Physics 742, Standard Model: Homework #8 Solution

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1 Physics 74, Standard Model: Homework #8 Solution Quark masses The isospin-violating ratio o quark masses is given as, Using Eq. (8.63) in the textbook, m u r m d. () m d + m u m =(m ± u + m d ) c, () m K =(m ± u + m s ) c, (3) m K =(m s + m d ) c. (4) Eq. () can be rewritten as, under the lowest order predictions o chiral perturbation theory, r chiral = m d + m s m s m u m d + m u = m K m K ± m ±. (5) Inserting the PDG values, m K = 497.6MeV, m K ± = 493.7MeV, m ± = 39.6MeV, we get, r chiral =.. (6) On the other hand, i we directly use quark mass by the PDG, m d =4.8MeV, m u =.3MeV, we get, The relative error between the two results is, which shows poor agreement as expected. (b) Given that r PDG =.35. (7) r PDG r chiral r chiral = 4.86%, (8) =e i a T a / e i /, (9)

2 where and = p p + p 6 + K + K p + p 6 K K C q A () 3, /3 Q = e /3 C A, () /3 we expand the e ective operator tr( Q Q) to second order in the meson ield, The second order terms in are " " Q Q = + i + i + # Q " i + i + # Q. () i Q Q + i i Q Q + # i Q = Q Q +4 Q Q Q. (3) By taking trace o the results above, we get, 4tr[ Q Q Q ] / + + K K +, (4) where in the last step, we have done explicit matrix manipulation and trice using and Q. This tells us that only K ± and ± receive an additional contribution to their masses and that the size o this contribution is same or them: m K = m ± K + ±, (5) m = m ± + ±. (6) The additional contribution can be easily determined by, =m ± m = 63MeV. (7) (c) Note that m K ± m ± = m K ± + m ± =m k ± m ± / (m s + m u ) (m d + m u )=m s m d. (8) Using the above results, we can rewrite the isospin-violating ratio o quark masses as

3 r cor = m d + m s m s + m d m d m u m d + m u Substituting corresponding numerical values, we obtain, = m K + m ± m K ± m m (9) r cor =.8. () The relative error now becomes roughly %, which shows better agreement than that in (a). K! e consider the process shown in Diagram A. The amplitude o this process is given by im = vū(e µ g µ P L V us )v s m ū u (e P L V ud )v d, () where we neglected momentum o +. Contracting indices above and identiying the two actors o interest (see equation in Georgi) : F h J µ L K+ i vū µ P L v s, () F h + J Lµ i ū µ µ P L v d, (3) the above amplitude can be rewritten in terms o the actors above, im = 4e V usv ud m F F. (4) In order to compute F, F, let us consider the current J L in equation in Georgi. For F, we only need the irst term in the ormula, By explicit matrix manipulation, the ollowing holds true: rom which, F is given as tr[(t it ) ] = + p, (5) For F, the second term in is relevant, Similarly one can show that F = h + (J Lµ ij Lµ) i = p (p +) µ (6) J µa L = itr(t a [,@ µ ]) (7) tr[(t 4 it 5 )[,@ µ ]] = p [(@µ K + ) K + (@ µ )] +, (8) where we omitted terms that will vanish when acting on the bra and ket o F.ehave 3

4 F = p h µ K + K µ ) K + i i = p (p K + + p )µ. (9) Putting all components together, the amplitude becomes im = Momentum conservation, p K + = p + + p Thereore the matrix element becomes and its square is ie V usv ud m p + (p K + + p ). (3) implies that p + p = p K + p + p = m K ± m ± m, (3) p + (p K + + p )=p + +p + p im = = m ± + m K ± m ± m = m K ± m. (3) ie V usv ud m ( m K m ± ), (33) M = e 4 m 4 V us V ud ( m K m ± ). (34) Note that there is no angular dependence in the above result and the associated decay width takes a particularly simple orm Using the ollowing kinematic conditions, ~p = = M 8 and putting corresponding numerical values, we inally get ~p m K ± (35) q E m, (36) E = m K m ± m ±, (37) m K ± =.8 4 MeV. (38) Now we want to compare the result to the PDG value. The lietime o K ± and the branching ratio or K +! + are, respectively, The partial width or this process is given by =.38 9 s, (39) BR(K +! + ) =.7%. (4) PDG = tot BR(K +! + ) = ~ BR(K+! + ) =. 4 MeV. (4) The PDG value and our result show airly good agreement with the relative error o 5%. 4

5 3 K! e (a) Let us veriy equation in Georgi. First one need to explicitly expand equation with T a = T 4 + it 5. Setting to zero all unnecessary ields in the matrix, There are only three surviving terms and the current J µ becoms J µ 3 p h + (K µ µ + + µ K ) K i = 3 p ( p +p + + p K ) µ. (4) To check the current conservation, q µ J µ = 3 p (p + + p p K ) µ ( p +p + + p K ) µ = 3 p (p + p p K +p K p p K p + + p p +). (43) In chiral limit where mesons are massless, this reduces to q µ J µ = 3 p (p K p p K p + + p p +) 6=. (44) (b) The kinetic term in the Lagrangian that we consider is L k = 4 tr[(@ µ )(@ µ )] (45) By expanding this term in the meson ield, we have ( L k = "@ 4 tr i µ + i + i ) 6 µ + i + i + i )# 6 (46) For the 4-meson interaction, we need to keep terms o order O( 4 ) in the expansion : " L (4) k = 4 tr 6 i + 6 = 4 apple tr = 4 tr i 3 (@ µ )(@ µ 3 )+ 4 i 3 # i (@ m u 3 )(@ µ ) 4 (@ µ )(@ µ ) apple 4 (@ ) i i (@ µ )(@ µ ) 3 (@ µ )(@ µ 3 ) 3 (@ ) 3. (47) In the last step, we used Eq. () in the problem set. As suggested in the problem, we will simpliy by setting all non-relevant meson ield to zero:! p + K + C A (48) K 5

6 The two terms in the last step o Eq. (47) can be explicitly expressed as tr[ (@ )] = [K ( + )+K (K + )], (49) tr[(@ ) 3 ]= 4 [K+ + (@ K )+K + (@ K + ) + K K + + (@ )+K + K (@ + )]. (5) Each term can be urther rewrite as, using integration by parts, tr[ (@ )] = K K + (@ ) + + (@ µ )(@ µ + )+ (@ + ) + K + (@ K ) + + (@ µ K )(@ µ + )+K (@ + ) (5) tr[(@ ) 3 ]= 4 K+ K (@ + )+K + (@ )+ + (@ K (K + )K + (K + K + ) = K+ K (@ µ )(@ µ + )+ (@ µ K )(@ µ + )+ + (@ µ K )(@ µ ). (5) In the chiral limit, these result in h + tr[ (@ )] K i = (p +p p + + p + + p K p K p + + p + ) = p p + p K p +, (53) h + tr[(@ ) 3 ] K i = (p p + p K p + p K p ). (54) Now we will use the above results to compute the current associated to Diagram B, where the incoming meson ield is shown in Diagram C. This current is given by p J µ q µ = q h + L (4) k K i, (55) where q is the momentum o K. Here, the irst bracket is the contribution o Diagram C, and the second is that o Diagram B. The overall sign is ixed in a way that the total current vanishes. Using the our results or L (4) k, q µj µ can be expressed as q µ J µ = 3 p (p K p + p p + p K p ). (56) This is the exactly same results in (a) but with the opposite sign. Thereore, by adding both contributions, the current q µ J µ is conserved in the chiral limit, q µ J µ =. (57) 6

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