Question 7.P.3: Find the Clebsch–Gordan coefficients associated with the add...

The addition of j_{1} =1 and j_{2} =1 is encountered, for example, in a two-particle system where the angular momenta of both particles are orbital.

The allowed values of the total angular momentum are between \left|j_{1}-j_{2}\right| \leq j\leq j_{1}+j_{2}; hence j = 0, 1, 2. To calculate the relevant Clebsch–Gordan coefficients, we need to find the basis vectors \left\{|j,m 〉\right\} , which are common eigenvectors of \hat{\vec{J}} ^{2}_{1} ,\hat{\vec{J}} ^{2}_{2} ,\vec{J}^{2} and \hat{J} _{z} , in terms of \left\{|1,1;m_{1} , m_{2} 〉\right\} .

Eigenvectors |j,m 〉 associated with j = 2

The state |2,2 〉 is simply given by

|2,2 〉=|1,1;1,1 〉;                                            (7.350)

the corresponding Clebsch–Gordan coefficient is thus given by 〈1,1;1,1|2,2 〉=1.

As for |2,1 〉, it can be found by applying J_{-} to |2,2 〉 and (J_{1-} +J_{2-} ) to |1,1;1,1 〉, and then equating the two results

J_{-} |2,2 〉=(J_{1-} +J_{2-} )|1,1;1,1 〉.                      (7.351)

This leads to

2\hbar |2,1 〉=\sqrt{2} \hbar \left(|1,1;1,0 〉+|1,1;0,1 〉\right)                             (7.352)

or to

|2,1 〉=\frac{1}{\sqrt{2} } \left(|1,1;1,0 〉+|1,1;0,1 〉\right);                            (7.353)

hence 〈1,1;1,0|2,1 〉 = 〈1,1;0,1|2,1 〉 = 1/\sqrt{2} . Using (7.353), we can find |2,0 〉 by applying J_{-} to |2,1 〉 and (J_{1-} +J_{2-} ) to [|1,1;1,0 〉 + |1,1;0,1 〉]:

J_{-} |2.1 〉=\frac{1}{\sqrt{2} }\hbar (J_{1-} +J_{2-} )[|1,1;1,0 〉 + |1,1;0,1 〉],                   (7.354)

which leads to

|2,0 〉=\frac{1}{\sqrt{6} }\left(|1,1;1,-1 〉 + 2|1,1;0,0 〉 + |1,1;-1,1 〉\right) ;             (7.355)

hence 〈1,1;1,-1|2,0 〉 = 〈1,1;-1,1|2,0 〉 =1/\sqrt{6} and 〈1,1;0,0|2,0 〉 =2/\sqrt{6}.

Similarly, by repeated applications of J_{-} and (J_{1-} +J_{2-} ), we can show that

|2,-1 〉=\frac{1}{\sqrt{2} }\left(|1,1;0,-1 〉 + |1,1;-1,0 〉\right) ,               (7.356)

|2,-2 〉 = |1,1;-1,-1 〉 ,                             (7.357)

with 〈1,1;0,-1|2,-1 〉 = 〈1,1;-1,0|2,-1 〉 = 1/\sqrt{2} and 〈1,1;-1,-1|2,-2 〉 = 1 .

Eigenvectors |j,m 〉 associated with j = 1

The relation

|1,m 〉=\sum\limits_{m_{1}=-1}^{1}\sum\limits_{m_{2}=-1}^{1} 〈1,1; m_{1},m_{2}|1,m 〉|1,1;m_{1},m_{2}〉                        (7.358)

leads to

|1,1 〉 = a|1,1;1,0 〉 +b|1,1;0,1 〉,                          (7.359)

where a=〈1,1;1,0|1,1 〉 and b=〈1,1;0,1|1,1 〉. Since |1,1 〉 ,|1,1;1,0 〉 and |1,1;0,1 〉 are all normalized, and since is orthogonal to [latex] |1,1;0,1 〉 and a and b are real, we have

〈1,1|1,1 〉 =a^{2} +b^{2} =1.                              (6.360)

Now, since 〈2,1|1,1 〉 =0, equations (7.353) and (7.359) yield

〈2,1|1,1 〉 =\frac{a}{\sqrt{2} } +\frac{b}{\sqrt{2} } =0.                  (7.361)

