For the π(spin 0) and the K^* (spin 1), show that (2.91) gives m(π) =mu+md-(3 a / mu md)

For the π(spin 0) and the K^* (spin 1), show that (2.91)...

For the $\pi$ (spin 0) and the $\mathrm{K}^*$ (spin 1), show that (2.91) gives $$ \begin{aligned} m(\pi) & =m_u+m_d-\left(3 a / m_u m_d\right) \\ m\left(\mathrm{~K}^*\right) & =m_u+m_s+\left(a / m_u m_s\right) . \end{aligned} $$ Calculate the masses of all the members of the $0^{-}$and $1^{-}$meson multiplets (Fig. 2.12) using $$ m_u=m_d=0.31, \quad m_s=0.48, \quad m_c=1.65, \quad a / m_u^2=0.16, $$ all in units of GeV. Compare your predictions with the meson masses listed in the particle data tables. Check that $$ (\rho-\pi)=\frac{m_s}{m_u}\left(\mathrm{~K}^*-\mathrm{K}\right)=\frac{m_c}{m_u}\left(\mathrm{D}^*-\mathrm{D}\right)=\frac{m_c m_s}{m_u^2}\left(\mathrm{~F}^*-\mathrm{F}\right), $$ where the meson names are used to denote their masses.

For the $\pi$ (spin 0) and the $\mathrm{K}^*$ (spin 1), show that (2.91) gives
$$
\begin{aligned}
m(\pi) & =m_u+m_d-\left(3 a / m_u m_d\right) \\
m\left(\mathrm{~K}^*\right) & =m_u+m_s+\left(a / m_u m_s\right) .
\end{aligned}
$$
Calculate the masses of all the members of the $0^{-}$and $1^{-}$meson multiplets (Fig. 2.12) using
$$
m_u=m_d=0.31, \quad m_s=0.48, \quad m_c=1.65, \quad a / m_u^2=0.16,
$$
all in units of GeV. Compare your predictions with the meson masses listed in the particle data tables.
Check that
$$
(\rho-\pi)=\frac{m_s}{m_u}\left(\mathrm{~K}^*-\mathrm{K}\right)=\frac{m_c}{m_u}\left(\mathrm{D}^*-\mathrm{D}\right)=\frac{m_c m_s}{m_u^2}\left(\mathrm{~F}^*-\mathrm{F}\right),
$$
where the meson names are used to denote their masses.

Key Concepts

Constituent Quark Model

This model treats hadrons as bound states of constituent quarks, which have effective masses that account for both the bare quark mass and the effects of strong interactions at low energies. The masses of mesons and baryons can be approximated as the sum of the constituent quark masses plus additional corrections from interactions, making it a useful framework for understanding hadron spectroscopy.

Hyperfine Interaction

The hyperfine interaction arises from the spin-spin coupling between constituent quarks and is analogous to the magnetic hyperfine splitting in atomic physics. In the context of mesons, the difference in mass between states having different spin configurations (for example, spin 0 vs spin 1) is explained by this interaction, which typically depends inversely on the product of the constituent quark masses. This mechanism helps account for the fine differences observed in meson mass spectra.

Meson Multiplet Structure

Mesons are classified according to their quantum numbers, such as total spin and parity, forming multiplets like the 0? (pseudoscalar) and 1? (vector) groups. The mass splitting within these multiplets can be understood through the interplay between the quark masses and their spin-dependent interactions. Establishing the relationships between masses in different multiplets or flavor combinations provides insights into the dynamics governing hadronic structures.

Flavor Dependence of Hyperfine Splitting

The strength of the hyperfine interaction in mesons depends on the masses of the constituent quarks, leading to a flavor-dependent splitting. By comparing mass differences for mesons with different quark compositions, one can extract ratios that reflect how the spin-spin interaction scales with the quark masses. This concept is crucial when testing the predictions of the quark model and verifying the relationships between the mass differences in various meson families.

Transcript

00:01 In this question we are given 9 variants which are spin 3 over 2 particles and we want to actually label them or draw them on a graph based on their strangeness value as well as they are charged.

00:24 So we have assuming we have a strangeness as our y -axis.

00:33 And charge to be our x -axis.

00:38 So i've drawn it like that because strangeness, our strangeness value range from 0 to minus 2.

00:46 So we have, here is the zero strangeness particles and then we have the charge particles.

00:53 We have zero charge, we have minus e, we have plus e, and finally we have plus 2e.

01:11 So for the first row, right, for the particles with zero strangeness, this is actually the lambda barion.

01:24 Right, so we are given that the lambda barion has four particles and they all have zero strangeness, but they have charges of plus 2e, plus e, 0 and minus e.

01:35 So they exist over here, here, here and here.

01:45 So there's four different particles and they all have mass of 1385 mv per c squared.

01:57 So just to write down what are the quark contents.

02:01 This is up up up, this is up, up down, up down, and down, down, down, down.

02:11 Now, for the next set of particles, that is the sigma barions.

02:24 Sorry, this is not 1385, right? this is 1 ,232, right? the 1385 is the sigma barions, actually.

02:38 And for these barions, they exist on the strangeness equals to minus 1 plane.

02:50 So over here where we have strangeness of minus 1...

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