Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 29
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Kleinert, Mod. Phys. Lett. A 4 , 2329 (1989) (http://www.physik.fu-berlin.de/~kleinert/199); Phys. Lett. B 236 , 315 (1990) (ibid.http/202).H. Kleinert, PATH INTEGRALS1032.2 Exact Solution for Free ParticleNote that the free-particle amplitude happens to be independent of the numberN + 1 of time slices. The amplitude (2.71) agrees, of course, with the Schrödingerresult (1.338) for D = 1.2.2.2Fluctuations around Classical PathThere exists another method of calculating this amplitude which is somewhat moreinvolved than the simple case at hand, but which turns out to be useful for the treatment of a certain class of nontrivial path integrals, after a suitable generalization.This method is based on all paths with respect to the classical path, i.e., all pathsare split into the classical pathxcl (t) = xa +xb − xa(t − ta ),tb − ta(2.72)along which the free particle would run following the equation of motionẍcl (t) = 0,(2.73)x(t) = xcl (t) + δx(t).(2.74)plus deviations δx(t):Since initial and final points are fixed at xa , xb , respectively, the deviations vanishat the endpoints:δx(ta ) = δx(tb ) = 0.(2.75)The deviations δx(t) are referred to as the quantum fluctuations of the particleorbit.
In mathematics, the boundary conditions (2.75) are referred to as Dirichletboundary conditions. When inserting the decomposition (2.74) into the action weobserve that due to the equation of motion (2.73) for the classical path, the actionseparates into the sum of a classical and a purely quadratic fluctuation termoM Z tb n 2dt ẋcl (t) + 2ẋcl (t)δ ẋ(t) + [δ ẋ(t)]22 taZ tbZZtbM tbM tb2dtẍcl δx +dtẋcl + M ẋδx − Mdt(δ ẋ)2=ta2 ta2 tataZ tZ tbbM=dtẋ2cl +dt(δ ẋ)2 .2 tataThe absence of a mixed term is a general consequence of the extremality propertyof the classical path,δA= 0.(2.76)x(t)=xcl (t)It implies that a quadratic fluctuation expansion around the classical actionAcl ≡ A[xcl ](2.77)1042 Path Integrals — Elementary Properties and Simple Solutionscan have no linear term in δx(t), i.e., it must start as1A = Acl +2ZtbtadtZδ2A0 + ...
.dt0 δx(t)δx(t )δx(t)δx(t )x(t)=xcl (t)tb0ta(2.78)With the action being a sum of two terms, the amplitude factorizes into the productof a classical amplitude eiAcl /h̄ and a fluctuation factor F0 (tb − ta ),(xb tb |xa ta ) =ZDx eiA[x]/h̄ = eiAcl /h̄ F0 (tb , ta ).(2.79)For the free particle with the classical actionAcl =ZtbtadtM 2ẋ ,2 cl(2.80)the function factor F0 (tb − ta ) is given by the path integralZF0 (tb − ta ) =Dδx(t) expih̄ZtbtadtM(δ ẋ)2 .2(2.81)Due to the vanishing of δx(t) at the endpoints, this does not depend on xb , xa butonly on the initial and final times tb , ta .
The time translational invariance reducesthis dependence further to the time difference tb −ta . The subscript zero of F0 (tb −ta )indicates the free-particle nature of the fluctuation factor. After inserting (2.72) into(2.80), we find immediatelyAcl =M (xb − xa )2.2 tb − ta(2.82)The fluctuation factor, on the other hand, requires the evaluation of the multipleintegralF0N (tb− ta ) =1q2πh̄i/MNYn=1Zdδxn∞−∞q2πh̄i/M expi NA ,h̄ fl(2.83)where ANfl is the time-sliced fluctuation actionANfl+1δxn − δxn−1M NX=2 n=1!2.(2.84)At the end, we have to take the continuum limitN → ∞, = (tb − ta )/(N + 1) → 0.H.
Kleinert, PATH INTEGRALS1052.2 Exact Solution for Free Particle2.2.3Fluctuation FactorThe remainder of this section will be devoted to calculating the fluctuation factor(2.83). Before doing this, we shall develop a general technique for dealing withsuch time-sliced expressions. Due to the frequent appearance of the fluctuatingδx-variables, we shorten the notation by omitting all δ’s and working only withx-variables.A useful device for manipulating sums on a sliced time axis such as (2.84) is thedifference operator ∇ and its conjugate ∇, defined by1∇x(t) ≡ [x(t) − x(t − )].1∇x(t) ≡ [x(t + ) − x(t)],(2.85)They are two different discrete versions of the time derivative ∂t , to which bothreduce in the continuum limit → 0:→0∇, ∇ −−−→ ∂t ,(2.86)if they act upon differentiable functions.
Since the discretized time axis with N + 1steps constitutes a one-dimensional lattice, the difference operators ∇, ∇ are alsocalled lattice derivatives.For the coordinates xn = x(tn ) at the discrete times tn we write1(x− xn ), n+11=(x − xn−1 ), n∇xn =N ≥ n ≥ 0,∇xnN + 1 ≥ n ≥ 1.(2.87)The time-sliced action (2.84) can then be expressed in terms of ∇xn or ∇xn as(writing xn instead of δxn )ANfl =N+1M XM NX(∇xn )2 =(∇xn )2 .2 n=02 n=1(2.88)In this notation,the limit → 0 is most obvious: The sum n goes into theR tbintegral ta dt, whereas both (∇xn )2 and (∇xn )2 tend to ẋ2 , so thatPANfl →ZtbtadtM 2ẋ .2(2.89)Thus, the time-sliced action becomes the Lagrangian action.Lattice derivatives have properties quite similar to ordinary derivatives.
