# Clebsch-Gordan Coefficient Recursion Derivation

I've got a question regarding a very particular recursion relation that Condon and Shortley use in their book "The Theory of Atomic Spectra," (TAS) which Racah then uses to derive the algebraic form of the Clebsch-Gordan coefficients (CGs).

The part that confuses me is buried in TAS chapters $$8^3$$ to $$9^3$$, and specifically I am confused concerning equation $$10^3 1$$

Condon and Shortley derive the properties of a generic vector operator T which has the commutation rules of an angular momentum operator. Namely, that:

$$[J_i , T_i] = 0$$

$$[J_i , T_j] = i\hbar T_k$$

$$[J_i , T_k] = -i\hbar T_j$$

Where each of these is valid in all permutations.

While doing this, in chapter $$9^3$$: "Dependence of the matrix T on ", they conjure up a symbol which they define as $$(\alpha j\vdots\ T \vdots \alpha'j)$$, as follows for j = j'

$$(\alpha j\vdots\ T \vdots \alpha'j) = \frac{(\alpha J m|\mathcal{T}|\alpha' J m+1)}{\sqrt{(j-m)(j+m+1)}} = \frac{(\alpha J m-1|\mathcal{T}|\alpha' J m)}{\sqrt{(j-m+1)(j+m)}}$$ Utilizing this shorthand, the authors proceed to find the "non-vanishing matrix components of T" in equation $$9^3 11$$:

$$(\alpha jm| \textbf{T} | \alpha'j+1m\pm 1) = \mp (\alpha j\vdots\ T \vdots \alpha'j+1)\frac{1}{2}\sqrt{(j\pm m+1)(j\pm m+2)}(\textbf{i} \pm i \textbf{j})$$ $$(\alpha jm| \textbf{T} | \alpha'j+1m) = (\alpha j\vdots\ T \vdots \alpha'j+1)\sqrt{(j+1)^2 -m^2} \textbf{k}$$ $$(\alpha jm| \textbf{T} | \alpha'jm\pm 1) = (\alpha j\vdots\ T \vdots \alpha'j+1)\frac{1}{2}\sqrt{(j\mp m)(j\pm m+1)}(\textbf{i} \pm i \textbf{j})$$ $$(\alpha jm| \textbf{T} | \alpha'jm) = (\alpha j\vdots\ T \vdots \alpha')m \textbf{k}$$ $$(\alpha jm| \textbf{T} | \alpha'j-1m\pm 1) = \pm (\alpha j\vdots\ T \vdots \alpha'j-1)\frac{1}{2}\sqrt{(j\mp m)(j\pm m-1)}(\textbf{i} \pm i \textbf{j})$$ $$(\alpha jm| \textbf{T} | \alpha'j-1m) = (\alpha j\vdots\ T \vdots \alpha'j-1)\sqrt{(j^2 - m^2 }\textbf{k}$$

On the following page, after mentioning that

"if $$T_x$$, $$T_y$$, and $$T_z$$ commute with $$\textbf{J}^2$$, the matrix $$(\alpha j\vdots T \vdots \alpha'j)$$ is diagonal in j; if they commute with A it is diagonal in $$\alpha$$. These remarks hold in particular for $$J_x$$, $$J_y$$, and $$J_z$$, for which

$$(\alpha j\vdots J \vdots \alpha'j') = \hbar \delta_{jj'} \delta_{\alpha \alpha'} "$$

On the very same page, the authors then attempt to derive $$(j_1 j_2 j\vdots J_1 \vdots j_1 j_2 j')$$ and $$(j_1 j_2 j\vdots J_2 \vdots j_1 j_2 j')$$ for the case j = j', and it is said that:

from $$9^3 11$$

$$j(j+1)\hbar (j_1 j_2 j \vdots J_{1} \vdots j_1 j_2 j) = (j_1 j_2 j m | \textbf{J}_1 \cdot \textbf{J} | j_1 j_2 jm)$$

But

$$\textbf{J}_2^2 = (\textbf{J} - \textbf{J}_1)^2 = \textbf{J}^2 - 2\textbf{J}_1 \cdot \textbf{J} + \textbf{J}_1^2;$$

$$\textbf{J}_1 \cdot \textbf{J} = \frac{1}{2} (\textbf{J}_1^2 - \textbf{J}_2^2+ \textbf{J}^2)$$

