Physics 513, Quantum Field Theory Homework 8 Due Tuesday, 11 th November 2003

Transcription

1 Physics 53 Quantum Field Theory Homework 8 Due Tuesday th November 3 Jacob Lewis Bourjaily Problem 4 We are to consider the problem of the creation of Klein Gordon particles by a classical source This process can be described by the Hamiltonian H H o + d 3 x j(x)φ(x) where H o is the Klein-Gordon Hamiltonian φ(x) is the Klein-Gordon filed and j(x) is a c-number scalar function Let us define the number λ by the relation d 3 p λ (π) 3 j (p) E p a) We are to show that the probability that the source creates no particles is given by [ P () {exp T i d 4 x j(x)φ I (x) Without loss of understanding we will denote φ φ I Almost entirely trivially we see that H I d 3 x j(x)φ(x) Therefore P () {exp T {exp T [ i dt H I (t ) [ i d 4 x j(x)φ(x) b) We are to evaluate the expression for P () to the order j and show generally that P () λ + O(λ ) First let us only consider the amplitude for the process We can make the naïve expansion T { exp [ i d 4 x j(x)φ(x) + i d 4 x j(x) φ(x) For every odd power of the expansion there will be at least one field φ Io that cannot be contracted from normal ordering and therefore will kill the entire term So only even terms will contribute to the expansion It should be clear that the amplitude will be of the form O(j )+O(j 4 ) Let us look at the O(j ) term That term is given by { ( ) } T ½ d 4 x j(x)φ(x) d 4 xd 4 y j(x)j(y) T {φ(x)φ(y)} d 4 xd 4 y j(x)j(y) D F (x y) d d 4 xd 4 4 p i y (π) 4 p m + iɛ e ip(x y) j(x)j(y) d 4 p (π) 4 d 4 x j(x)e ipx d 4 y j(y)e ipy }{{}}{{} j (p) j (p) d 4 p (π) 4 j (p) i p m + iɛ d 3 p dp (π) 3 (π) j (p) i p m + iɛ i p m + iɛ

2 JACOB LEWIS BOURJAILY We know how to evaluate the integral dp (π) j (p) i dp p m + iɛ (π) j (p) i (p ) Ep + iɛ dp (π) j (p) i (p E p )(p + E p ) The function has a simple pole at p E p with the residue i j (p) p E p i j (p) p E p E p We know from elementary complex analysis that the contour integral is πi times the residue at the pole Therefore d 3 p dp (π) 3 (π) j (p) i d 3 p m + iɛ p (π) 3 j (p) E p λ Because we now know the amplitude to the first order of λ (or rather the second order of j) we have shown as desired that P () ½λ + λ + O(λ ) c) We must represent the term computed in part (b) as a Feynman diagram and show that the whole perturbation series for P () in terms of Feynman diagrams is precisely P () e λ The term computed in part (b) can be represented by faf λ It has two points (neither originated by the source) and a time direction specified (not to be confused with charge or momentum) We can write the entire perturbation series as [ P () {exp T i d 4 x j(x)φ(x) +faf+ To get the series we must figure out the correct symmetry factors If one begins with n vertices then n of them must be chosen as in ; there are n/ n ways to do this After that each one of the in vertices must be paired with one of the out vertices; you can do this n! ways So the symmetry factor for the term with n uninteracting propagators is +faf+ S(n) n n! We may now compute the probability explicitly faf P () + ( ) ( λ) n n n! n ( n ( e λ/) ) ( λ/) n n! P () e λ

3 PHYSICS 53: QUANTUM FIELD THEORY HOMEWORK 8 3 d) Let us now compute the probability that the source creates one particle of momentum k First we should perform this computation to O(j) and then to all orders using the same trick as in part (c) to sum the series Let us calculate the amplitude that a particle is created with the explicit momentum k T { φ k exp [ i i d 4 x j(x) a k i d 4 x j(x) a k i d 4 x j(x) i d 4 x (π) 4 i j (k) Ek d 4 x j(x)φ(x) j(x) Ek e ikx d 3 p ( ap (π) 3 e ipx + a pe ipx) Ep d 3 p (π) 3 a pe ipx Ep d 3 p (π) 3 e ipx (π) 3 δ (3) (p k) Ep Now the probability of creating such a particle is the modulus of the amplitude P ( k ) j(x) E k We can compute the probability that a particle is created with any momentum by simply integrating over all the possible k This yields d 3 k P () (π) 3 j (x) λ E k Therefore in Feynman graphs xfef i λ The entire perturbation in Feynman diagrams is therefore P () xfef +faf+ i λe λ/ P () λe λ e) We are to show that the probability of producing n particles is given by a Poisson distribution From part (d) above we know that each creation vertex on the Feynman diagram must be multiplied by i λ Now because each of the final products are identical and there are n! ways of arranging them the symmetry factor in each case is n! The probability is approximated by P (n) λn n! Like we have done before to get the correct probability we must take into account the probability that no particle is created Therefore P (n) λn e λ n! f) We must show that a poisson distribution given above with parameter λ has a norm of an expectation value of λ and a variance of λ First let us compute the norm of the distribution function n λ n n! eλ e λ n λ n n! e λ e λ

