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My question is about the relation between the members of the members of a multiplet, for example the baryon-Decuplet or one of the Baryon-Octets. By what transformation are they related? Are they related by a SU(3) transformation $e^{i\phi_aT_a}$, where $T_a$ are the 8 generators of the SU(3)? Or are they related by members of the algebra of SU(3), i.e. by raising and lowering operators formed by the generators themselves (for example $T_1 \pm i T_2$)? If the latter is true, why do I always read, the multiplet is a representation of SU(3)? It should rather be a representation of the algebra $su(3)$ of SU(3). I would be so grateful, if you could help me!

taxus1
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2 Answers2

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Are they related by a SU(3) transformation $e^{i\phi_aT_a}$, where $T_a$ are the 8 generators of the SU(3)? Or are they related by members of the algebra of SU(3), i.e. by raising and lowering operators formed by the generators themselves (for example $T_1 \pm i T_2$) Why do I always read "the multiplet is a representation of SU(3)"? It should rather be a representation of the algebra su(3) of SU(3).

A: both. Your cooked up a false dichotomy. The algebra elements, the generators, act on the multiplets of a given dimensionality (octet, decuplet, etc,...), and, consequently, their exponentials, the group elements, likewise act on those as well, never moving you out of these multiplets. They can, in general, connect all particles to all particles. However, the multiplet bases corresponding to particles are organized by eigenvalues of algebra generators.

The crucial point you ignored is that each particle is an eigenstate of the Casimir invariants of the algebra, and also chosen generators, in your case Isospin and Hypercharge, (Cartan subalgebra); so, as you noticed, raising and lowering operators for such eigenvalues move you from state to state, unlike a generic linear combination of generators, which mix your particle states. It is easiest to monitor this for the octet and decupet you mentioned, but, when you feel conceptually swamped, look at just isospin, that contains all elements of your puzzlement.

enter image description here enter image description here

On the horizontal axes, you have $I_3$ eigenstates, so the irrep is organized by su(2), the algebra. You move them among themselves by $I_{\pm}$. By contrast, a generic su(2) transformation on them will give you "mush", a linear combination of those, of indistinct $I_3$, just like a generic SU(2) transformation.

This is regardless of the dimensionality of the multiplet, technically a property of the algebra. I.e. you use 8×8 or 10×10 matrices to achieve that in the respective multiplets. (It happens that you can bypass multiplication by 8×8 matrices in favor of commutation of 3×3 matrices, but let's not get there...)

That's all; I assume you have not been introduced to the Cartan-Weyl basis, the more organized group-algebraic codification of this picture.

Cosmas Zachos
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  • Thanks a lot for the detailed answer! It almost solved my problem. If both, the group elements as well as the raising and lowering operators relate the vectors of a multiplet, can we state an equality like $T_- = U (\pi)$ where $T_-$ is made from the algebra and $U(\pi)$ is a group element? – taxus1 Jan 11 '23 at 08:08
  • For isospin SU(2), $\exp(i\pi \sigma_1~/2) = i \sigma_1 $, not $\sigma_-$, but, acting on $|1/2, 1/2\rangle$, a proton, it has essentially the same effect. – Cosmas Zachos Jan 11 '23 at 16:37
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If you have a representation of algebra $su(3)$ of dimension $d$, and the carrier space is spanned by $\{\vert\psi_a\rangle, a=1,\ldots,d\}$, then the same set spans the carrier space for the $d$-dimensional irrep of group $SU(3)$ by exponentiation.

It is clear that, since elements in the algebra will not take you out of the carrier space, repeated powers of these, as obtained through exponentiation, will also not take you out of the space.

ZeroTheHero
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  • Thanks a lot for your answer! It helped me to understand that both, the generators as well as the SU(2)-elements allow me to move around within a multiplet. However what are the transformation relating to members of a multiplet? Generator matrices or SU(2)-matrices? – taxus1 Jan 11 '23 at 08:10
  • @taxus both. In the analogous case of rotations, any hermitian generator $L_k$ AND the group element $U(\theta_k)=\exp(i\theta_k L_k)$ will connect states since $\exp(i\theta_k L_k)=\sum_a (i\theta_k L_k)^a/a!$. Whereas the action of a generator $L_k$ will connect states with $\Delta m=0, \pm 1$, the action of a finite group element through repeated action of $L_k$ will in general connect all $m$ states (although the details depends on the rotation angle $\theta_k$). – ZeroTheHero Jan 11 '23 at 09:28
  • Thus, it is not $L_y=U(\theta)$ but $U(\theta)=\exp(i\theta L_y)$. The same holds for $\lambda\in su(3)$ and the group transformation $U(\beta)=\exp(i\beta\lambda)$ where (say) $\lambda=T_-+T_+$ so $\lambda$ is hermitian. – ZeroTheHero Jan 11 '23 at 09:34