Clebsch-Gordon Coefficients

As we have seen, the operator group $J_1^{\,2}$ , $J_2^{\,2}$ , $J^{\,2}$ , and $J_z$ is incompatible with the group $J_1^{\,2}$ , $J_2^{\,2}$ , $J_{1\,z}$ , and $J_{2\,z}$ . This means that if the system is in a simultaneous eigenstate of the former group then, in general, it is not in a simultaneous eigenstate of the latter. In other words, if the quantum numbers $j_1$ , $j_2$ , $j$ , and $m$ are known with certainty then a measurement of the quantum numbers $m_1$ and $m_2$ will give a range of possible values. We can use the completeness relation (6.20) to write

$$\vert j_1,j_2;j,m\rangle = \sum_{m_1}\sum_{m_2} \langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle \vert j_1,j_2;m_1,m_2\rangle.$$ (6.22)

Thus, we can write the eigenkets of the first group of operators as a weighted sum of the eigenkets of the second set. The weights, $\langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle$ , are called the Clebsch-Gordon coefficients. If the system is in a state where a measurement of $J_1^{\,2}, J_2^{\,2}, J^{\,2}$ , and $J_z$ is bound to give the results $j_1\,(j_1+1)\,\hbar^{\,2}, j_2\,(j_2+1)\,\hbar^{\,2}, j\,(j+1)\,\hbar^{\,2}$ , and $j_z\,\hbar$ , respectively, then a measurement of $J_{1\,z}$ and $J_{2\,z}$ will give the results $m_1\,\hbar$ and $m_2\,\hbar$ , respectively, with probability $\vert\langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle\vert^{\,2}$ .

The Clebsch-Gordon coefficients possess a number of very important properties. First, the coefficients are zero unless

$$m = m_1 + m_2.$$ (6.23)

To prove this, we note that

$$(J_z - J_{1\,z} - J_{2\,z})\, \vert j_1,j_2; j, m\rangle =0.$$ (6.24)

Forming the inner product with $\langle j_1, j_2; m_1, m_2\vert$ , we obtain

$$(m-m_1-m_2) \,\langle j_1, j_2; m_1, m_2\vert j_1,j_2; j, m\rangle=0,$$ (6.25)

which proves the assertion. Thus, the $z$ -components of different angular momenta add algebraically. So, an electron in an $l=1$ state, with orbital angular momentum $\hbar$ , and spin angular momentum $\hbar/2$ , projected along the $z$ -axis, constitutes a state whose total angular momentum projected along the $z$ -axis is $3\,\hbar/2$ . What is uncertain is the magnitude of the total angular momentum.

Second, the coefficients vanish unless

$$\vert j_1-j_2\vert \leq j \leq j_1+j_2.$$ (6.26)

We can assume, without loss of generality, that $j_1\geq j_2$ . We know, from Equation (6.23), that for given $j_1$ and $j_2$ the largest possible value of $m$ is $j_1+j_2$ (because $j_1$ is the largest possible value of $m_1$ , etc.). This implies that the largest possible value of $j$ is $j_1+j_2$ (because, by definition, the largest value of $m$ is equal to $j$ ). Now, there are $(2\,j_1+1)$ allowable values of $m_1$ , and $(2\,j_2+1)$ allowable values of $m_2$ . Thus, there are $(2\,j_1+1)\,(2\,j_2+1)$ independent eigenkets, $\vert j_1,j_2; m_1,m_2\rangle$ , needed to span the ket space corresponding to fixed $j_1$ and $j_2$ . Because the eigenkets $\vert j_1, j_2; j, m\rangle$ span the same space, they must also form a set of $(2\,j_1+1)\,(2\,j_2+1)$ independent kets. In other words, there can only be $(2\,j_1+1)\,(2\,j_2+1)$ distinct allowable values of the quantum numbers $j$ and $m$ . For each allowed value of $j$ , there are $2\,j+1$ allowed values of $m$ . We have already seen that the maximum allowed value of $j$ is $j_1+j_2$ . It is easily seen that if the minimum allowed value of $j$ is $j_1-j_2$ then the total number of allowed values of $j$ and $m$ is $(2\,j_1+1)\,(2\,j_2+1)$ . In other words [59],

$$\sum_{j=j_1-j_2,j_1+j_2} (2\,j+1) = \left[(1+j)^2\right]_{j=j_1-j_2-1}^{j=j_1+j_2}=(2\,j_1+1)\,(2\,j_2+1).$$ (6.27)

This proves our assertion.

