# A Proof - gauge transformation of yang mills field strength

#### spaghetti3451

In Yang-Mills theory, the gauge transformations

$$\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})\psi$$

and

$$A^{a}_{\mu} \to A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c}$$

induce the gauge transformation

$$F_{\mu\nu}^{a} \to F_{\mu\nu}^{a} - f^{abc}\theta^{b}F_{\mu\nu}^{c}.$$

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Proof:

Let us begin with the definition of the field-strength tensor in terms of the covariant derivative $D_{\mu} = \partial_{\mu}-iA_{\mu}^{a}T^{a}_{\bf R}$:

$$F^{a}_{\mu\nu}T^{a}_{\bf R} = i\left[D_{\mu},D_{\nu}\right].$$

In order to determine the gauge transformation of $F^{a}_{\mu\nu}T^{a}_{\bf R}$, we must first determine the gauge transformation of $D_{\mu}D_{\nu}\psi$. It is rather easy to show that

$$D_{\mu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}\psi$$

and hence

$$D_{\mu}D_{\nu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}D_{\nu}\psi.$$

Therefore, we now have

$$F_{\mu\nu}^{a}T^{a}_{\bf R} = i\left[D_{\mu},D_{\nu}\right]$$

$$F_{\mu\nu}^{a}T^{a}_{\bf R} \to i(1 \pm i\theta^{b}T^{b}_{\bf R})\left[D_{\mu},D_{\nu}\right]$$

$$F_{\mu\nu}^{a}T^{a}_{\bf R} \to (1 \pm i\theta^{b}T^{b}_{\bf R})F_{\mu\nu}^{a}T^{a}_{\bf R}$$

$$F_{\mu\nu}^{a}T^{a}_{\bf R} \to F_{\mu\nu}^{a}T^{a}_{\bf R} \pm i\theta^{b}T^{b}_{\bf R}F_{\mu\nu}^{a}T^{a}_{\bf R}$$

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In the last line of the derivation, I don't have a third term needed to form a structure constant. Where have I gone wrong?

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#### samalkhaiat

$$D_{\mu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}\psi$$

and hence

$$D_{\mu}D_{\nu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}D_{\nu}\psi.$$
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In the last line of the derivation, I don't have a third term needed to form a structure constant. Where have I gone wrong?
The second equation is not correct because the covariant derivative $D_{\mu}$ does not pass through the gauge function $\theta^{a} (x)$.

#### spaghetti3451

Consider the proof of $D_{\mu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}\psi$ (under the gauge transformation $\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})\psi$ of the Yang-Mills Lagrangian).

Skip this proof if you want to. The point of including this tedious proof is to obtain the important result that

$\left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)(1 \pm i\theta^{b}T^{b}_{\bf R}) = (1 \pm i\theta^{a}T^{a}_{\bf R})(\partial_{\mu}-iA_{\mu}^{a}T^{a}_{\bf R})$

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$D_{\mu}\psi \to \left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)\left(1 \pm i\theta^{d}T^{d}_{\bf R}\right)\psi$

$= \left(\partial_{\mu} - iA_{\mu}^{a}T_{\bf R}^{a} \mp i(\partial_{\mu}\theta^{a})T_{\bf R}^{a} \mp if^{abc}A_{\mu}^{b}\theta^{c}T_{\bf R}^{a}\right)\left(1 \pm i\theta^{d}T^{d}_{\bf R}\right)\psi$

$= \partial_{\mu}\psi - iA_{\mu}^{a}T_{\bf R}^{a}\psi \mp i(\partial_{\mu}\theta^{a})T_{\bf R}^{a}\psi \mp if^{abc}A_{\mu}^{b}\theta^{c}T_{\bf R}^{a}\psi \pm i\partial_{\mu}(\theta^{d}T^{d}_{\bf R}\psi) \pm A_{\mu}^{a}T^{a}_{\bf R}\theta^{d}T^{d}_{\bf R}\psi.$