Oneonly has to be careful in distinguishing ∇ and ∇. For example, they allow for auseful operation summation by parts which is analogous to the integration by parts.Recall the rule for the integration by partsZtbta˙ = g(t)f (t)tb −dtg(t)f(t)taZtbtadtġ(t)f (t).(2.90)1062 Path Integrals — Elementary Properties and Simple SolutionsOn the lattice, this relation yields for functions f (t) → xn and g(t) → pn :N+1Xn=1+1pn ∇xn = pn xn |N−0NX(∇pn )xn .(2.91)n=0This follows directly by rewriting (2.35).For functions vanishing at the endpoints, i.e., for xN +1 = x0 = 0, we can omitthe surface terms and shift the range of the sum on the right-hand side to obtainthe simple formula [see also Eq.
(2.47)]N+1Xn=1pn ∇xn = −NX(∇pn )xn = −n=0N+1X(∇pn )xn .(2.92)n=1The same thing holds if both p(t) and x(t) are periodic in the interval tb − ta , so thatp0 = pN +1 , x0 = xN +1 . In this case, it is possible to shift the sum on the right-handside by one unit arriving at the more symmetric-looking formulaN+1Xn=1pn ∇xn = −N+1X(∇pn )xn .(2.93)n=1In the time-sliced action (2.84) the quantum fluctuations xn (=δxˆ n ) vanish at theends, so that (2.92) can be used to rewriteN+1Xn=1(∇xn )2 = −NXn=1xn ∇∇xn .(2.94)In the ∇xn -form of the action (2.88), the same expression is obtained by applyingformula (2.92) from the right- to the left-hand side and using the vanishing of x0and xN +1 :NX(∇xn )2 = −n=0N+1Xn=1xn ∇∇xn = −NXn=1xn ∇∇xn .(2.95)The right-hand sides in (2.94) and (2.95) can be written in matrix form as−−NXn=1NXn=1xn ∇∇xn ≡ −xn ∇∇xn ≡ −with the same N × N -matrix∇∇ ≡ ∇∇ ≡12NXxn (∇∇)nn0 xn0 ,n,n0 =1NXxn (∇∇)nn0 xn0 ,−21 0 ...
01 −2 1 . . . 0...00(2.96)n,n0 =10000...0 0 . . . 1 −210 0 ... 01 −2.(2.97)H. Kleinert, PATH INTEGRALS1072.2 Exact Solution for Free ParticleThis is obviously the lattice version of the double time derivative ∂t2 , to which itreduces in the continuum limit → 0. It will therefore be called the lattice Laplacian.A further common property of lattice and ordinary derivatives is that they canboth be diagonalized by going to Fourier components.
When decomposingx(t) =Z∞−∞dωe−iωt x(ω),(2.98)and applying the lattice derivative ∇, we find1 −iω(tn +)e− e−iωtn x(ω)−∞Z ∞1=dωe−iωtn (e−iω − 1) x(ω).−∞∇x(tn ) =Z∞dω(2.99)Hence, on the Fourier components, ∇ has the eigenvalues1 −iω(e− 1).(2.100)In the continuum limit → 0, this becomes the eigenvalue of the ordinary timederivative ∂t , i.e., −i times the frequency of the Fourier component ω.
As a reminderof this we shall denote the eigenvalue of i∇ by Ω and havei(i∇x)(ω) = Ω x(ω) ≡ (e−iω − 1) x(ω).(2.101)For the conjugate lattice derivative we find similarlyi(i∇x)(ω) = Ω x(ω) ≡ − (eiω − 1) x(ω),(2.102)where Ω is the complex-conjugate number of Ω, i.e., Ω ≡ Ω∗ . As a consequence, theeigenvalues of the negative lattice Laplacian −∇∇≡−∇∇ are real and nonnegative:i −iωi1(e− 1) (1 − eiω ) = 2 [2 − 2 cos(ω)] ≥ 0.(2.103)Of course, Ω and Ω̄ have the same continuum limit ω.When decomposing the quantum fluctuations x(t) [=δx(t)]ˆinto their Fouriercomponents, not all eigenfunctions occur. Since x(t) vanishes at the initial timet = ta , the decomposition can be restricted to the sine functions and we may expandx(t) =Z0∞dω sin ω(t − ta ) x(ω).(2.104)The vanishing at the final time t = tb is enforced by a restriction of the frequenciesω to the discrete valuesπmπmνm =.(2.105)=tb − ta(N + 1)1082 Path Integrals — Elementary Properties and Simple SolutionsThus we are dealing with the Fourier seriesx(t) =∞Xm=1s2sin νm (t − ta ) x(νm )(tb − ta )(2.106)with real Fourier components x(νm ).
A further restriction arises from the fact thatfor finite , the series has to represent x(t) only at the discrete points x(tn ), n =0, . . . , N + 1. It is therefore sufficient to carry the sum only up to m = N and toexpand x(tn ) asx(tn ) =NXm=1s2sin νm (tn − ta ) x(νm ),N +1(2.107)√where a factor has been removed from the Fourier components, for convenience.The expansion functions are orthogonal,N2 Xsin νm (tn − ta ) sin νm0 (tn − ta ) = δmm0 ,N + 1 n=1(2.108)N2 Xsin νm (tn − ta ) sin νm (tn0 − ta ) = δnn0N + 1 m=1(2.109)and complete:(where 0 < m, m0 < N + 1). The orthogonality relation follows by rewriting theleft-hand side of (2.108) in the formN+1Xiπ(m − m0 )iπ(m + m0 )2 1Reexpn − expnN +12N +1N +1n=0("#"#),(2.110)with the sum extended without harm by a trivial term at each end.
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