4.7: The general problem

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The problem of adding two arbitrary angular moment ${\bf J}_1$ and ${\bf J}_2$ amounts to finding a unitary transformation from the set of basis vectors defined by the tensor products of the individual eigenstates of $J_1^2$ and $J_{1z}$ and $J_2^2$ and $J_{2z}$ to the eigenstates of $J^2$ and $J_z$, where

$\begin{displaymath} {\bf J}= {\bf J}_1 + {\bf J}_2 \end{displaymath}$

The individual eigenstates of 1 and 2 satisfy

$\displaystyle J_1^2\vert j_1\;\;m_1\rangle$	$\textstyle =$	$\displaystyle j_1(j_1+1)\hbar^2\vert j_1\;\;m_1\rangle$
$\displaystyle J_{1z}\vert j_1\;\;m_1\rangle$	$\textstyle =$	$\displaystyle m_1\hbar\vert j_1\;\;m_1\rangle$
$\displaystyle J_2^2\vert j_2\;\;m_2\rangle$	$\textstyle =$	$\displaystyle j_2(j_2+1)\hbar^2\vert j_2\;\;m_2\rangle$
$\displaystyle J_{2z}\vert j_2\;\;m_2\rangle$	$\textstyle =$	$\displaystyle m_2\hbar\vert j_2\;\;m_2\rangle$

We may also define raising and lowering operators according to

$\displaystyle J_{1\pm}\vert j_1\;\;m_1\rangle$	$\textstyle =$	$\displaystyle \hbar\sqrt{j_1(j_1+1)-m_1(m_1\pm 1)}\vert j_1\;\;m_1\pm 1\rangle$
$\displaystyle J_{2\pm}\vert j_2\;\;m_2\rangle$	$\textstyle =$	$\displaystyle \hbar\sqrt{j_2(j_2+1)-m_2(m_2\pm 1)}\vert j_2\;\;m_2\pm 1\rangle$

We then define the tensor product basis vectors as

$\begin{displaymath} {\vert j_1\;\;m_1;j_2\;\;m_2\rangle}= \vert j_1\;\;m_1\rangl... ...2\rangle \;\;\;\;\;\;\;\;m_1=-j_1,...,j_1,\;\;m_2=-j_2,...,j_2 \end{displaymath}$

and we seek a transformation to a set basis set denoted ${\vert J\;\;M\rangle}$ that satisfies

$\displaystyle J^2{\vert J\;\;M\rangle}$	$\textstyle =$	$\displaystyle J(J+1)\hbar^2{\vert J\;\;M\rangle}$
$\displaystyle J_z{\vert J\;\;M\rangle}$	$\textstyle =$	$\displaystyle M\hbar{\vert J\;\;M\rangle}$
$\displaystyle J_{\pm}{\vert J\;\;M\rangle}$	$\textstyle =$	$\displaystyle \hbar\sqrt{J(J+1)-M(M\pm 1)}\vert J\;\;M\pm 1\rangle$

The method of obtaining the transformation is simply to expand the new basis vectors in terms of the old:

$\begin{displaymath} {\vert J\;\;M\rangle}= \sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2}^... ...\;m_2\rangle}{\langle j_1\;\;m_1;j_2\;\;m_2\vert}J\;\;M\rangle \end{displaymath}$

The coefficients

$\begin{displaymath} {\langle j_1\;\;m_1;j_2\;\;m_2\vert}{J\;\;M\rangle} \end{displaymath}$

are the general Clebsch-Gordan coefficients. In principle, they can be determined by the programmatic procedure outlined in the last section applied to the arbitrary angular momenta. Note that

$\begin{displaymath} {\langle j_1\;\;m_1;j_2\;\;m_2\vert}{J\;\;M\rangle}=0 \end{displaymath}$

unless $m_1+m+2=M$, which restricts the summations in the above expansion considerably.

Although a general formula exists for Clebsch-Gordon coefficients (see below), let us first examine some of the properties of the coefficients that are useful in constructing the unitary transformation:

1.

The Clebsch-Gordon coefficients are real:

$\begin{displaymath} {\langle j_1\;\;m_1;j_2\;\;m_2\vert}{J\;\;M\rangle}= {\langle J\;\;M}{\vert j_1\;\;m_1;j_2\;\;m_2\rangle} \end{displaymath}$

2.

