Homework & Soluton Prof. Lee, Hyun Mn Contrbutors Park, Ye J(yej.park@yonse.ac.kr) Lee, Sung Mook(smlngsm0919@gmal.com) Cheong, Dhong Yeon(dhongyeoncheong@gmal.com) Ban, Ka Young(ban94gy@yonse.ac.kr) Ro, Tae Gyu(shxorb34@gmal.com) - 1 -
1 HW Lecture1 1 Derve Euler-Lagrange eq. for L 1 µφ µ φ m φ λ 4! φ4 S d 4 xl(φ, µ φ) [ δs d 4 x φ δφ + ] ( µ φ) δ( µφ) [{ } { }] d 4 x φ µ δφ + µ ( µ φ) ( µ φ) δφ so, as the total dervatve term vanshes on the boundares wth δφ 0 then, we can wrte φ µ ( µ φ) 0 µ µ φ + m φ + 1 3! λφ3 0 ( + m )φ + 1 3! λφ3 0 Where are the feld from? [Research "Representaton of Lorentz group"] The Lorentz algebra could be wrtten as SO(3, 1) that s same wth SU() L SU() R. It s a knd of Le algebra. The Lorentz algebra s defned as J ± J ± K where J s a generator of rotaton and K s a generator of boost. Ther representaton s labeled by angular momentum, j, where j 0, 1, 1, 3,. And the dmenson of the representaton (j, j + ) s (j + 1)(j + + 1). And f consderng the scalar feld, we get the values of j and j + could be both 0 snce the dmenson s 1 and spn s equal to zero. Thus that s labeled by (0, 0). In the same way, Drac feld s four-dmenson and ts spn s 1 and ts representaton s a drect sum of ( 1, 0) and (0, 1 ), whch s called The Weyl spnor representaton(the left-handed Weyl spnor and the rght-handed Weyl spnor, respectvely). Vector feld havng spn-1 belongs to the ( 1, 1 ) representaton, whch s also called as the gauge feld. 3 Why "δs0" gves the Largest Contrbuton when 0 lmt? From the path ntegral formulaton, the probablty ampltude s provded by the formula Prob all possble paths Ths problem s hghly related to statonary phase method. Here s the smple revew of ths. e S - -
Consder, I lm dx e λf(x) λ when f(x) has a global mnmum at x x 0,.e f (x 0 ) 0. Then the domnant contrbutons to the above ntegral, as λ wll come from the ntegraton regon around x 0, snce t the largest value on the ntegral regon, and the value exponentally decays on the other regon. Formally, we may expand f(x) about ths pont: Snce f (x 0 ) 0, ths becomes: f(x) f(x 0 ) + f (x 0 )(x x 0 ) + 1 f (x 0 )(x x 0 ) + f(x) f(x 0 ) + 1 f (x 0 )(x x 0 ) Insertng the expanson nto the expresson for I gves [ ] lm λ e λf(x 0) dx e λ f (x 0 )(x x 0 ) π 1/ lm λ λf e λf(x 0) (x 0 ) Ths approxmaton s known as the statonary phase or saddle pont approxmaton. formula also holds for the magnary case, I dx e λf(x) Ths snce when λ s very large, phases change very rapdly as the value of exponent s large, hence they wll add ncoherently, varyng between constructve and destructve addton at dfferent tmes. Our case s a trval applcaton of ths formula, whch s f(x) replaced wth S, and λ replaced wth 1/. - 3 -
