# Proof of angular momentum on $3j$ Wigner state equal to zero

I've been trying to prove the well-known identity for the following angular momentum state: $$$$|\Psi\rangle = \sum_{m_1,m_2,m_3}\begin{pmatrix} j_1&j_2&j_3\\m_1&m_2&m_3 \end{pmatrix}|j_1 m_1\rangle|j_2 m_2\rangle|j_3 m_3\rangle=|0\ 0\rangle.$$$$ My idea was to calculate $$\langle J^2\rangle$$ on $$|\Psi\rangle$$ and wish for luck it comes out zero, implying that $$J=\sqrt{\langle J^2\rangle}=0$$. This is what I've got for now: $$$$\begin{split} \langle J^2\rangle = \langle\Psi|J^2|\Psi\rangle= \sum_{\substack{m_1,m_2,m_3 \\ m'_{1},m'_{2},m'_{3}}}\begin{pmatrix} j_1&j_2&j_3\\m^{'}_{1}&m^{'}_{2}&m^{'}_{3} \end{pmatrix}\begin{pmatrix} j_1&j_2&j_3\\m_1&m_2&m_3 \end{pmatrix}\times\langle{j_1 m^{'}_{1}|\langle j_2 m'_{2}|\langle j_3 m'_{3}} |J^2| j_1 m_1\rangle| j_2 m_2\rangle| j_3 m_3\rangle \end{split}$$$$ I am somewhat confused now, because I assumed from the definition of 3j symbol, that $$\vec{J}_1+\vec{J}_2=\vec{J}_3$$, so $$J^2$$ should produce on this state $$j_3(j_3+1)$$. But this doesn't lead to the desired result. I'd be very grateful for a hint.

• You seem to be adding three momenta to produce a fourth, not two to produce a third. So, it's not clear why you think $J_1 + J_2 = J_3$. Can you provide a citation/reference from which you are copying down this "well-known" result?
– hft
Commented Nov 8, 2022 at 20:17
• This is a problem 1.13 from "From Nucleons to Nucleus" by J. Suchonen, chapter 1. It has also been given in a few books among standard properties of 3j-symbol like orthogonality & completeness and raised in the post physics.stackexchange.com/questions/451706/… (without rigorous proof though). Commented Nov 8, 2022 at 20:44

The simplest way is to verify that $$J_\pm$$ acting on your state gives $$0$$. The only $$J$$-state killed by $$J_\pm$$ is $$\vert 00\rangle$$.

The 3j’s are defined by your equation, i.e. they are the coefficient needed to combine $$j_1,j_2$$ and $$j_3$$ so they give a scalar. In the vector notation this would be $$\vec J_1+\vec J_2+\vec J_3=0$$. In particular, this implies $$m_1+m_2+m_3=0$$.

The notation $$\vec J_1+\vec J_2=\vec J_3$$ is usually used to couple using Clebsch-Gordan coefficients. Since the only way to get $$J=0$$ from $$J_3$$ is by coupling with another $$J_3$$, the 3js can be understood as related to the double coupling $$(j_1\otimes j_2)\otimes j_3\to 0$$.

To be explicit, consider the scalar $$\vert 00\rangle$$ constructed using the double coupling $$(j_1\otimes j_2)\otimes j_3\to 0$$, which contains the product of CGs $$C_{j_1m_1;j_2m_2}^{j_3m_3} C_{j_3m_3;j_3,-m_3}^{00} =\frac{1}{\sqrt{2j_3+1}}(-1)^{j_3-m_3} C_{j_1m_1;j_2m_2}^{j_3,m_3}\, =\chi \left(\begin{array}{ccc} j_1 &j_2&j_3\\ m_1&m_2&m_3\end{array}\right)\, , \tag{1}$$ where $$\chi$$ is some phase. It should then be no surprise to find the relation $$C_{j_1m_1;j_2m_2}^{j_3m_3}=(-1)^{-j_1+j_2-m_3}\sqrt{2j_3+1} \left(\begin{array}{ccc}j_1&j_2&j_3\\ m_1&m_2&-m_3\end{array}\right)$$ where $$\chi=(-1)^{-j_1+j_2+j_3}$$ and independent of the magnetic quantum numbers. Eq.(1) is the a physicist’s proof or your original claim.

Note finally that, for CGs, we must have $$m_1+m_2=m_3$$, in contradistinction with the similar relation to 3j’s.

I've been trying to prove the well-known identity for the following angular momentum state: $$$$|\Psi> = \sum_{m_1,m_2,m_3}\begin{pmatrix} j_1&j_2&j_3\\m_1&m_2&m_3 \end{pmatrix}|j_1 m_1>|j_2 m_2>|j_3 m_3>=|0\ 0>.$$$$

This is the correct expression for how to add three angular momenta together in order to form a fourth (total) angular momentum state with zero squared angular momentum and zero z-axis angular momentum.

I am somewhat confused now, because I assumed from the definition of 3j symbol, that $$\vec{J}_1+\vec{J}_2=\vec{J}_3$$

It is not clear to which definition you are referring. You may be getting confused with the definition of the Clebsch-Gordan coefficients. For example, per Wikipedia, "The CG coefficients are defined so as to express the addition of two angular momentum in terms of a third... The 3-j symbols, on the other hand, are coefficients with which three angular momentum must be added so that the resultant is zero." (Emphasis Added.)

Anyways, the properties of the 3j-symbol tell you that you must have $$M = m_1 + m_2 + m_3 = 0$$, therefore we are at least in the right total $$S_z$$ subspace. We also have from the 3j-symbol properties that $$J = j_1 + j_2 + j_3$$ is an even integer.

The rest of the result can be taken as the definition of the 3j-symbol. But, if you want to prove it you can act with $$J^2 = (J_1 + J_2 + J_3)^2$$ and show that the result is zero.