2
$\begingroup$

I know that Wigner $3j$-symbols have certain symmetry factors arising by exchange of two columns within one symbol.

But what happens if you have two 3j symbols and do an exchange like this:

$ \left(\,\begin{matrix}% j_1&j_2&J\\% m_1&m_2&-M_J %6 \end{matrix}\,\right)% \left(\,\begin{matrix}% J&j_3&j\\% M_J&m_3&-m_j %6 \end{matrix}\,\right)% \Rightarrow \left(\,\begin{matrix}% j_3&j_2&J\\% m_3&m_2&-M_J %6 \end{matrix}\,\right)% \left(\,\begin{matrix}% J&j_1&j\\% M_J&m_1&-m_j %6 \end{matrix}\,\right)%$

So I did an exchange of $j_1$ with $j_3$ and since that operation is an exchange between two 3j symbols, does there also arise a symmetry factor $(-1)^{something}$ ?

That is an example of three body coupling, where the first two states of the bra $j_1$ and $j_2$ are coupled to $J$ and then $J$ and the third state $j_3$ are coupled to $j$.

$\endgroup$
  • $\begingroup$ Somewhat related: Adding 3 electron spins. The behaviour of multiple additions of angular momenta under particle exchange need not be particularly pretty. $\endgroup$ – Emilio Pisanty Jun 4 at 7:29
  • $\begingroup$ @EmilioPisanty Thanks, but here I am not dealing with electrons. It's about Nuclei and there is not written how the symmetry behaviour is under exchange of the coupling order $\endgroup$ – Armani42 Jun 4 at 7:35
1
$\begingroup$

In general, states with total angular momentum $J$ obtained from the triple couplings $j_1j_2j_3$ can be obtained in a number of ways.

A first is to couple $j_1j_2$ to $j_{12}$, and couple this to $j_3$ to get $j$, giving states symbolically denoted by $\vert (j_1j_2)j_{12}j_3;jM\rangle$. A second is to couple first $j_2j_3$ to $j_{23}$, and then $j_1$ to this to get $j$, with states denoted $\vert j_1(j_2j_3)j_{23};jM\rangle$.

The states $\vert j_1(j_2j_3)j_{23};JM\rangle$ are not linearly independent from the $\vert (j_1j_2)j_{12}j_3;JM\rangle$. In fact, the overlap is related to a $6j$ coefficient: $$ \langle (j_1j_2)j_{12}j_3;jM\vert j_1(j_2j_3)j_{23};jM\rangle =(-1)^{j_1+j_2+j_3+J} \sqrt{(2j_{12}+1)(2j_{23}+1)} \left\{\begin{array}{ccc} j_1&j_2&j_{12}\\ j_3&j&j_{23}\end{array}\right\}\, . $$ and the overlap need not be $\pm 1$.

Your specific case seem to have $j_{12}=j_{23}=J$. I think you should follow that route to understand the effects of your permutation $j_1\leftrightarrow j_3$, i.e. write states in $\vert (j_1j_2)Jj_3;jm\rangle$ basis, with CG and all, permute $j_1$ and $j_3$, and then take the overlap of the resulting state $P_{13}\vert (j_1j_2)Jj_3;jm\rangle$ with your original basis $\vert (j_1j_2)Jj_3;jm\rangle$ to get the effect of the permutation. There are equalities of the type $$ \sum_{e\epsilon} (-1)^{2e}\sqrt{(2c+1)(2d+1)} C_{b\beta;d\delta}^{e\epsilon}C^{e\epsilon}_{f\varphi;c\gamma} \left\{\begin{array}{ccc} a&b&c\\ e&f&d\end{array}\right\}= C_{a\alpha;b\beta}^{c\gamma}C_{a\alpha;f\varphi}^{d\delta} $$ which can be found in

Varshalovich, D.A., Moskalev, A.N. and Khersonskii, V.K.M., 1988. Quantum theory of angular momentum.

to help you sort this out. The most explicit example of this procedure I could find is in this paper:

Rowe, D.J., Sanders, B.C. and De Guise, H., 1999. Representations of the Weyl group and Wigner functions for SU (3). Journal of Mathematical Physics, 40(7), pp.3604-3615.

where the effect of permutations is examined, although in a slightly different context.

$\endgroup$
  • $\begingroup$ Thank you, that helped me a lot! $\endgroup$ – Armani42 Jun 20 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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