A combination of (7.360) and (7.361) leads to a=-b=\pm 1/ \sqrt{2} . The signs of a and b have yet to be found. The phase convention mandates that coefficients like 〈j_{1} ,j_{2} ;j_{1} ,(j-j_{1} )|j,j 〉 must be positive. Thus, we have a=1/\sqrt{2} and b=-1/\sqrt{2}, which when inserted into (7.359) give

|1,1 〉=\frac{1}{\sqrt{2} }\left(|1,1;1,0 〉 - |1,1;0,1 〉\right) .                 (7.362)

This yields 〈1,1;1,0|1,1 〉 = \frac{1}{2} and 〈1,1;0,1 |1,1 〉 = -\frac{1}{2} .

To find |1,0 〉 we proceed as we did above when we obtained the states |2,1 〉 , |2,0 〉,...,|2,-2 〉 by repeatedly applying J_{-} on |2,2 〉. In this way, the application of J_{-} on |1,1 〉 and (J_{1-} +J_{2-} ) on [|1,1;1,0 〉 - |1,1;0,1 〉],

J_{-} |1,1 〉=\frac{1}{2}(J_{1-} +J_{2-} )[|1,1;1,0 〉 - |1,1;0,1 〉]                      (7.363)

gives

\sqrt{2} \hbar |1,0 〉=\frac{\sqrt{2} \hbar}{2} [|1,1;1,-1 〉 - |1,1;-1,1 〉] ,                        (7.364)

or

|1,0 〉=\frac{1}{\sqrt{2} }\left(|1,1;1,-1 〉 - |1,1;-1,1 〉\right) ,                    (7.365)

with 〈1,1;1,-1|1,0 〉 =\frac{1}{\sqrt{2}} and 〈1,1;-1,1 |1,0 〉 =-1/\sqrt{2} .

Similarly, we can show that

|1,-1 〉=\frac{1}{\sqrt{2} }\left(|1,1;0,-1 〉 - |1,1;-1,0 〉\right) ;                     (7.366)

hence 〈1,1;0,-1|1,-1 〉 =1/\sqrt{2} and 〈1,1;-1,0|1,-1 〉 =-1/\sqrt{2} .

Eigenvector |0,0 〉 associated with j = 0

Since

|0,0 〉=a|1,1;1,-1 〉 +b|1,1;0,0 〉 +c|1,1;-1,1 〉,                       (7.367)

where a=|1,1;1,-1|0,0 〉, b= |1,1;0,0|0,0 〉, and c= |1,1;-1,1|0,0 〉 are real, and since the states |0,0 〉,|1,1;1,-1 〉,|1,1;0,0 〉, and |1,1;-1,1 〉 are normal, we have

〈0,0|0,0 〉 = a^{2}+b^{2}+c^{2}=1.                          (7.368)

Now, combining (7.355), (7.365), and (7.367), we obtain

〈2,0|0,0 〉 = \frac{a}{\sqrt{6}} +\frac{2b}{\sqrt{6}} +\frac{c}{\sqrt{6}} =0,                            (7.369)

〈1,0|0,0 〉 = \frac{a}{\sqrt{2}} -\frac{c}{\sqrt{2}}=0.                             (7.370)

Since a is by convention positive, we can show that the solutions of (7.368), (7.369), and (7.370) are given by a=1/\sqrt{3} ,b=-1/ \sqrt{3} ,c=1/\sqrt{3} , and consequently

|0,0 〉=\frac{1}{\sqrt{3} } \left(|1,1;1,-1 〉 - |1,1;0,0 〉 + |1,1;-1,1 〉\right) ,           (7.371)

with 〈1,1;1,-1|0,0 〉 = 〈1,1;-1,1|0,0 〉 =1/\sqrt{3} and 〈1,1;0,0|0,0 〉 = -1/\sqrt{3} .

Note that while the quintuplet states |2,m 〉 (with m = \pm 2,\pm 1,0) and the singlet state |0,0 〉 are symmetric, the triplet states |1,m 〉 (with m =\pm 1,0) are antisymmetric under space inversion.