1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
    Commented Dec 8, 2022 at 4:32
  • $\begingroup$ @Jagerber48 Thank you so much! I will take a look in that book. $\endgroup$
    – Hsu Bill
    Commented Dec 12, 2022 at 1:20

1 Answer 1

1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.