QCD lecture 04 - Magnetic momenta of baryons in quark model Proton and neutron magnetic momenta for symmetric wave function 1 Symmetric wave function of the proton could guessed. As three particles with spin 2 mix 1 into particle with spin 2 either as doublet with mixed symmetry in the first two indeces or mixed asymmetry: 2 ⊗ 2 ⊗ 2 = 4S ⊕ 2m;s ⊕ 2m;a: Clearly to get symmetric function for juudi component of the proton, it must be cou- pled only to 2m;s. The full symmetric proton wave fuction can be obtain then by cyclic permutations: 1 h i jp; "i = p juudi j("# + #") " −2 ""#i + cyclic permutations : 18 Assigning magnetic moment of quark q in state jq "i value µq we can write for proton magnetic moment: 3 4µ − µ µ = [(µ − µ + µ ) + (−µ + µ + µ ) + 4(µ + µ − µ )] = u d : p 18 u u d u u d u u d 3 This can be further evolved using relation between spin and magnetic moment of fermion 1 with spin 2 : Qqe µq = : 2mq and expressing in terms of nuclear magneton µN = e=(2mp). Assuming the same masses for u and d quarks, i.e. µN = e=(6mu), we get: e 2 1 µp = 4 + = 3µN : 6mu 3 3 Neutron magnetic momentum can be quickly obtained using isospin symmetru (u d): 4µ − µ 1 2 µ = d u = −4 − µ = −2µ : n 3 3 3 N N The prediction of this simple additive quark model is quite close to the measured values of proton µp µp = 2:793 : : : µN ; µn = −1:913 : : : µN ; ) = 1:460 :::: µn Ratio of proton and neutron magnetic momenta is described even better: µ 4µ − µ 4Q − Q 3 p = u d = u d = − : µn 4µd − µu 4Qd − Qu 2 From measured values of proton and neutron magnetic momenta we can extract mag- netic momenta of quarks u and d: 4µ + µ 4µ + µ µ = p n = 1:852µ ; µ = n p = −0:972µ : u 5 N d 5 N From the ratio of quark magnetic momenta we can extract the difference in mass between quarks: µ m m u = −2 d ; ) d = 0:952: µd mu mu Proton and neutron magnetic momenta for antisymmetric wave function In analogy with symmetric case, we can guess the anti-symetric proton wave function. In this case the juudi has to be combined with 2m;a: 1 h i jp; "i = p juudi j("# − #") "i + cyclic permutations : 6 Magnetic moment is then 3 µ = [µ − µ + µ − µ + µ + µ ] = µ : p 6 u u d u u d d Neutron is than µn = µu: And the ration is µ µ 1 p = d = − : µn µu 2 Magnetic momenta of other baryons 1 Magnetic momenta of other baryons from basic spin 2 octet could be easily guessed from the proton moment for all states except Λ0 and Σ0. We can guess the wave function of Σ0 from the wave function of Σ+ + 1 h i Σ ; " = p juusi j("# + #") " −2 ""#i + cyclic permutations 18 0 + by using lower operator E−α3 as Σ is in the same isospin multiplet as Σ . In such state, we will get symmetric mixture of quarks u and d and such symmetric state should be combined again with spin 2m;s: 0 1 h i Σ ; " = p j(ud + du)si j("# + #") " −2 ""#i + cyclic permutations : 36 Baryon Λ0 will contain than the asymmetric combination of ud and du states. This should be mixed with 2m;a to get fully symmetric wave function: 0 1 h i Λ ; " = p j(ud − du)si j("# − #") "i + cyclic permutations : 12 And for magnetic moment of Λ0 we have 3 md µΛ0 = (4µs) = µs = µd : 12 ms Plugging in measured value of magnetic moment µΛ0 = −0:613±0:004µN we get for strange mass quark: µd ms = md = 1:59md; or ms = 1:51mu: µΛ0 We have now all parameters of the model fully constrained. We can then check the prediction for magnetic momentum of Σ+. Replacing quark d with s in the formula for proton moment we get 4µu − µs µ + = = 2:674µ ; Σ 3 N which is in good agreement with measured value of µΣ+ = 2:458 ± 0:010µN . Magnetic moment of Σ0 is then 3 · 2 2µu + 2µd − µs µ 0 = (µ − µ + µ − µ + µ + µ + 4µ + 4µ − 4µ ) = ; Σ 36 u d s u d s u d s 3 numerically 3:704 − 1:944 + 0:613 2:373 µ 0 = µ = µ = 0:791µ : Σ 3 N 3 N N However, this quantity is not directly measurable. PDG, gives Σ0 ! Λ transition moment. In reality, the observed Λ0 and Σ0 particles are mixture of the above jΛ0i and jΣ0i states. The mixing angle # = −0:014 ± 0:004 rad is small and we can write: 0 0 0 0 0 0 Λ = Λ + # Σ ; Σ = −# Λ + Σ :.