Question 7.P.4: (a) Find the total spin of a system of three spin 1/2 partic...

(a) To add j_{1} =\frac{1}{2} ,j_{2} =\frac{1}{2}, and j_{3} =\frac{1}{2}, we begin by coupling j_{1} and j_{2} to form j_{12} =j_{1} +j_{2} , where \left|j_{1} -j_{2} \right| \leq j_{12} \leq \left|j_{1} +j_{2} \right| ; hence j_{12} =0,1. Then we add j_{12} and j_{3} ; this leads to \left |j_{12} -j_{3} \right| \leq j \leq \left|j_{12} +j_{3} \right| or j=\frac{1}{2} ,\frac{3}{2} .

We are going to denote the joint eigenstates of \hat{\vec{J} }^{2}_{1} ,\hat{\vec{J} }^{2}_{2} ,\hat{\vec{J} }^{2}_{3} ,\hat{\vec{J} }^{2}_{12} ,\hat{\vec{J} }^{2} , and J_{z} by |j_{12} ,j,m 〉 and the joint eigenstates of \hat{\vec{J} }^{2}_{1} ,\hat{\vec{J} }^{2}_{2} ,\hat{\vec{J} }^{2}_{3} \hat{J} _{1_{z} } ,\hat{J} _{2_{z} } , and \hat{J} _{3_{z} } by |j_{1},j_{2},j_{3};m_{1} ,m_{2},m_{3} 〉; since j_{1}=j_{2}=j_{3}=\frac{1}{2} and m_{1}=\pm \frac{1}{2} ,m_{2}=\pm \frac{1}{2} ,m_{3}=\pm \frac{1}{2}, we will be using throughout this
problem the lighter notation |j_{1},j_{2},j_{3};\pm ,\pm ,\pm 〉 to abbreviate |\frac{1}{2} ,\frac{1}{2} ,\frac{1}{2} ;\pm \frac{1}{2} ,\pm \frac{1}{2} ,\pm \frac{1}{2} 〉.

In total there are eight states |j_{12} ,j,m 〉 since (2j_{1}+1)(2j_{2}+1)(2j_{3}+1)=8. Four of these correspond to the subspace j=\frac{3}{2} ;|1,\frac{3}{2} ,\frac{3}{2} 〉,|1,\frac {3}{2} ,\frac{1}{2} 〉,|1,\frac{3}{2} ,-\frac{1}{2} 〉, and |1,\frac{3}{2} ,-\frac{3}{2} 〉. The remaining four belong to the subspace j=\frac{1}{2} :|0,\frac{1}{2} ,\frac{1}{2} 〉,|0,\frac{1}{2} ,-\frac{1}{2} 〉,|1,\frac{1}{2} ,\frac{1}{2} 〉, |1,\frac{1}{2} ,-\frac{1}{2} 〉. To construct the states |j_{12} ,j,m 〉 in terms of |j_{1},j_{2},j_{3};\pm ,\pm ,\pm 〉, we are going to consider the two subspaces j=\frac{3}{2} and j=\frac{1}{2} separately.

Subspace j=\frac{3}{2}

First, the states |1,\frac{3}{2} ,\frac{3}{2} 〉 and |1,\frac{3}{2} ,-\frac{3}{2} 〉 are clearly given by

|1,\frac{3}{2} ,\frac{3}{2} 〉=|j_{1},j_{2},j_{3};+ ,+ ,+ 〉,         |1,\frac{3}{2} ,-\frac{3}{2} 〉=|j_{1},j_{2},j_{3};- ,- ,- 〉.        (7.372)

To obtain |1,\frac{3}{2} ,\frac{1}{2} 〉, we need to apply, on the one hand, \hat{J}_{-} on |1,\frac{3}{2} ,\frac{3}{2} 〉 (see (7.220)),

