In higher dimensional representations of $\mathfrak{su(3)}$, what will be the quadratic Casimir operator? Is it same as in lower dimensions or different?


I suspect you have not asked what you wanted to. The operator form of all Casimir operators is the same as lionelbrits points out, and SU(3) has two independent ones, unlike SU(2), which only has one, the quadratic one. The two are a quadratic and a cubic, included here for completeness.

However, the eigenvalues of these operators vary with (irreducible) representation, and, in fact, serve to distinguish among them and label/characterize them cf. Pais (1966).

All states in a given irreducible representation assume the same value for each Casimir operator, which serves as the identity in a space with the dimension of that representation. This is because states in a given representation are connected by the action of the generators of the Lie algebra, and all generators commute with the Casimir operators.

For SU(3), the quadratic Casimir is $\hat{C_2}=\sum_k \hat{F_k} \hat{F_k}$, and the cubic one is $ \hat{C_3}=\sum_{jkl}d_{jkl} \hat{F_j} \hat{F_k} \hat{F_l}$. The F̂s are the 8 normalized generators of this Lie algebra.

The irreducible representations of SU(3) are denoted in the Dynkin basis by $D(p,q)$, consisting of $p$ quarks and $q$ antiquarks (in Young tableaux, $p$ is the number of single-box columns and $q$ the number of double-box columns): they have dimension $d(p,q)=\frac{1}{2}(p+1)(q+1)(p+q+2)$.

For example, for the triplet representation, $D(1,0)$, the eigenvalue of $\hat{C}_2$ is 4/3, and of $\hat{C}_3$, 10/9.

More generally, for generic irrep $D(p,q)$, the eigenvalue of $\hat{C}_2$ is $(p^2+q^3+3p+3q+pq)/3$, which I suspect was what you were really asking, above.

N.B. An aside, strictly speaking: The eigenvalue ("anomaly coefficient") of the cubic one, $\hat{C}_3$, is$(p-q)(3+p+2q)(3+q+2p)/18$, an odd function under the interchange p↔q. Consequently, it vanishes for real representations, p=q, such as the adjoint, $D(1,1)$, i.e. both this cubic Casimir and anomalies vanish for the octet, the 27, the 64, etc..

  • $\begingroup$ How can I compute the quadratic Casimir for $u(n)$ and $u(m|n)$ (unitary Lie and superLie algebras)? How can I label highest-weight states of these algebras? $\endgroup$ – QGravity Jan 13 '18 at 16:40
  • $\begingroup$ @QGravity : Straightforward, but long answer: Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Graduate Texts in Mathematics, 222 (2nd ed.), Springer, ISBN 978-3319134666 $\endgroup$ – Cosmas Zachos Jan 13 '18 at 16:55

The algebraic form of the quadratic casimir $T^2$ depends only on the structure constants, and are therefore the same in any representation. As in $SU(2)$, of course, it's matrix form is representation dependent.

Edit: To see why this is the case, suppose you constructed $T^2$ out of other elements of the algebra, and you did this for say, the fundamental representation. You then had to show that $\left[T^2, T_i\right]=0$ for all $T_i$. Of course, you can only use the structure constants, i.e., $[T_i, T_j] = i f_{ijk} T_k$ and $\{T_i, T_j\} = \frac{1}{N} \delta_{ij} + d_{ijk} T_k$ to show this. But these don't depend on the representation. They are what define the algebra, i.e., what make it $\mathfrak{su}(N)$.

  • $\begingroup$ can you suggest me any material for proof of what you have said " The algebraic form of the quadratic T2 casimir depends only on the structure constants, and are therefore the same in any representation" $\endgroup$ – Ipsit Panda Dec 5 '14 at 13:12
  • $\begingroup$ Ummm the Lie algebra does not depend on the representation, but the anticommutator does... In fact, the anticommutator expression you wrote only holds for the fundamental. $\endgroup$ – Cosmas Zachos Oct 11 '17 at 23:26

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