Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 52

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(3.181)(xb + xa )j +3 ω (tb − ta ) − 2 tanω222MωThis result could also have been obtained more directly by taking the potential plusa constant-current term in the action−ZtbtadtM 2 2ω x − xj ,2(3.182)tb − ta 2j .2Mω 2(3.183)and by completing it quadratically to the form−ZtbtaMjdt ω 2 x −2Mω 22+2343 External Sources, Correlations, and Perturbation TheoryThis is a harmonic potential shifted in x by −j/Mω 2 . The time evolution amplitudecan thus immediately be written down as(xb tb |xa ta )j=constωsiMωMωexp2πih̄ sin ω(tb − ta )2h̄ sin ω(tb − ta )("2 2 #jjcos ω(tb − ta )(3.184)×xb −+ xa −Mω 2Mω 2!ji tb − ta 2jxa −+j .−2 xb −h̄ 2Mω 2Mω 2Mω 2=In the free-particle limit ω → 0, the result becomes particularly simple:(xb ta |xa ta )j=const0i M (xb − xa )2= qexph̄ 2 tb − ta2πih̄(tb − ta )/M 1i 1(xb + xa )(tb − ta )j −(tb − ta )3 j 2 .

(3.185)× exph̄ 224M"1#As a cross check, we verify that the total exponent is equal to i/h̄ times the classicalactionAj,cl =ZtbtaM 2dtẋ + jxj,cl ,2 j,cl(3.186)calculated for the classical orbit xj,cl (t) connecting xa and xb in the presence of theconstant current j. This satisfies the Euler-Lagrange equationẍj,cl = j/M,(3.187)which is solved byjt − tajxj,cl (t) = xa + xb − xa −(tb − ta )2(t − ta )2 .+2Mtb − ta 2M(3.188)Inserting this into the action yieldsAj,clM (xb − xa )2 1(tb − ta )3 j 2+ (xb + xa )(tb − ta )j −=,2 tb − ta224M(3.189)just as in the exponent of (3.185).Let us remark that the calculation of the oscillator amplitude (xa tb |xa t)jω in(3.168) could have proceeded alternatively by using the orbital separationx(t) = xj,cl (t) + δx(t),(3.190)where xj,cl (t) satisfies the Euler-Lagrange equations with the time-dependent sourcetermẍj,cl (t) + ω 2 xj,cl (t) = j(t)/M,(3.191)H.

Kleinert, PATH INTEGRALS2353.7 Time Evolution Amplitude at Fixed Path Averagerather than the orbital separation of Eq. (3.7),x(t) = xcl (t) + δx(t),where xcl (t) satisfied the Euler-Lagrange equation with no source. For this inhomogeneous differential equation we would have found the following solution passingthrough xa at t = ta and xb at t = tb :1 Z tb 0sin ω(t − ta )sin ω(tb − t)+ xb+dt Gω2 (t, t0 )j(t0 ).xj,cl (t) = xasin ω(tb − ta )sin ω(tb − ta )M ta(3.192)The Green function Gω2 (t, t0 ) appears now at the classical level. The separation(3.190) in the total action would have had the advantage over (3.7) that the sourcecauses no linear term in δx(t).

Thus, there would be no need for a quadratic completion; the classical action would be found from a pure surface term plus one halfof the source part of the actionAcltbMM 2=dt(ẋcl,j − ω 2x2j,cl ) + jxj,cl =xj,cl ẋj,cl ta22taZ tbZ tbj1Mdtxj,cl jx−ẍj,cl − ω 2xj,cl +++dt2 j,clM2 tataZM1 tb=dtxj,cl (t)j(t).(x ẋ − xa ẋa ) +x=xj,cl2 b b2 taZtb(3.193)Inserting xj,cl from (3.192) and Gω2 (t, t0 ) from (3.36) leads once more to the exponentin (3.168). The fluctuating action quadratic in δx(t) would have given the samefluctuation factor as in the j = 0 -case, i.e., the prefactor in (3.168) with no furtherj 2 (due to the absence of a quadratic completion).3.7Time Evolution Amplitude at Fixed Path AverageAnother interesting quantity to be needed in Chapter 15 is the Fourier transform ofthe amplitude (3.184):(xb tb |xa ta )xω0= (tb − ta )Z∞−∞dj −ij(tb −ta )x0 /h̄e(xb tb |xa ta )jω .2πh̄(3.194)This is the amplitude for a particle to run from xa to xb along restricted paths whoseRtemporal average x̄ ≡ (tb − ta )−1 ttab dt x(t) is held fixed at x0 :(xb tb |xa ta )xω0=ZiDx δ(x0 − x̄) exph̄ZtbtaMdt (ẋ2 − ω 2 x2 ) .2(3.195)This property of the paths follows directly from the fact that the integral over theRtime-independent source j (3.194) produces a δ-function δ((tb − ta )x0 − ttab dt x(t)).Restricted amplitudes of this type will turn out to have important applications laterin Subsection 3.25.1 and in Chapters 5, 10, and 15.2363 External Sources, Correlations, and Perturbation TheoryThe integral over j in (3.194) is done after a quadratic completion in Aj − j(tb −ta )x0 with Aj of (3.181):#"1ω(tb − ta )Aj − j(tb − ta )x0 =(j − j0 )2 + Ax0 , (3.196)3 ω (tb − ta ) − 2 tan22Mωwithj0 =Mω 2ω(tb − ta ) − 2 tan ω(tb2−ta )#"ω(tb − ta )(xb + xa ) ,ω(tb − ta ) x0 − tan2andAx0"Mωω(tb − ta )i ω(tb − ta )x0 − tan=− h(xa + xb )ω(tb −ta )22 ω(tb −ta )−2 tan 2#2.With the completed quadratic exponent (3.196), the Gaussian integral over j in(3.194) can immediately be done, yielding(xb tb |xa ta )xω0= (xb tb |xa tavuu)tiMω 3 /2πh̄ω(tb − ta ) − 2 tan ω(tb2−ta )i x0expA.h̄(3.197)If we set xb = xa and integrate over xb = xa , we find the quantum-mechanicalversion of the partition function at fixed x0 :ω(tb − ta )/2i1exp − (tb − ta )Mω 2 x20 .

