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I tried to use this online CG calculator to find the singlet state in the $SU(3)$ relation: $$ \textbf{3}~\otimes~\bar{\textbf{3}} =\textbf{1}~\oplus~\textbf{8} $$ Taking the basis of $\bf 3$ and $\bar{\bf 3}$ as $|r\rangle,|g\rangle,|b\rangle$ and $|\bar r\rangle,|\bar g\rangle,|\bar b\rangle$, respectively, this calculator gives the following CG coefficients: $1/\sqrt 3, -1/\sqrt 3, 1/\sqrt 3$. I know however that the singlet is supposed to be: $$ |0\rangle = \frac{|\bar rr\rangle + |\bar gg\rangle + |\bar bb\rangle}{\sqrt 3} $$ i.e. without a minus sign. Where did I go wrong?

Qmechanic
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ersbygre1
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  • Maybe that's connected to the minus sign in front of SU(2)'s $\bar d$? https://physics.stackexchange.com/q/360090/ – ersbygre1 Jul 08 '21 at 12:18

1 Answers1

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This depends very much on the phases of the states. Basically, it could be that $\vert \text{your } \bar d\rangle$ is $- \vert \text{their } \bar d\rangle$, as you suggest.

The difficulty is there is non standard phases for states so it's quite difficult to use someone else's table without first doing some checks.

IIRC this code uses for basis states the Gelfand-Tseitlin basis (which is pretty much the standard basis for $\mathfrak{su}(n)$ calculations), and the matrix elements for any irreps are actually given in a paper by the same authors. Soooo... what you need to do is construct the $(1,0)$ and the $(0,1)$ matrices (or $\boldsymbol{3}$ and $\boldsymbol{\bar 3}$ if you prefer this sadistic notation of HEP), compare the signs of the paper with your own sign convention, and infer the relative phases between their $\vert \bar r\rangle, \vert \bar g\rangle$ and $\vert \bar d\rangle$ with your states. (Constructing the $(1,0)$ isn't strictly necessary but if you can't get your $(1,0)$ matrix elements to agree with theirs you're in trouble.)

The lesson here is that, for anything beyond $\mathfrak{su}(2)$, phases have to be checked very very carefully and it's not so easy to just use existing codes as tools just "off-the-shelf".

Aside: if you need another source, there this recent C++ code:

https://gitlab.com/tdytrych/su3lib

which will give you reduced CGs (have to be multiplied by an $\mathfrak{su}(2)$ CG). The code is an upgrade of an older code and is based on the gold standard for $\mathfrak{su}(3)$:

Draayer JP, Akiyama Y. Wigner and Racah coefficients for SU 3. Journal of Mathematical Physics. 1973 Dec;14(12):1904-12.

It must be run locally though, it is also based on the GT basis, and there's no $\mathfrak{su}(n)$ version that I'm aware. The online code you're using has a different way of disentangling outer multiplicities (i.e. repeated representations); the symmetries of the Draayer code are described in the JMP paper.

ZeroTheHero
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  • Thanks for the advice concerning the phases. You mentioned constructing (1,0) matrices: do you mean the 8 generators in that representation? And check whether they are the Gell-Mann matrices for (1,0)? – ersbygre1 Jul 08 '21 at 12:51
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    Yes. Constructing the matrix elements of the simple generators should be enough. What you need to do is check if, by changing the phase of $\vert \text{your } \bar d\rangle$, you can reproduce the matrix elements of that paper by Alex. The tricky part will be for $(0,1)$, where extra phases will appear. This is because there are many "natural" way of constructing $(0,1)$, one of which is from the antisymmetric part of $(1,0)\otimes(1,0)$, another one is by directly conjugating the variable in $(1,0)$; both lead to different signs. – ZeroTheHero Jul 08 '21 at 12:56
  • I'm sorry, I don't really understand. I've got my (0,1) generators defined as the negative c.c. of the 1/2*Gell-Mann generators, but I cannot see how to extract any information out of that paper in order to compare: https://homepages.physik.uni-muenchen.de/%7Evondelft/Papers/ClebschGordan/Alex2010.pdf Where's the connection of all those weights to the generators? – ersbygre1 Jul 08 '21 at 13:13
  • The problem is you defined your $(0,1)$ as the negative $cc$ of the Gell-Mann. That's a choice of phase in your basis states, i.e. you have made some consistent choices of phases in your $\vert \bar{\text{state}}\rangle$ with the result that your $(0,1)$ matrix elements are as you state. The negative signs enter per your convention, and this might not be where the negative sign enter in the GT basis. – ZeroTheHero Jul 08 '21 at 13:33
  • soo... you will need to go to the GT basis, and use the matrix elements defined in this basis to see if the negative signs occur in the same places as you (presumably they will not), and then find a way to make everything agree by flipping signs of some of the GT states. Presumably you only need to do this for $\vert \bar{d}\rangle$ to make everything work but I haven't done this myself so I can't quite guarantee the result. – ZeroTheHero Jul 08 '21 at 13:36
  • Thank you, you've given me a lot of new information. I simply can't not accept this answer. If you have time for one last question, do you know a reference where to find the GT conventions? – ersbygre1 Jul 08 '21 at 13:53
  • It's in the Alex paper on which the code is based. I can't remember if it's an appendix or in the main text. Alex A, Kalus M, Huckleberry A, von Delft J. A numerical algorithm for the explicit calculation of SU (N) and SL (N, C) Clebsch–Gordan coefficients. Journal of Mathematical Physics. 2011 Feb 7;52(2):023507. – ZeroTheHero Jul 08 '21 at 14:33