(a) Find the Clebsch–Gordan coefficients associated with the coupling of the spins of the electron and the proton of a hydrogen atom in its ground state.
(b) Find the transformation matrix which is formed by the Clebsch–Gordan coefficients. Verify that this matrix is unitary
In their ground states the proton and electron have no orbital angular momenta. Thus, the total angular momentum of the atom is obtained by simply adding the spins of the proton and electron.
This is a simple example to illustrate the general formalism outlined in this section. Since j_{1} =\frac{1}{2} and j_{2} = \frac{1}{2} , j has two possible values j = 0, 1. When j = 0, there is only a single state |j,m 〉=|0,0 〉; this is called the spin singlet. On the other hand, there are three possible values of m = -1, 0, 1 for the case j = 1; this corresponds to a spin triplet state |1,-1 〉,|1,0 〉,|1,1 〉.
From (7.121),
|j,m 〉=\left(\sum\limits_{m_{1}=-j_{1} }^{j_{1}}\sum\limits _{m_{2}=-j_{2} }^{j_{2}}|j_{1},j_{2};m_{1},m_{2} 〉〈j_{1},j_{2}; m_{1} ,m_{2} | \right) |j,m 〉,
=\sum\limits_{m_{1}m_{2}} 〈j_{1},j_{2};m_{1},m_{2}|j,m 〉 |j_{1},j_{2};m_{1},m_{2} 〉. (7.121)
we can express the states |j,m 〉 in terms of |\frac {1}{2},\frac{1}{2};m_{1},m_{2} 〉 as follows:
|j,m 〉=\sum\limits_{m_{1}=-1/2}^{1/2} \sum\limits_{m_{2} =-1/2}^{1/2} 〈\frac{1}{2},\frac{1}{2};m_{1},m_{2}|j,m 〉 |\frac{1}{2},\frac{1}{2};m_{1},m_{2} 〉, (7.159)
which, when applied to the two cases j = 0 and j = 1, leads to
|0,0 〉=〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|0,0 〉\left |\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}\right\rangle +〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|0,0 〉\left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}\right\rangle , (7.160)
|1,1 〉=〈\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}|1,1 〉\left |\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}\right\rangle , (7.161)
|1,0 〉=〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|1,0 〉 \left |\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}\right\rangle +〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|1,0 〉 \left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}\right\rangle , (7.162)
|1,-1 〉=〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},-\frac{1}{2}|1,-1 〉 \left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},-\frac{1}{2}\right\rangle . (7.163)
To calculate the Clebsch–Gordan coefficients involved in (7.160)–(7.163), we are going to adopt two separate approaches: the first approach uses the recursion relations (7.150) and (7.155),
\sqrt{(j\mp m)(j\pm m+1)} 〈j_{1},j_{2};m_{1},m_{2}|j,m\pm 1 〉=\sqrt{(j_{1}\pm m_{1})(j_{1}\mp m_{1}+1)} 〈j_{1},j_{2}; m_{1}\mp 1,m_{2}|j,m〉
+\sqrt{(j_{2}\pm m_{2})(j_{2}\mp m_{2}+1)} 〈j_{1},j_{2}; m_{1},m_{2}\mp 1|j,m〉. (7.150)
\sqrt{(j\mp m+1)(j\pm m)} 〈j_{1},j_{2};m_{1},m_{2}|j,m 〉
=\sqrt{(j_{1}\pm m_{1})(j_{1}\mp m_{1}+1)} 〈j_{1},j_{2}; m_{1}\mp 1,m_{2}|j,m\mp 1 〉
+\sqrt{(j_{2}\pm m_{2})(j_{2}\mp m_{2}+1)} 〈j_{1},j_{2}; m_{1},m_{2}\mp 1|j,m\mp 1 〉. (7.155)
while the second uses the algebra of angular momentum.