We can simplify the third and fifth terms to obtain

$\mp i(\partial_{\mu}\theta^{a})T_{\bf R}^{a}\psi \pm i\partial_{\mu}(\theta^{d}T^{d}_{\bf R}\psi) = \pm i\theta^{a}T^{a}_{\bf R}\partial_{\mu}\psi$

and we can simplify the fourth and sixth terms using the commutation relation of the ${\bf su}(N)$ algebra to obtain

$\mp if^{abc}A_{\mu}^{b}\theta^{c}T_{\bf R}^{a}\psi \pm A_{\mu}^{a}T^{a}_{\bf R}\theta^{d}T^{d}_{\bf R}\psi = \pm \theta^{a}A_{\mu}^{b}(if^{abc}T^{c}_{\bf R} + T^{b}_{\bf R}T^{a}_{\bf R})\psi = \pm \theta^{a}T^{a}_{\bf R}A^{b}_{\mu}T^{b}_{\bf R}\psi$

Therefore, the gauge transformation of the covariant derivative $D_{\mu}\psi$ simplifies to

$D_{\mu}\psi \to \partial_{\mu}\psi-iA^{a}_{\mu}T^{a}_{\bf R}\psi \pm i\theta^{a}T^{a}_{\bf R}\partial_{\mu}\psi \pm \theta^{a}T^{a}_{\bf R}A^{b}_{\mu}T^{b}_{\bf R}\psi$
$= (1\pm i\theta^{a}T^{a}_{\bf R})(\partial_{\mu}-iA^{b}_{\mu}T^{b}_{\bf R})\psi$
$= (1\pm i\theta^{a}T^{a}_{\bf R})D_{\mu}\psi$

#### spaghetti3451

We can compute the gauge transformation of $D_{\mu}D_{\nu}\psi$ similarly.

$D_{\mu}D_{\nu}\psi \to \left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)(1 \pm i\theta^{b}T^{b}_{\bf R})D_{\nu}\psi$

Tracing the computation in my previous post, we find that

$\left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)(1 \pm i\theta^{b}T^{b}_{\bf R}) = (1 \pm i\theta^{a}T^{a}_{\bf R})(\partial_{\mu}-iA_{\mu}^{a}T^{a}_{\bf R})$

Therefore, the gauge transformation of $D_{\mu}D_{\nu}\psi$ is given by

$D_{\mu}D_{\nu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}D_{\nu}\psi.$

#### spaghetti3451

I do not see how I have passed the covariant derivative $D_{\mu}$ through the gauge function $\theta(x)$.

#### dextercioby

Homework Helper
That $\pm$ everywhere is tiring for the eye, plus that's simply wrong to use. So choose a sign to begin with and respect it in the sequel.

#### samalkhaiat

We can compute the gauge transformation of $D_{\mu}D_{\nu}\psi$ similarly.

$D_{\mu}D_{\nu}\psi \to \left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)(1 \pm i\theta^{b}T^{b}_{\bf R})D_{\nu}\psi$

Tracing the computation in my previous post, we find that

$\left(\partial_{\mu} - i(A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c})T_{\bf R}^{a}\right)(1 \pm i\theta^{b}T^{b}_{\bf R}) = (1 \pm i\theta^{a}T^{a}_{\bf R})(\partial_{\mu}-iA_{\mu}^{a}T^{a}_{\bf R})$

Therefore, the gauge transformation of $D_{\mu}D_{\nu}\psi$ is given by

$D_{\mu}D_{\nu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}D_{\nu}\psi.$
The problem you are asking is very simple, and you can do it if you use simple and neat expressions. Let me show you one method. And you will see that even a school kid can follow the algebra. Okay, I will work with matrix-valued fields and functions, i.e., fields and functions taking value in the Lie algebra of the gauge group. These are the connection $\mathbb{A}_{\mu}(x) = A_{\mu}^{a}(x)T^{a}$ and the field tensor $\mathbb{F}_{\mu\nu}(x) = F_{\mu\nu}^{a}T^{a}$. The $T^{a}$’s are arbitrary (anti-hermitian) matrix representation of the Lie algebra $[T^{a} , T^{b}] = C^{abc}T^{c}$ . Of course the field tensor is defined by

$$\mathbb{F}_{\mu\nu}(x) = \partial_{\mu}\mathbb{A}_{\nu} - \partial_{\nu}\mathbb{A}_{\mu} + [\mathbb{A}_{\mu} , \mathbb{A}_{\nu}] . \ \ \ \ \ (1)$$

