I understand the standard theory of angular momentum and the reps. of SU(2). I found there are some zero elements in CG coefficient table. I can derive them by using the recurrence relations between CG coefficients.But I am wondering is there any physical picture rather than mathematical derivation behind that. Namely, I am wondering is there any perspectives due to symmetry or whatever to argue the zero elements in CG table?

CG coefficient table from Griffiths' QM textbook


2 Answers 2


This is a cute question. You have inadvertently opened an interesting can of worms.

Why some CGs have zeroes is an ongoing research topic. In some cases the reason is trivial. The CGs $C^{jm_3}_{j_1m_1;j_2m_2}$ are $0$ when the triangularity condition $\vert j_1-j_2\vert \le j \le j_1+j_2$ is not met. There is also the condition on the magnetic number: the CG is $0$ unless $m_1+m_2=m_3$.

There are also some symmetry considerations. Since $$ C^{lm_3}_{j_1m_1;j_2m_2}=(-1)^{j_1+j_2-j_3}C^{j,-m_3}_{j_1,-m_1;j_2,-m_2} \tag{1} $$ it follows that $C^{l0}_{l_10;l_20}=0$ whenever $l_1+l_2-l_3$ is odd. Note that Eq.(1) is valid in general but, if $m_i=0$, then the angular momenta must be integers hence the specialization from $j_i\to l_i$ in $C^{l0}_{l_10;l_20}=0$. This is the case of your question. These CGs are $0$ because of parity arguments, which can be seem using combination of spherical harmonics.

Indeed: $$ Y^{l_1}_{m_1}(\theta,\varphi)Y^{l_2}_{m_2}(\theta,\varphi) =\sum_{l_3} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}}C^{l0}_{l_10;l_20}C^{lm_3}_{l_1m_1;l_2m_2} Y^{l_3}_{m_3}(\theta,\varphi)\, . $$ The parity of a spherical harmonics $Y^{l_1}_{m_1}(\theta,\varphi)$ is $(-1)^{l_1}$ so the left hand side has parity $(-1)^{l_1+l_2}$ hence the right hand side must have the same parity for all $m_1,m_2,m_3$. The parity requirement is then enforced by the CG $C^{l0}_{l_10;l_20}$.

More interestingly, there are "accidental" zeros, the origin of which are not always clear. To declutter the notation I will use $$ C^{jm_3}_{j_1m_1;j_2m_2}:= \left\langle\begin{array}{cc|c} j_1&j_2&j_3\\ m_1&m_2&m_3\end{array}\right\rangle $$

In Table 1 of

Srinivasa Rao, K., 1985. A note on the classification of the zeros of angular momentum coefficients. Journal of Mathematical Physics, 26(9), pp.2260-2261,

some accidental zeroes are listed and explained in terms of subgroup chains.


Heim, T.A., Hinze, J. and Rau, A.R.P., 2009. Some classes of ‘nontrivial zeroes’ of angular momentum addition coefficients. Journal of Physics A: Mathematical and Theoretical, 42(17), p.175203.

the authors show that CGs of the type $$ \left\langle\begin{array}{cc|c} d^2 & 2 d^2-d & d^2 \\ \frac{1}{2} (d-1) (d+2) & \frac{1}{2} (1-d) (d+2) & 0\end{array}\right\rangle $$ are $0$ for $d=1,2,3$. The $d=3$ is may be due to the vanishing of some quadrupole coupling in nuclear physics, but it's not clear if there's a physics-only reason why this coupling should vanish.


Brudno, S. and Louck, J.D., 1985. Nontrivial zeros of weight 1 3 j and 6 j coefficients: Relation to Diophantine equations of equal sums of like powers. Journal of Mathematical Physics, 26(9), pp.2092-2095.

the authors show that CGs of the type $$ \left\langle\begin{array}{cc|c} \frac{u+x}{2} & \frac{v+y}{2} & \frac{1}{2} (u+v+x+y-2) \\ \frac{x-u}{2} & \frac{y-v}{2} & \frac{1}{2} (x+y-u-v)\end{array}\right\rangle $$ are $0$ whenever the integers $x,y,u,v$ satisfy $(x+u)(y-v)=(x-u)(y+v)$. A non-trivial instance of this is $x=2,y=16,u=1,v=8$, which gives $$ \left\langle\begin{array}{cc|c} \frac{3}{2} & 12 & \frac{25}{2} \\ \frac{1}{2} & 4 & \frac{9}{2}\end{array}\right\rangle=0\, . $$ There is apparently no physics-only reason for this type of zeroes. These papers are just a sample of literature on the topic.

Disclaimer: this answer was NOT AI-generated.

  • $\begingroup$ the disclaimer felt like a punch to the gut, honestly. Is this the future of PSE? $\endgroup$ Jun 6 at 21:13
  • $\begingroup$ @AccidentalFourierTransform I hope not. $\endgroup$ Jun 6 at 21:46
  • $\begingroup$ Chatbots are nowhere near as circumspect; the are more likely to hedge and weasel... But they might learn, in time, to fake-use this disclaimer? $\endgroup$ Jun 8 at 19:46

Indeed, such zeros are the analog of nodes in continuous antisymmetric functions.

A sufficient condition for zeros for the special $m_1=m_2=m_3=0$ cases is that $j_1+j_2+j_3$ be a an odd integer, but this is not necessary, as you may see in the 2×2 table. The symmetry properties of the 3-j symbols are very rich; see, e.g., $j_1=j_2=3/2$, $j_3=2$, $m_1=m_2=1/2$, $m_3=1$.

Ultimately, such zeros are due to antisymmetry in the spherical basis. When you compose two spins, starting from the highest possible $m_3$, symmetric, and descending by a lowering operator, you do not change this symmetry at the lower rung, so there must be a new, antisymmetric, state orthogonal to the original symmetric one, and so on. Symmetric irreps intercalate with the antisymmetric ones, and the antisymmetric ones necessarily vanish at zero m s, for example.

An easy way to see this is in the Clebsch resolution of the composition (tensoring) of two three-vectors, $\vec a$ and $\vec b$, resolving to a symmetric traceless rank 2-tensor, an antisymmetric axial vector, and a symmetric scalar, of spins 2; 1;0, respectively, $$ T_{ij}={a_ib_j + a_j b_i\over 2}-\delta_{ij} {a_kb_k\over 3} ,\\ A_{ij}= {a_ib_j - a_j b_i\over 2} =\epsilon^{ijk}\left (\epsilon_{krs}{a_rb_s - a_s b_r\over 2} \right ) ,\\ S_{ij}= {\delta_{ij}\over 3} a_kb_k , $$ all mutually orthogonal.

It is evident that $A_{33}=0$, as you see in your 1×1 CG table.


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