[NOTES/QM-17004] Recurrence Relations for CG Coefficients

Using ladder operators


Restrictions on total $J M $ values

The restrictions $1)-4)$, given above, on the allowed values of the total angular momentum $J$ will be derived by considering the matrix elements $\matrixelement{j_1j_2m_1m_2}{J_z}{JM}$,$\matrixelement{j_1j_2m_1m_2} {J_z}{JM}$ and $\matrixelement{j_1j_2m_1m_2}{J_\pm}{JM}$ and by repeated use of \eqref{E5} and \eqref{E6}

Proof of $M=m_1+m_2$.

The first result is easy to prove. Since $J_z=J^{(1)}_z + J^{(2)}_z$, taking the matrix element and using the properties \begin{eqnarray} J_z\ket{JM} &=& M\hbar \ket{JM} \\ (J^{(1)}_z + J^{(2)}_z)\ket{j_1m_1j_2m_2}& =& (m_1+m_2)\hbar \ket{j_1m_1j_2m_2} \end{eqnarray} we obtain \begin{equation} \matrixelement{JM}{(J^{(1)}_z + J^{(2)}_z -J_z)}{j_1m_1j_2m_2} =0 \label{E11} \end{equation} Therefore, \begin{equation} ( m_1+m_2 -M )\innerproduct{JM}{j_1m_1j_2m_2} =0 \label{E12} \end{equation} Thus if $M\ne m_1+m_2,$ the Clebsch Gordon coefficient $\innerproduct{JM} {j_1j_2m_1m_2}$ has to be zero. In other words, a nonzero value of $ \innerproduct{JM}{j_1j_2m_1m_2}$ is possible only when \begin{equation} M = m_1 + m_2 \label{E13} \end{equation}

Range of $J$ values.

The results will be derived by considering the matrix elements $\matrixelement{j_1m_1j_2m_2}{J_z}{JM}$ and \linebreak $\matrixelement{j_1m_1,j_2m_2}{J_ \pm}{JM}$. Note that there is one relation between three the variables $m_1m_2,M$. Hence we need the Clebsch Gordon coefficients for all allowed values of $M$ and $m_1 $which vary in the range $M=-J,\cdots,J$ and $m_1=-j_1,\cdots,j_1$. We will now argue that these can all be related to single coefficient $ \innerproduct{JJ}{;j_1J-j_1}.$ They can all be related to $\innerproduct{j_1j_1,j_2(J-j_1)}{JJ}$ \begin{description}

• Use \eqref{qm-lec-17003;E5} with $M=J, m_1=j_1$ \begin{eqnarray} { 0 \times \innerproduct{j_1m_1,j-2m_2}{J(M + 1)}} \nonumber &&\\ &=& \innerproduct{j_1(j_1-1), j_2 m_2}{JJ} \sqrt{2j_1} \nonumber\\ && + \innerproduct{j_1j_1,j_2(m_2-1)}{JM} \sqrt{j_2(j_2+1)-m_2(m_2-1)} \label{E32} \end{eqnarray}
• Use \eqref{qm-lec-17003;E6} with $M=J, m_1=j_i$  \begin{eqnarray} { \sqrt{2J} \innerproduct{;j_1m_2}{J(J-1)}}\nonumber &&\\ &=& \innerproduct{j_1(j_1+1),j_2 m_2}{JM} \times 0 \nonumber\\ && + \innerproduct{j_1j_1, j_2(m_2+1)}{JM} \sqrt{j_2(j_2+1)-m_2(m_2+1)} \label{E33} \end{eqnarray}
• Use \eqref{qm-lec-17003;E5} with $M=J-1,m_1=j_1$ \begin{eqnarray} { \sqrt{2J} \innerproduct{;j_1m_2}{JJ}} \nonumber &&\\ &=& \innerproduct{;(j_1-1)m_2}{J(J-1)} \sqrt{2j_1} \nonumber\\ && + \innerproduct{j_1j_1,j_2 (m_2-1)}{J(J-1)} \sqrt{(j_2(j_2+1) -m_2(m_2-1)} \label{E34} \end{eqnarray}
• Use \eqref{qm-lec-17003;E6} with $M=J, m_1=j_1-1$ \begin{eqnarray} { \sqrt{2J} \innerproduct{j_1(j_1-1), j_2m_2}{J(J-1)}}\nonumber &&\\ &=& \innerproduct{j_1j_1,j_2 m_2}{JJ} \times 0 \nonumber\\ && + \innerproduct{j_1(j_1-1),j_2 m_2+1}{JM} \sqrt{j_2(j_2+1)-m_2(m_2+1)} \label{E35} \end{eqnarray}
• Use \eqref{qm-lec-17003;E6} with $M=J-1,m_1=j_1$ \begin{eqnarray} { \sqrt{(J+M)(J- M+1)} \innerproduct{j_1j_2m_1m_2}{J(J-2)}}\nonumber \\ &=& \innerproduct{j_1(j_1+1),j_2 m_2}{JM} \sqrt{j_1(j_1+1)-m_1(m_1+1)} \nonumber\\ && + \innerproduct{j_1j_2j_1 (m_2+1)}{JM} \sqrt{(j_2(j_2+1)-m_2(m_2+1)} \label{E36} \end{eqnarray}. We can continue in this fashion. We see that the Clebsch Gordon coefficient for different pairs of values of $M,m_1$ are known in terms of a single coefficient for $M=J,m_1=j_i$ as follows .                                                     