Using this relation gives,

$$(\gamma j_1 j_2 j\vdots J_2 \vdots \gamma j_1 j_2 j) = \frac{j_1 (j_1+1)-j_2(j_2+1)+j(j+1)}{2j(j+1)}\hbar$$

"and the corresponding element of $$J_2$$ is obtained by interchanging subscripts 1 and 2"

Ok! So from here on out I'm fine. Utilizing that equation and moving forward to derive CG doesn't pose as much of a problem as the idea that I can derive

$$j(j+1)\hbar (j_1 j_2 j \vdots J_{1} \vdots j_1 j_2 j) = (j_1 j_2 j m | \textbf{J}_1 \cdot \textbf{J} | j_1 j_2 jm)$$

and

$$(\gamma j_1 j_2 j\vdots J_2 \vdots \gamma j_1 j_2 j) = \frac{j_1 (j_1+1)-j_2(j_2+1)+j(j+1)}{2j(j+1)}\hbar$$

from the relations $$9^3 11$$.

I have no idea how they do this. If anyone is familiar with Condon and Shortley's text and understands this step please help!

All help is greatly appreciated!

• what is this $9^3 11$ stuff? Sep 20 '18 at 1:43
• In the actual book in question, this is how they dictate chapter, subsection, and equations, for instance: Chapter 9, subsection 3 is $9^3$, and the 11th numbered equation in that chapter is denoted $9^3 11$ Does this make sense? I don't personally think so, but in the off chance that I managed to reach someone who had studied out of the same text, I wanted them to be able to refer to it. Sep 20 '18 at 1:51
• This notation is a world’s first for me... Sep 20 '18 at 2:48
• I have to agree with you. The whole thing is confusing to me. But Racah and Zare both reference it, in addition to the many early literature publications about LS and SS coupling. It's a very weird thing to read, but I'm positive that it makes more sense than I can see. That's why I thought I'd just put it out there and see if anyone knows. Sep 20 '18 at 3:18
• @ZeroTheHero : No. The notation $\,n^m\,$ means Chapter $\,m\,$ Section $\,n$. See here : imgur.com/a/4c2dChE Sep 20 '18 at 13:30

My original question was how to derive the following equation: $$(\gamma j_1 j_2 j\vdots J_2 \vdots \gamma j_1 j_2 j) = \frac{j_1 (j_1+1)-j_2(j_2+1)+j(j+1)}{2j(j+1)}\hbar$$

From the relations: $$(\alpha j\vdots\ T \vdots \alpha'j) = \frac{(\alpha J m|\mathcal{T}|\alpha' J m+1)}{\sqrt{(j-m)(j+m+1)}} = \frac{(\alpha J m-1|\mathcal{T}|\alpha' J m)}{\sqrt{(j-m+1)(j+m)}}$$

where $$\mathcal{T} = T_x -iT_y$$

and

\begin{align*} &(\alpha jm| \textbf{T} | \alpha'j+1m\pm 1) = \mp (\alpha j\vdots\ T \vdots \alpha'j+1)\frac{1}{2}\sqrt{(j\pm m+1)(j\pm m+2)}(\textbf{i} \pm i \textbf{j})\\ &(\alpha jm| \textbf{T} | \alpha'j+1m) = (\alpha j\vdots\ T \vdots \alpha'j+1)\sqrt{(j+1)^2 -m^2} \textbf{k}\\ &(\alpha jm| \textbf{T} | \alpha'jm\pm 1) = (\alpha j\vdots\ T \vdots \alpha'j+1)\frac{1}{2}\sqrt{(j\mp m)(j\pm m+1)}(\textbf{i} \pm i \textbf{j})\\ &(\alpha jm| \textbf{T} | \alpha'jm) = (\alpha j\vdots\ T \vdots \alpha')m \textbf{k}\\ &(\alpha jm| \textbf{T} | \alpha'j-1m\pm 1) = \pm (\alpha j\vdots\ T \vdots \alpha'j-1)\frac{1}{2}\sqrt{(j\mp m)(j\pm m-1)}(\textbf{i} \pm i \textbf{j})\\ &(\alpha jm| \textbf{T} | \alpha'j-1m) = (\alpha j\vdots\ T \vdots \alpha'j-1)\sqrt{(j^2 - m^2 }\textbf{k}\\ \end{align*}

In the case where $$\textbf{T}$$ is a general operator satisfying the canonical angular momentum commutation relations with the generic angular momentum operator $$\textbf{J}$$, namely these:

\begin{align*} &[J_i , T_i] = 0\\ &[J_i , T_j] = i\hbar T_k\\ &[J_i , T_k] = -i\hbar T_j\\ \end{align*}

Where standard permutation of the indices applies.