\hat{J}_{\pm }|j_{12} ,j,m 〉=\hbar \sqrt{j(j+1)-m(m\pm 1)} |j_{12} ,j,m \pm 1〉,                    (7,220)

\hat{J}_{-}|1,\frac{3}{2} ,\frac{3}{2} 〉=\hbar \sqrt{\frac{3}{2}\left(\frac{3}{2}+1\right) -\frac{3}{2}\left(\frac{3}{2}-1\right) } |1,\frac{3}{2} ,\frac{1}{2} 〉=\hbar \sqrt{3} |1,\frac{3}{2} ,\frac{1}{2} 〉,           (7.373)

and, on the other hand, apply (\hat{J}_{1-}+\hat{J}_{2-} +\hat {J}_{3-}) on |j_{1},j_{2},j_{3};+ ,+ ,+ 〉 (see (7.221) to (7.223)).

\hat{J}_{1_{\pm } }|j_{1},j_{2},j_{3};m_{1},m_{2},m_{3} 〉=\hbar \sqrt{j_{1}(j_{1}+1)-m_{1}(m_{1}\pm 1)} |j_{1},j_{2},j_{3};(m_{1}\pm 1),m_{2},m_{3} 〉 ,          (7.221)

\hat{J}_{2_{\pm } }|j_{1},j_{2},j_{3};m_{1},m_{2},m_{3} 〉=\hbar \sqrt{j_{2}(j_{2}+1)-m_{2}(m_{2}\pm 1)} |j_{1},j_{2},j_{3}; m_{1},(m_{2}\pm 1),m_{3} 〉 ,          (7.222)

\hat{J}_{3_{\pm } }|j_{1},j_{2},j_{3};m_{1},m_{2},m_{3} 〉=\hbar \sqrt{j_{3}(j_{3}+1)-m_{3}(m_{3}\pm 1)} |j_{1},j_{2},j_{3}; m_{1},m_{2},(m_{3}\pm 1) 〉 .         (7.223)

This yields

(\hat{J}_{1-}+\hat{J}_{2-}+\hat{J}_{3-})|j_{1},j_{2},j_{3};+ ,+ ,+ 〉=\hbar \left(|j_{1},j_{2},j_{3};- ,+ ,+ 〉+|j_{1},j_{2},j_{3};+ ,- ,+ 〉+|j_{1},j_{2},j_{3};+ ,+ ,- 〉\right) ,               (7.374)

since \sqrt{\frac{1}{2}(\frac{1}{2}+1)-\frac{1}{2}(\frac{1}{2}-1)} =1. Equating (7.373) and (7.374) we infer

|1,\frac{3}{2} ,\frac{1}{2} 〉=\frac{1}{\sqrt{3} } \left(|j_{1}, j_{2},j_{3};- ,+ ,+ 〉+|j_{1},j_{2},j_{3};+ ,- ,+ 〉+|j_{1},j_{2},j_{3};+ ,+ ,- 〉\right).         (7.375)

Following the same method—applying \hat{J}_{-} on |1,\frac{3}{2} ,\frac{1}{2} 〉 and (\hat{J}_{1-}+ \hat{J}_{2-}+\hat{J}_{3-}) on the right-hand side of (7.375) and then equating the two results—we find

|1,\frac{3}{2} ,-\frac{1}{2} 〉=\frac{1}{\sqrt{3} } \left(|j_{1}, j_{2},j_{3};+ ,-,- 〉+|j_{1},j_{2},j_{3};- ,+ ,- 〉+|j_{1},j_{2},j_{3};- ,- ,+ 〉\right).       (7.376)

Subspace j=\frac{1}{2}

We can write |0,\frac{1}{2} ,\frac{1}{2} 〉 as a linear combination of |j_{1},j_{2},j_{3};+ ,+ ,- 〉 and |j_{1},j_{2},j_{3};- ,+ ,+ 〉:

|0,\frac{1}{2} ,\frac{1}{2} 〉=\alpha |j_{1},j_{2},j_{3};+ ,+,- 〉+\beta |j_{1},j_{2},j_{3};- ,+,+ 〉.                (7.377)

