0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – hft
    Commented Nov 8, 2022 at 20:17
  • $\begingroup$ 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). $\endgroup$ Commented Nov 8, 2022 at 20:44

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.)


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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.