Andre Schneider P622

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1 Andre Schneder P6 Probem Set #0 March, 00 Srednc 7. Suppose that we have a theory wth Negectng the hgher order terms, show that Souton Knowng β(α and γ m (α we can wrte β(α =b α O(α 3 (. γ m (α =c α O(α (. m(µ = Keepng ony the hghest order terms n α n α(µ m(µ. (.3 α(µ dm = ( αc O(α m (.4 dα = ( α b O(α. (.5 dα we obtan Thus, or dα α = b α = b n µ n µ = (b α. (.6 dm = c αm dm m = c αd(α n m = c ( n. (.7 b b α n m = c b n (α. (.8 If the mts of the ntegra are from m(µ and m(µ on the RHS and α(µ and α(µ on the LHS n m(µ n m(µ = c b (n α(µ n α(µ (.9 whch mpes that m(µ = c/b α(µ m(µ. (.0 α(µ

2 Srednc 8. Consder ϕ 4 theory L = ϕ µ ϕ µ ϕ mm ϕ 4 λλ µ ϕ 4, (. n d =4 dmensons. Compute the beta functon to O(λ, the anomaous dmenson of m to O(λ, and the anomaous dmenson of ϕ to O(λ. Souton Rewrtng the Lagrangan as L = µ ϕ 0 µ ϕ 0 m 0 ϕ 0 4 λ 0ϕ 4 0, (. where the new feds and coupngs are reated to the od ones by ϕ 0 = ϕ / ϕ, (.3 m 0 = m / ϕ m, (.4 λ 0 = λ λ ϕ, (.5 and snce the s now ony cance the nfntes they can be wrtten as the foowng sums ϕ = m = λ = n= n= n= a n (λ n, (.6 b n (λ n, (.7 c n (λ n. (.8 From the resuts to Π( and V 4 (,, 3, 4 to O(λ (probems 4.5 and 6. we have that Thus, chosng the s to cance ony the nfntes we obtan A =0, (.9 B = λ (4π ( µ n, (.0 m C = 3λ (4π 3 ( µ n. (. m a (λ =O(λ, (. b (λ = λ (4π O(λ, (.3 c (λ = 3λ (4π O(λ. (.4 Frst we consder the nvarance of λ 0 wth respect to µ. We start defnng G(λ, n( λ ϕ = G n (λ n (.5 n= we have that n λ 0 = G(λ, n λ n µ. (.6

3 Dferentatng wth respect to n µ and requrng that λ 0 s ndependent of t mpes Thus, where the beta functon s gven by 0= d n λ 0 G(λ, dλ = λ dλ λ = ( λg (λ λg (λ dλ λ.... (.7 dλ = λ β(λ (.8 β(λ =λ dg dλ (λ = λ d dλ (c (λ a (λ O(λ 3 = 3λ (4π O(λ3. (.9 In the equaton above G (λ was obtaned expandng n( λ ϕ to frst order n λ and comparng wth n= G n(λ/ n. Now we consder the nvarance of m 0 wth respect to µ. Defnng we have that M(λ, n( / m / ϕ = n= M n (λ n (.0 n m 0 = M(λ, n m. (. Dferentatng wth respect to n µ and requrng that λ 0 s ndependent of t mpes 0= d n m 0 M(λ, dλ = λ dm m ( M = (λ M (λ dλ... dm m. (. Thus, usng resuts (.8 and (.9, we obtan the anomaous dmenson of m γ m (λ ( dm M m = (λ M (λ... (λ β(λ = λ dm (λ O(λ dλ = λ d dλ (b (λ a (λ O(λ = λ (4π O(λ. (.3 In the equaton above M (λ was obtaned expandng n(m / ϕ / to frst order n λ and comparng wth n= M n(λ/ n. To obtan the anomaous dmenson of ϕ we consder the propagator n the MS normazaton scheme, ( = d 4 xe x 0 Tϕ(xϕ(0 0 (.4 3