First approach: using the recursion relations
First, to calculate the two coefficients 〈\frac{1}{2},\frac{1}{2}; \pm \frac{1}{2},\mp \frac{1}{2}|0,0 〉 involved in (7.160), we need, on the one hand, to substitute j=0, m=0, m_{1}=m_{2}= \frac{1}{2} into the upper sign relation of (7.150):
〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|0,0 〉=-〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|0,0 〉. (7.164)
On the other hand, the substitution of j = 0 and m = 0 into (7.125)
\sum\limits_{m_{1}m_{2}}= 〈j_{1},j_{2};m_{1},m_{2}|j,m 〉^{2} =1 . (7.125)
yields
〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|0,0 〉^{2} + 〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|0,0 〉^{2} =1 (7.165)
Combining (7.164) and (7.165) we end up with
〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|0,0 〉=\pm \frac{1}{\sqrt{2} } . (7.166)
The sign of 〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|0,0 〉 has to be positive because, according to the phase convention, the coefficient 〈j_{1},j_{2};j_{1},(j-j_{1})|j,j 〉 is positive; hence
〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|0,0 〉=\frac{1}{\sqrt{2} } . (7.167)
As for 〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|0,0 〉, its value can be inferred from (7.164) and (7.167):
〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|0,0 〉=-\frac{1}{\sqrt{2} } . (7.168)
Second, the calculation of the coefficients involved in (7.161) to (7.163) goes as follows. The orthonormalization relation (7.125) leads to
〈\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}|1,1 〉^{2}=1, 〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},-\frac{1}{2}|1,-1 〉^{2}=1 , (7.169)
and since 〈\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}|1,1 〉 and 〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},-\frac{1}{2}|1,-1 〉 are both real and positive, we have
〈\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}|1,1 〉=1, 〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},-\frac{1}{2}|1,-1 〉=1 . (7.170)
As for the coefficients 〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac {1}{2}|1,0 〉 and 〈\frac{1}{2},\frac{1}{2};-\frac{1}{2}, \frac{1}{2}|1,0 〉, they can be extracted by setting j=1,m =0,m_{1}=\frac{1}{2},m_{2}=-\frac{1}{2} and j=1,m=0, m_{1}=-\frac{1}{2},m_{2}=\frac{1}{2}, respectively, into the lower sign case of (7.155):
\sqrt{2} 〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|1,0 〉 =〈\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}|1,1 〉 , (7.171)
\sqrt{2} 〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|1,0 〉 =〈\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}|1,1 〉 . (7.172)
Combining (7.170) with (7.171) and (7.172), we find
〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|1,0 〉 =〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|1,0 〉 =\frac{1}{\sqrt{2} } . (7.173)
Finally, substituting the Clebsch–Gordan coefficients (7.167), (7.168) into (7.160) and (7.170), and substituting (7.173) into (7.161) to (7.163), we end up with
|0,0 〉=-\frac{1}{\sqrt{2} } \left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}\right\rangle +\frac{1}{\sqrt{2} } \left|\frac{1}{2},\frac {1}{2};\frac{1}{2},-\frac{1}{2}\right\rangle , (7.174)
|1,1 〉=\left|\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}\right \rangle , (7.175)
|1,0 〉=\frac{1}{\sqrt{2} } \left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}\right\rangle +\frac{1}{\sqrt{2} } \left|\frac{1}{2},\frac {1}{2};\frac{1}{2},-\frac{1}{2}\right\rangle , (7.176)
|1,-1 〉=\left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},-\frac{1}{2} \right\rangle . (7.177)
Note that the singlet state |0,0 〉 is antisymmetric, whereas the triplet states |1,-1 〉,|1,0 〉, and |1,1 〉 are symmetric.
