To prove the Photon-Decoupling Identity for colour-ordered Yang-Mills amplitudes:

$$0= A(1,2,3,...,n)+A(2,1,3,...,n)+...+A(2,3,...,1,n) \tag{1}$$

I know I must use $(2)$, which expresses the pure gluon tree amplitudes (Yang-Mills amplitudes) in the following (colour-decomposed) form:

$$\mathcal{A}_n =g^{n-2}\sum_{non-cyc\\ perms} Tr[T^{a_1} ... T^{a_n}]A(1,...,n) \tag{2}$$

$A(1,...,n)$ represents the colour-ordered amplitude.

I'm told that:

If an external leg is a photon, the amplitude on $(2)$ must vanish, and that we can see this by looking at the standard form of the Lagrangian

$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \tag{3}$$

where $F_{\mu\nu}$ corresponds to

$$F_{\mu\nu}= \partial_\mu A_\nu - \partial_\nu A_\mu -\frac{ig}{\sqrt{2}}[A_\mu , A_\nu] \tag{4}$$

I'm also told that I should set one of the legs of the diagram as a photon and that this corresponds to setting:

$$(A_\mu)^i_j= A_\mu \delta^i_j \tag{5}$$

which decouples, since the interaction terms involve commutators. Setting $T^{a_1}= \mathbb{1} $ (unitary matrix)

What I don't understand:

Why must I turn/consider one of the legs of the Feynman representation to be a photon? Why am I allowed to do this??

Why does equation $(5)$ correspond to considering a leg to be a photon?

I know photons and gluons don't couple, but I don't understand why it is so mathematically: What does it mean by "it decouples since interaction terms involve commutators"?

  • $\begingroup$ Are you following a reference? $\endgroup$
    – Qmechanic
    Commented Apr 18, 2020 at 12:11
  • $\begingroup$ There is no reference as it is part of a solved homework from a lecturer. This is all there is to it. $\endgroup$ Commented Apr 18, 2020 at 12:17

1 Answer 1


If you consider a $U(N)$ gauge theory, the Lie algebra is equivalent to that of $U(1)\times SU(N)$, and you can use the name "photon" for the gauge field proportional to the $U(1)$ generator in the Lie algebra, and "gluon" for the gauge field proportional to the $SU(N)$ generator (do not confuse these names with the actual photons and gluons in the Standard Model, it is just common terminology to call photon any abelian gauge field and gluon any non-abelian one).

In more details, the $U(N)$ generators are complex $N\times N$ matrices with the additional constraint of being hermitean: $(T^a)^\dagger = T^a$, with $a = 1, \dots , N^2$. You can organize the basis of hermitean matrices in such a way that $N^2-1$ of them are trace-less matrices, and these will be the generators of $SU(N)$ and the remaining one is the identity matrix, which is the generator of the abelian subgroup $U(1)$.

Just to be clear, let me spell out the relation between your two-indices $i,j$ notation, and my one-index $a$ notation is:

$(A_\mu)^i_j = A^a_\mu (T^a)^i_j$

namely each $T^a$ is an $N\times N$ matrix, and the indices $i,j = 1,\dots,N$ are the matrix indices, while $a = 1,\dots , N^2$ (or $N^2 -1$ after we separate the identity) is an index that runs over the basis in space of matrices. So the gauge field that you write in your equation (5) is precisely the one in which the matrix $T^a$ is the identity $N\times N$ matrix.

As you correctly pointed out, the self-interactions between the gauge fields involve the commutator of the corresponding $T^a$ generators. So since our photon is the one whose $T^a$ is the identity matrix, and since the identity matrix commutes with any other matrix, the photon is going to drop from any self interaction.


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