4 and the bare propagator 0 ( = d 4 xe x 0 Tϕ 0 (xϕ 0 (0 0, (.5 whch shoud be ndependent of µ. The two are reated by Tang the ogartm and dferentatng wth respect to n µ, we obtan 0= d n 0 ( = d n ϕ ( 0 ( = ϕ (. (.6 ( n µ dλ λ dm (. (.7 m The anomaous dmenson γ ϕ (λ of ϕ s the frst RHS n the equaton above, that s, whch s of order O(λ. γ ϕ (λ n ϕ = a (λ a (λ a (λ... (.8 3 Srednc 5. Derve the fermon-oop correcton to the scaar propagator by worng through ( ( ( (η, η, J exp g d 4 x 0 (η, η, J (3. J(x η α (x η α (x and show that t has a reatve mnus sgn reatve to the case of the scaar oop. Fgure : Feynman dagram for the fermon-oop correcton to the scaar propagator. Souton The fermon-oop correcton for a scaar propagator (see fgure ( subjected to the nteracton L Yu = gϕψγψ, (3. where γ can be γ 5 or the dentty I matrx, can be obtaned notng that 0 (η, η, J = exp d 4 xd 4 yη(xs(x yη(y exp d 4 xd 4 yj(x (x yj(y. (3.3 where S(x y = (x y = d 4 p ( /p me p(x y (π 4 p m, (3.4 ɛ d 4 e (x y (π 4 M ɛ (3.5 4

5 are the free fed propagators. Thus, expandng both 0 (η, η, J and exp constrant (0, 0, 0 = we have that (η, η, J = P f =0 P f! V =0 g V! ( ( d 4 x J(x L Yu ( η α (x Pf d 4 yd 4 zη(ys(y zη(z V s=0 J(x, η, α(x ( γ P s! η α (x η α (x wth the normazaton V Ps d 4 vd 4 wj(v (v wj(w (3.6 we obtan the partton functon to a orders. However, snce we are ony nterested on the process descrbed n fgure ( we ony need to tae nto connected dagrams that have two externa scaars and two vertces, that s, V =, P s = and P f =. Thus, to obtan the one fermon oop correcton (FL to the propagator we have to evauate FL (η, η, J = (g!!! = g 3» d 4 x» J(x η γ «d 4 x α(x η α (x J(x η β (x γ» d 4 yd 4 zη(ys(y zη(z d 4 y d 4 z η(y S(y z η(z» d 4 vd 4 wj(v (v wj(w» d 4 v d 4 w J(v (v w J(w d 4 xd 4 x J(x J(x d 4 vd 4 wj(v (v wj(w» η γ α(x η α (x η β (x γ η β (x d 4 yd 4 zη(ys(y zη(z! η β (x d 4 v d 4 w J(v (v w J(w d 4 y d 4 z η(y S(y z η(z. (3.7 Now, snce the functona dervatves wth /η(x and /η(x must act on dfferent ntegras so that we have a cosed fermon oop, the ast equaty becomes FL (η, η, J = g d 4 xd 4 x γs(x xγs(x x 6 J(x J(x» d 4 vd 4 wj(v (v wj(w d 4 v d 4 w J(v (v w J(w. (3.8 The mnus sgn came from the odd number of antcommutaton reatons between the feds η, η and ther functona dervatves. The factor of two came from the two possbe ways the fermon functona dervatves can act on the two ntegras. Note that factor of /6 reduces to / when we consder that the functona dervatves /J(x and /J(x have to act on dfferent ntegras so that we have a cosed fermon oop. Thus the one fermon-oop correcton s Π FL = g d 4 xd 4 x γs(x xγs(x x. (3.9 Foowng the same scheme we have for the oop correcton of a scaar oop " «# SL (η, η, J = (g d 4 3» 4 x d 4 vd 4 wj(vs(v wj(w! J(x 4!» = g 5 d 4 xd 4 x J(x J(x d 4 vd 4 wj(v (v wj(w d 4 v d 4 w J(v (v w J(w» J(x J(x J(x J(x d 4 v d 4 w J(v (v w J(w d 4 v d 4 w J(v (v w J(w = g d 4 xd 4 x (x x (x x 6 J(x J(x» d 4 vd 4 wj(v (v wj(w d 4 v d 4 w J(v (v w J(w. (3.0 From the frst to the second equaty a factor of 4! arses from the number of ways we can rearrange the four v and w ntegras. From the second to the thrd equaty a factor of arses from the two possbe ways the functona dervatves can act on the doube and trpe prme ntegras. As n the 5