Next we consider the state $\ket{j_1m_1,j_2m_2}$ with $m_1=j_1$ and $m_2=j_2$. In this state $M$ has the highest value $j_1+j_2$. All these coefficients can be fixed in terms of a single coefficient $\innerproduct{j_1j_1,j_2m_2=(j-j_1)}{JM}$ which is non zero only if $j-j_1$ lies in between $-j_1$ and $j_1$. Thus \begin{equation} -j_2 \le j-j_1 \le j_2 \label{E22} \end{equation} By repeating the above steps with $j_1$ and $j_2$ interchanged we would get \begin{equation} -j_1 \le j-j_2 \le j_1 \label{E23} \end{equation} These two conditions, \eqref{E22} and \eqref{E23}, are equivalent to the requirement $J >|j_1-j_2|.$ Since maximum possible value of $M=m_1+m_2$ is $j_1+j_2$ we must have $J< j_1+j_2$. Thus the total angular momentum is constrained to lie between $|j_1-j_2|$ and $j_1+j_2$. The three numbers $ J,j_1,j_2$ should be such that they satisfy triangle inequalities \begin{equation} |j_1-j_2| \le J \le j_1+j_2 . \label{E24} \end{equation} It should be remarked that \eqref{E22}-\eqref{E24}, written as conditions on $j_1$ and $j_2$, are equivalent to each of the following two alternate forms \begin{equation} |J-j_2| \le j_1 \le J+j_2 \qquad \mbox{\rm and}\qquad |J-j_1| \le j_2 \le J + j_1. \label{E25} \end{equation}.

$J$ takes values in steps of 1 

The range of $J$, the total angular momentum value, has been determined; the minimum value being $|j_1-j_2|$ and the maximum value is $(j_1+j_2)$. Are all integral, and half integral values, allowed in this range allowed ? We must fix which values of $J$ are allowed and which ones are not allowed? The state $\ket{j_1j_2m_1m_2}$, when both $m_1$ and $m_2$ have their maximum allowed values $j_1$ and $j_2$, the value of $M=m_1+m_2$ is maximum and equal to $j_1+j_2$. There is only one such state and this must correspond to $J=j_1+j_2$. \begin{equation} \ket{J=(j_1+j_2), M=(j_1+j_2)} = \ket{j_1m_1=j_1,j_2m_2=j_2} \label{E38} \end{equation} The next value of $M=j_1+j_2-1$ corresponds to two linearly independent states corresponding to

1. $m_1=j_1-1 , m_2=j_2$, and
2. $m_1=j_1, m_2=j_2-1$

What are the corresponding $J$ values? One combination of these two states must be the $M=j_1+j_2-1$ partner of the state \eqref{E38}; and the other linear combination can only correspond to the next value $J =j_1+j_2-1$.Continuing in this way, we now consider $M=j_1+j_2-2$ which will come from three sets of $m_1,m_2$ values,{\it i.e.},

1. $m_1=j_1-2, m_2=j_2$
2. $m_1=j_1-1, m_2=j_2-1$, and,
3. $m_1=j_1, m_2=j_2-2$

Of the three states $\ket{j_1m_1,j_2m_2}$ corresponding the above values two linear combinations will correspond to the $J$ values $j_1+j_2$ already found; a third linear combination must therefore correspond to the value $J=j_1+j_2-2$. Proceeding in this fashion we see that the successive $J$ values differ by one. {\it How are we sure that all the $J$ values have been correctly identified ?}. We will now count the number of states in two different ways to confirm the conclusion that $J$ takes all the values in the allowed range from $|j_1-j_2|$ to $j_1+j_2.

Count in two ways to cross check. 

Thus we have two set of orthonormal bases \begin{equation} \Big\{\ket{j_1m_1,j_2m_2}\Big| m_1=-j_1,\cdots,j_1, m_2=-j_2, j_2\cdots \Big\}\label{E26} \end{equation} and \begin{equation} \Big\{\ket{j_1j_2;JM}\Big| J=|j_1-j_2|,\cdots, j_1+j_2,M=-J,\cdots,J \Big\}\label{E27} \end{equation} It is easily seen that the total number of vectors in two bases are equal. The number of elements in the set \eqref{\TheFile;E26}) is $(2j_1+1)(2j_2+1)$ and in the set \eqref{\TheFile;E27} is also the same \begin{equation} \sum_{J=|j_1-j_2|}^{j_1+j_2} (2J+1) = (2j_1+1)(2j_2+1). \label{E28} \end{equation} The two bases in \eqref{E26} and \eqref{E27} are orthonormal and hence the transformation connecting the two must be a unitary transformation and the Clebsch Gordon coefficients must satisfy the relations \begin{eqnarray} \sum_{JM} \innerproduct{j_1j_2;m_1m_2}{j_1j_2JM}\innerproduct{JM} {j_1j_2;m_1^\prime m_2^\prime} &=& \delta_{m_1m_1^\prime} \delta_{m_2m_2^\prime} \label{E29} \\ \sum_{m_1,m_2} \innerproduct{JM}{j_1j_2;m_1m_2}\innerproduct{j_1j_2;m_1m_2} {j_1j_2J^\prime M^\prime} &=& \delta_{JJ^\prime} \delta_{MM^\prime}\label{E30} \end{eqnarray} These relations can also be seen as a consequence of the completeness formula such as \begin{equation} \sum_{JM} \ket{JM}\bra{JM} = \hat{I} \label{E31} \end{equation} and a similar relation for the vectors $\ket{j_1j_2;m_1m_2}.$ The Clebsch Gordon coefficients for addition of angular momenta are tabulated. Two such tables for $j_2=\frac{1}{2}$ and $j_2=1$ are given on the next page.