4 4 JACOB LEWIS BOURJAILY The expectation value for the number created is simply nλ n E(n) n! e λ λe λ λ n (n )! λe λ n n m λ m m! λe λ e λ λ To compute the variance we will use the relation V ar(n) E(n ) E(n) Let us compute E(n ) E(n ) n λn n! e λ k Knowing this it is clear that λe λ n n λn (n )! λe λ ((n ) + ) λn (n )! n [ λe λ (n ) λn (n )! + n λe λ e λ + λe λ λ + λ e λ λ + λ n n λ n (n )! λ n (n )! V ar(n) λ + λ λ λ n ] λ n (n )! Problem 44 The cross section for scattering of an electron by the Coulomb field of a nucleus can be computed to lowest order without quantizing the electromagnetic field We will treat the field as a given classical potential A µ (x) The interaction Hamiltonian is then H I d 3 x e ψγ µ ψa µ where ψ(x) is the usual quantized Dirac field a) We must show that the T -matrix element for an electron scatter to off a localized classical potential is given to the lowest order by p f it p i ieū(p f )γ µ u(p i ) õ(p f p i ) where õ is the Fourier transform of A µ We may compute this contribution directly p f it p i d 4 x p f T {H I (x)} p i ie d 4 x A µ p f T { ψ(x)γ µ ψ(x)} p i ie d 4 x A µ p f ψ(x)γ µ ψ(x) p i ie d 4 x A µ (x)u s (p f )γ µ u s (p i )e ix(p f p i) ieu s (p f )γ µ u s (p i ) d 4 x A µ (x)e ix(p f p i ) ieu s (p f )γ µ u s (p i )õ(p f p i )

5 PHYSICS 53: QUANTUM FIELD THEORY HOMEWORK 8 5 b) If A µ (x) is time independent its Fourier transform contains a delta function of energy We therefore define p f it p i im (π)δ(e f E i ) Given this definition of M we must show that the cross section for scattering off a timeindependent localized potential is given by v i E i d 3 p f (π) 3 E f (π)δ(e f E i ) M(p i p f ) From class we know that we can represent an incoming wave packet with momentum p i in the z-direction and impact parameter b by the relation d 3 p i ψ b (π) 3 e ibp i ψ(p i ) p i Epi The probability of interaction given an impact parameter is then P (b) d3 p f (π) 3 p f it ψ b E f d3 p f (π) 3 E f d3 p f (π) 3 E f Therefore d b P (b) d 3 p i d 3 k (π) 6 e ib(pi k) ψ(p i )ψ (k) p f it p i p f it k Epi E k d 3 p i d 3 k e ib(pi k) ψ(p i )ψ (k)(π) δ(e f E pi )δ(e f E k )M(p i p f )M(k p f ) Epi E k (π) 6 d3 p f (π) 3 d b d3 pd 3 k e ib(p k) E f (π) 6 ψ(p)ψ (k)(π) δ(e f E p )δ(e f E k )M(p p f )M(k p f ) Ep E k d 3 pd 3 k d3 p f (π) 3 E f d3 p f (π) 3 E f v i (π) d3 p f (π) 3 E f v i (π) ψ(p)ψ (k) Ep E k (π) δ () (p k )δ(e f E p )δ(e f E k )M(p p f )M(k p f ) (π) 6 d 3 pd 3 k (π) 3 ψ(p)ψ (k) Ep E k δ () (p k )δ(p z k z )δ(e f E p )M(p p f )M(k p f ) d 3 p (π) 3 ψ(p) δ(e f E p ) M(p p f ) E p With a properly normalized wave function this reduces directly to (allow me to apologize for the inconsistency with notation It is hard to keep track of The incoming momentum p has energy E i ) d 3 p f v i E i (π) 3 (π)δ(e f E i ) M(p i p f ) E f Now let us try to write an expression for /dω d 3 p f (π) 3 (π)δ(e f E i ) M v i E i E f p f dp f dω (π) δ(p p) M v i E f E i v f dω p (π) 4vi M E i dω 6π M Therefore we have that dω 6π M

6 6 JACOB LEWIS BOURJAILY c) We will now specialize to the non-relativistic scattering of a Coulomb potential (A Ze/4πr) We must show that in this limit dω α Z 4m v 4 sin 4 (θ/) Let us first take the Fourier transform of the Coulomb potential à µ (k) Ze d 3 r eikr 4π r Ze 4π 4π k õ(k) Ze k From part (a) above we calculated that M ieu s (p f )γ µ u s (p)ãµ(p f p) ie Z (p f p) us (p f )γ u s (p) In the nonrelativistic limit E >> p so we may approximate that Therefore our amplitude becomes u s (p f )γ u s (p) u s (p f )u s (p) Eδ s s M ie Z s (p f p) Eδs From part (b) we may compute /dω directly dω 4Z e 4 E6 6π (p f p) 4 Z α E p 4 ( cosθ) Z α E 4p 4 sin 4 (θ/) Z α 4E v 4 sin 4 (θ/) In the nonrelativistic limit we have that E m Therefore we may conclude as desired that dω α Z 4m v 4 sin 4 (θ/)