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? $\endgroup$ 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. $\endgroup$ Jun 30, 2019 at 18:03
  • $\begingroup$ Oh I see! Yes indeed, the original version of my answer had a typo, thanks for correcting. $\endgroup$ Jun 30, 2019 at 21:15
  • 2
    $\begingroup$ Thank you for this excellent answer. $\endgroup$ Jul 1, 2019 at 4:26

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.