Question 1.10.18: Problem 1.18 Consider the bound state of two quarks having t...

(a) Consider the two quarks to move circularly, much like the electron and proton in a hydrogen atom; then we can write the force between them as

\mu \frac{\upsilon ^2}{r}=\frac{dV(r)}{dr}=k,              (1.240 )

where \mu =m/2  is the reduced mass and V(r) is the potential. From the Bohr quantization condition of the orbital angular momentum, we have

L=\mu \upsilon r=n\hbar .           (1.241)

Multiplying (1.240) by (1.241), we end up with \mu ^2\upsilon ^3=n\hbar k,  which yields the (quantized) speed of the relative motion for the two-quark system:
\upsilon _n=\left(\frac{\hbar k}{\mu ^2} \right) ^{1/3}n^{1/3}.              (1.242)

The radius can be obtained from (1.241), r_n=n\hbar /(\mu \upsilon _n);  using (1.242), this leads to

r _n=\left(\frac{\hbar ^2}{\mu k} \right) ^{1/3}n^{2/3}.              (1.243)

We can obtain the total energy of the relative motion by adding the kinetic and potential energies:

E _n=\frac{1}{2}\mu \upsilon ^2_n+kr_n=\frac{3}{2} \left(\frac{\hbar ^2k^2}{\mu } \right) ^{1/3}n^{2/3}.             (1.244)

In deriving this relation, we have used the relations for \upsilon_n  and r_n  as given by (1.242) by (1.243), respectively.

The angular frequency of the radiation generated by a transition from n to m is given by

\omega _{nm}=\frac{E_n-E_m}{\hbar}=\frac{3}{2} \left(\frac{k^2}{\mu \hbar} \right) ^{1/3} \left(n^{2/3}-m^{2/3}\right).               (1.245)

(b) Inserting n = 1, \hbar c\simeq 0.197  GeV  fm, \mu c^2=mc^2/2=1  GeV,  and  k=0.5  GeV  fm^{-1}

into (1.242) to (1.244), we have

\upsilon _1=\left(\frac{\hbar ck}{(\mu c^2)^2} \right)^{1/3}c\simeq \left(\frac{0.197  GeV  fm \times 0.5  GeV  fm^{-1}}{(1  GeV)^2} \right)^{1/3} c=0.46c,                   (1.246)

where c is the speed of light and

r _1=\left(\frac{(\hbar c)^2}{\mu c^2k} \right)^{1/3}\simeq \left(\frac{(0.197  GeV  fm )^2}{1  GeV \times 0.5  GeV  fm^{-1}} \right)^{1/3} =0.427  fm,                  (1.247)