Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 30
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Being of thegeometric type, this can be calculated right away. For m = m0 the sum adds up to1, while for m 6= m0 it becomes001 − eiπ(m−m ) eiπ(m−m )/(N +1)2 1Re− (m0 → −m0 ) .0 )/(N +1)iπ(m−mN +121−e"#(2.111)The first expression in the curly brackets is equal to 1 for even m − m0 6= 0; whilebeing imaginary for odd m − m0 [since (1 + eiα )/(1 − eiα ) is equal to (1 + eiα )(1 −e−iα )/|1 − eiα |2 with the imaginary numerator eiα − e−iα ].
For the second term thesame thing holds true for even and odd m + m0 6= 0, respectively. Since m − m0 andm + m0 are either both even or both odd, the right-hand side of (2.108) vanishes form 6= m0 [remembering that m, m0 ∈ [0, N + 1] in the expansion (2.107), and thus in(2.111)]. The proof of the completeness relation (2.109) can be carried out similarly.H. Kleinert, PATH INTEGRALS1092.2 Exact Solution for Free ParticleInserting now the expansion (2.107) into the time-sliced fluctuation action (2.84),the orthogonality relation (2.108) yieldsANfl =N+1M XM NX(∇xn )2 =x(νm )Ωm Ωm x(νm ).2 n=02 m=1(2.112)Thus the action decomposes into a sum of independent quadratic terms involvingthe discrete set of eigenvalues11πmΩm Ωm = 2 [2 − 2 cos(νm )] = 2 2 − 2 cosN +1,(2.113)and the fluctuation factor (2.83) becomes1F0N (tb − ta ) = q2πh̄i/MNYNYn=1Zdxn∞−∞q2πh̄i/MiMΩm Ωm [x(νm )]2 .×exph̄ 2m=1(2.114)Before performing the integrals, we must transform the measure of integration fromthe local variables xn to the Fourier components x(νm ).
Due to the orthogonalityrelation (2.108), the transformation has a unit determinant implying thatNYdxn =n=1NYdx(νm ).(2.115)m=1With this, Eq. (2.114) can be integrated with the help of Fresnel’s formula (1.334).The result isNY11q.F0N (tb − ta ) = q2πh̄i/M m=1 2 Ωm Ωm(2.116)To calculate the product we use the formula5N Ym=11 + x2 − 2x cosx2(N +1) − 1mπ=.N +1x2 − 1(2.117)Taking the limit x → 1 givesNYm=12 Ωm Ωm =NYm=12 1 − cosmπ= N + 1.N +1(2.118)The time-sliced fluctuation factor of a free particle is therefore simply51F0N (tb − ta ) = q,2πih̄(N + 1)/MI.S.
Gradshteyn and I.M. Ryzhik, op. cit., Formula 1.396.2.(2.119)1102 Path Integrals — Elementary Properties and Simple Solutionsor, expressed in terms of tb − ta ,1F0 (tb − ta ) = q.2πih̄(tb − ta )/M(2.120)As in the amplitude (2.71) we have dropped the superscript N since this final resultis independent of the number of time slices.Note that the dimension of the fluctuation factor is 1/length. In fact, one mayintroduce a length scale associated with the time interval tb − ta ,l(tb − ta ) ≡q2πh̄(tb − ta )/M,and writeF0 (tb − ta ) = √1il(tb − ta )(2.121).(2.122)With (2.120) and (2.82), the full time evolution amplitude of a free particle (2.79)is again given by (2.71)i M (xb − xa )2(xb tb |xa ta ) = qexp.h̄ 2 tb − ta2πih̄(tb − ta )/M"1#(2.123)It is straightforward to generalize this result to a point particle moving throughany number D of Cartesian space dimensions.
If x = (x1 , . . . , xD ) denotes the spatialcoordinates, the action isMA[x] =2Ztbtadt ẋ2 .(2.124)Being quadratic in x, the action is the sum of the actions for each component.Hence, the amplitude factorizes and we find1(xb tb |xa ta ) = qD2πih̄(tb − ta )/Mi M (xb − xa )2,exph̄ 2 tb − ta"#(2.125)in agreement with the quantum-mechanical result in D dimensions (1.338).It is instructive to present an alternative calculation of the product of eigenvaluesin (2.116) which does not make use of the Fourier decomposition and works entirelyin configuration space. We observe that the productNY2 Ωm Ωm(2.126)m=1is the determinant of the diagonalized N × N -matrix −2 ∇∇.
This follows fromthe fact that for any matrix, the determinant is the product of its eigenvalues.H. Kleinert, PATH INTEGRALS1112.2 Exact Solution for Free ParticleThe product (2.126) is therefore also called the fluctuation determinant of the freeparticle and writtenNYm=12 Ωm Ωm ≡ detN (−2 ∇∇).(2.127)With this notation, the fluctuation factor (2.116) readsi−1/2h1.F0N (tb − tb ) = qdetN (−2 ∇∇)2πh̄i/M(2.128)Now one realizes that the determinant of 2 ∇∇ can be found very simply from theexplicit N × N matrix (2.97) by induction: For N = 1 we see directly thatdetN =1 (−2 ∇∇) = |2| = 2.(2.129)For N = 2, the determinant is2 −1 = 3.detN =2 (− ∇∇) =−12 2(2.130)A recursion relation is obtained by developing the determinant twice with respectto the first row:detN (−2 ∇∇) = 2 detN −1 (−2 ∇∇) − detN −2 (−2 ∇∇).(2.131)With the initial condition (2.129), the solution isdetN (−2 ∇∇) = N + 1,(2.132)in agreement with the previous result (2.118).Let us also find the time evolution amplitude in momentum space.
