# Selection rules on quadratic terms

Let's assume I have to find expectation value of $$z^2$$ in the $$|l_im_i\rangle$$ state.

Can I use the selection rules in this way?

$$\langle 10|z^2|10\rangle$$

$$=\langle 10|z \color{red}{|10\rangle\langle10|} z|10\rangle$$ $$=\color{blue}{\langle 10|z|10\rangle}\color{green}{\langle10|z|10\rangle}$$

Since, we have $$l_1 = l_2$$ and $$m_1= m_2$$ , $$\langle 10|z^2|10\rangle$$ will be zero. We would need $$l_2 = l_1 \pm 1$$ to have nonzero expectation value.

My question is how to write the expectation value of $$\langle 10|z^2|10\rangle$$, so I can easily apply the selection rules. Any other kind of examples would help me too. I feel like I have done the completeness (the red part) relation wrong while I was breaking the $$z^2$$.

• You must insert a complete set of states in lieu of your red states. You already noted that the red states you used take you nowhere (0) so you must move to $m=\pm1$ ones instead. What does the Wigner-Eckart theorem steer you to do? Commented Aug 15, 2019 at 16:14
• Correction, I meant $l\pm1$, as you indicated. The point is $z \sim T^{(1)}_0$ and Kronecker-multiplying two $|10\rangle$s never yields a $|10\rangle$, but, instead, $\sqrt{2/3}~|20\rangle -1/\sqrt{3} ~ |00\rangle$. Commented Aug 15, 2019 at 21:28
• There is my problem Cosmas Zachos. I need to better understand how to manipulate completeness there. Could you give me an example for the problem so I can do it? Commented Aug 16, 2019 at 0:12

$$\langle 10|z^2|10\rangle = \langle 10|z| 1\!\! 1 |z|10\rangle \\ =\langle 10|z \color{red}{\sum_{l=0}^{\infty}\sum_{m=-l}^{l} |lm\rangle\langle lm|} z|10\rangle\\ =\color{blue}{\langle 10|z|20\rangle}\color{blue}{\langle 20|z|10\rangle} +\color{green}{\langle10|z|00\rangle}\color{green}{\langle00|z|10\rangle} ,$$ Since z is a vector operator like l=1 , m=0, and the only surviving Clebsch expansion is $$|10;10\rangle= \sqrt{2/3}|20\rangle -\sqrt{1/3} |00\rangle ,$$ by use of the Wigner-Eckart theorem...