Quantum Field Theory III

Transcription

1 Quantum Feld Theory III Prof. Erck Wenberg February 16, Lecture 9 Last tme we showed that f we just look at weak nteractons and currents, strong nteracton has very good SU() SU() chral symmetry, and there s also pretty good approxmate SU(3) SU(3) symmetry. But now we know the Lagrangan for QCD If we have some m f = 0 then there s symmetry L QCD = q f (D/ m f )q f 1 4 F a µf aµ (1) q L f U Lq L f, qr f U Rq R f () However why s the symmetry SU(N) nstead of U(N)? One of the U(1) symmetry just corresponds to U L = U R U(1) whch just counts the number of u s and d s. The other U(1) symmetry U L = U R seems to be absent, so t must correspond to a Goldstone boson. So n SU() SU() chral theory we should have a fourth Goldstone boson n addton to the three pons. Now the mass of π s about MeV, and the next lght pseudoscalar boson whch has zero strangeness s η whch has mass 549 MeV, whch s too large to be also a Goldstone bosons. nythng else would be heaver and not plausble. So ths s known as the U(1) problem. It s a long story to a resoluton of ths problem. Let s suppose we have a spontaneously broken gauge theory. Frst let s consder gauge propagators, and we use photon for example. The poton propagator s a seres of The frst term s just (g µ q + λqµ q ) q (3) Now the second dagram wthout the legs s, n general Π µ = (q g µ q µ q )Π(q ) (4) whose form s dctated by current conservaton, thus Ward dentty. So f we put n the legs we wll get (g αµ q + λqα q µ ) ( q (q g µ q µ q )(Π) ) (g β q + λq q β ) q = (g αβ q qα q β ) q Π (5) Note that the second term n both of the propagators dot nto the mddle term to gve zero, so they are rrelevant. We can see the thrd term wll be (g αβ q qα q β ) q Π (6) 1

2 Now we can sum the seres q = q (g αβ qα q β ) Π n + (λ 1) qα q β q (g αβ qα q β ) q n=0 1 1 Π(q ) +... q (7) Because ths s QED the second term n the propagator does not matter, so t s just 1/q. ny pole n ths expresson must come from Π(q ). Say f then the pole wll be at µ and t wll be approxmately mass squared. Let s consder broken U(1) gauge theory Π(q ) = µ q + f(q ) (8) L = 1 4 F µ + D µ φ V ( ϕ ) (9) and the potental has a mnmum at ϕ = /. Let s choose ϕ 1 = and ϕ = 0 then we can defne our feld as ϕ = 1 [ + χ(x) + ψ(x)] (10) then from the potental V there wll be no mass term for ψ, but a mass term for χ. The covarant dervatve wll be Dµ ϕ = µ χ + µ ψ e µ ψ + e µ ( + χ) (11) So the absolute value square s D µ ϕ = 1 [ ( µ χ e µ ψ) + ( µ ψ + e µ + e µ χ) ] (1) Let s extract from ths term the terms quadratc n the felds, and wrte down the Lagrangan for the exctaton felds ψ and χ L = 1 4 F µ + 1 ( µχ) + 1 ( µψ) + 1 e µ + e µ µ ψ V (χ, ψ) +... (13) Now apart from the usual terms we have two pecular terms Now let s look at contrbutons to photon self-energy. The sum s e g µ, eq µ (14) ( g µ q q µ q ) e q (15) So the correcton s n the form of a pole wth photon mass m = e. Note here the mass comes from two very pecular terms and not from ordnary feld mass term.

