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I've been reading about gauge field theories, and I keep coming across the generators of the $SU(3)$ and $SU(2)$ Symmetry Groups. I read that each generator corresponds to a gauge boson, but I'm struggling with the conceptual interpretation of the so call "generators". What is the physical meaning of the generators which appear to be matrices?

Any help to further my understanding is greatly appreciated.

Qmechanic
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K Ferreira
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  • Related: http://physics.stackexchange.com/q/156415/2451 , http://physics.stackexchange.com/q/133758/2451 – Qmechanic Oct 05 '16 at 20:08

1 Answers1

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What is the physical meaning of the generators which appear to be matrices?

Basically, they physically represent infinitesimally small operations, for example,

  1. angular momentum as the generator of rotations,

  2. linear momentum as the generator of translations,

  3. electric charge being the generator of the U(1) symmetry group of electromagnetism,

  4. the color charges of quarks are the generators of the SU(3) color symmetry in quantum chromodynamics,

They are very useful, due to their simplicity, in checking commutation relations, related to the Lie Algebra of any particular group.

The Pauli matrices, multiplied by $i$, are the basis for $su(2)$, i.e. the Lie algebra involved, not the group.

${\displaystyle {\begin{aligned}\sigma _{1}=\sigma _{x}&={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\\\sigma _{2}=\sigma _{y}&={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\\\sigma _{3}=\sigma _{z}&={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}\,.\end{aligned}}}$

However, as you probably know, they are incorporated in an exponential function when we need to express say, a rotation through an angle of 30 degrees.

Given a $SO(3; 1)$ representation, one may try to construct a representation of $SO(3; 1)$, the identity component of the Lorentz group, by using the exponential mapping. If $X$ is an element of $SO(3; 1)$ in the standard representation, then

${\displaystyle \Lambda =e^{iX}\equiv \sum _{n=0}^{\infty }{\frac {(iX)^{n}}{n!}}}$

is a Lorentz transformation by general properties of Lie algebras.

EDIT From a comment by WetSavannaAnimal aka Rod Vance

One thinks of Lie algebras of Lie groups as Lie algebras over real fields, so that $i$ is not thought of as a scalar in the field: the co-efficients of superpositions in the Lie Algebra are always real.

  • Good answer: a couple of pedantic points: the Pauli matrices multiplied by i that are the basis for $\mathfrak{su}(2)$, i.e. the Lie algebra, not the group and one thinks of Lie algebras of Lie groups as Lie algebras over real fields, so that $i$ is not thought of as a scalar in the field: the co-efficients of superpositions in the LA are always real. – Selene Routley Oct 05 '16 at 21:31
  • After thinking about this for a while recently, I thought it might be good to post my ideas to get feedback just in case I'm mistaken (which I very well may be). Is it the case that the $SU(3)$ generators in the phase transformation cause a change in the color charge of a quark? This idea, if correct, can be extended to $SU(2)$. – K Ferreira Apr 01 '17 at 04:32