Orthogonality:

$\begin{displaymath} \sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2}^{j_2} {\langle J\;\;M}... ...m_1;j_2\;\;m_2\vert}J'\;\;M'\rangle = \delta_{JJ'}\delta_{MM'} \end{displaymath}$

This can be seen by recognizing that

$\begin{displaymath} \langle J\;\;M\vert J'\;\;M'\rangle = \delta_{JJ'}\delta_{MM'} \end{displaymath}$

so that if we insert an identity operator in the inner product in the form

$\begin{displaymath} I = \sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2}^{j_2} {\vert j_1\;\;m_1;j_2\;\;m_2\rangle}{\langle j_1\;\;m_1;j_2\;\;m_2\vert} \end{displaymath}$

the orthogonality relation results. A similar orthogonality relation is

$\begin{displaymath} \sum_{J=\vert j_1-j_2\vert}^{j_1+j_2}\sum_{M=-J}^{J} {\langl... ...\;m_1';j_2\;\;m_2'\rangle = \delta_{m_1 m_1'}\delta_{m_2 m_2'} \end{displaymath}$

which can also be proved starting from

$\begin{displaymath} {\langle j_1\;\;m_1;j_2\;\;m_2\vert}j_1\;\;m_1';j_2\;\;m_2'\rangle = \delta_{m_1 m_1'}\delta_{m_2 m_2'} \end{displaymath}$

and inserting identity in the form

$\begin{displaymath} I = \sum_{J=\vert j_1-j_2\vert}^{j_1+j_2}\sum_{M=-J}^{J} {J\;\;M\rangle}{\langle J\;\;M}\vert \end{displaymath}$

Note that the label $J$ is not a fixed label like $j_1$ and $j_2$. This is because different total $J$ values can result. The minimum value of $J$ is clearly $\vert j_1-j_2\vert$, while its maximum is $j_1+j_2$, and we need to sum over these in the completeness relation.

3.

Recursion relation:

$\displaystyle \sqrt{J(J+1)-M(M\pm 1)}{\langle j_1\;\;m_1;j_2\;\;m_2\vert}\vert J\;\;M\pm 1\rangle$	$\textstyle =$	$\displaystyle \sqrt{j_1(j_1+1)-m_1(m_1\mp 1)}\langle j_1\;\;m_1\pm 1;j_2\;\;m_2\vert{J\;\;M\rangle}$

	$\textstyle +$	$\displaystyle \sqrt{j_2(j_2+1)-m_2(m_2\mp 1)}\langle j_1\;\;m_1;j_2\;\;m_2\mp 1\vert{J\;\;M\rangle}$

which can be derived starting from

$\begin{displaymath} J_{\pm} = J_{1\pm} + J_{2\mp} \end{displaymath}$

and taking matrix elements of both sides between the new and old basis vectors:

$\begin{displaymath} {\langle j_1\;\;m_1;j_2\;\;m_2\vert}J_{\pm}{\vert J\;\;M\ran... ...angle j_1\;\;m_1;j_2\;\;m_2\vert}J_{2\pm}{\vert J\;\;M\rangle} \end{displaymath}$

On the left side, $J_{\pm}$ acts on ${\vert J\;\;M\rangle}$ to produce the term on the left in the recursion relation. On the right, the operators $J_{1\pm}$ and $J_{2\pm}$ operate to the left as $J^{\dagger}_{1\pm}$ and $J^{\dagger}_{2\pm}$. However,

$\displaystyle J^{\dagger}_{1\pm}$	$\textstyle =$	$\displaystyle J_{1\mp}$
$\displaystyle J^{\dagger}_{2\pm}$	$\textstyle =$	$\displaystyle J_{2e\mp}$

and hence produce the opposite action as the $J_{\pm}$ on the left.

Finally, the general formula for the Clebsch-Gordan coefficients is

$\displaystyle {\langle j_1\;\;m_1;j_2\;\;m_2\vert}{J\;\;M\rangle}$	$\textstyle =$	$\displaystyle \delta_{m_1+m_2,M}$
	$\textstyle \times$	$\displaystyle \left[(2J+1){(s-2J)!(s-2j_2)!(s-2j_1)! \over (s+1)!} (j_1+m_1)!(j_1-m_1)!(j_2+m_2)!(j_2-m_2)!(J+M)!(J-M)!\right]^{1/2}$

	$\textstyle \times$	$\displaystyle \sum_{\nu}(-1)^{\nu} {1 \over \nu!(j_1+j_2-J-\nu)!(j_1-m_1-\nu)!(j_2+m_2-\nu)!(J-j_2+m_1+\nu)!(J-j_1-m_2+\nu)!}$

where $s=j_1+j_2+J$, and the $\nu$ summation runs over all values for which all of the factorial arguments are greater than or equal to 0.

This formula is rather cumbersome to work with, so it is useful to deduce some special cases. These are as follows:

i.

$\begin{displaymath} \langle j_1\;\;j_1;j_2\;\;j_2\vert J\;\;J\rangle = 1 \end{displaymath}$

ii.

if $m_1=\pm j_1$ or $m_2=\pm j_2$ and $M=\pm J$, then

$\displaystyle {\langle j_1\;\;m_1;j_2\;\;m_2\vert}J\;\;J\rangle$	$\textstyle =$	$\displaystyle \langle j_1\;\;-m_1;j_2\;\;-m_2\vert J\;\;-J\rangle$

	$\textstyle =$	$\displaystyle (-1)^{j_1-m_1} \sqrt{{(2J+1)!(j_1+j_2-J)! \over (j_1+j_2+J+1)!(J+j_1-j_2)!(J-j_1+j_2)!}} \sqrt{{(j_1+m_1)!(j_2+m_2)! \over (j_1-m_1)!(j_2-m_2)!}}$

iii.

