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 $z^2$.

  • 4
    $\begingroup$ 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? $\endgroup$ Commented Aug 15, 2019 at 16:14
  • $\begingroup$ 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, $\sqrt{2/3}~|20\rangle -1/\sqrt{3} ~ |00\rangle$. $\endgroup$ Commented Aug 15, 2019 at 21:28
  • $\begingroup$ 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? $\endgroup$
    – user193422
    Commented Aug 16, 2019 at 0:12

1 Answer 1


$$\langle 10|z^2|10\rangle = \langle 10|z| 1\!\! 1 |z|10\rangle \\ =\langle 10|z \color{red}{\sum_{l=0}^{\infty}\sum_{m=-l}^{l} |lm\rangle\langle lm|} z|10\rangle\\ =\color{blue}{\langle 10|z|20\rangle}\color{blue}{\langle 20|z|10\rangle} +\color{green}{\langle10|z|00\rangle}\color{green}{\langle00|z|10\rangle} ,$$ 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...


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