# Peeter Joot's (OLD) Blog.

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## 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.)]

# Motivation.

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).

# References

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