6 fermon oop case the factor of /6 reduces to / when we consder that the functona dervatves /J(x and /J(x have to act on dfferent ntegras, that s Π SL = g d 4 xd 4 x (x x (x x. (3. whch s the same as that of the fermon oop correcton Π FL except for a factor of. 4 Srednc 5.3 Consder mang ϕ a scaar rather than a pseudoscaar, so that the Yuawa nteracton s L Yu = gϕψψ. In ths case renormazabty requres us to add a term L ϕ 3 = 6 κκϕ 3, as we as a near term n ϕ to cance the tadpoes. Fnd the one-oop contrbutons to the renormazng factors for ths theory n the MS scheme. Souton Lagrangan The Lagrangan for ths theory s L = L 0 L L 0 = Ψ/ Ψ mψψ µ ϕ µ ϕ M ϕ (4. L = g gϕψψ 4 λλϕ 4 6 κκϕ 3 Y Yϕ L ct (4. L ct = ( Ψ Ψ/ Ψ ( m mψψ ( ϕ µ ϕ µ ϕ ( M M ϕ (4.3 The Scaar Propagator To obtan the one oop contrbutons to ths theory we start wth the correctons to the scaar propagator shown n fgure (. Fgure : Feynman dagrams for the one-oop correctons to the scaar propagator. 6

7 The contrbuton of the frst dagram n fgure ( s gven by Π Ψoop ( = ( (g ( d 4 (π 4 Tr S(/ / S(/ (4.4 where S(/p = /p m p m ɛ. (4.5 Snce the denomnator of the propagator s just a number the trace nsde the ntegra can be evauated notng that Tr( / / m( / m = 4( m The denomnators can be combned usng Feynman s formua ( m m = supressg the ɛs, where q = x and D = x( x m. Changng the ntegraton varabes from to q = x we have that where Π Ψoop ( = 4g 0 dx 4N. (4.6 0 dx (q D (4.7 d 4 q N (π 4 (q D, (4.8 N = q x( x m ( xq = q D ( xq. (4.9 Movng to d =4 spacetme dmensons and mang g g µ / we obtan Π Ψoop ( = 4g µ 0 dx dx d 4 q N (π 4 (q D = 4g µ d 4 q q D 0 (π 4 (q D = g 6 4π dxd 0 D n(d/µ = g 6m 4π dxd D n(d/µ 0 = g 6m 4π fnte (4.0 where µ =4πe γ µ and n the ast ne ony the dvergent terms are expct snce we are nterested n the MS scheme. The contrbuton of the second dagram, from probem 4.5 of the prevous homewor set, s Π ϕ 4 = λm 6π fnte (4. where the fnte terms were not wrten expctey snce we are worng n the MS scheme. The thrd dagram correcton s Π ϕ 3( = (κ ( d 4 ( ( (π 4 (4. where (p = p M ɛ. (4.3 7

8 A smar ntegra was obtaned n probem 6.. Usng the resut from there Π ϕ 3( = The ast dagram gves the contrbuton of the counterterms, that s, κ 6π fnte. (4.4 Π ct ( = ( ϕ ( M M. (4.5 Snce we want the sum of a one-oop contrbutons to the scaar propagator to be fnte we must have whch mpes that The Fermon Propagator g 6m 4π λm 6π κ 6π ( ϕ ( M M = 0 (4.6 g ϕ = 4π (4.7 M = ( λ 6π g 6m 4π M κ 6π M. (4.8 Now we move on to the one-oop correctons to the fermon propagator. The Feynman dagrams for ths case can be seen on fgure (3. Fgure 3: Feynman dagrams for the one-oop correctons to the scaar propagator. The contrbuton of the frst dagram n fgure (3 w be the same as the one for the theory wth pseudoscaar ϕ: ( Σ oop (/p = (g d 4 S( (π 4 /p / ( Σ oop (/p = g 6π ( /p m fnte. (4.9 The dveregent terms cance f addng Σ oop (/p wth eads to a fnte resut. Thus, Σ ct (/p = ( Ψ /p ( m m (4.0 g Ψ = 6π (4. m = g 8π. (4. 8

