Solving the Puzzle of ${M_{D^*}-M_{D}\over M_{D_s^*}-M_{D_s}}$ $\simeq$ ${M_{B^*}-M_{B}\over M_{B_s^*}-M_{B_s}}$ $\simeq$ $1$

Details Solving the Puzzle of ${M_{D^*}-M_{D}\over M_{D_s^*}-M_{D_s}}$ $\simeq$ ${M_{B^*}-M_{B}\over M_{B_s^*}-M_{B_s}}$ $\simeq$ $1$ Solving the Puzzle of ${M_{D^*}-M_{D}\over M_{D_s^*}-M_{D_s}}$ $\simeq$ ${M_{B^*}-M_{B}\over M_{B_s^*}-M_{B_s}}$ $\simeq$ $1$

1997-01-15

arxiv.org/abs/hep-ph/9701295v1

Physics

High Energy Physics

High Energy Physics - Phenomenology

10 pages, Latex, 5 figures

Scientific paper

The commonly used Hamiltonian of the chromomagnetic hyperfine splitting is inversely proportional to the product of the masses of two constituent quarks composing the meson. So it is expected to have $(M_{D^*}-M_{D})/(M_{D_s^*}-M_{D_s})$ $\simeq$ $(M_{B^*}-M_{B})/(M_{B_s^*}-M_{B_s})$ $\simeq$ $1.6$, when the constituent quark masses $m_{u,d}=0.33$ GeV and $m_s=0.53$ GeV are used. However, the experimental results show that the above ratios are very close to 1. We solve this puzzle by employing the Hamiltonian recently proposed by Scora and Isgur.

Affiliated with

Also associated with

No associations

Rating

Solving the Puzzle of ${M_{D^*}-M_{D}\over M_{D_s^*}-M_{D_s}}$ $\simeq$ ${M_{B^*}-M_{B}\over M_{B_s^*}-M_{B_s}}$ $\simeq$ $1$ does not yet have a rating. At this time, there are no reviews or comments for this scientific paper.

If you have personal experience with Solving the Puzzle of ${M_{D^*}-M_{D}\over M_{D_s^*}-M_{D_s}}$ $\simeq$ ${M_{B^*}-M_{B}\over M_{B_s^*}-M_{B_s}}$ $\simeq$ $1$, we encourage you to share that experience with our LandOfFree.com community. Your opinion is very important and Solving the Puzzle of ${M_{D^*}-M_{D}\over M_{D_s^*}-M_{D_s}}$ $\simeq$ ${M_{B^*}-M_{B}\over M_{B_s^*}-M_{B_s}}$ $\simeq$ $1$ will most certainly appreciate the feedback.

Rate now

Comments { 0 }

Profile ID: LFWR-SCP-O-119576