This question is important because Racah ($$\textit{Phys. Rev.}$$ $$\textbf{62}$$ 438 1942) and by extension, Zare (Angular Momentum 1988) utilize this result (among others) to determine the algebraic expression for a general Clebsch-Gordon coefficient (CGc). Being able to calculate any CGc is useful when applying the Wigner-Eckhart theorem to spherical tensor operators such as the Dipole and Quadrupole operators that are important in spectroscopic transitions in Chemistry and physics.

After staring at Condon & Shortley's $$\textit{Theory of Atomic Spectra}$$ (TAS) for an embarrassing number of days, I believe that I've figured it out. So here goes:

This entire proof is based around finding the form of the matrix elements of this generic operator $$\textbf{T}$$ between two states, so we must first familiarize ourselves with the allowed values and selection rules that govern angular momenta, and which will facilitate this solution:

As a consequence of the commutation rules above and the general commutation rule $$[\textbf{A} \cdot \textbf{B}, \textbf{C}] = \textbf{A} \cdot [\textbf{B},\textbf{C}]- [\textbf{C}, \textbf{A}] \cdot \textbf{B}$$

\begin{align*} &[\textbf{T}_1 \cdot \textbf{T}_2, \textbf{J}] = \textbf{T}_1 \cdot [\textbf{T}_2,\textbf{J}]- [\textbf{J},\textbf{T}_1] \cdot \textbf{T}_2\\ &= \textbf{T}_1 (-i \hbar \textbf{T}_2) - (-i \hbar \textbf{T}_1 \textbf{T}_2)\\ &= -i \hbar \textbf{T}_1 \textbf{T}_2 + i \hbar \textbf{T}_1 \textbf{T}_2\\ &= 0 \\ \end{align*}\\

Such that any products of two $$\textbf{T}$$ operators commute with $$\textbf{J}$$. We then expand the commutation relations $$[\textbf{J}^2 , [\textbf{J}^2,\textbf{T}]]$$ in two ways to obtain a useful equality:

\begin{align*} &[\textbf{J}^2,\textbf{T}] = -2 i\hbar(\textbf{J} \times \textbf{T} - i\hbar \textbf{T})\\ \end{align*}\\

such that

\begin{align*} &[\textbf{J}^2 , [\textbf{J}^2,\textbf{T}]] = 2\hbar^2 (\textbf{J}^2 \textbf{T} + \textbf{T} \textbf{J}^2) - 4\hbar^2 \textbf{J} (\textbf{J} \cdot \textbf{T}) \\ \end{align*}\\

We also know that because $$[\textbf{J}^2,\textbf{T}] = (\textbf{J}^2\textbf{T}-\textbf{T}\textbf{J}^2)$$, this is equivalent to:

\begin{align*} &[\textbf{J}^2 , (\textbf{J}^2\textbf{T}-\textbf{T}\textbf{J}^2)] = \textbf{J}^4 \textbf{T} - 2\textbf{J}^2 \textbf{T}\textbf{J}^2 +\textbf{T} \textbf{J}^4 \\ \end{align*}\\

Such that

\begin{align*} &\textbf{J}^4 \textbf{T} - 2\textbf{J}^2 \textbf{T}\textbf{J}^2 +\textbf{T} \textbf{J}^4 =2\hbar^2 (\textbf{J}^2 \textbf{T} + \textbf{T} \textbf{J}^2) - 4\hbar^2 \textbf{J} (\textbf{J} \cdot \textbf{T})\\ \end{align*}\\