I will write the gauge group element (or the transformation matrix) as $U(x) = e^{\Theta (x)}$, with $\Theta (x) = \theta^{a}(x)T^{a}$ and $\theta^{a}(x)$ being the arbitrary gauge functions. When I consider infinitesimal transformation, I will write $U(x) = 1 + \Theta (x)$. On any Lie algebra valued function, the covariant derivative is defined by $$D_{\mu}\Theta (x) = \partial_{\mu} \Theta + [ \mathbb{A}_{\mu} , \Theta ] . \ \ \ \ \ \ (2)$$

Now, under a finite gauge transformation $U(x)$, the gauge field transforms in-homogeneously as $$\mathbb{A}_{\mu} \to U^{-1} \mathbb{A}_{\mu} U + U^{-1}\partial_{\mu}U . \ \ \ \ \ (3)$$ This means that the gauge field (or the connection) transforms in the adjoint representation of the corresponding global symmetry group, i.e., when the transformation matrix $U$ is constant. Infinitesimally, i.e., to first order in $\Theta$, Eq(3) becomes

$$\mathbb{A}_{\mu} \to \mathbb{A}_{\mu} + \partial_{\mu}\Theta + [\mathbb{A}_{\mu} , \Theta ] .$$ Let’s rewrite this in the language of infinitesimal variation or change $\delta \mathbb{A}_{\mu}$ and use the definition of covariant derivative $$\delta \mathbb{A}_{\mu} = \partial_{\mu}\Theta + [\mathbb{A}_{\mu} , \Theta ] = D_{\mu} \Theta . \ \ \ \ (4)$$ From this, you can easily obtain the following commutator equation

$$[D_{\mu} , D_{\nu}] \Theta = D_{\mu} \delta\mathbb{A}_{\nu} - D_{\nu} \delta\mathbb{A}_{\mu} . \ \ \ \ \ (5)$$ Now, we have every things needed to solve your problem. That is, given the transformation $\delta\mathbb{A}_{\nu}$ we would like to find the corresponding infinitesimal transformation of the field tensor, i.e. $\delta \mathbb{F}_{\mu\nu}$. Applying the variation symbol $\delta$ on both sides of Eq(1) and rearranging the terms, we obtain

$$\delta \mathbb{F}_{\mu\nu} = \partial_{\mu} \delta\mathbb{A}_{\nu} + [\mathbb{A}_{\mu} , \delta\mathbb{A}_{\nu} ] - \partial_{\nu} \delta\mathbb{A}_{\mu} - [\mathbb{A}_{\nu} , \delta\mathbb{A}_{\mu} ] .$$ Using the definition of the covariant derivative, the above equation becomes

$$\delta \mathbb{F}_{\mu\nu} = D_{\mu} \delta\mathbb{A}_{\nu} - D_{\nu} \delta\mathbb{A}_{\mu} . \ \ \ \ \ \ \ (6)$$ Now, using Eq(5), equation (6) becomes

$$\delta \mathbb{F}_{\mu\nu} = [D_{\mu} , D_{\nu}] \Theta . \ \ \ \ \ \ \ \ (7)$$

So, we need to calculate the commutator on the RHS of (7). To do that, we only need to first find the action of $D_{\mu}D_{\nu}$ on $\Theta$ and then the commutator can be constructed by interchanging the indices. From the definition of the covariant derivative, we obtain

$$D_{\mu}D_{\nu} \Theta = D_{\mu} \left( \partial_{\nu} \Theta + [\mathbb{A}_{\nu} , \Theta ] \right) .$$ Now, on the RHS, apply the definition of $D_{\mu}$ to get

$$D_{\mu}D_{\nu} \Theta = \partial_{\mu} \left( \partial_{\nu} \Theta + [\mathbb{A}_{\nu} , \Theta ] \right) + [\mathbb{A}_{\mu} , \partial_{\nu} \Theta + [ \mathbb{A}_{\nu} , \Theta ] ] .$$

