There is a simple formula for the irreducible characters of SU(2), $$ \chi^{(j)} (\theta) = \frac{\sin (j+\frac{1}{2})\theta}{\sin \frac{1}{2}\theta}, $$ where the irreps have dimensionality $( 2 j +1)\times(2j+1)$.

Is there a similar formula for the characters of the irreps of SU(3)?


1 Answer 1


Yes, in fact there is! The formula for the character of the irreducible representation of $SU(3)$ with highest weight $(p,q)$ is

\begin{align} \chi^{p,q}(\theta, \phi) = e^{i \theta (p+2q)}\sum\limits_{k=q}^{p+q}\sum\limits_{l=0}^q e^{-3i(k+l) \theta/2} \left(\frac{\sin((k-l+1)\phi/2)}{\sin (\phi/2)} \right) \end{align}

For example, one can check that the defining representation (highest weight $(1,0)$) has

\begin{align} \chi^{1,0}(\theta, \phi) &= e^{i \theta}\sum\limits_{k=0}^{1}\sum\limits_{l=0}^0 e^{-3i(k+l) \theta/2} \left(\frac{\sin((k-l+1)\phi/2)}{\sin (\phi/2)} \right) \\ &= e^{i \theta}+e^{i(\phi-\theta)/2} + e^{-i(\phi+\theta)/2} \end{align}

which as expected is just the trace of some $U \in SU(3)$ in terms of two independent eigenvalues, where the three eigenvalues are related to each other by $\det U =1$.

An equivalent formula is also given in Baaquie's paper (Equations 3.1 and 3.2):

\begin{align} \chi^{p,q}(x,y) &= -\frac{i}{s(x,y)} \big( - e^{i ((p+1) y - (q+1) x)} + e^{i ((p+1) x - (q+1) y)} \\ &+ e^{-i (p+1) (x+y)} (e^{-i (q+1) x} - e^{-i (q+1) y}) + e^{i (q+1) (x+y)}(e^{i (p+1) y} - e^{i (p+1) x}) \big) \\ s(x,y) &= 8 \sin \left(\frac{x-y}{2}\right) \sin \left(\frac{1}{2} (x+2 y)\right) \sin \left(\frac{1}{2} (2 x+y)\right) \end{align}

This actually took me longer to sort out than is ideal, so here I document some things to note in case there is confusion for others in future.

  • A derivation of $\chi^{p,q}(\theta, \phi) $ can be found in Chapter 10.15 of Greiner - Quantum Mechanics: Symmetries (2nd revised edition). There is a typo in the final quoting of the result - in Equation 10.121 the first $\varphi$ should instead be a $\psi$. If anyone knows a a mathematics textbook that explicitly writes down the $SU(3)$ result, I'd appreciate a message about it!
  • The characters of $Gl(n,\mathbb{C})$ are the Schur polynomials, then one needs to specialize to the subgroup $SU(n)$. This is effectively what is carried out in the cited derivation, although written more from the perspective of physicists and specialized to the case $n=3$.
  • The result is also quoted here. Typos also appear: in Equations 5.65 and 5.82, $m_{12}$ is almost surely meant to be $m_{13}$.
  • On notation: irreducible representations of $SU(n)$ each have a corresponding Young Diagram with at most $n-1$ rows. One then chooses to uniquely label the irreducible representation according $n-1$ non-negative integers $(q_1,\dots, q_{n-1})$ where $q_i$ is the number of boxes in row $i$ in the diagram (counting from the top down), or one labels the irreducible representation according to $(p_1,\dots,p_{n-1}) := (q_1-q_2,q_2-q_3, \dots,q_{n-2}-q_{n-1},q_{n-1})$, where now $p_i$ gives the total number of columns of length $i$ in the diagram. The $h_{ij}$'s and $m_{ij}$'s that appear respectively in Greiner and Prakash's derivations correspond to the $q_i$'s, whereas the $p_i$'s correspond to the labeling in for example Brian Hall's book (that book uses $(p_1,p_2) = (m_1,m_2)$). I have converted the formula to use the $p_i$ labeling in my answer above, which to be explicit corresponds to $(p,q) = (p_1,p_2)$ and for example the $(p+1)(q+1)(p+q+2)/2$ dimension formula for the irreducible representations. Baaquie's paper uses the convention (2, 1) for (p, q) of the fundamental representation, so you need to subtract 1 from each of his weights to get the more standard convention that I used. I have already done this in the formula above.

Edit: The formula is now correctly described as applying to SU(3) in particular, and not for SU(n) in general.

  • 1
    $\begingroup$ I really appreciate all the details you have provided (I assume you updated the Wikipedia article also!). One clarification though: the formula you have written is for SU(3) and not SU(n), right? $\endgroup$ Commented Jun 30, 2019 at 11:07
  • $\begingroup$ No problem! Yes I updated the wikipedia article. Yes, it's all for $SU(3)$ specifically. For $SU(n)$ there would need to be $n-1$ integers to specify the irreducible representations. $\endgroup$ Commented Jun 30, 2019 at 18:03
  • $\begingroup$ Oh I see! Yes indeed, the original version of my answer had a typo, thanks for correcting. $\endgroup$ Commented Jun 30, 2019 at 21:15
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    $\begingroup$ Thank you for this excellent answer. $\endgroup$ Commented Jul 1, 2019 at 4:26

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