(a) Consider the two quarks to move circularly, much like the electron and proton in a hydrogen atom; we can write the force between them as
\mu \frac{\upsilon ^{2} }{r} =\frac{dV(r)}{dr} =k, (6.292)
where μ = m/2 is the reduced mass. From the Bohr quantization condition of the orbital angular momentum, we have
L=\mu \upsilon r=n\hbar , (6.293)
Multiplying (6.292) by (6.293), we end up with \mu ^{2} \upsilon ^{3} =n\hbar k which yields the speed of the relative motion of the two-quark system:
\upsilon _{n} =\left(\frac{n\hbar k}{\mu ^{2}} \right) ^{1/3} . (6.294)
The radius can be obtained from (6.293), r_{n} =n\hbar /(\mu \upsilon _{n} ); using (6.294) this leads to
r_{n} =\left(\frac{n^{2} \hbar ^{2} }{\mu k} \right) ^{1/3}. (6.295)
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{n^{2} \hbar ^{2}k^{2} }{\mu } \right) ^{1/3} . (6.296)
In deriving this we have used the relations for \upsilon _{n} and r_{n} as given by (6.294) and (6.295),
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\hbar } \left(\frac{k^{2}}{\mu \hbar } \right) ^{1/3} \left(n^{2/3} -m^{2/3} \right) . (6.297)
(b) The radial equation is given by (6.57):
-\frac{\hbar ^{2} }{2M} \frac{d^{2} U_{nl} (r)}{dr^{2} } + V_{eff}(r) U_{nl} (r)= E_{n} U_{nl} (r), (6.57)
-\frac{\hbar ^{2} }{2\mu } \frac{d^{2} U_{nl} (r)}{dr^{2} } +\left[kr+\frac{l(l+1)\hbar ^{2} }{2Mr^{2}} \right] U_{nl} (r)= E_{n} U_{nl} (r), (6.298)
where U_{nl} (r)=rR_{nl} (r) . Since we are dealing with l = 0, we have
-\frac{\hbar ^{2} }{2\mu } \frac{d^{2} U_{n0} (r)}{dr^{2} } +kr U_{n0} (r)= E_{n} U_{n0} (r) , (6.299)
which can be reduced to
\frac{d^{2} U_{n0} (r)}{dr^{2} }-\frac{2\mu k}{\hbar ^{2}} \left(r-\frac{E}{k} \right) U_{n0} (r)=0. (6.300)
Making the change of variable x=(2\mu k/\hbar ^{2})^{1/3} (r-E/k) , we can rewrite (6.300) as
\frac{d^{2}\phi _{n}(x) }{dx^{2} } -x\phi _{n}(x)=0 . (6.301)
We have already studied the solutions of this equation in Chapter 4; they are given by the Airy functions Ai(x):\phi (x)=BAi(x) .The bound state energies result from the zeros of Ai(x). The boundary conditions on U_{nl} of (6.301) are U_{nl}(r=0) =0 and U_{nl}(r\rightarrow +\infty )=0 . The
second condition is satisfied by the Airy functions, since Ai(x \rightarrow +\infty )=0. The first condition corresponds to \phi [-(2\mu k/\hbar ^{2})^{1/3} E/k]=0 or to Ai[-(2\mu k/\hbar ^{2})^{1/3} E/k]=Ai(R_{n} )=0, where R_{n} are the zeros of the Airy function.
The boundary condition U_{nl}(r=0)=0 then yields a discrete set of energy levels which can be expressed in terms of the Airy roots as follows:
Ai\left[-\left(\frac{2\mu k}{\hbar ^{2}} \right) ^{1/3} \frac{E}{k} \right] =0\Longrightarrow -\left(\frac{2\mu k}{\hbar ^{2}} \right) ^{1/3} \frac{ E_{n}}{k} = R_{n}; (6.302)
hence
E_{n}=-\left(\frac{\hbar ^{2}k ^{2}}{2\mu } \right) ^{1/3} R_{n}. (6.303)
The radial function of the system is given by R_{n0} (r)=(1/r) U_{n0} (r)=(B_{n}/r)Ai(x) or
R_{n0} (r)=\frac{B_{n}}{r} Ai(x)=\frac{B_{n}}{r} Ai\left[\left (\frac{2\mu k}{\hbar ^{2}} \right) ^{1/3} r+R_{n} \right] . (6.204)
The energy expression (6.303) has the same structure as the energy (6.296) derived from the Bohr model E^{B}_{n} =\frac{3}{2} (n^{2}\hbar ^{2}k^{2}/\mu )^{1/3} ; the ratio of the two expressions is
\frac{E_{n}}{E^{B}_{n}} =-\frac{2}{3}\frac{R_{n}}{(2n^{2} )^{1/3} } . (6.305)
(c) In the following calculations we will be using k=15Ge V fm^{-1},\mu c^{2}=m/2=2.2GeV and \hbar c=197.3 Me Vfm. The values of the four lowest energy levels corresponding to the expression E^{B}_{n} =\frac{3}{2} (n^{2}\hbar ^{2} k^{2}/\mu )^{1/3} , derived from the Bohr model, are
E^{B}_{1} =\frac{3}{2}\left(\frac{\hbar ^{2}k^{2}}{\mu } \right) ^{1/3} =2.38 GeV , E^{B}_{2} =2^{2/3} E^{B}_{1} =3.77 GeV , (6.306)
E^{B}_{3} =3^{2/3} E^{B}_{1} =4.95 GeV , E^{B}_{4} =4^{2/3} E^{B}_{1} =5.99 GeV . (6.307)
Let us now calculate the exact energy levels. As mentioned in Chapter 4, the first few roots of the Airy function are given by R_{1}=-2.338, R_{2}=-4.088, R_{3}=-5.521, R_{4}=-6.787, so we can immediately obtain the first few energy levels:
E_{1}=\left(\frac{\hbar ^{2}k^{2}}{2\mu } \right) ^{1/3} R_{1} =2.94GeV, E_{2}=\left(\frac{\hbar ^{2} k^{2}}{2\mu } \right) ^{1/3}R_{2}=5.14GeV, (6.308)
E_{3}=\left(\frac{\hbar ^{2}k^{2}}{2\mu } \right) ^{1/3}R_{3} =6.95GeV, E_{4}=\left(\frac{\hbar ^{2}k^{2}}{2\mu } \right) ^{1/3}R_{4}=8.54GeV. (6.309)