Kinetic lagrangian for non-abelian gauge theories

[patrice]

hello everyone!

i've got a very special question on the gauge invariance (gauge group: SU(N), non-abelian) of the kinetic term in the lagrangian. is it invariant for any representation? i only know, that for the fundamental rep. this is true.

to be more specific:

the usual kinetic term is -\frac{1}{4} F_{\mu \nu}^{a}F^{\mu \nu a}

arbitrary rep.: F_{\mu \nu} =F_{\mu \nu}^a T^a

the field tensor transforms like this: F_{\mu \nu}'=U F_{\mu \nu} U^{-1}, where U(x)=\exp(i \theta^a (x) T^a). (in an arbitrary representation with generators T^a)

in the fundamental representation of SU(N), i.e. T^a=\frac{\lambda^a}{2}, the lagrangian is invariant under gauge transformations, which can easily be proved.

and now for the adjoint representation of SU(N): (T^a)_{ij}=-if^{aij}
(note: the generators are here (N^2 -1)\times (N^2-1)-matrices.) does anyone know, or is anyone smart enough :-) to proof, that this also holds for this representation?

thank you, I'm looking forward to any answer!

patrice

 

[samalkhaiat]

hello everyone!

i've got a very special question on the gauge invariance (gauge group: SU(N), non-abelian) of the kinetic term in the lagrangian. is it invariant for any representation? i only know, that for the fundamental rep. this is true.

to be more specific:

the usual kinetic term is -\frac{1}{4} F_{\mu \nu}^{a}F^{\mu \nu a}

arbitrary rep.: F_{\mu \nu} =F_{\mu \nu}^a T^a

the field tensor transforms like this: F_{\mu \nu}'=U F_{\mu \nu} U^{-1}, where U(x)=\exp(i \theta^a (x) T^a). (in an arbitrary representation with generators T^a)

in the fundamental representation of SU(N), i.e. T^a=\frac{\lambda^a}{2}, the lagrangian is invariant under gauge transformations, which can easily be proved.

and now for the adjoint representation of SU(N): (T^a)_{ij}=-if^{aij}
(note: the generators are here (N^2 -1)\times (N^2-1)-matrices.) does anyone know, or is anyone smart enough :-) to proof, that this also holds for this representation?

thank you, I'm looking forward to any answer!

patrice

Hi,

Each representation of the gauge group has its own (invariant) Lagrangian. The gauge fields do not carry the fundamental representation! The ( N^{2}-1 ) gauge fields transform in the adjoint representation. So, I am afraid, your question is meaningless.

regards

sam

 

[patrice]

hi sam!

thank you for your answer! it was a great help for me. i remember my professor saying that the gauge fields transform in the adjoint representation. up to now it's not yet clear to me, how the gauge transformation "prescription" A_{\mu}'=UA_\mu U^{-1}+\frac{i}{g} U (\partial _\mu U^{-1}) is connected with the generators of the adjoint representation of Lie group SU(N), namely the structure constants of the corrensponding Lie Algebra.

Does A_\mu= A_\mu ^a T^a not mean (which one follows from the prescription above), that A_\mu is Lie Algebra valued for the representation of our choice for the matter multiplett \psi, even the fundamental rep?

by the way: before i read your post i wanted to say, that the problem is solved, because F_{\mu \nu}^a is independent from the representation and so the Lagrangian is too. but now I'm unsure...

i found a maybe good article on adjoint endomorphisms at wikipedia which i will try to understand later on. but now for the moment my head is full :) I'm going to tell you, if i could get wise from that arcticle..

best regards

patrice

 

[samalkhaiat]

hi sam!

thank you for your answer! it was a great help for me. i remember my professor saying that the gauge fields transform in the adjoint representation. up to now it's not yet clear to me, how the gauge transformation "prescription" A_{\mu}'=UA_\mu U^{-1}+\frac{i}{g} U (\partial _\mu U^{-1}) is connected with the generators of the adjoint representation of Lie group SU(N), namely the structure constants of the corrensponding Lie Algebra.

Does A_\mu= A_\mu ^a T^a not mean (which one follows from the prescription above), that A_\mu is Lie Algebra valued for the representation of our choice for the matter multiplett \psi, even the fundamental rep?

by the way: before i read your post i wanted to say, that the problem is solved, because F_{\mu \nu}^a is independent from the representation and so the Lagrangian is too. but now I'm unsure...

i found a maybe good article on adjoint endomorphisms at wikipedia which i will try to understand later on. but now for the moment my head is full :) I'm going to tell you, if i could get wise from that arcticle..

best regards

patrice

Very soon, maybe tomorow, I will post some useful details for you. As for Wikipedia, I have nothing good to say about it. Try Textbooks instead.

regards

sam

 

[patrice]

hello

sorry for not posting that long. yes, indeed this wikipedia article is a little short and cannot help me with my problem.
neither could the textbooks that i tried: peskin & schroeder (introduction to quantum field theory), ryder (qft) and bjorken drell (relativistic quantum field theory)

do you know or does anyone know a textbook which gives a more detailed discussion on representations of gauge potentials? that would be really nice.

thanks a lot

patrice

 

[Avodyne]

In an arbitrary rep, the kinetic term is {\rm Tr}\,F^{\mu\nu}F_{\mu\nu} (up to an overall constant).

Try Srednicki's book (google to find a free draft version online).

 

[patrice]

thank you, avodyne! i'll try this book. it really seems to be well structured and quite instructive.

patrice

 

[schieghoven]

Hi, you seem to be mostly on the right track...

... up to now it's not yet clear to me, how the gauge transformation "prescription" A_{\mu}'=UA_\mu U^{-1}+\frac{i}{g} U (\partial _\mu U^{-1}) is connected with the generators of the adjoint representation of Lie group SU(N), namely the structure constants of the corrensponding Lie Algebra.

Does A_\mu= A_\mu ^a T^a not mean (which one follows from the prescription above), that A_\mu is Lie Algebra valued for the representation of our choice for the matter multiplett \psi, even the fundamental rep?

Gauge theory posits a symmetry described by the transformation rule \psi \rightarrow U \psi, where U belongs to a gauge group G, and \psi belongs to a vector space which is referred to as the fundamental representation of G. The covariant derivative D_{\mu} \psi transforms in the same way (in other words, it also belongs to the fundamental representation), and this can be used to show gauge invariance of the kinetic term for \psi in the Lagrangian, as you seemed to refer to in your original post. The adjoint representation is the vector space of quantities which transform as Z \rightarrow UZU^{-1}; the connection A_\mu is of this type for *global* gauge transformations, while the field strength F_{\mu\nu} is an even better example, since it transforms this way even under local gauge transformations. A_\mu belongs to this space as a consequence of its definition in the covariant derivative. For your question, yes, both A_\mu and F_{\mu\nu} take values in the Lie algebra of G.

I'm not entirely sure, but perhaps your question in your original post asks what changes if we substitute G for one of its other representations? If so, the answer is nothing, because G can be whatever group you like; the fundamental representation is still the space which holds \psi, and A and F will be in the adjoint representation of that space.

Good luck with the studies!
Dave