Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

On conditions for Clebsh-Gordan coefficients to be zero

Posted by peeterjoot on November 23, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]


In section 28.2 of the text [1] is a statement that the Clebsh-Gordan coefficient

\begin{aligned}\left\langle{{m_1 m_2}} \vert {{jm}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.1)

unless m = m_1 + m_2. It appeared that it was related to the operation of J_z, but how exactly wasn’t obvious to me. In tutorial today we hashed through this. Here’s the details lying behind this statement

Recap on notation.

We are taking an arbitrary two particle ket and decomposing it utilizing an insertion of a complete set of states

\begin{aligned}{\left\lvert {jm} \right\rangle} = \sum_{m_1' m_2'} \Bigl({\left\lvert {j_1 m_1'} \right\rangle} {\left\lvert {j_2 m_2'} \right\rangle}{\left\langle {j_1 m_1'} \right\rvert} {\left\langle {j_2 m_2'} \right\rvert}\Bigr){\left\lvert {jm} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.2)

with j_1 and j_2 fixed, this is written with the shorthand

\begin{aligned}{\left\lvert {j_1 m_1} \right\rangle} {\left\lvert {j_2 m_2} \right\rangle} &= {\left\lvert {m_1 m_2} \right\rangle} \\ {\left\langle {j_1 m_1} \right\rvert} {\left\langle {j_2 m_2} \right\rvert} {\left\lvert {jm} \right\rangle} &= \left\langle{{m_1 m_2}} \vert {{jm}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(2.3)

so that we write

\begin{aligned}{\left\lvert {jm} \right\rangle} = \sum_{m_1' m_2'} {\left\lvert {m_1' m_2'} \right\rangle} \left\langle{{m_1' m_2'}} \vert {{jm}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.5)

The J_z action.

We have two ways that we can apply the operator J_z to {\left\lvert {jm} \right\rangle}. One is using the sum above, for which we find

\begin{aligned}J_z {\left\lvert {jm} \right\rangle} &= \sum_{m_1' m_2'} J_z {\left\lvert {m_1' m_2'} \right\rangle} \left\langle{{m_1' m_2'}} \vert {{jm}}\right\rangle \\ &= \hbar \sum_{m_1' m_2'} (m_1' + m_2') {\left\lvert {m_1' m_2'} \right\rangle} \left\langle{{m_1' m_2'}} \vert {{jm}}\right\rangle \\ \end{aligned}

We can also act directly on {\left\lvert {jm} \right\rangle} and then insert a complete set of states

\begin{aligned}J_z {\left\lvert {jm} \right\rangle} &=\sum_{m_1' m_2'} {\left\lvert {m_1' m_2'} \right\rangle}{\left\langle {m_1' m_2'} \right\rvert}J_z {\left\lvert {jm} \right\rangle} \\ &=\hbar m\sum_{m_1' m_2'} {\left\lvert {m_1' m_2'} \right\rangle}\left\langle{{m_1' m_2'}} \vert {{jm}}\right\rangle \\ \end{aligned}

This provides us with the identity

\begin{aligned}m\sum_{m_1' m_2'} {\left\lvert {m_1' m_2'} \right\rangle}\left\langle{{m_1' m_2'}} \vert {{jm}}\right\rangle = \sum_{m_1' m_2'} (m_1' + m_2') {\left\lvert {m_1' m_2'} \right\rangle} \left\langle{{m_1' m_2'}} \vert {{jm}}\right\rangle \end{aligned} \hspace{\stretch{1}}(3.6)

This equality must be valid for any {\left\lvert {jm} \right\rangle}, and since all the kets {\left\lvert {m_1' m_2'} \right\rangle} are linearly independent, we must have for any m_1', m_2'

\begin{aligned}(m - m_1' - m_2') \left\langle{{m_1' m_2'}} \vert {{jm}}\right\rangle {\left\lvert {m_1' m_2'} \right\rangle} = 0\end{aligned} \hspace{\stretch{1}}(3.7)

We have two ways to get this zero. One of them is a m = m_1' + m_2' condition, and the other is for the CG coeff \left\langle{{m_1' m_2'}} \vert {{jm}}\right\rangle to be zero whenever m \ne m_1' + m_2'.

It’s not a difficult argument, but one that wasn’t clear from a read of the text (at least to me).


[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.


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