Cornell University, Department of Physics May 2, 207 PHYS 4444, Particle physics, HW # 9, due: 4/3/207, :40 AM Question : Making invariant Consider a theory where the symmetry is SU(3) SU(2) U() and we have the following complex scalars Q(3, 2) x, R(T, 3) 2, H( 3, 2). (). What should be the values of x and T such that we can write a term of the form QRH. Give all possible values. Answer: First, the U() charge should add up to zero, so x = 2 = 3. Second, for SU(3) we note that 3 3 = 8 +. If we combine Q and H into a, we need T =. If we combine Q and H into an 8, we need T = 8 = 8 since the adjoint representation is real. 2. Repeat the above now looking for a term of the form QRH. Answer: In this case, we need x = 2 ( ) =. For SU(3), we note that 3 3 = 3 + 6. If we combine Q and H into a 3, then we need T = 3. If we combine them into a 6, then we need T = 6. Question 2: Gluon self interaction Using the definition F µν a = µ A ν a ν A µ a gf abc A µ b Aν c. (2) show that the gluon self interaction is given by L = gf abc ( µ A aν )A µ b Aν c 4 g2 (f abc A µ b Aν c )(f ade A dµ A eν ). (3)
Answer: The gluon kinetic term is given by 4 F a µν F a,µν = 4 ( µ A ν a ν A µ a gf abc A µ b Aν c ) ( µ A aν ν A aµ gf ade A dµ A eν ) = 4 ( µ A ν a ν A µ a) ( µ A aν ν A aµ ) 4 g2 f abc f ade A µ b Aν c A dµ A eν 4 [ gf adea dµ A eν µ A ν a + gf ade A dµ A eν ν A µ a gf abc A µ b Aν c µ A aν + gf abc A µ b Aν c ν A aµ ] (4) = 4 ( µ A ν a ν A µ a) ( µ A aν ν A aµ ) 4 g2 f abc f ade A µ b Aν c A d,µ A e,ν + gf abc A µ b Aν c µ A a,ν For the three-point interactions, I have used the antisymmetry of f abc and freedom to relabel colour and Lorentz indices to combine the four terms into one. Question 3: β function In class we discussed the idea that coupling constants run, that is, that higher order corrections can be absorbed into the definition of the coupling constant. For example, consider the cross section for e + e µ + µ. In the CM frame and to leading order, this is given by (up to normalization factors) σ = α2 E 4, (5) where E is the energy of the electron. Higher order effects change this result. Most of the effect can be absorbed into the running of α, that is σ = α(µ)2 E 4, (6) where µ E is the energy scale in the problem, and α(µ) is a running coupling constant that satifies the differential equation: α log(µ) = β(α), (7) where the beta function can be calculated to the desired precision in perturbation theory. In QED at one loop with only electrons in the loop we have. Verify that the solution of the beta function equation is β(α) = Bα 2, B = 2 3π. (8) α(µ ) = α(µ 2 ) + B log 2 ( µ2 µ ). (9)
Answer: In our case we have a diffrential equation of the form Bα 2 = α t with t log µ. The solution for this is α(t) = () Bt + C where C is a constant that need to be determined from the initial conditions. Knowing α(t) at one point we then get C = B log µ + α(µ) Since we know that C is the same for any µ we conclude that And thus B log µ + α(µ ) = B log µ 2 + α(µ 2 ) α(µ ) = α(µ 2 ) + B log ( µ2 µ (0) (2) (3) ). (4) 2. Use α(m e ) /37.0 to calculate α(m Z ) where m Z 9 GeV. Answer: α(m Z ) 37.0 + 2 ( ) 3π log me 34.5 (5) m Z 3. Find the Landau pole, that is, find µ where α(µ). What can you say about this infinity? Answer: We have it when ( ) µ α(m e ) = B log m e µ = m e exp[/(αb)] m e 0 280 0 280 MeV (6) And we see that this number is much above the Planck scale, so above the UV cutoff of QED, so it is not important. Measurements found that α(m Z ) /28. The reason for the disagreement with your result above is that there are other particles in the loop beside the electrons. The generalization of eq. (9) is α(µ ) = α(µ 2 ) + B ( ) Q 2 i NC i mi log. (7) µ i where the sum over i is the sum of all the charged particles with mass below µ, i.e. we assume µ 2 {m i } < µ. N i C = (3) for leptons (quarks), and Q i is the electric charge. 3
4. Give a physical argument why we only sum over particles with mass less than µ. Answer: Because only at distances smaller than /m are pair production important. 5. Use the physical masses and charges of the known fermions q = : m l (0.5, 00, 777) MeV, q = 2/3 : m u (0.3,.4, 74) GeV, q = /3 : m d (0.3, 0.4, 4.2) GeV, (8) and calculate α(m Z ). How close is your number to the measured value? (Note that m u, m d and m s are much larger than what you find in PDG, because the correct quark masses to use here are valence quark masses, which is not what the PDG gives.) Answer: Numerically, we get /α(m Z ) 28.3 clearly very good compared to the measured value. The small deviation may be due to higher order effects. We now move to QCD where the beta function is more complicated ( B = 2n ) f 3 2π (9) where n f is the number of quark flavors with masses below the relevant scale. Below we use the input α s (m Z ) 0.2 6. Note that the sign of the beta function changes sign depending on the number of flavors. How many flavors are needed to change the sign of the beta function? Answer: n f = 33/2, that is the sign change when the number of flavor move from 6 to 7. 7. Sketch the shape of the function α s (µ) for µ between and 0 4 GeV for (i) a theory with fewer flavors, and (ii) a theory with more flavors than this critical value. Use log scale for µ. Answer: For simplicity, we assume all flavours to be lighter than GeV so we don t have to worry about threshold effects. We then get the following plots: 8. Estimate Λ QCD, that is, the scale where α s =. For simplicity, you can neglect all quark masses but that of the top, that is, you can use n f = 5 in Eq. (9). Answer: We need to use n f = 5, since the top is heavier than our starting point, m Z, we have B = 23/(6π).22 and we get B log(λ QCD /m Z ) /α s (m Z ) (20) 4
and thus ( ) α s Λ QCD m Z exp m Z e 6.0 223 MeV. (2) B 9. The proton mass is very roughly m P 3Λ QCD. Can you tell if the mass of the proton would be lighter or heavier if we did not have the third generation, assuming the same measured value of α s (m Z )? Answer: With one less flavor B would be larger and thus Λ QCD would be larger, and thus the proton will be heavier. 5
Αs 0.35 Αs 0.6 n f =5 0.30 n f =25 0.4 0.25 0.2 0.20 0.0 0.5 5 0 50 00 Μ 500 000 6 5 0 50 00 Μ 500 000