,hx4 (z1 )x4 (z2 )iω = 72z1−2z22 + 24z1−4 z24 + 9,hx4 (z1 )x4 (z2 )x4 (z3 )iω = 27 · 8z1−2 z22 + 63 · 32z1−2z2−2 z34(3B.1)+ 351 · 8z1−2 z32 + 9 · 8z1−4 z24 + 63 · 32z1−4 z22 z32 + 369 · 8z1−4z34+ 27 · 8z2−2 z32 + 9 · 8z2−4 z34 + 27.The cumulants arehx4 (z1 )x4 (z2 )iω,c = 72z1−2z22 + 24z1−4 z24 ,444hx (z1 )x (z2 )x (z3 )iω,c =288(7z1−2z2−2 z34(3B.2)+9z1−2 z32+7z1−4 z22 z32+10z1−4 z34 ).The powers of z show by how many steps the intermediate states have been excited.
They determinethe energy denominators in the formulas (3.515) and (3.516). Apart from a factor (g/4)n and aH. Kleinert, PATH INTEGRALSAppendix 3B Energy Shifts for gx4 /4-Interaction351factor 1/(2ω)2n which carries the correct length scale of x(z), the energy shifts ∆E = ∆1 E0 +∆2 E0 + ∆3 E0 are thus found to be given by∆1 E0 = 3,11∆2 E0 = − 72 · + 24 ·,241 11 11 11 1= 333 · 4.∆3 E0 = 288 7 · · + 9 · · + 7 · · + 10 · ·2 42 24 24 4(3B.3)Between excited states, the calculation is somewhat more tedious and yields4hx4 (z)iω = 6n2 + 6n + 3,4hx (z1 )x (z2 )iω,c =+++444(3B.4)(16n + 96n + 212n + 204n + 72)z1−2 z22(n4 + 10n3 + 35n2 + 50n + 24)z1−4 z24(n4 − 6n3 + 11n2 − 6n)z14 z2−4(16n4 − 32n3 + 20n2 − 4n)z12 z2−2 ,6543432(3B.5)2hx (z1 )x (z2 )x (z3 )iω,c = [(16n + 240n + 1444n + 4440n + 7324n + 6120n + 2016)× (z1−2 z2−2 z34 +z1−4 z22 z32 )+ (384n5 + 2880n4 + 8544n3 + 12528n2 + 9072n + 2592)z1−2z32+ (48n5 + 600n4 + 2880n3 + 6600n2 + 7152n + 2880)z1−4z34+ (16n6 − 144n5 + 484n4 − 744n3 + 508n2 − 120n)z14z2−2 z3−2+ (−48n5 + 360n4 − 960n3 + 1080n2 − 432n)z14z3−4+ (16n6 + 48n5 + 4n4 − 72n3 − 20n2 + 24n)z12 z2−4 z32+ (−384n5 + 960n4 − 864n3 + 336n2 − 48n)z12 z3−2+ (16n6 − 144n5 + 484n4 − 744n3 + 508n2 − 120n)z12z22 z3−4+ (16n6 + 48n5 + 4n4 − 72n3 − 20n2 + 24n)z1−2 z24 z3−2 ].(3B.6)From these we obtain the reduced energy shifts:∆1 E0 = 6n2 + 6n + 3,∆2 E0 = −(16n4 + 96n3 + 212n2 + 204n + 72) ·−(n4 + 10n3 + 35n2 + 50n + 24) ·−(n4 − 6n3 + 11n2 − 6n) · −14−(16n4 − 32n3 + 20n2 − 4n) ·= 2 · (34n3 + 51n2 + 59n + 21),654(3B.7)1214−12(3B.8)321212∆3 E0 = [(16n + 240n + 1444n + 4440n + 7324n + 6120n + 2016) · ( ·+(384n5 + 2880n4 + 8544n3 + 12528n2 + 9072n + 2592) · 12 ·+(48n5 + 600n4 + 2880n3 + 6600n2 + 7152n + 2880) ·14·+(16n6 − 144n5 + 484n4 − 744n3 + 508n2 − 120n) ·+(−48n5 + 360n4 − 960n3 + 1080n2 − 432n) · 14 · 1414·12+(16n6 − 144n5 + 484n4 − 744n3 + 508n2 − 120n) ·+(16n6 + 48n5 + 4n4 − 72n3 − 20n2 + 24n) · 12 · −12 ]12·14+(16n6 + 48n5 + 4n4 − 72n3 − 20n2 + 24n) ·+(−384n5 + 960n4 − 864n3 + 336n2 − 48n) ·= 4 · 3 · (125n4 + 250n3 + 472n2 + 347n + 111).1212··14+1412· )141412(3B.9)3523 External Sources, Correlations, and Perturbation TheoryAppendix 3CRecursion Relations for PerturbationCoefficients of Anharmonic OscillatorBender and Wu24 were the first to solve to high orders recursion relations for the perturbationcoefficients of the ground state energy of an anharmonic oscillator with a potential x2 /2 + gx4 /4.Their relations are similar to Eqs.
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