(3.198)Zωx0 = q2h̄2πh̄(tb − ta )/Mi sin[ω(tb − ta )/2]As a check we integrate this over x0 and recover the correct Zω of Eq. (2.404).We may also integrate over both ends independently to obtain the partitionfunctionZωopen,x0 =vuutiω(tb − ta )exp − (tb − ta )Mω 2 x20 .sin ω(tb − ta )2h̄(3.199)Integrating this over x0 and going to imaginary times leads back to the partitionfunction Zωopen of Eq.

(2.405).3.8External Source in Quantum-Statistical Path IntegralIn the last section we have found the quantum-mechanical time evolution amplitudein the presence of an external source term. Let us now do the same thing for thequantum-statistical case and calculate the path integral(xbh̄β|xa 0)jω=Z(1Dx(τ ) exp −h̄Z0h̄βM 2dτ(ẋ + ω 2 x2 ) − j(τ )x(τ )2). (3.200)This will be done in two ways.H. Kleinert, PATH INTEGRALS2373.8 External Source in Quantum-Statistical Path Integral3.8.1Continuation of Real-Time ResultThe desired result is obtained most easily by an analytic continuation of thequantum-mechanical results (3.23), (3.168) in the time difference tb − ta to an imaginary time −ih̄(τb − τa ) = −ih̄β.

This gives immediately(xb h̄β|xa 0)jω=sM2πh̄2 βsωh̄β1exp − Aext[j] ,sinh ωh̄βh̄ e(3.201)with the extended classical Euclidean oscillator actionjjjAexte [j] = Ae + Ae = Ae + A1,e + A2,e ,(3.202)where Ae is the Euclidean actionAe =hiMω(x2b + x2a ) cosh ωh̄β − 2xb xa ,2 sinh βh̄ω(3.203)while the linear and quadratic Euclidean source terms areAj1,eZ τb1dτ [xa sinh ω(h̄β − τ ) + xb sinh ωτ ]j(τ ),=−sinh ωh̄β τa(3.204)andAj2,e = −1MZ0h̄βdτZ0τdτ 0 j(τ ) Gω2 ,e (τ, τ 0 )j(τ 0 ),(3.205)where Gω2 ,e (τ, τ 0 ) is the Euclidean version of the Green function (3.36) with Dirichletboundary conditions:sinh ω(h̄β − τ> ) sinh ωτ<ω sinh ωh̄βcosh ω(h̄β − |τ − τ 0 |) − cosh ω(h̄β − τ − τ 0 )=,2ω sinh ωh̄βGω2 ,e (τ, τ 0 ) =(3.206)satisfying the differential equation(−∂τ2 + ω 2 ) Gω2 ,e (τ, τ 0 ) = δ(τ − τ 0 ).(3.207)It is related to the real-time Green function (3.36) byGω2 ,e (τ, τ 0 ) = i Gω2 (−iτ, −iτ 0 ),(3.208)the overall factor i accounting for the replacement δ(t − t0 ) → iδ(τ − τ 0 ) on theright-hand side of (3.148) in going to (3.207) when going from the real time t tothe Euclidean time −iτ .

The symbols τ> and τ< in the first line (3.206) denote thelarger and the smaller of the Euclidean times τ and τ 0 , respectively.2383 External Sources, Correlations, and Perturbation TheoryThe source terms (3.204) and (3.205) can be rewritten as follows:Aj1,e = −ioMω nhxb (e−βh̄ω Ae − Be ) xa (e−βh̄ω Be − Ae ) ,sinh ωh̄β(3.209)andAj2,e1 Z h̄β Z h̄β 0 −ω|τ −τ 0 |=−dτdτ ej(τ )j(τ 0 )4Mω 00hiMω+eβh̄ω (A2e + Be2 ) − 2Ae Be .2 sinh ωh̄β(3.210)We have introduced the Euclidean versions of the functions A(ω) and B(ω) inEqs.

(3.173) and (3.174) ash̄β1Ae (ω) ≡ iA(ω)|tb −ta =−ih̄β =dτ e−ωτ j(τ ),(3.211)Mω 0Z h̄β1Be (ω) ≡ iB(ω)|tb −ta =−ih̄β =dτ e−ω(h̄β−τ ) j(τ ) = −e−βh̄ω Ae (−ω). (3.212)Mω 0ZFrom (3.201) we now calculate the quantum-statistical partition function. Setting xb = xa = x, the first term in the action (3.202) becomesAe =Mω2 sinh2 (ωh̄β/2)x2 .sinh βh̄ω(3.213)If we ignore the second and third action terms in (3.202) and integrate (3.201) overx, we obtain, of course, the free partition functionZω =1.2 sinh(βh̄ω/2)(3.214)In the presence of j, we perform a quadratic completion in x and obtain a sourcedependent part of the action (3.202):Aje = Ajfl,e + Ajr,e,(3.215)where the additional term Ajr,e is the remainder left by a quadratic completion.

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