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In Assa Auerbach's book on page 166, he describes the construction of a bosonic representation of $SU(N)$ where the generators $S^{mn} \rightarrow b^\dagger_m b_n$. I'm a bit confused about the constraint in cases $N\ne 2$:

$$ n_b= \sum_{n}^N b^\dagger_n b_n = NS $$

where $S$ is the spin representation and $n_b$ is the total number of bosons representing that spin. Clearly, in $SU(2)$, $S$ takes on positive integer and half integer values, so that $NS = (0, 1,2 ...)$.

What about in $N>2$? Wikipedia shows that the adjoint irrep of $SU(3)$ has $16/3$ which is not a half integer. Basically I'm asking what the allowed values of $S$ are for a given $N$, and how to work this out?

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  • $\begingroup$ What is the meaning of $16/3$? Is it a Casimir eigenvalue? $\endgroup$ – ZeroTheHero Jan 15 at 0:13
  • $\begingroup$ Yes it's a sum over the square of all 8 generators $\endgroup$ – Adam B Jan 15 at 0:35
  • $\begingroup$ ? The quadratic Casimir for SU(3) is 4/3 for the triplet, 3 for the adjoint (octet), etc. What, exactly, are you insinuating? $\endgroup$ – Cosmas Zachos Jan 15 at 1:47
  • $\begingroup$ Related Casimir formulas. $\endgroup$ – Cosmas Zachos Jan 15 at 2:39
  • $\begingroup$ Which Wikipedia page? $\endgroup$ – Qmechanic Jan 15 at 5:59
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Jordan's (1935) nifty map is a Lie algebra isomorphism of generator matrices (and the vectors they act on) to Fock bilinears (and the Fock states they operate on). It expedites some representation theory, as Schwinger illustrated on SU(2) — after a hiatus of 17 years.

As you presumably already saw, for SU(2), $$ {\vec J} \equiv {\mathbf a}^\dagger \cdot\frac{ {\vec \sigma } } {2} \cdot {\mathbf a}^{\,} ~, \qquad n \equiv \sum_{i=1}^2 a^\dagger_i a^{\,}_i \equiv {\mathbf a}^\dagger \cdot {\mathbf a}^{\,} ~ , $$ so the quadratic Casimir operator is $$ {\vec J} \cdot {\vec J} = \frac{n}{2} \left ( \frac{n}{2}+1\right )~. $$

You may thus identify n with 2j, and count aggregate modes, whose net number the Js cannot change. They automatically separate (reduce) the subspaces characterized by a given common j=n/2. That is, they connect only states $a_1^{\dagger ~ k} a_2^{\dagger ~ l}|0\rangle$ among themselves, sharing a common $k+l=n=2j$. You may call this "projection" if you focus on a j of interest to you.

On to SU(3), now with three fundamental oscillators, so their dot products range over indices from 1 to 3 (N for general SU(N)). The conventionally normalized generators are the eight $$ F_a \equiv {\mathbf a}^\dagger \cdot\frac{ { \lambda_a } } {2} \cdot {\mathbf a} ~, $$ where the $\lambda_a$ s are the Gell—Mann matrices, obeying the Fierz identity where $$\lambda^{a ~~i} _j \lambda^{a~~k} _l = 2 \delta^i_l \delta^k _l -\frac{2}{3} \delta^i_j \delta^k _l ~. $$

The quadratic Casimir then evaluates to (Einstein summation convention over repeated indices!) $$ {\mathbf a}^\dagger \cdot\frac{ { \lambda_a } } {2} \cdot {\mathbf a}~~~ {\mathbf a}^\dagger \cdot\frac{ { \lambda_a } } {2} \cdot {\mathbf a}=-\frac{1}{6} {\mathbf a}^\dagger \cdot {\mathbf a}~~{\mathbf a}^\dagger \cdot {\mathbf a}+\frac{1}{2}a_i^\dagger a_k a^\dagger_k a_i \\ =\frac{1}{3} n(n+3)~. $$

You may recognize this as the quadratic Casimir $(p+3p)/3$ of p symmetrized quarks' representation, D(p,0), so n is quark number, without any antiquarks—no columns in the respective Young tableau. The dimension d(p,0) of these representations is (p+1)(p+2)/2 ; hence p=1 for the fundamental triplet, with Casimir eignevalue 4/3. (Your 16/3 is for Gell—Mann matrix bilinears, which are here normalized by a 2 each, to become SU(3) generators.)

To reassure yourself, inspect states for p=2, so d=3, so C=10/3. Further, p=3, so d=10, so C=6, etc.

NB The adjoint, D(1,1) is not in this symmetrized collection--it has an antiquark as well, hence the telltale Γ Young tableau.

For basics of the direct SU(N) generalization, see Iachello, or Greiner-Mueller ...

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  • $\begingroup$ So then the number of bosons can still take any integer $n_b = (0,1,2,3..)$ in $SU(N)$ given the proper representation? I've seen papers varying $n_b$ and $N$ independently (e.g. looking at limits where one or the other is sent to infinity, or they are both sent to infinity with $n_b/M$ held fixed), and this is the origin of my question-- (1) is this valid (are they actually independent)? and (2) then this process is actually taking limits on the representation of SU(N)? I can provide an example paper if it helps to answer my question $\endgroup$ – Adam B Jan 15 at 19:43
  • $\begingroup$ Yes. $n_b$ can take arbitrary integral values and characterizes a rep... albeit not uniquely (which did not come up here.... that's why you use more, higher, Casimirs). N characterizes a group. The two are conceptually independent. What is M? $\endgroup$ – Cosmas Zachos Jan 15 at 20:20
  • $\begingroup$ sorry typo, M=N -- the paper uses M and I am going back and forth between notation! $\endgroup$ – Adam B Jan 15 at 20:24

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