Fgure 1: Homework 1.3 reference(schwartz "Quantum feld theory and Standard model") HW Lecture 1 Show dq 1... dq N e 1 qaq+jq ( (π) N Let s look at the 1 dmensonal case, whch s expressed as dx e 1 ax +Jx det(a) Ths s a smple Gaussan ntegral, whch can be evaluated as dx e 1 ax +Jx dx e a (x J a ) e J a ) 1/ e 1 JA 1 J ( π a ) 1 e J a Now, f we expand ths to an N N matrx, the exponent n the ntegral takes the form 1 q A j q j + J q - 4 -
Usng the same analogy wth the 1D case, we get 1 ( ( ) ) ( q A 1 J A j q j By ntroducng q q A 1 q, we can evaluate 1 q A j q j as, 1 qt A q 1 qt S T ( SAS T ) S q ( ) ) A 1 J + 1 j J A 1 j J j q 1 T D q 1 ) (d 1 q 1 + d q + + d N q N Where S s an orthogonal transformaton matrx, and D s the dagonalzed matrx. As the matrx s dagonalzed, we get the followng relaton Hence, we get det(a) d 1 d N [ dq 1 dq N e 1 (π) qt Aq+Jq N d 1 d N [ (π) N det(a) ] 1 e 1 J 1 A 1 J ] 1 e 1 J 1 A 1 J From Show Free feld operator φ(x) d 3 p (π) 3 (a pe p x + (a pe p x )), wth [a p, a p ] (π)3 δ 3 ( p p ) 0 T (φ(x)φ(y) 0 φ 0 ( x, t) a p 0 0, 0 a p 0. d 4 p e p (x y) (π) 4 p m + ɛ d 3 (π) 3 1 ωk [a k (t)e kx + a k (t)ekx ] d 3 k 1 d 3 k 1 1 0 φ 0 (x 1 )φ 0 (x ) 0 (π) 3 (π) 3 0 a k1 a k ω1 ω 0 e (k x k 1 x 1 ) d 3 k 1 d 3 k 1 1 (π) 3 (π) 3 (π) 3 δ 3 ( k 1 k )e (k x k 1 x 1 ) ω1 ω d 3 k 1 (π) 3 e k(x x 1 ) ω k 0 T {φ 0 (x 1 )φ 0 (x )} 0 0 φ 0 (x 1 )φ 0 (x ) 0 θ(t 1 t ) + 0 φ 0 (x )φ 0 (x 1 ) 0 θ(t t 1 ) d 3 k 1 (π) 3 [e k(x x 1 ) θ(t 1 t ) + e k(x 1 x ) θ(t t 1 )] ω k d 3 k 1 (π) 3 [e k( x 1 x ) e ωkτ θ(τ) + e k( x 1 x ) e ωkτ θ( τ)] ω k d 3 k 1 (π) 3 e k( x 1 x ) [e ωkτ θ( τ) + e ωkτ θ(τ)] ω k - 5 -
Lemma Proof. e ωkτ θ( τ) + e ωkτ ω k θ(τ) lm ɛ 0 π dω ω ω k + ɛeωτ 1 ω ωk + ɛ 1 [ω (ω k ɛ)][ω ( ω k + ɛ)] [ 1 ω k 1 ω (ω k ɛ) 1 ω ( ω k + ɛ) ] dω ω (ω k ɛ) eωτ πe ωkτ θ( τ) + O(ɛ) dω ω ( ω k + ɛ) eωτ πe ωkτ θ(τ) + O(ɛ) Puttng t together, we fnd lm ɛ 0 d 3 k 1 (π) 3 e k( x 1 x ) ω k ω k π dω ω ω k + ɛeωτ Remark. k 0 k + m. The propagatng feld can be off-shell.. Ths s a classcal Green s functon for the Klen-Gordon equaton : ( + m )D F (x, y) δ 4 (x y) 3 HW Lecture3 1 Show that the Drac equaton, (γ µ µ m) ψ 0 mples the Klen-Gordon equaton, ( + m ) ψ 0 (γ ν ν + m)(γ µ µ m)ψ (γ µ γ ν µ ν + m )ψ 0 Note that γ µ γ ν µ ν 1 γ{γµ, γ ν } 1 gµν µ ν µ µ Therefore, each compoents of ψ solves the Klen-Gordon equaton, ( + m )ψ 0 Usng the followng relatons F 0 E, F j ε jk B k construct the Maxwell equatons (note that the Banch dentty s [µ F λν] 0). By Banch dentty, F jk + j F k + k F j 0-6 -
