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For $SU(2)$, we can contract two spin-1/2 indices and they break apart into two irreducible representation as (math notation) - $$ 2\otimes2 = 3\oplus 1 $$ that is the triplet and the singlet sector. The triplet and singlet projection operators are given as $$ P_t = \vec{S}_1.\vec{S}_2 + \frac{3}{4}\mathbb{I}\\ P_s = -\vec{S}_1.\vec{S}_2 + \frac{1}{4}\mathbb{I}.\\ $$ Now, for the general $SU(N)$ scenario, we have $$ N\otimes \bar{N} = (N^2-1) \oplus 1 $$ where $N^2-1$ is the adjoint representation and $1$ is the singlet representation. What is the projector onto these sectors in terms of $\vec{T}_1.\vec{T}_2$ where $T_1$ and $T_2$ are the vectors with elements being the elements of the lie algebra of $SU(N)$. I have been using approaches using the Young Tableaux but I am not getting anywhere. It seems like there is an easy way to do this. What is that?

Qmechanic
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CondensedChatter
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1 Answers1

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Indeed.

Review for su(2), in your notation, where $\vec S$ are the normalized 3-vector generators, so $\vec \sigma /2$ for the fundamental and anti fundamental, so you have $$ \vec{S}_1\cdot \vec{S}_2=\tfrac{1}{2}((\vec S_1+\vec S_2)^2-\vec S_1^2- \vec S_2^2)= -3/4 + \tfrac{1}{2}((\vec S_1+\vec S_2)^2, $$ hence $=-3/4+0= -3/4$ for the singlet (spinless) and $= -3/4+1=1/4$ for the triplet (adjoint), as detailed in most QM texts.

Hence, these eigenvalues plug into the projectors below to confirm they are that, $$ P_s = -\vec{S}_1\cdot \vec{S}_2 + \tfrac{1}{4}\mathbb{I},\\ P_a = \vec{S}_1\cdot \vec{S}_2 + \tfrac{3}{4}\mathbb{I}\qquad \leadsto \\ P_sP_a=0; \qquad P_s+P_a={\mathbb I}. $$ The two projector relations on the last line serve as checks of the above, but could determine the spin/quadratic Casimir of the adjoint, if you were oblivious of it!!

Now repeat this for su(3), for normalized 8-vector generators $\vec F$ ($= \vec\lambda /2$ for the 3 and its conjugate), and quadratic Casimir 4/3 for the triplet and 2 for the adjoint. (Note this differs from the 3 of Wikipedia, because of the different normalization involved here, different by a factor of 3/2, the ratio of rep indices. As mentioned above, you don't really need it, as the Casimir for the fundamental will suffice.)

Consequently, $$ \vec F_1\cdot \vec F_2= -4/3 ~~~~\hbox {for the singlet, and}\\ \vec F_1\cdot \vec F_2= -4/3 +2/2=-1/3 ~~~~\hbox {for the adjoint}, $$ hence $$ P_s = -\vec{F}_1\cdot \vec{F}_2 - \tfrac{1}{3}\mathbb{I},\\ P_a = \vec{F}_1\cdot \vec{F}_2 + \tfrac{4}{3}\mathbb{I}\qquad \leadsto \\ P_sP_a=0; \qquad P_s+P_a={\mathbb I}. $$ $P_a$ projects out the singlet, and $P_s$ projects out the adjoint. $P_sP_a$ is a quadratic polynomial in the $F\cdot F$ with roots at the right places, -1/3 and -4/3.

You may extend this for the su(N) algebra, given the normalizations, in, e.g., this, $$ P_s = -\vec{T}_1\cdot \vec{T}_2 - \tfrac{(N-1)^2-2}{2N}\mathbb{I},\\ P_a = \vec{T}_1\cdot \vec{T}_2 + \tfrac{N^2-1}{2N}\mathbb{I}\qquad \leadsto \\ P_sP_a=0; \qquad P_s+P_a={\mathbb I}. $$

Edit in response to last comment (geeky): Bypassing the charge conjugation stunt, here is the su(3) case displaying the P&S conventions you use. Start with $3\otimes 3= 6\oplus \bar 3$, which is more similar to the su(2) one! Using "=" signs to indicate obvious "up to a total-vs-uncoupled 9×9 matrix basis changes", you have $ 2\vec F_1\cdot \otimes \vec F_2 = \tfrac{4}{3} {\mathbb I}_{3} +\tfrac{10}{3} {\mathbb I}_{6} - 2\cdot \tfrac{4}{3} {\mathbb I}_{9} ~~\implies ~~ \vec F_1\cdot \otimes \vec F_2 =-\tfrac{2}{3} P_{\bar 3}+ \tfrac{1}{3} P_6. $ With the condition $P_{\bar 3}+P_6={\mathbb I}_9$, it yields $P_{\bar 3}=(- \vec F_1\cdot \otimes \vec F_2+1/3); ~~~P_6=( \vec F_1\cdot \otimes \vec F_2 +2/3$).

Now on to $3\otimes \bar 3= 8\oplus 1$. As you wish, call the 9×9 matrix $ \vec F_1\cdot \otimes (-\vec F_2)^* \equiv G$. As above, $2G= 0 {\mathbb I}_{1} + 3 {\mathbb I}_{8} - 2\cdot \tfrac{4}{3} {\mathbb I}_{9} ~~\implies ~~ G=-\tfrac{4}{3} P_ 1+ \tfrac{1}{6} P_8.$ As before, it implies $P_1=\tfrac{2}{3}(-G +1/6); ~~~P_8=\tfrac{2}{3}(G +4/3)$. So your comment was trenchant and my glib scaling in the answer was off! See how $P_1 +P_8={\mathbb I}_9$ is satisfied now!

This adjusts/morphs mutatis mutandis to the $su(N)$ case you confirmed, $P_1=\tfrac{2}{N}(-G +1/2N); ~~~~P_{N^2-1}=\tfrac{2}{N}(G +(N^2-1)/2N)$.
Cosmas Zachos
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