QCD lecture 04 - Magnetic momenta of in model

Proton and magnetic momenta for symmetric

1 Symmetric wave function of the could guessed. As three with 2 mix 1 into with spin 2 either as doublet with mixed in the first two indeces or mixed : 2 ⊗ 2 ⊗ 2 = 4S ⊕ 2m,s ⊕ 2m,a. Clearly to get symmetric function for |uudi 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 |p, ↑i = √ |uudi |(↑↓ + ↓↑) ↑ −2 ↑↑↓i + cyclic permutations . 18

Assigning of quark q in state |q ↑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 1 with spin 2 : Qqe µq = . 2mq

and expressing in terms of nuclear magneton µN = e/(2mp). Assuming the same for u and d , i.e. µN = e/(6mu), we get: e  2 1 µp = 4 + = 3µN . 6mu 3 3

Neutron magnetic momentum can be quickly obtained using 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 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 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 |uudi has to be combined with 2m,a: 1 h i |p, ↑i = √ |uudi |(↑↓ − ↓↑) ↑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 , ↑ = √ |uusi |(↑↓ + ↓↑) ↑ −2 ↑↑↓i + cyclic permutations 18 0 + by using lower operator E−α3 as Σ is in the same isospin 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 , ↑ = √ |(ud + du)si |(↑↓ + ↓↑) ↑ −2 ↑↑↓i + cyclic permutations . 36 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 , ↑ = √ |(ud − du)si |(↑↓ − ↓↑) ↑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 |Λ0i and |Σ0i states. The ϑ = −0.014 ± 0.004 rad is small and we can write:

0 0 0 0 0 0 Λ = Λ + ϑ Σ , Σ = −ϑ Λ + Σ .