I am studying about Wigner-Eckart theorem, and I have a question about the reduced matrix element.

Wigner-Eckart theorem: (I follow the terms as in Wikipedia, https://en.wikipedia.org/wiki/Wigner%E2%80%93Eckart_theorem)

For a spherical tensor, $T^{(k)}_q$, qth component of rank k tensor operator, and states $|j',m'\rangle,|j,m\rangle$, the matrix element of that tensor can be expressed as:

$$\langle j',m'|T^{(k)}_q|j,m\rangle=\langle j'||T^{(k)}_q||j\rangle\langle j'm'k,q|j,m\rangle$$

Now I consider this tensor in the hyperfine states basis, that is, $|F,m_F\rangle,|F',m'_{F}\rangle$, then according to the Wigner-Eckart theorem the matrix element can be expressed as:

$$\langle F',m'_{F}|T^{(k)}_q|F,m_F\rangle=\langle F'||T^{(k)}_q||F\rangle\langle F'm'_Fk,q|F,m_F\rangle$$

I have seen several data discussing about the dipole transition strength by Daniel A. Steck (https://steck.us/alkalidata/rubidium87numbers.1.6.pdf bottom of pg,6--top of pg7). In those data, they use the rank 1 spherical tensor. In those two pages, they use Wigner-Eckart theorem first:

$$\langle F,m_{F}|er_q|F',m'_F\rangle=\langle F||e\mathbf{r}||F'\rangle\langle Fm_F|F'1m'_Fq\rangle$$

Then they simplify the reduced matrix element further:

$$\langle F||e\mathbf{r}||F'\rangle=\langle J||e\mathbf{r}||J'\rangle\times(Wigner\ 6-j\ symbol)$$

Why can they use Wigner-Eckart theorem on reduced matrix element? That is the most confusing point in my mind.

  • $\begingroup$ The expansion involving the 6j symbol is not a direct application of the Wigner Eckart theorem but instead involves other angular momentum combining rules. See Steck’s textbook quantum and atom optics for many more details. $\endgroup$
    – Jagerber48
    Dec 8, 2022 at 4:32
  • $\begingroup$ @Jagerber48 Thank you so much! I will take a look in that book. $\endgroup$
    – Hsu Bill
    Dec 12, 2022 at 1:20

1 Answer 1


For atomic reduced matrix elements, when dealing with $jj$ coupling the tensor acts on only part of the state so you have to decouple the states to the $LS$ scheme, compute the matrix element in this basis, and then recouple to $jj$. Thus the matrix element in the $jj$ scheme can be expressed as a spatial matrix element expressed in terms of $L$ multiplied by some recoupling coefficient (that’s the 6j-symbol) that deals with all this coupling-recoupling business.

Here the notation is not quite the same as your $F$ is my $j$ and that kind of stuff but that’s the basic idea: your tensor acts only on the $J$ part of the state so you need to decouple-recouple various J’s, I’s and F’s to change from the matrix elements between the two schemes.

  • $\begingroup$ Thank you for your comment! $\endgroup$
    – Hsu Bill
    Dec 12, 2022 at 1:21

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