Second approach: using angular momentum algebra
Beginning with j = 1, and since |1,1 〉 and |\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2} 〉 are both normalized, equation (7.161) leads to
|\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}|1,1 〉^{2} =1 . (7.178)
From the phase convention, which states that 〈j_{1},j_{2};j,(j-j_{1})|j,j 〉 must be positive, we see that |\frac{1}{2},\frac {1}{2};\frac{1}{2},\frac{1}{2}|1,1 〉 =1, and hence
|1,1 〉=\left|\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2}\right \rangle . (7.179)
Now, to find the Clebsch–Gordan coefficients in |1,o 〉, we simply apply \hat{J} _{-} on |1,1 〉 :
\hat{J} _{-} |1,1 〉=(\hat{J} _{1-}+\hat{J} _{2-})\left|\frac{1}{2} ,\frac{1}{2};\frac{1}{2},\frac{1}{2}\right\rangle, (7.180)
which leads to
|1,0 〉=\frac{1}{\sqrt{2} } \left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}\right\rangle +\frac{1}{\sqrt{2} } \left|\frac{1}{2},\frac {1}{2};\frac{1}{2},-\frac{1}{2}\right\rangle , (7.181)
hence 〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|1,0 〉=1/ \sqrt{2} and 〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac {1}{2}|1,0 〉=1/\sqrt{2} . Next, applying \hat{J} _{-} on (7.181), we get
|1,-1 〉=\left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},-\frac{1}{2} \right\rangle . (7.182)
Finally, to find |0,0 〉, we proceed in two steps: first, since
|0,0 〉=a\left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2} \right\rangle +b\left|\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2} \right\rangle, (7.183)
where a=〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|0,0 〉 and b=〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2} |0,0 〉 , a combination of (7.181) with (7.183) leads to
〈0,0|1,0 〉=\frac{a}{\sqrt{2} } +\frac{b}{\sqrt{2} } =0; (7.184)
second, since |0,0 〉 is normalized, we have
〈0,0|0,0 〉=a^{2} +b^{2} =1. (7.185)
Combining (7.184) and (7.185), and since 〈\frac{1}{2},\frac{1}{2} ;\frac{1}{2},-\frac{1}{2}|0,0 〉 must be positive, we obtain a=〈\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}|0,0 〉=-1/ \sqrt{2} and b=〈\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}|0,0 〉=1/\sqrt{2} . Inserting these values into (7.183) we obtain
|0,0 〉=-\frac{1}{\sqrt{2} } \left|\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2}\right\rangle +\frac{1}{\sqrt{2} } \left|\frac{1}{2},\frac {1}{2};\frac{1}{2},-\frac{1}{2}\right\rangle. (7.186)
(b) Writing (7.174) to (7.177) in a matrix form:
\left(\begin{matrix} |0,0 〉 \\ |1,1 〉 \\ |1,0 〉 \\ |1,-1 〉 \end {matrix} \right) =\left(\begin{matrix} 0 & 1/\sqrt{2} & -1/\sqrt{2} & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right) \left(\begin{matrix} |\frac{1}{2},\frac{1}{2};\frac {1}{2},\frac{1}{2} 〉 \\ |\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2} 〉 \\ |\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2} 〉 \\ |\frac{1}{2}, \frac{1}{2};-\frac{1}{2},-\frac{1}{2} 〉 \end{matrix} \right), (7.187)
we see that the elements of the transformation matrix
U=\left(\begin{matrix} 0 & 1/\sqrt{2} & -1/\sqrt{2} & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 0 & 1 \end {matrix} \right), (7.188)
which connects the \left\{|j,m 〉\right\} vectors to their \left\{|j_{1},j_{2};m_{1},m_{2} 〉\right\} counterparts, are given by the Clebsch–Gordan coefficients derived above. Inverting (7.187) we obtain
\left(\begin{matrix} |\frac{1}{2},\frac{1}{2};\frac{1}{2},\frac{1}{2} 〉 \\ |\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2} 〉 \\ |\frac{1}{2},\frac{1}{2};-\frac{1}{2},\frac{1}{2} 〉 \\ |\frac{1}{2},\frac{1}{2};-\frac{1}{2},-\frac{1}{2} 〉 \end{matrix} \right) =\left(\begin{matrix} 0 & 1 & 0 & 0 \\ 1/\sqrt{2} & 0 & 1/\sqrt{2} & 0 \\ -1/\sqrt{2} & 0 & 1/ \sqrt{2} & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right)\left(\begin {matrix} |0,0 〉 \\ |1,1 〉 \\ |1,0 〉 \\ |1,-1 〉 \end{matrix} \right). (7.189)
From (7.187) and (7.189) we see that the transformation matrix U is unitary; this is expected since U^{-1} =U^{\dagger }.