Third, the sum of the modulus squared of all of the Clebsch-Gordon coefficients is unity: that is,

$$\sum_{m_1}\sum_{m_2} \vert\langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle\vert^{\,2} =1.$$ (6.28)

This assertion is proved as follows:

$$\begin{multline} \langle j_1, j_2; j, m\vert j_1, j_2; j, m\rangle =\\ [0.5ex] \... ...ert\langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle\vert^{\,2} =1, \end{multline}$$

where use has been made of the completeness relation (6.20).

Finally, the Clebsch-Gordon coefficients obey two recursion relations. To obtain these relations, we start from

$$J^{\pm} \vert j_1,j_2;j,m\rangle = (J_1^\pm + J_2^\pm )\sum_{m_1'... ..., j_2; m_1', m_2'\vert j_1, j_2; j, m\rangle \vert j_1, j_2; m_1', m_2'\rangle,$$ (6.29)

where $J^\pm = J_x\pm {\rm i}\,J_y$ , et cetera, Making use of the well-known properties of the ladder operators, $J^\pm$ , $J_1^\pm$ , and $J_2^\pm$ , which are specified by analogy with Equations (4.55)-(4.56), we obtain

$$\begin{multline} \sqrt{j\,(j+1)- m\,(m\pm 1)}\, \vert j_1,j_2;j,m\pm 1\rangle =\... ...2'\,(m_2'\pm 1)}\, \vert j_1, j_2; m_1', m_2'\pm 1\rangle\right). \end{multline}$$

Taking the inner product with $\langle j_1, j_2; m_1, m_2\vert$ , and making use of the orthonormality property of the basis eigenkets, we get the desired recursion relations:

$$\begin{multline} \sqrt{j\,(j+1)- m\,(m\pm 1)}\,\langle j_1, j_2; m_1, m_2\vert j... ...mp 1)}\,\langle j_1, j_2; m_1, m_2\mp 1\vert j_1,j_2;j, m\rangle. \end{multline}$$

It is clear, from the absence of complex coupling coefficients in the previous relations, that we can always choose the Clebsch-Gordon coefficients to be real numbers. This is convenient, because it ensures that the inverse Clebsch-Gordon coefficients, $\langle j_1, j_2; j, m\vert j_1, j_2; m_1, m_2\rangle$ , are identical to the Clebsch-Gordon coefficients. In other words,

$$\langle j_1, j_2; j, m\vert j_1, j_2; m_1, m_2\rangle = \langle j_1, j_2; m_1, m_2\vert j_1, j_2; j, m\rangle.$$ (6.30)

The inverse Clebsch-Gordon coefficients are the weights in the expansion of the $\vert j_1,j_2; m_1,m_2\rangle$ in terms of the $\vert j_1, j_2; j, m\rangle$ :

$$\vert j_1,j_2; m_1,m_2\rangle = \sum_{j}\sum_m \langle j_1, j_2; j, m\vert j_1, j_2; m_1, m_2\rangle \vert j_1,j_2; j,m\rangle.$$ (6.31)

It turns out that the recursion relations (6.32), together with the normalization condition (6.28), are sufficient to completely determine the Clebsch-Gordon coefficients to within an arbitrary sign (multiplied into all of the coefficients). This sign is fixed by convention. [To be more exact, each Clebsch-Gordon sub-table associated with a specific value of $j$ (see later) is undetermined to an arbitrary sign. It is conventional to give the Clebsch-Gordon coefficient with the largest value of $m$ a positive sign.] The easiest way of demonstrating this assertion is by considering a specific example.