If $j_1+j_2=J$,

$\begin{displaymath} {\langle j_1\;\;m_1;j_2\;\;m_2\vert}{J\;\;M\rangle}= \sqrt{{... ...{(J+M)!(J-M)! \over (j_1+m_1)!(j_1-m_1)!(j_2+m_2)!(j_2-m_2)!}} \end{displaymath}$

\(\displaystyle J_1^2\vert j_1\;\;m_1\rangle\)	\(\textstyle =\)	\(\displaystyle j_1(j_1+1)\hbar^2\vert j_1\;\;m_1\rangle\)
\(\displaystyle J_{1z}\vert j_1\;\;m_1\rangle\)	\(\textstyle =\)	\(\displaystyle m_1\hbar\vert j_1\;\;m_1\rangle\)
\(\displaystyle J_2^2\vert j_2\;\;m_2\rangle\)	\(\textstyle =\)	\(\displaystyle j_2(j_2+1)\hbar^2\vert j_2\;\;m_2\rangle\)
\(\displaystyle J_{2z}\vert j_2\;\;m_2\rangle\)	\(\textstyle =\)	\(\displaystyle m_2\hbar\vert j_2\;\;m_2\rangle\)

\(\displaystyle J_{1\pm}\vert j_1\;\;m_1\rangle\)	\(\textstyle =\)	\(\displaystyle \hbar\sqrt{j_1(j_1+1)-m_1(m_1\pm 1)}\vert j_1\;\;m_1\pm 1\rangle\)
\(\displaystyle J_{2\pm}\vert j_2\;\;m_2\rangle\)	\(\textstyle =\)	\(\displaystyle \hbar\sqrt{j_2(j_2+1)-m_2(m_2\pm 1)}\vert j_2\;\;m_2\pm 1\rangle\)

\(\displaystyle J^2{\vert J\;\;M\rangle}\)	\(\textstyle =\)	\(\displaystyle J(J+1)\hbar^2{\vert J\;\;M\rangle}\)
\(\displaystyle J_z{\vert J\;\;M\rangle}\)	\(\textstyle =\)	\(\displaystyle M\hbar{\vert J\;\;M\rangle}\)
\(\displaystyle J_{\pm}{\vert J\;\;M\rangle}\)	\(\textstyle =\)	\(\displaystyle \hbar\sqrt{J(J+1)-M(M\pm 1)}\vert J\;\;M\pm 1\rangle\)

\(\displaystyle \sqrt{J(J+1)-M(M\pm 1)}{\langle j_1\;\;m_1;j_2\;\;m_2\vert}\vert J\;\;M\pm 1\rangle\)	\(\textstyle =\)	\(\displaystyle \sqrt{j_1(j_1+1)-m_1(m_1\mp 1)}\langle j_1\;\;m_1\pm 1;j_2\;\;m_2\vert{J\;\;M\rangle}\)

	\(\textstyle +\)	\(\displaystyle \sqrt{j_2(j_2+1)-m_2(m_2\mp 1)}\langle j_1\;\;m_1;j_2\;\;m_2\mp 1\vert{J\;\;M\rangle}\)

\(\displaystyle J^{\dagger}_{1\pm}\)	\(\textstyle =\)	\(\displaystyle J_{1\mp}\)
\(\displaystyle J^{\dagger}_{2\pm}\)	\(\textstyle =\)	\(\displaystyle J_{2e\mp}\)

\(\displaystyle {\langle j_1\;\;m_1;j_2\;\;m_2\vert}{J\;\;M\rangle}\)	\(\textstyle =\)	\(\displaystyle \delta_{m_1+m_2,M}\)
	\(\textstyle \times\)	$\displaystyle \left[(2J+1){(s-2J)!(s-2j_2)!(s-2j_1)! \over (s+1)!} (j_1+m_1)!(j_1-m_1)!(j_2+m_2)!(j_2-m_2)!(J+M)!(J-M)!\right]^{1/2}$

	\(\textstyle \times\)	$\displaystyle \sum_{\nu}(-1)^{\nu} {1 \over \nu!(j_1+j_2-J-\nu)!(j_1-m_1-\nu)!(j_2+m_2-\nu)!(J-j_2+m_1+\nu)!(J-j_1-m_2+\nu)!}$

\(\displaystyle {\langle j_1\;\;m_1;j_2\;\;m_2\vert}J\;\;J\rangle\)	\(\textstyle =\)	\(\displaystyle \langle j_1\;\;-m_1;j_2\;\;-m_2\vert J\;\;-J\rangle\)

	\(\textstyle =\)	$\displaystyle (-1)^{j_1-m_1} \sqrt{{(2J+1)!(j_1+j_2-J)! \over (j_1+j_2+J+1)!(J+j_1-j_2)!(J-j_1+j_2)!}} \sqrt{{(j_1+m_1)!(j_2+m_2)! \over (j_1-m_1)!(j_2-m_2)!}}$