3 Now let s wrte the feld n another form ϕ(x) = 1 ( + ρ(x))e ξ(x)/ (16) For small felds ths s essentally the same felds we used before, but thngs dffer n hgher order terms. Now the covarant dervatve s So the absolute value square s D µ ϕ = µ ϕ + e µ ϕ = 1 [ µ ρ + µ ξ ( + ρ D µ ϕ = 1 ( µρ) + 1 ( µξ + e µ ) ( + ρ ) ( )] + ρ + e µ e ξ/ (17) ) = 1 ( µρ) + e ( µ + 1 e µξ) ( 1 + ρ ) (18) Now we can redefne the feld B µ = µ + 1 e µξ (19) Ths s essentally a gauge transformaton, so F µ s not changed. Now the Lagrangan wll look lke L = 1 4 F µ + 1 ( µρ) + e ( Bµ B µ 1 + ρ ) + ρ V ( + ρ) (0) Now the feld ξ totally dsappears, and we have a massve vector feld B µ wth mass m B = e coupled to a massve scalar feld. If we count the degree of freedom we can see that the degree of freedom s = 4. Now f we had a massless vector and two massve scalars we wll have degree of freedom + = 4, so the degree of freedom matches. What happens here s that the Goldstone boson got eaten by the gauge boson. What we have done above s essentally a gauge transformaton. It s lke takng a gauge condton, but nstead of requrng µ µ = 0 we requre ϕ to be real. Ths s how we elmnated ξ and make the gauge boson massve. Ths s why we obtan a mass term whch volates gauge nvarance: because we have chosen a gauge. Now let s generalze the Yang-Mlls. ϕ l are real scalars wth vacuum expectaton value ϕ l = l, and we have a gauge group G wth acton on the ϕ s δϕ l = β a t a lm ϕ m (1) Note ϕ l forms a representaton of the gauge group and t mght be reducble. Smlar to the above we mpose a gauge condton ϕ l (t a lm m) = 0 () Ths s to requre that the feld s orthogonal to the VEV. Now the covarant dervatve s D µ ϕ l = µ ϕ l a µt a lm ϕ m = µ ϕ l a µt a lm ( m + ϕ m) (3) nd the square s D µ ϕ l = ( µ ϕ l ) + a µ b [ µ t a lm ( m + ϕ m) ] [ t b ln ( n + ϕ n)] ( µ ϕ l )ta lm ( m + ϕ m) a µ (4) 3

4 Now we use the gauge condton to smplfy the above expresson and we can extract the quadratc terms L = 1 4 F µ + 1 ( µϕ l ) 1 a µ b µ(t a lm m)(t b ln n) V (ϕ ) +... (5) If we wrte the mass term for the gauge feld as 1/ a µ b µ(µ ) ab then the mass matrx s (µ ) ab = (t a lm m)(t b ln n) (6) Suppose some lnear combnaton of generators c a t a s unbroken, then c a t a lm m = 0 and ths corresponds to a zero egenvector of ths mass matrx. Conversely f t has a zero egenvector µ ab c ac b = 0 then we know that c a t a lm m = 0 and ths corresponds to an unbroken symmetry. gan the countng of degree of freedom obvously works. Now we have a bunch of massless or massve vector bosons and some massve scalars wth no Goldstone bosons. Ths gauge choce s called untarty gauge. Ths s because all the partcles appearng n ths gauge are physcal partcles. If we work out the gauge propagator n ths gauge we wll get [ g µ k m + kµ k ] V m (7) V Note that the second term s problematc and we need to cancel them to get renormalzaton to work n loop dagrams. So ths gauge s not a very good choce f we work on renormalzablty. So let s work n R ξ gauge. For smplcty we work n U(1) theory. φ = 1 (φ 1 + φ ) = 1 ( + h(x) + ϕ(x)) (8) Remember we do perturbaton theory by add a gauge-fxng term to the Lagrangan and add a ghost Lagrangan. The gauge-fxng term for R ξ gauge s and the ghost Lagrangan wll be L ghost = c(δg)c = c [ µ µ ξe ( + h) ] c = c The quadratc terms n the full Lagrangan wll be G = 1 ξ ( µ µ eξϕ) (9) [ ξm (1 + h ) ] c (30) L = 1 ( µh) 1 m h h + 1 ( µϕ) 1 m ξϕ 1 4 F µ 1 ξ ( µ µ ) + 1 m µ + m µ µ ϕ + eϕ µ µ + c [ ξm ] (31) c Now the two terms n the second lne wll add up to be a total dervatve of the gauge fxng term and we can throw them away. The propagator for the vector boson s [ k m g µ k ] µk k ξm (1 ξ) (3) 4

5 The h propagator s just a usual massve propagator, so as ϕ and c. Now when we do perturbaton theory we have three propagators that depend on ξ, namely µ, ϕ, c. However ξ s an arbtrary varable and we expect that all the ξ factors should cancel n our loop calculatons, and ths s an advantage n our calculaton. Ths s called renormalzable gauge, because renormalzaton s easy n ths gauge. If we take the lmt ξ, the vector propagator goes to (g µ k m kµ k ) m (33) The propagators for ϕ and c go to zero, because they are unphyscal partcles. However the couplng between c and h goes to nfnty. nd we have a loop of ghosts and external h legs then the dagram wll gve a fnte contrbuton. So ths s lke untarty gauge wth some funny nteractons,.e. nteracton between arbtrary number of h bosons. These show up n the untarty gauge f we calculate the path ntegral and we wll get some δ 4 (0) terms whch correspond to these nteractons. 5