Since |0,\frac{1}{2} ,\frac{1}{2} 〉 is normalized, while |j_{1},j_{2},j_{3};+ ,+ ,- 〉 and |j_{1},j_{2},j_{3};- ,+ ,+ 〉 are orthonormal, and since the Clebsch–Gordan coefficients, such as α and β, are real numbers, equation (7.377) yields

\alpha ^{2} +\beta ^{2} =1.                    (7.378)

On the other hand, since 〈1,\frac{3}{2} ,\frac{1}{2}|0,\frac{1}{2} ,\frac{1}{2} 〉=0, a combination of (7.375) and (7.377) leads to

\frac{1}{\sqrt{3} } (\alpha +\beta )=0\Longrightarrow \alpha =-\beta .                             (7.379)

A substitution of \alpha =-\beta into (7.378) yields \alpha =-\beta=1/\sqrt{2} , and substituting this into
(7.377) we obtain

|0,\frac{1}{2} ,\frac{1}{2} 〉=\frac{1}{\sqrt{2} } \left(|j_{1},j_{2} ,j_{3};+ ,- ,- 〉-|j_{1},j_{2},j_{3};- ,+ ,+ 〉\right) .                      (7.380)

Following the same procedure that led to (7.375)—applying \hat {J}_{-} on the left-hand side of (7.380) and (\hat{J}_{1-} +\hat{J}_{2-}+\hat{J}_{3-}) on the right-hand side and then equating the two results—we find

|0,\frac{1}{2} ,-\frac{1}{2} 〉=\frac{1}{\sqrt{2} } \left(-|j_{1}, j_{2},j_{3};+ ,- ,- 〉+|j_{1},j_{2},j_{3};- ,- ,+ 〉\right).                  (7.381)

Now, to find |0,\frac{1}{2} ,\frac{1}{2} 〉, we may write it as a linear combination of |j_{1},j_{2},j_{3};+ ,+ ,- 〉,|j_{1},j_{2} ,j_{3};+ ,- ,+ 〉, and |j_{1},j_{2},j_{3};- ,+ ,+ 〉:

|1,\frac{1}{2} ,\frac{1}{2} 〉=\alpha |j_{1},j_{2},j_{3};+ ,+ ,- 〉+\beta |j_{1},j_{2},j_{3};+ ,- ,+ 〉+\gamma |j_{1},j_{2},j_{3};- ,+ ,+ 〉.          (7.382)

This state is orthogonal to |0,\frac{1}{2} ,\frac{1}{2} 〉, and hence \alpha =\gamma ; similarly, since this state is also orthogonal to |1,\frac{3}{2} ,\frac{1}{2} 〉, we have \alpha +\beta +\gamma =0, and hence 2\alpha +\beta =0 or \beta =-2\alpha =-2\gamma .

Now, since all the states of (7.382) are orthonormal, we have \alpha ^{2} +\beta ^{2} +\gamma ^{2} =1, which when combined with \beta =-2\alpha =-2\gamma leads to \alpha =\gamma =-1/\sqrt{6} and \beta =2/ \sqrt{6} . We may thus write (7.382) as

|1,\frac{1}{2} ,\frac{1}{2} 〉=\frac{1}{\sqrt{6} } \left(-|j_{1}, j_{2},j_{3};+ ,+ ,- 〉+2|j_{1},j_{2},j_{3};+ ,- ,+ 〉-|j_{1},j_{2},j_{3} ;- ,+ ,+ 〉\right) .                (7.383)

