I seem to be missing something regarding why Yang-Mills theories are Lorentz invariant quantum mechanically.

Start by considering QED. If we just study the physics of a massless $U(1)$ gauge field then by the usual Wigner little group classification one-particle states form representations of the little group for $k^\mu=(k,0,0,k)$ which is just $ISO(2)$. As is standard, we assume that our states only transform non-trivially under the $SO(2)$ subgroup of $ISO(2)$ so that states are labeled by their momentum and their helicity, $h$. That is, under an $SO(2)$ rotation by angle $\theta$ we have $U(\theta)|k,h\rangle=e^{ih\theta}|k,h\rangle$. Correspondingly, the creation operator $a^\dagger_h(k)$ transforms as $U(\theta)a^\dagger_h(k)U(\theta)^\dagger=e^{ih\theta}a^\dagger_h(k)$. Because we assumed that the states transform trivially under the other $ISO(2)$ generators, the same equations will hold true if we replace $U(\theta)$ by $U(W)$ where $W$ is any member of the $ISO(2)$ group.

It's then well known that if we try to write down a field operator $A_\mu$ in the usual manner, we will find that the transformation properties of the creation and annihilation operators force $A_\mu$ to transform under $U(W)$ as $U(W)A_\mu U(W)^\dagger=W_\mu{}^\nu A_\nu+\partial_\nu\Omega$ for Lorentz transformation $W$ and some irrelevant function $\Omega$. This is the standard argument by which Lorentz invariance is found to demand gauge invariance for massless particles.

My question then is what happens when you consider instead a non-Abelian gauge theory? I'd assume the same story holds and all gauge bosons just acquire an internal index, say $a$. That is, I thought states would be labeled like $|k,h,a\rangle$ and field operators would look like $A_\mu^a$ and transform as $U(W)A_\mu^aU(W)^\dagger=W_\mu{}^\nu A_\nu^a+\partial_\nu\Omega^a$ for some function $\Omega^a$.

This seems to be wrong, though, since the YM lagrangian is not invariant under the above and it is instead only invariant under full non-Abelian gauge transformations, $A\to U^{-1}(A+d)U$.

The only way out I can see is that single particle non-Abelian states may change under $U(W)$ as $U(W)|k,h,a\rangle=e^{ih\theta}\sum_b D(W)_{ab}|k,h,b\rangle$, where $D(W)_{ab}$ is a unitary matrix which is rotating the internal index on the gauge states. Previously, I'd been assuming that $D(W)_{ab}$ is trivial. If this is the proper transformation, I could see the $A_\mu^a$ operator inheriting the more familiar non-Abelian gauge transformation behavior.

However, naively it seems wrong that a Lorentz rotation will rotate the gauge indices of states since this would seem to be a mix of spacetime and internal transformations and hence ruled out by Coleman-Mandula. An internal non-Abelian transformation generator $V(X)$ should act on our state as $V(X)|k,h,a\rangle=\sum_b D_{ab}|k,h,b\rangle$ for some unitary matrix $D_{ab}$ and hence $V(X)U(\theta)|k,h,a\rangle\neq U(\theta)V(X)|k,h,a\rangle$ so there would be symmetry generators which do not commute with Poincare.

In summary, my questions are:

1) How does the non-Abelian gauge operator change under Lorentz transformations?

2) Does a one-particle non-Abelian spin-1 state $|k,p,a\rangle$ indeed transform as $U(W)$ as $U(W)|k,h,a\rangle=e^{ih\theta}\sum_b D(W)_{ab}|k,h,b\rangle$ with $D(W)_{ab}$ a non-trivial element of the non-Abelian group? If so, why doesn't this violate Coleman-Mandula?

  • $\begingroup$ I found a curious paper, where the "gap" is resolved, by using some "deformed Lorentz transformations" applying only on the non-physical (scalar, longitudinal) degrees of freedom. Abelian and non-abelian cases are studied (see for instance formulae $(3.2), (3.7)$. $\endgroup$
    – Trimok
    Commented Sep 16, 2014 at 11:00
  • 2
    $\begingroup$ Just a comment: what needs to be Lorentz invariant are the observables, not the operators I'd say. The observable are usually defined as some trace over the internal (gauge) degrees of freedom. No operator is invariant under a gauge-transform: they are covariant... but the trace is of course. Perhaps it helps in your case. $\endgroup$
    – FraSchelle
    Commented Sep 17, 2014 at 5:39
  • $\begingroup$ Yeah, I'm entirely fine with the fact that the operators are not Lorentz invariant. My issue is that the Lagrangian is not invariant under the naive transformations above. Since the Lagrangian is what appears in S-matrix calculations it needs to be Lorentz invariant for the theory to be Lorentz invariant, I believe. $\endgroup$
    – user26866
    Commented Sep 17, 2014 at 16:52
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    $\begingroup$ Gauge theories have Ward identities. So yes, your amplitude might naively have spacetime indices, but when you contract it with the polarization vectors of the gluons, the spurious dependence is supposed to vanish (transverse modes, little group, etc). $\endgroup$
    – Siva
    Commented Sep 17, 2014 at 20:35

2 Answers 2


I think that the problematic part here is the notion of what demands what. For example, you state

This is the standard argument by which Lorentz invariance is found to demand gauge invariance for massless particles.

but I am not completely sure if I agree with this, or at least to the interpretation you are carrying with it.

