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

Question

I've been trying to prove the well-known identity for the following angular momentum state: \begin{equation} |\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. \end{equation} 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{equation} \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} \end{equation} 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.

Answer 1

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.

Answer 2

I've been trying to prove the well-known identity for the following angular momentum state: \begin{equation} |\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>. \end{equation}

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.)

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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.