Finally, applying \hat {J}_{-} on the left-hand side of (7.383) and (\hat{J}_{1-} +\hat{J}_{2-}+\hat{J}_{3-}) on the right-hand side and equating the two results, we find

|1,\frac{1}{2} ,-\frac{1}{2} 〉=\frac{1}{\sqrt{6} } \left(|j_{1}, j_{2},j_{3};+ ,- ,- 〉-2|j_{1},j_{2},j_{3};- ,+ ,- 〉+|j_{1},j_{2},j_{3} ;- ,- ,+ 〉\right) .         (7.384)

(b) Since we have three different (nonidentical) particles, their spin angular momenta mutually commute. We may thus write their Hamiltonian as \hat{H}=-(\epsilon _{0}/\hbar ^{2})(\hat {\vec{S} }_{1}+\hat{\vec{S} }_{2} ).\hat{\vec{S} }_{3}. Due
to this suggestive form of \hat{H}, it is appropriate, as shown in (a), to start by coupling \hat {\vec{S} }_{1} with \hat {\vec{S} }_{2} to obtain \hat{\vec{S} }_{12}= \hat{\vec{S} }_{1}+\hat{\vec{S} }_{2}, and then add \hat {\vec{S} }_{12} to \hat {\vec{S} }_{3} to generate the total spin: \hat{\vec{S} }=\hat{\vec{S} }_{12}+\hat {\vec{S} }_{3}. We may thus write \hat{H} as

\hat{H}=-\frac{\epsilon _{0}}{\hbar ^{2}} \left(\hat{\vec{S} }_{1}+\hat{\vec{S} }_{2}\right) .\hat{\vec{S} }_{3}=- \frac{\epsilon _{0}}{\hbar ^{2}}\hat{\vec{S} }_{12}.\hat{\vec{S} }_{3}=-\frac {\epsilon _{0}}{2\hbar ^{2}}\left(\hat{S}^{2}-\hat{S}^{2}_{12}- \hat{S}^{2}_{3} \right) ,         (7.385)

since \hat{\vec{S} }_{12}.\hat{\vec{S} }_{3}=\frac{1}{2}[(\hat{\vec{S} }_{12}+\hat{\vec{S} }_{3})^{2} -\hat{S}^{2}_{12}-\hat{S}^{2}_{3}]. Since the operators \hat{H},\hat {S}^{2},\hat{S}^{2}_{12} , and \hat{S}^{2}_{3} mutually commute, we may select as their joint eigenstates the kets |s_{12} ,s,m 〉; we have seen in (a) how to construct these states. The eigenvalues of \hat{H} are thus given by

=-\frac{\epsilon _{0}}{2}\left[s(s+1)-s_{12}(s_{12}+1)-\frac{3}{4}\right]|s_{12} ,s,m 〉 ,                               (7.386)

since s_{3} =\frac{1}{2} and \hat{S}^{2}_{3} |s_{12} ,s,m 〉=\hbar ^{2}s_{3}(s_{3}+1) |s_{12} ,s,m 〉=(3\hbar ^{2}/4)|s_{12} ,s,m 〉 .

As shown in (7.386), the energy levels of this system are degenerate with respect to m, since they depend on the quantum numbers s and s_{12} but not on m:

E_{s_{12},s} =-\frac{\epsilon _{0}}{2}\left[s(s+1)-s_{12}(s_{12}+1)-\frac{3}{4}\right].                  (7.387)

For instance, the energy E_{s_{12},s} =E_{1,3/2} =-\epsilon _{0}/2 s fourfold degenerate, since it corresponds to four different states: |s_{12} ,s,m 〉=|1 ,\frac{3}{2},\pm \frac{3}{2} 〉 and |1 ,\frac{3}{2},\pm \frac{1}{2} 〉. Similarly, the energy E_{0,1/2} =0 is twofold degenerate; the corresponding states are |0 ,\frac{1}{2},\pm \frac{1}{2} 〉 .Finally, the energy E_{1,1/2} =\epsilon _{0} is also twofold degenerate since it corresponds to |1 ,\frac{1}{2},\pm \frac{1}{2} 〉.