KLEIN-GORDON EQUATION AND PROPAGATOR (A REVIEW) The relativistic equation relating a particles energy E, three momentum. E 2 = (p) 2 + m 2.

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1 KLEIN-GORDON EQUATION AND PROPAGATOR A REVIEW The relativistic equation relating a particles energy E, three momentum p, and mass m is E 2 = p 2 + m 2. Note that and c have been set equal to. The relativistic wave equation for a free particle of spin zero is found by making the replacements E i / t and p i in Eq.. and applying the resulting operator to the free particle wave function φx. The resulting equation 2 φx t 2 2 x φx + m 2 φx = φx + m 2 φx = 0 2 is called the Klein-Gordon equation for a free particle. Note that µ = 0,, 2, 3, and x = x 0 = t, x, x 2, x 3 To find the Klein-Gordon equation for a spin zero particle interacting with an electro-magnetic field, replace µ by µ + iqa µ where q is the charge of the particle, A µ is the is the four dimensional vector potential. ψx +iqa µ ψx +iq Aµ ψx q 2 A µ A µ ψx+m 2 ψx = 0. 3 Date: July 26, 204.

2 2 KLEIN-GORDON EQUATION AND PROPAGATORA REVIEW Introduce ˆV = +iqa µ + iq Aµ q2 A µ A µ. 4 Then the Klein-Gordon equation is written ψx + m 2 ψx = ˆV ψx. 5 Define the Klein-Gordon propagator F x y by 2 F x y x x Fx y + m 2 F x y = δ 4 x y. 6 Write F x y as a Fourier transform, i.e., F x y = d 4 k 2π 4 Fkexp ik x y 7 where the four-dimensional momentum vector k = k 0, k, k 2, k 3. Substitute Eq. 7 into Eq. 6, and find d 4 k 2π 4 Fk exp ik x y = d δ 4 4 k x y = exp ik x y 8 2π 4 So F k =. 9 A solution to Eq. 9 is F k = /k 2 m 2, but this is not the most general solution. The general solution is F k = /[k 0 2 k 2 m 2 ]+C δk 0 +C 2 δk 0 + 0

3 KLEIN-GORDON EQUATION AND PROPAGATORA REVIEW 3 This can be verified by multiplying Eq. 0 by [k 0 2 k 2 m 2 ], and using k 0 δk 0 = 0 and k 0 + δk 0 + = 0. Eq. 9 is then recovered. Use partial fractions to show k 0 2 k 2 m = 2 2 k 0 k 2 + m 2 k 0 +, so F k = 2 2 k 0 k C δk C 2 δk The integral I below is part of the integral in Eq. 8 and will be evaluated using reasonable physical constraints, which in turn will determine the constants C and C 2 : where I = and + I = + dk 0 F k exp ik 0 x 0 y 0 = I + I 2 3 dk 0 2 k 0 k 2 + m 2 k 0 + exp ik 0 x 0 y 0, 4 I 2 = + dk 0 [C δk 0 +C 2 δk 0 + ] exp ik 0 x 0 y 0. 5

4 4 KLEIN-GORDON EQUATION AND PROPAGATORA REVIEW To evaluate I using contour integration, introduce the closed contour below where ɛ is a small constant and M is a large real value of k 0, which eventually : from M to ɛ along the real k 0 axis 2 from ɛ to + ɛ along a semi-circle of radius ɛ, which is centered at, and is below the k 0 3 from + ɛ to + ɛ along the k 0 axis 4 from + ɛ to + + ɛ along a semi-circle of radius ɛ, which is centered at +, and is below the k 0 5 from + + ɛ to +M along the k 0 axis 6 +M to M along a semi-circle S M of radius M, which is centered at the origin of the k 0 axis and is below the This contour will be referred to as contour C. There are no poles inside the contour, so the integral around the contour is zero. When x 0 y 0 > 0 and M, the integral vanishes on S M. I is identified

5 KLEIN-GORDON EQUATION AND PROPAGATORA REVIEW 5 as a Cauchy principle value. So I = Resk 0 = + πiresk 0 = = 2 exp i x 0 y 0 exp+i x 0 y 0 Hx 0 y 0 where H is the unit step function defined by Hx 0 y 0 = if x 0 y 0 > 0 and Hx 0 y 0 = 0 if x 0 y 0 < 0. Integrate I 2 over the contour C and get 6 I 2 = C exp i x 0 y 0 +C 2 exp+i x 0 y 0 Hx 0 y 0 Use these results in Eq. 3 and find 7 I = 2 k 2 + m + C 2 exp i x 0 y 0 Hx 0 y k 2 + m + C 2 2 exp+i x 0 y 0 Hx 0 y 0 8 The first term represents a Klein-Gordon particle of positive energy traveling forward in time from y to x. The second term represents a Klein-Gordon particle of negative energy traveling forward in time. This is rejected by Feynman as being unphysical, so eliminate the second term by setting C 2 equal to /2. Next, evaluate I when x 0 y 0 < 0 by introducing the closed contour below:

6 6 KLEIN-GORDON EQUATION AND PROPAGATORA REVIEW from M to ɛ along the k 0 axis 2 from ɛ to + ɛ along a semi-circle of radius ɛ, which is centered at, and is above the k 0 3 from + ɛ to + ɛ along the k 0 axis 4 from + ɛ to + + ɛ along a semi-circle of radius ɛ, which is centered at +, and is above the k 0 5 from + + ɛ to +M along the q 0 axis 6 +M to M along a semi-circle S M of radius M, which is centered at the origin of the k 0 axis and is above the k 0 This contour will be referred to as contour C2. There are no poles inside the contour, so the integral around the contour is zero. When x 0 y 0 < 0 and M, the integral vanishes on S M. We identify I as a Cauchy principle value, so I = Resk 0 = + πiresk 0 = = 2 exp i x 0 y 0 exp+i x 0 y 0 Hy 0 x 0. 9

7 KLEIN-GORDON EQUATION AND PROPAGATORA REVIEW 7 Then I = 2 k 2 + m + C 2 exp i x 0 y 0 Hy 0 x k 2 + m + C 2 2 exp+i x 0 y 0 Hy 0 x 0 = 2 k 2 + m + C 2 exp+i y 0 x 0 Hy 0 x k 2 + m + C 2 2 exp i y 0 x 0 Hy 0 x 0 20 The second term represents a Klein-Gordon particle of positive energy traveling forward in time from x to y. The first term represents a Klein- Gordon particle of negative energy traveling forward in time. This term is rejected as unphysical and is eliminated by setting C equal to /2. Combining these results, Eq. 2 becomes F k = 2 k 0 πiδk0 + 2 k 0 + k 2 + m + 2 πiδk The theory of generalized functions proves that /x±iɛ = /x πiδx. Thus F k = 2 k 0 + iɛ k 0 +. iɛ 22 An alternative expression for F k is F k = k 0 2 k 2 m 2 + iɛ. 23