9 p p p p p p Fgure 4: Feynman dagrams for the one-oop correctons to the fermon scaar vertex. The Fermon-Scaar Vertex The Feynman dagrams for the one-oop correcton to a fermon-scaar vertex can be seen on fgure (4. The contrbuton of the frst dagram s ( V Y (p,p = (g 3 = g 3 = g 3 3 d 4 S( (π 4 /p / S(/p / ( d 4 (/p / m( /p / m (π 4 ((p m ((p m ( M df 3 d 4 q (π 4 where, wth the transformaton q = x p x p, N (q D 3 (4.3 N =(/q x /p ( x /p m( /q x /p ( x /p m (4.4 D = x ( x p x ( x p x x p p (x x m x 3 M. (4.5 Rewrtng N = q Ñ (near n q, where Ñ =( x /p ( x /p m(x /p ( x /p m. (4.6 Snce the terms that are near n q ntegrate to zero and ony the frst term s dvergent, the contrbuton of V Y,oop (p,p to the propagator n the MS scheme s (see secton 5 n Srednc V Y (p,p= The contrbuton of the second dagram s nu snce V Y (p,p= ( 3 6 κ(g = 6 κg s fnte (the argest term s of order Thus, we obtan The Three-Scaar Vertex g3 8π fnte. (4.7 d 4 (π 4 S(/ (( (( d 4 (π 4 S(/ (( (( (4.8 d 4 (π 4 5 and ths ntegra converges. g = g 8π. (4.9 The Feynman dagrams for the one-oop correcton to a three-scaar vertex can be seen on fgure (5. 9

10 3 3 3 Fgure 5: Feynman dagrams for the one-oop correctons to the three-scaar vertex. Not a possbe permutatons were drawn. The contrbuton of the frst dagram to the one-oop correcton of the three scaar vertex s nu snce ( 3 V 3 (,, 3 = (κ 3 d 4 (π ( 4 (( (( (4.30 s fnte. The second dagram contrbuton s ( 3 V 3 (,, 3 = ( (g 3 = g 3 df 3 d 4 q (π 4 where, from secton 6 n Srednc, d 4 (π S(/ S(/ 4 / S(/ / N d 4 q D 3 q N (π 4 q D 3 d 4 (π S(/ S(/ 4 / 3 S(/ / (4.3 q = x x, (4.3 q = x x 3 3, (4.33 N =3mTr/q/q = mq, (4.34 N =3mTr/q/q = mq, (4.35 D = x 3 x x 3x x x 3 m, (4.36 D = x x x x 3 3 x x 3 m, (4.37 f we eep ony the dvergent terms. Snce ony terms proportona to q 3 and q can dverge and Tr/a/b/c = 0 the ony term that s eft s proportona to 3mTr/q/q = mq. Thus, The thrd dagram contrbuton s V 3 (,, 3 = ( (κ( λ = 3λκ 6π usng resut (3.9 from Srednc. Addng a contrbutons we obtan V 3 (,, 3 = mg3 8π = 3mg3 π. (4.38 d 4 (π ( 4 (( (( (( (4.39 κ = 3λ 6π 3mg3 κπ (4.40 0