A simpleFourier transform of initial and final positions according to the rule (2.37) yields(pb tb |pa ta ) =ZD−ipb xb /h̄d xb eZdD xa eipa xa /h̄ (xb tb |xa ta )2= (2π)D δ (D) (pb − pa )e−ipb (tb −ta )/h̄ .2.2.4(2.133)Finite Slicing Properties of Free-Particle AmplitudeThe time-sliced free-particle time evolution amplitudes (2.70) happens to be independent of the number N of time slices used for their calculation. We have pointedthis out earlier for the fluctuation factor (2.119).
Let us study the origin of thisindependence for the classical action in the exponent. The difference equation ofmotion−∇∇x(t) = 0(2.134)1122 Path Integrals — Elementary Properties and Simple Solutionsis solved by the same linear functionx(t) = At + B,(2.135)as in the continuum. Imposing the initial conditions givesxcl (tn ) = xa + (xb − xa )n.N +1(2.136)The time-sliced action of the fluctuations is calculated, via a summation by partson the lattice [see (2.91)]. Using the difference equation ∇∇xcl = 0, we findN+1XAcl = M(∇xcl )22n=1M=2N +1xcl ∇xcl n=0(2.137)−NXn=0xcl ∇∇xcl!N +1MM (xb − xa )2==.x ∇x 2 cl cl n=02 tb − taThis coincides with the continuum action for any number of time slices.In the operator formulation of quantum mechanics, the -independence of theamplitude of the free particle follows from the fact that in the absence of a potentialV (x), the two sides of the Trotter formula (2.26) coincide for any N.2.3Exact Solution for Harmonic OscillatorA further problem to be solved along similar lines is the time evolution amplitudeof the linear oscillatorDpi(xb tb |xa ta ) =DxexpA[p, x]2πh̄h̄ZiA[x] ,=Dx exph̄Z0Z(2.138)with the canonical actionA[p, x] =Ztbta1 2 Mω 2 2p −x ,dt pẋ −2M2(2.139)M 2(ẋ − ω 2x2 ).2(2.140)!and the Lagrangian actionA[x] =ZtbtadtH.
Kleinert, PATH INTEGRALS2.3 Exact Solution for Harmonic Oscillator2.3.1113Fluctuations around Classical PathAs before, we proceed with the Lagrangian path integral, starting from the timesliced form of the actionAN = +1 hiM NX(∇xn )2 − ω 2 x2n .2 n=1(2.141)The path integral is again a product of Gaussian integrals which can be evaluatedsuccessively. In contrast to the free-particle case, however, the direct evaluationis now quite complicated; it will be presented in Appendix 2B. It is far easier toemploy the fluctuation expansion, splitting the paths into a classical path xcl (t) plusfluctuations δx(t).
The fluctuation expansion makes use of the fact that the actionis quadratic in x = xcl + δx and decomposes into the sum of a classical partAcl =ZtbtaM 2(ẋcl − ω 2x2cl ),2(2.142)M[(δ ẋ)2 − ω 2 (δx)2 ],2(2.143)dtand a fluctuation partAfl =Ztbtadtwith the boundary conditionδx(ta ) = δx(tb ) = 0.(2.144)There is no mixed term, due to the extremality of the classical action. The equationof motion isẍcl = −ω 2 xcl .(2.145)Thus, as for a free-particle, the total time evolution amplitude splits into a classicaland a fluctuation factor:(xb tb |xa ta ) =ZDx eiA[x]/h̄ = eiAcl /h̄ Fω (tb − ta ).(2.146)The subscript of Fω records the frequency of the oscillator.The classical orbit connecting initial and final points is obviouslyxcl (t) =xb sin ω(t − ta ) + xa sin ω(tb − t).sin ω(tb − ta )(2.147)Note that this equation only makes sense if tb − ta is not equal to an integer multipleof π/ω which we shall always assume from now on.66For subtleties in the immediate neighborhood of the singularities which are known as causticphenomena, see Notes and References at the end of the chapter, as well as Section 4.8.1142 Path Integrals — Elementary Properties and Simple SolutionsAfter an integration by parts we can rewrite the classical action Acl asAcl =ZtbtadttbiMMhxcl (−ẍcl − ω 2 xcl ) + xcl ẋcl .ta22(2.148)The first term vanishes due to the equation of motion (2.145), and we obtain thesimple expressionAcl =M[x (t )ẋ (t ) − xcl (ta )ẋcl (ta )].2 cl b cl b(2.149)Sinceω[x − xa cos ω(tb − ta )],sin ω(tb − ta ) bω[x cos ω(tb − ta ) − xa ],ẋcl (tb ) =sin ω(tb − ta ) bẋcl (ta ) =(2.150)(2.151)we can rewrite the classical action asAcl =2.3.2hiMω(x2b + x2a ) cos ω(tb − ta ) − 2xb xa .2 sin ω(tb − ta )(2.152)Fluctuation FactorWe now turn to the fluctuation factor.
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