Now we know that $$\textbf{J} \cdot \textbf{T}$$ commutes with $$\textbf{J}$$, and if we take the matrix element of this expression between two states $$\langle \alpha j m|$$ and $$|\alpha' j' m'\rangle$$ where $$j \neq j'$$, this term ($$\textbf{J} (\textbf{J} \cdot \textbf{T}$$) vanishes, leaving behind the following as we take $$j(j+1)$$ to be the eigenvalue of $$\textbf{J}^2$$, and $$j^2(j+1)^2$$ to be the eigenvalue of $$\textbf{J}^4$$:

\begin{align*} &\langle \alpha j m|(\textbf{J}^4 \textbf{T} - 2\textbf{J}^2 \textbf{T}\textbf{J}^2 +\textbf{T} \textbf{J}^4)|\alpha' j' m'\rangle =\langle \alpha j m|(2\hbar^2 (\textbf{J}^2 \textbf{T} + \textbf{T} \textbf{J}^2) - 4\hbar^2 \textbf{J} (\textbf{J} \cdot \textbf{T}))|\alpha' j' m'\rangle\\ &= \hbar^4[j^2(j+1)^2-2j(j+1)j'(j'+1)+j'^2(j'+1)^2]\langle \alpha j m|\textbf{T}|\alpha' j' m'\rangle = 2\hbar^4[j(j+1)+j'(j'+1)]\langle \alpha j m|\textbf{T}|\alpha' j' m'\rangle \end{align*}\\

And from some rearrangement of this, we can find that in order to have a non-vanishing $$\langle \alpha j m|\textbf{T}|\alpha' j' m'\rangle$$ we must have $$j'-j = 0,\pm1$$

A more important result for us is that if we take the matrix element between states with the same value of j instead, we get a different expression, where the value

$$\hbar^4[j^2(j+1)^2-2j(j+1)j(j+1)+j^2(j+1)^2] = 0$$

the value

$$(\textbf{J}^2 \textbf{T} + \textbf{T} \textbf{J}^2) = 2(\textbf{J}^2 \textbf{T})$$

and we retain the term $$4\hbar^2 \textbf{J} (\textbf{J} \cdot \textbf{T})$$ This gives us, upon dotting with $$\langle \alpha j m|$$ and $$|\alpha' j m'\rangle$$,

$$j(j+1)\hbar^2\langle \alpha j m|\textbf{T}|\alpha' j m'\rangle = \langle \alpha j m|\textbf{J}|\alpha j m'\rangle \langle \alpha j m'|\textbf{J} \cdot \textbf{T} |\alpha' j m'\rangle$$

For any operator T, including when $$\textbf{T} = \textbf{S}$$, or $$\textbf{T} = \textbf{J}_1$$, the latter of which we will use.

Using this, we look at the case of the z-components of the matrix $$\textbf{T} = \textbf{J}_1$$ and $$\textbf{J}$$ Using what we've just found, we can write:

$$j(j+1)\hbar^2\langle j_1 j_2 j m|\textbf{J}_{1z}|j_1 j_2 j m\rangle = \langle j_1 j_2 j m|\textbf{J}_z|j_1 j_2 j m\rangle \langle j_1 j_2 j m|\textbf{J} \cdot \textbf{J}_1 |j_1 j_2 j m\rangle$$

We now make use of only one of the equations $$9^3 11$$ in my original question from Condon and Shortley

$$(j_1 j_2 j m| \textbf{T} | j_1 j_2 j m) = (j_1 j_2 j \vdots\ T \vdots j_1 j_2 j)m \textbf{k}$$

Where $$(j_1 j_2 j m|$$ and $$| j_1 j_2 j m)$$ are meant to be $$\langle j_1 j_2 j m|$$ and $$| j_1 j_2 j m\rangle$$, respectively.

A quick word on the derivation of this expression is warranted here. It is derived as follows: Define $$T_-$$ and $$J_-$$ as $$T_x - iT_y$$ and $$J_x - iJ_y$$, respectively. Then

\begin{align*} &[J_-^\dagger,T_-] = [J_x +iJ_y,T_-] = [J_-,T_-]+2i[J_y,T_x-iT_y] =2\hbar T_z\\ &2\hbar T_z = J_-^\dagger T_- - T_-J_-^\dagger \\ \end{align*}\\

Next, recall that $$m = m'$$ is the requirement for a nonvanishing matrix coefficient of $$T_z$$ because $$T_z$$ commutes with $$J_z$$. For the matrix coefficient where $$j = j'$$, we then have:

