Feynman Path Integrals in Quantum Mechanics

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2.3.1 The Free ParticleIt might be good to see a kernel once in its very explicit form. Maybe it will even behelpfull in later sections. For the free particle case we simply have to put L = mẋ 2 /2 in(9) and insert this in (8). I do not write down the calculations, since it is basically solvinga set of Gaussian integrals and can certainly be found in any of the references.[ ] −1/2 2πi¯h(tb − t a )K(b, a) =exp im(x b − x a ) 2m2¯h(t b − t a )(11)2.4 Classical LimitIn the classical limit, i.e. when ¯h → 0, only paths that lie very close to the classical onegive significant contribution to the probability amplitude.One can see this in the following way: Let’s look at any two neighbouring paths x(t) andx ′ (t) which contribute to the path integral (6). We can write x ′ (t) = x(t) + η(t), withη(t) small. For the action we get:∫S[x ′ (t)] = S[x(t) + η(t)] = S[x(t)] +η(t) δS[x(t)]δx(t) dt + O(η2 ) (12)The contribution of the two paths to the path integral is:(A = e iS[x(t)]/¯h + e iS[x′ (t)]/¯h ≃ e iS[x(t)]/¯h 1 + exp ī ∫hη(t) δS[x(t)]δx(t) dt ), (13)where we have neglected terms of order η 2 . The phase difference between the contributionsis approximately ¯h −1 ∫ (η(t)δS[x(t)]/δx(t))dt. Even if δS gets very small, in the limit¯h → 0 the phase difference is huge and consequently on average destructive interferencebetween neighbouring paths occurs! There is an exception for the classical path. SinceδS at x c (t) vanishes, the action is of the form S[x c + η] = S[x c ] + O(η 2 ) and even for verysmall values of ¯h constructive interference results.3 Equivalence to Schrödingers Operator NotationIn the classical formulation of Quantum Mechanics the Schrödinger equation is postulated.In this section we will show, that it is possible to derive the Schrödinger equation fromthe postulates made in part 2.3.1 Derivation of the Schrödinger EquationLet’s start with Eq. (10) for the wave function. We set t 1 = t and t 2 = t + ɛ and use forthe kernel the expression derived in (9). In words, for a short time interval ɛ the actionis approximately ɛ times the lagrangian for this interval. We getΨ(x, t + ɛ) =∫ ∞−∞(1A expɛ ī ( x − yh L ɛ, x + y )), t Ψ(y, t)dy (14)2Observe that this equation is not exact. But it needs only be true to first order to ɛ, since(5) still holds 6 ; i.e. we shall derive the Schrödinger equation assuming (14) to be true infirst order to ɛ. This can be done in the following way 7 : write the Lagrangian in the formL = mẋ 2 /2 − V (x, t) and substitute y = x + η. The integration over y becoms an integral6 For a finite time interval T the number of factors is T/ɛ. If an error of order ɛ 2 is made in each factor,the resulting error is of order ɛ and vanishes in the limit7 Intermediate steps are left out here. The are found in [3]4
over η. The resulting integrals are Gaussian and can be solved explicitely. Expanding 8 tofirst order in ɛ and second order to η leeds then to the equationΨ + ɛ ∂Ψ∂t = Ψ − iɛ¯h V Ψ − ¯hɛ ∂ 2 Ψ2im ∂x 2 . (15)(15) is true in first order to ɛ, if Ψ satisfies [defining H = − ¯h22m ∂x+ V ] the differential2Equation−¯h ∂Ψ= HΨ , (16)i ∂twhich is the Schrödinger equation!3.2 Schrödinger Equation for the KernelThe kernel K(2, 1) introduced by (6) is only defined for t 2 > t 1 . Since K(2, 1), consideredas a function of the variables 2, is a special wave function it is clear that (16) is satisfied:−¯hi∂Ψ∂t 2K(2, 1) − H 2 K(2, 1) = 0 for t 2 > t 1 (17)Where H 2 acts on the variables 2 only.If we define K(2, 1) = 0 for t 2 < t 1 , (17) is also valid for t 2 < t 1 but K(2, 1) getsdiscontinous at t 2 = t 1 . One can show, that the kernel satisfies−¯hi∂Ψ∂t 2K(2, 1) − H 2 K(2, 1) = −¯h i δ(x 2 − x 1 )δ(t 2 − t 1 ) (18)Remark:We introduced K(2, 1) as the probability amplitude. Another possibility is to start fromthe usual formulation of Quantum Mechanics. The kernel can then be defined as theGreen’s function for the Schrödinger equation. This means: one defines K(2, 1) to bethe solution of Eq. (18). This is equivalent 9 to the following definition in the betterknown Dirac representation: K(2, 1) = 〈x 2 |Û(t)|x 1〉, where Û(t) = e−iĤt/¯h as long as theHamiltonian is time-independent.Finally we remark that K(2, 1) only depends on the time difference (t 2 − t 1 ) [as longas the Hamiltonian is time-independent].4 Pertubation TheoryThe idea of Pertubation theory in the path integral formulation of Quantum Mechanicsis quite simple. We look at a particle moving in some potential 10 V (x, t). If we insert (4)in (6) and expand the exponential, we getK V (2, 1) = K 0 (2, 1) + K 1 (2, 1) + K 2 (2, 1) + · · · , where (19)∫ 2[ ( ∫ īt2mẋ 2 )]K 0 (2, 1) = exp1 h t 12 dt Dx(t) (20)K 1 (2, 1) = − ī ∫ 2[ ( ∫ īt2mẋ 2 )] ∫ t2exph 1 h t 12 dt V [x(s), s]ds Dx(t) (21)t 1K 2 (2, 1) = − 1 ∫ 2[ ( ∫ īt2mẋ 2 )] [∫ t2] 22¯h 2 exp1 h t 12 dt V [x(s), s]ds Dx(t) (22).8 Demanding this expansion to be true in the limit ɛ → 0, the promised condition in A appears, with the result A = 2πi¯hɛ 1/2.9mThis follows from ”normal” Quantum Mechanics. Discussed in [4], [5]10 The index V in Eq. (19) shall just symbolize that the particle is moving in some potential V (x, t)t 1∂ 25
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Feynman Path Integrals in Quantum Mechanics

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