This can be rewritten as

$$D_{\mu}D_{\nu} \Theta = \partial_{\mu} \partial_{\nu} \Theta + \partial_{\mu} [ \mathbb{A}_{\nu} , \Theta ] + [\mathbb{A}_{\mu} , \partial_{\nu}\Theta ] + [\mathbb{A}_{\mu} , [\mathbb{A}_{\nu} , \Theta ]] .$$

Now, obtain another equation by interchanging $\mu \leftrightarrow \nu$ and subtract it from the above equation

$$[D_{\mu} , D_{\nu}] \Theta = [\partial_{\mu}\mathbb{A}_{\nu} - \partial_{\nu}\mathbb{A}_{\mu} , \Theta ] + [\mathbb{A}_{\mu} , [\mathbb{A}_{\nu} , \Theta]] - [\mathbb{A}_{\nu} , [\mathbb{A}_{\mu} , \Theta]] .$$ And finally, we use the Jacobi identity

$$[\mathbb{A}_{\mu} , [\mathbb{A}_{\nu} , \Theta]] - [\mathbb{A}_{\nu} , [\mathbb{A}_{\mu} , \Theta]] = [[\mathbb{A}_{\mu} , \mathbb{A}_{\nu}] , \Theta ] ,$$ to arrive at

$$[D_{\mu} , D_{\nu}] \Theta = [\partial_{\mu}\mathbb{A}_{\nu} - \partial_{\nu}\mathbb{A}_{\mu} + [\mathbb{A}_{\mu} , \mathbb{A}_{\nu}], \Theta ] ,$$ which is

$$[D_{\mu} , D_{\nu}] \Theta = [\mathbb{F}_{\mu\nu} , \Theta ] . \ \ \ \ \ \ \ \ \ (8)$$ So, from (7) and (8), you find the answer to your problem

$$\delta \mathbb{F}_{\mu\nu} = [\mathbb{F}_{\mu\nu} , \Theta ] , \ \ \ \ \ \ \ \ \ \ (9)$$

or

$$\mathbb{F}_{\mu\nu} \to \mathbb{F}_{\mu\nu} + [\mathbb{F}_{\mu\nu} , \Theta ] . \ \ \ \ \ (10)$$

This is nothing but the infinitesimal version of the following finite transformation $$\mathbb{F}_{\mu\nu} \to U^{-1} \mathbb{F}_{\mu\nu} U .$$

In terms of the components of the field tensor $F^{a}_{\mu\nu}$, Eq(9) becomes

$$\delta F^{c}_{\mu\nu}T^{c} = F^{a}_{\mu\nu}\theta^{b} [T^{a} , T^{b}] = C^{abc}F^{a}\theta^{b}T^{c} .$$ Thus $$\delta F^{c}_{\mu\nu} = C^{abc}F^{a}_{\mu\nu}\theta^{b} .$$
So, the lesson from this is: use neat and simple expressions, and don't let your equations look like tangled spaghetti.