\begin{align*} &\langle \alpha j m| 2\hbar T_z | \alpha j m \rangle = \langle \alpha j m|J_-^\dagger T_-| \alpha j m \rangle - \langle \alpha j m|T_-J_-^\dagger| \alpha j m \rangle\\ &\langle \alpha j m| 2\hbar T_z | \alpha j m \rangle = \sqrt{(j+m)(j-m+1)} \langle \alpha j m-1|T_-| \alpha j m \rangle - \sqrt{(j-m)(j+m+1)}\langle \alpha j m|T_-| \alpha j m+1 \rangle \\ &\langle \alpha j m| 2\hbar T_z | \alpha j m \rangle = \sqrt{(j+m)(j-m+1)}\sqrt{(j+m)(j-m+1)} ( \alpha j m \vdots T \vdots \alpha j m ) - \sqrt{(j-m)(j+m+1)}\sqrt{(j-m)(j+m+1)}( \alpha j m \vdots T \vdots \alpha j m ) \\ &\langle \alpha j m| 2\hbar T_z | \alpha j m \rangle = (j+m)(j-m+1) ( \alpha j m \vdots T \vdots \alpha j m ) -(j-m)(j+m+1)( \alpha j m \vdots T \vdots \alpha j m )\\ &\langle \alpha j m| 2\hbar T_z | \alpha j m \rangle = 2m\hbar ( \alpha j m \vdots T \vdots \alpha j m )\\ \end{align*}\\

Substitution of $$(j_1 j_2 j m| \textbf{T} | j_1 j_2 j m) = (j_1 j_2 j \vdots\ T \vdots j_1 j_2 j)m \textbf{k}$$ then leads to:

$$j(j+1)\hbar^2(j_1 j_2 j \vdots\ J_1 \vdots j_1 j_2 j)m = \langle j_1 j_2 j m|\textbf{J}_z|j_1 j_2 j m\rangle \langle j_1 j_2 j m|\textbf{J} \cdot \textbf{J}_1 |j_1 j_2 j m\rangle$$

Next, the term $$\langle j_1 j_2 j m|\textbf{J}_z|j_1 j_2 j m\rangle$$ is expanded as its' definition, $$m\hbar$$, where we have used the fact that $$\textbf{J}_{1z}$$ and $$\textbf{J}_{z}$$ commute, such that $$m' = m$$. We now have:

$$j(j+1)\hbar^2(j_1 j_2 j\vdots\ J_1 \vdots j_1 j_2 j)m = m\hbar \langle j_1 j_2 j m|\textbf{J} \cdot \textbf{J}_1 |j_1 j_2 j m\rangle$$

or

$$j(j+1)\hbar(j_1 j_2 j\vdots\ J_1 \vdots j_1 j_2 j) = \langle j_1 j_2 j m|\textbf{J} \cdot \textbf{J}_1 |j_1 j_2 j m\rangle$$

The very last step is to realize that because $$\textbf{J} = \textbf{J}_1 + \textbf{J}_2$$, we can write:

\begin{align*} &\textbf{J}_2^2 = (\textbf{J}-\textbf{J}_1)^2 = \textbf{J}^2-2\textbf{J}_1 \cdot\textbf{J} +\textbf{J}_1^2\\ &\textbf{J}_1 \cdot\textbf{J} = \frac{1}{2} (\textbf{J}_1^2 -\textbf{J}_2^2 +\textbf{J}^2)\\ \end{align*}\\

Such that if we now explicitly compute the expectation value of the RHS we can get the following:

\begin{align*} &j(j+1)\hbar(j_1 j_2 j\vdots\ J_1 \vdots j_1 j_2 j) = \hbar^2 \frac{j_1(j_1+1)-j_2(j_2+1)+j(j+1)}{2}\\ &=(j_1 j_2 j\vdots\ J_1 \vdots j_1 j_2 j) = \hbar \frac{j_1(j_1+1)-j_2(j_2+1)+j(j+1)}{2j(j+1)}\\ \end{align*}\\

And we're finished.

• I had intended to wait for someone to come along and answer my question, but in the days that followed, found the answer myself. And in the case of the Relativistic Lagrangian for the EM field I attempted to credit another person with the answer (@bolbteppa), but was downvoted because they (strangely) decided to answer the post in a comment. I could not force them to post it again as an answer. In these cases, what should I have done? And I do not mean this condescendingly, I would really like to know the proper conduct in these scenarios. Sep 22 '18 at 23:30
• I will, though, de-select my own answers as the "best" ones on both posts, I can see how that may be poor conduct. Sep 22 '18 at 23:31
• May I ask why you use Racah as your source? There are much “cleaner” derivations of these results elsewhere. Sep 24 '18 at 3:42
• Originally the only reason that I used Racah was because Zare credited him with the algebraic determination of the CGcs. Throughout researching this question and reading these comments I have started to learn about these other derivations. Sep 24 '18 at 15:15