Question 7.P.6: Work out the coupling of the isospins of a pion–nucleon syst...

Since the isospin of a pion meson is 1 and that of a nucleon is \frac {1}{2}, the total isospin of a pion– nucleon system can be obtained by coupling the isospins t_{1} =1 and t_{2} =\frac{1}{2}. The various values of the total isospin lie in the range \left|t_{1} -t_{2} \right| <T<t_{1} +t_{2} ; hence they are given by T=\frac{3}{2},\frac{1}{2}.

The coupling of the isospins t_{1} =1 and t_{2} = \frac{1}{2} is analogous to the addition of an orbital angular momentum l = 1 and a spin \frac {1}{2}; the expressions pertaining to this coupling are listed in (7.206) to (7.211).

\left|\frac{3}{2},\frac{3}{2}\right\rangle =\left|1,\frac{1}{2};1, \frac{1}{2}\right\rangle ,                          (7.206)

\left|\frac{3}{2},\frac{1}{2}\right\rangle =\sqrt{\frac{2}{3} } \left|1,\frac{1}{2};0,\frac{1}{2}\right\rangle +\frac{1}{\sqrt{3} } \left |1,\frac{1}{2};1,-\frac{1}{2}\right\rangle ,                        (7.207)

\left|\frac{3}{2},-\frac{1}{2}\right\rangle =\frac{1}{\sqrt{3} }\left|1,\frac{1}{2};-1,\frac{1}{2}\right\rangle + \sqrt{\frac{2}{3} }\left |1,\frac{1}{2};0,-\frac{1}{2}\right\rangle ,                        (7.208)

\left|\frac{3}{2},-\frac{3}{2}\right\rangle =\left|1,\frac{1}{2}; -1,-\frac{1}{2}\right\rangle .                                   (7.209)

\left|\frac{1}{2},\frac{1}{2}\right\rangle =\sqrt{\frac{2}{3} } \left|1,\frac{1}{2};1,-\frac{1}{2}\right\rangle -\frac{1}{\sqrt{3} } \left |1,\frac{1}{2};0,\frac{1}{2}\right\rangle ,                      (7.210)

\left|\frac{1}{2},-\frac{1}{2}\right\rangle =\frac{1}{\sqrt{3} } \left|1,\frac{1}{2};0,-\frac{1}{2}\right\rangle – \sqrt{\frac{2}{3} }\left |1,\frac{1}{2};-1,\frac{1}{2}\right\rangle .                      (7.211)

Note that there are three different π-mesons:

|1,1 〉=|\pi ^{+} 〉,           |1,0 〉=|\pi ^{0} 〉,           |1,-1 〉=|\pi ^{-} 〉,                   (7.388)

and two nucleons, a proton and a neutron:

\left|\frac{1}{2},\frac{1}{2}\right\rangle=|p 〉,         \left|\frac{1}{2},-\frac{1}{2}\right\rangle =|n 〉.                          (7.289)

By analogy with (7.206) to (7.211) we can write the states corresponding to T=\frac{3}{2} as

\left|\frac{3}{2},\frac{3}{2}\right\rangle=|1,1 〉\left|\frac{1}{2}, \frac{1}{2}\right\rangle=|\pi ^{+} 〉|p 〉 ,                        (7.390)

\left|\frac{3}{2},\frac{1}{2}\right\rangle =\sqrt{\frac{2}{3} } |1,0 〉\left|\frac{1}{2},\frac{1}{2}\right\rangle+\frac{1}{\sqrt{3} }|1,1 〉\left|\frac{1}{2},-\frac{1}{2}\right\rangle=\sqrt{\frac{2}{3} }|\pi ^{0} 〉|p 〉+\frac{1}{\sqrt{3} }|\pi ^{+} 〉 |n 〉 ,      (7.391)

\left|\frac{3}{2},-\frac{1}{2}\right\rangle =\frac{1}{\sqrt{3} } |1,-1 〉\left|\frac{1}{2},\frac{1}{2}\right\rangle+\sqrt{\frac{2}{3} }|1,0 〉\left|\frac{1}{2},-\frac{1}{2}\right\rangle=\frac{1}{\sqrt{3} }|\pi ^{-} 〉|p 〉+\sqrt{\frac{2}{3} }|\pi ^{0} 〉|n 〉 ,      (7.392)

\left|\frac{3}{2},-\frac{3}{2}\right\rangle=|1,-1 〉\left|\frac{1}{2},-\frac{1}{2}\right\rangle=|\pi ^{-} 〉|n 〉 ,                  (7.393)

and those corresponding to T=\frac{1}{2} as

\left|\frac{1}{2},\frac{1}{2}\right\rangle=\sqrt{\frac{2}{3} }|1,1 〉\left|\frac{1}{2},-\frac{1}{2}\right\rangle-\frac{1}{\sqrt{3} }|1,0 〉\left |\frac{1}{2},\frac{1}{2}\right\rangle=\sqrt{\frac{2}{3} }|\pi ^{+} 〉|n 〉-\frac{1}{\sqrt{3} }|\pi ^{0} 〉 |p 〉 ,      (7.394)

\left|\frac{1}{2},-\frac{1}{2}\right\rangle=\frac{1}{\sqrt{3} } |1,0 〉\left|\frac{1}{2},-\frac{1}{2}\right\rangle-\sqrt{\frac{2}{3} }|1,-1 〉\left|\frac{1}{2},\frac{1}{2}\right\rangle=\frac{1}{\sqrt{3} }|\pi ^{0} 〉|n 〉-\sqrt{\frac{2}{3} } |\pi ^{-} 〉|p 〉 .     (7.395)