0) 0 F jk + j F k0 + k F 0j 0 0 F jk + j F k0 + k F 0j 0 (usng η γα F α F γ ) 0 ɛ jkm B m + j E k k E j 0 (usng ɛ abc ɛ abd δc d ) 0 δn m B m + ɛ jkn ( j E k k E j ) 0 then, we can fnd 0 δ m n B m ɛ jkn ( j E k k E j ) 0 δ m n B m ɛ jkn k E j 0 B n ɛ jkn k E j ths equaton means, B t E : Faraday s Law, j, k 0) let 1, j, k 3 F jk + j F k + k F j 0 ɛ jka B a + j ɛ ka B a + k ɛ ja B a 0 1 ɛ 3a B a + ɛ 3a B a + 3 ɛ 1a B a 0 α B α α B α 0 lkewse, explctly calculate other cases, then we can fnd, B 0 There s smlar problem n "An ntroducton to quantum feld theory Ch. pb..(a) (Peskn Schroeder)" 3 Classcal electromagnetsm (wth no sources) follows from the acton S d 4 x ( 1 ) 4 F µνf µν, where F µν µ A ν ν A µ Derve Maxwells equatons as the Euler-Lagrange equatons of ths acton, treat ng the components A µ (x) as the dynamcal varables. Wrte the equatons n standard form by dentfyng E F 0 and ɛ jk B k F j L 1 4 F µνf µν 1 4 µa ν ν A µ ν A µ µ A ν 1 ( µa ν )( µ A ν ) + 1 ( µ A µ ) - 7 -
So, let s derve a equaton of moton. and, Lemma µ A ν 0 ( µ A ν ) ( µ A ν ) + ( α A α )η µν ( µ A ν ) ( α A α ) µ A ν ( β A α )η αβ δ µ ρ δ ν αη αβ η µν usng ths lemma, we can get µ ( µ A ν ) µ µ A ν + µ α A α η µν µ µ A ν + ν α A α µ µ A ν + ν µ A µ µ ( µ A ν ν A µ ) µ F µν so, 1. µ, ν 0, µ ( µ A ν ) µ A ν µ F µν 0 F 0 0 E 0 E 0. µ µ, ν j, 0 µ F µ 0 F 0 + j F j E t jɛ jk B k E t + jɛ jk B k E t + B - 8 -
4 Calculate the ampltude M of the followng dagrams usng Feynman rules. p 3 p 4 k 1 p 1 + p 3 p 3 p 4 k k p 3 p 4 k 1 k 1 p 1 p k 1 p 1 k 1 p p 1 p p 1 p k 1 p 1 + p 4 [Dagram 3-1] [Dagram 3-] [Dagram 3-3] [Dagram 3-4] p 1 p Dagram 3-1) The S-matrx n the momentum space s wrtten as, S M(π) 4 δ (4) (p 1 p ) Then usng the Feynman rules, one can evaluate the matrx element M, whch s actually the ampltude. d 4 k M (π) 4 k m ( λ)(π) 4 δ (4) (p 1 p ) }{{} ntegral of large k part k 3 dk 1 k k 0 Therefore the ampltute goes to nfnty, the mass renormalzaton s needed. Dagram 3-) M ( λ) d 4 k (π) 4 d 4 k 1 (π) 4 k1 m + ɛ k m + ɛ (π) 4 δ (4) (k + p 4 k 1 p )(π) 4 δ (4) (p 3 + k 1 p 1 k ) ( λ) d 4 k 1 (π) 4 k1 m + ɛ (k 1 + p 3 p 1 ) m + ɛ δ(4) (p 3 + p 4 p 1 p ) By evaluatng the large k 1 ntegral part, one can get Dagram 3-3) M k1dk 3 1 1 k1 4 log(k) k0 M ( λ) d 4 k 1 (π) 4 k1 m + ɛ (k 1 p 1 p ) m + ɛ δ(4) (p 3 + p 4 p 1 p ) - 9 -
Dagram 3-4) M ( λ) d 4 k 1 (π) 4 k1 m + ɛ (k 1 p 1 p 4 ) m + ɛ δ(4) (p 3 + p 4 p 1 p ) Clearly, ths dgerves. Therfore, we need to ntroduce the renormalzaton of quartc couplng lambda. Usng the same method and summng up all contrbutons, we could get the total λ ordered ampltude. - 10 -