11 Fgure 6: Feynman dagrams for the one-oop correctons to the four-scaar vertex. Not a possbe permutatons were drawn. The Four-Scaar Vertex The Feynman dagrams for the one-oop correcton to the four-scaar vertex can be seen on fgure (6. The contrbuton of the frst dagram s V 4 (,, 3, 4 = g 4 d 4 (π 4 Tr S(/ S(/ / S(/ / S(/ / / 3 ( 5 permutatons of, 3 and 4 = 3g4 π (4.4 usng the resut from secton 5 n Srednc. The resut s the same as n the case of a pseudoscaar ϕ. Ths happens because the ony dvergent part n the ntegra above s the one proportona to ( n the numerator and the four γ 5 s from the pseudoscaar nteracton w not change ths factor. The contrbutons of the second and thrd dagrams are V 4 (,, 3, 4 =κ 4 V 4 (,, 3, 4 = κ λ d 4 ( (π 4 (( (( (( 3 ( permutatons of, and 3, (4.4 d 4 ( (π 4 (( (( 3 4 ( 5 permutatons of, 3 and 4. (4.43 Both are fnte and, therefore, do not contrbute to the vertex correcton n the MS scheme. The contrbuton of the fourth dagram s V 4 (,, 3, 4 = λ = 3λ 6π usng the resut from secton 3 n Srednc. Thus, we have 5 Srednc 5. d 4 ( (π 4 (( ( permutatons of, 3 and 4 (4.44 (4.45 λ = 3g4 λπ 3λ 6π. (4.46 Compute the one-oop contrbutons to the anomaous dmenson of m, M, Ψ and ϕ. Souton Lagrangan The Lagrangan for the Yuawa theory L = Ψ Ψ/ Ψ m mψψ ϕ µ ϕ µ ϕ M M ϕ g g µ / ϕψγ 5 Ψ 4 λλ µ ϕ 4 (5.

12 can be rewrtten as where L = Ψ 0 / Ψ 0 mψ 0 Ψ 0 µ ϕ 0 µ ϕ 0 M 0 ϕ 0 g 0ϕ 0 Ψ 0 γ 5 Ψ 0 4 λ 0ϕ 4 0 (5. Choosng the s to cance the nfntes we have ( a n (λ Ψ = n = g 6π fnte n= ( b n (λ g m = n, = 8π fnte ϕ = M = g = λ = n= n= n= n= n= Ψ 0 = / Ψ Ψ, (5.3 m 0 = m Ψ m, (5.4 ϕ 0 = ϕ / ϕ, (5.5 M 0 = / M / ϕ M, (5.6 g 0 = g ϕ / Ψ µ/ g, (5.7 λ 0 = λ ϕ λ. (5.8 c n (λ n g, = 4π d n (λ n, = e n (λ n = g 8π f n (λ n = whch mpes that, to frst order, Ψ 0 = g 3π m 0 = g 6π ϕ 0 = g 8π ( λ M 0 = 3π g 8π g 0 = 5g 6π λ 0 = ( fnte ( λ 6π g m π M ( fnte ( 3λ 6π 3g π λ ( fnte ( fnte (5.9 (5.0 (5. (5. (5.3 (5.4 Ψ, (5.5 m, (5.6 ϕ, (5.7 ( m M M, (5.8 µ / g, (5.9 ( 3λ 6π g π 3g λπ µ λ. (5.0 To obtan the anomaous dmensons we mae use of the foowng resuts (secton 5 Srednc: dg = g β g(g, λ (5. dλ = λ β λ(g, λ (5.

13 Anomaous dmenson of m The anomaous dmenson of m s found from where s the anomaous dmenson of m. Anomaous dmenson of M 0= d n m 0 = d = g 8π The anomaous dmenson of M s found from 0= d n M 0 = d ( = 3π ( λ 3π g 8π dλ g 4π g 6π dg n m γ m (5.3 γ m dm m = g 6π (5.4 ( m M n M ( m M dg g π m M (γ M γ m γ M (5.5 where, eepng ony the O( 0 terms, we obtan γ M M dm = λ 3π g 8π ( m M (5.6 s the anomaous dmenson of M. Anomaous dmenson of Ψ The anomaous dmenson of the fed Ψ s found from s the anomaous dmenson of Ψ. γ Ψ = d n Ψ = d = g 6π g 3π dg = g 3π (5.7 3

14 Anomaous dmenson of ϕ The anomaous dmenson of ϕ s found from 0= d n ϕ = d g 8π = g dg 4π = g 8π (5.8 s the anomaous dmenson of ϕ. 4