Taking a look at Weinberg's book on QFT[1], Chapter 5.9, I see a different way of looking at this issue. Weinberg notices that there is no consistent way of constructing an operator for the spin-1 massless field. The equivalent to your statement is that if we are to forcefully construct that operator, it has to be equivalent to itself in addition to a derivative of some other operator.

Now, in my interpretation (and I don't think that at the time of this reply there is an unanimous and unambiguous interpretation of this) is that the statement of gauge invariance is that of a single massless spin-1 field is only defined up to a sum of a divergence, and the important part is that this is a statement which is true even in the absence of matter coupled to the field.

My understanding of your question leads me to think that you are putting at the same footing two distinct notions:

  1. The intrinsic gauge invariance of a single massless spin-1 field, which is required to make sense of the corresponding operator.
  2. The gauge invariance of an interacting theory with matter coupled to a massless spin-1 field.

I point out this distinction because point 2. is the one that forces you to put a collection of massless spin-1 fields in the Adjoint representation, such that the kinetic term of matter with covariant derivatives is invariant under a gauge transformation of matter fields, whereas point 1. tells you that each of the massless spin-1 fields has an intrinsic gauge invariance where each is identical to itself plus the addition of a divergence.

I then propose you an interpretation that would solve this issue, and it goes like this:

  • a) We know that in order to make sense of spin-1 massless fields, their operators have to enjoy a gauge invariance of the form $$\hat A_\mu \to \hat A_\mu + \partial_\mu \hat \Omega $$ as such we allow for the classical counterpart to enjoy this freedom from the first place (point 1.).
  • b) Due to the previous point, we notice that theories of local $U(1)$ symmetries of the matter fields $$ \psi \to e^{i \theta(x)} \psi $$ coupled to matter $$\mathcal L = i \overline \psi \gamma \cdot \partial \psi + \overline \psi \gamma \cdot A \psi $$ are immediately invariant as the new terms arising from the derivative terms, the divergence of the phase $\theta$, can be gauged away into the field $A_\mu$.
  • c) For non-abelian theories, while it's feasible possible to construct them without being interacting with matter, their motivation comes from empirical evidence where they interact with matter. In such formulations we know that the theory is invariant if there is a collection of massless spin-1 fields that transform under the adjoint transformation (this is point 2.).
  • d) Furthermore, because of the presence of derivatives, there will still be a divergence of a lie-algebra valued quantity (the phases) which can be gauged away by the collection of the massless spin-1 fields due to their intrinsic gauge invariance (point 1.).

In some sense you can read point c) as being what some people related closely to the global (a lot of grains of salt in this term) part of the gauge transformation, while d) the local part. In the end the theory works because the requirement for an interacting gauge theory plays along nicely with the quantum mechanical requirements for writing down the corresponding massless spin-1 fields!

Would this help you? I have thought about similar issues (I wrote, for example, [2] even though it's not direclty related to your issue) and I've come to believe that the interpretation I constructed and provided above solves a lot of conceptual problems in gauge theories (namely the simple Lie group type).

So to answer your questions explicitly:

  1. Each field operator (meaning each of the $a$, using your index notation) in the non-abelian cause transforms just as the abelian case under Lorentz transformations, that is with an inhomogeneous derivative term.
  2. The collection of fields (all the $a$) transforms in accordance with Coleman-Mandula as it transforms under the adjoint representation, plus each $a$ has the "freedom" to be defined up to a derivative. This freedom plays nicely in presence of matter where derivatives of phases appear, and in absence of matter can be used to write each operator in a convenient gauge.


[1] Quantum Theory of Fields - Vol 1, S. Weinberg

[2] Conceptual Challenges of Gauge Symmetry, Miguel Crispim Romão https://www.academia.edu/5659587/Conceptual_Challenges_of_Gauge_Symmetry

  • 1
    $\begingroup$ I think that what is very important to note is that if we have only helicity 1 particles (massless), we don't need to work with this redundant gauge fields. We can only dealing with the F's fields. The requirement for this fields describe a 1 helicity particle fix the F field in terms of creation and anihilation operators. The reason to work with the gauge fields is that local interactions build out of F fields are not long-ranged. $\endgroup$
    – Nogueira
    Commented Oct 22, 2016 at 0:36

Disclaimer: I am not sure about this answer, but nevertheless I'm quite confident about it (and would like to be corrected if it is incorrect to learn something).

The one-particle states refer to asymptotic states at infinity where interactions are supposed to switch-off, that is the gauge coupling is sent to zero. In this limit a $U(1)^N$ gauge symmetry survives and everything works as for N copies of non-interacting photons.

As an aside comment, notice that your question is relevant only for non-confining (non-abelian) gauge theories since otherwise the gauge bosons aren't asymptotic one particle states. For example, your question applies for $W$'s but not for gluons in the SM.


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