#### Geonaut

@samalkhaiat great answer! You just saved me from a hopeless mess of math, thank you! I'm currently trying to do the final step which is to prove that $$tr[F^a_{\mu\nu}\frac{\sigma^a}{2} F^{a\mu\nu}\frac{\sigma^a}{2}]$$ is gauge invariant using $$\delta F^a_{\mu \nu} = -C^{abc} F^c_{\mu \nu} \theta^b$$, but I'm having trouble with it. My approach looks like this: $$tr[(F^a_{\mu\nu}\frac{\sigma^a}{2} + \delta F^a_{\mu \nu}\frac{\sigma^a}{2})(F^{a\mu\nu}\frac{\sigma^a}{2} + \delta F^{a\mu\nu}\frac{\sigma^a}{2})]$$ = $$tr[(F^a_{\mu \nu}\frac{\sigma^a}{2} - C^{abc}F^c_{\mu\nu}\theta^b\frac{\sigma^a}{2})(F^{a\mu\nu}\frac{\sigma^a}{2} - C^{abc}F^{c\mu\nu}\theta^b\frac{\sigma^a}{2})]$$ = $$tr[(F^1_{\mu\nu}\frac{\sigma^1}{2} + F^2_{\mu\nu}\frac{\sigma^2}{2} +F^3_{\mu\nu}\frac{\sigma^3}{2} - C^{123}F^3_{\mu\nu} \theta^2 \frac{\sigma^1}{2} -C^{132}F^2_{\mu\nu}\theta^3 \frac{\sigma^1}{2} -C^{213}F^3_{\mu\nu}\theta^1\frac{\sigma^2}{2} - \ldots)(F^{1\mu\nu}\frac{\sigma^1}{2} +\ldots)]$$. I know that the trace of the terms involving different pauli matrices multiplying each other all vanish, and I can get the terms that look like $$tr[(F^a_{\mu\nu}\frac{\sigma^a}{2}\delta F^{a\mu\nu}\frac{\sigma^a}{2})]$$ to vanish, since, for example: $$-C^{123}F^1_{\mu\nu}F^{3\mu\nu}\theta^2 - C^{321}F^3_{\mu\nu}F^{1\mu\nu}\theta^2 = 0$$. However, the terms that look like $$\delta F^a_{\mu\nu} \frac{\sigma^a}{2} \delta F^{a\mu\nu}\frac{\sigma^a}{2}$$ are not going away!!! I know I'm making a serious mistake somewhere, but I guess I'm bad at math because I'm having a lot of trouble finding it.

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#### PeterDonis

Mentor
@Mr. Chiappone please stop using all bold. That's the online equivalent of shouting.

#### Geonaut

@Mr. Chiappone please stop using all bold. That's the online equivalent of shouting.
Sure, I'm not sure why that ended up being all in bold, but I fixed it.

#### PeterDonis

Mentor
I'm not sure why that ended up being all in bold, but I fixed it.
Ok, thanks! Sometimes the post editor does do weird things with fonts.

#### Geonaut

Ok, thanks! Sometimes the post editor does do weird things with fonts.
No problem!

#### samalkhaiat

$$tr[F^a_{\mu\nu}\frac{\sigma^a}{2} F^{a\mu\nu}\frac{\sigma^a}{2}]$$
This is wrong. Do not use the same repeated indices in both factors. The correct expression is this $$L \equiv \mbox{Tr} \left( F^{a}_{\mu\nu}t^{a} F^{b \mu\nu} t^{b} \right).$$ Now, $F^{a}_{\mu\nu}F^{b \mu\nu}$ is a number, so you can pull it out of the trace: $$L = F^{a}_{\mu\nu}F^{b \mu\nu} \ \mbox{Tr} (t^{a}t^{b}) .$$ But $\mbox{Tr}(t^{a}t^{b}) = \frac{1}{2} \delta^{ab}$. So, your expression becomes $$L = \frac{1}{2} F^{a}_{\mu\nu} F^{a \mu\nu}.$$ Now use $\delta (AB) = A (\delta B) + B (\delta A)$ to obtain $$\delta L = F^{a}_{\mu\nu} \ \delta F^{a \mu\nu} = F^{a}_{\mu\nu}F^{b \mu\nu} \theta^{c} C^{abc} .$$ This is zero because $C^{abc} = - C^{bac}$.

However, the terms that look like $$\delta F^a_{\mu\nu} \frac{\sigma^a}{2} \delta F^{a\mu\nu}\frac{\sigma^a}{2}$$ are not going away!!!
Infinitesimal transformation are linear in the group parameter $\theta$. So $\delta F \ \delta F \propto \theta^{2} \approx 0$.

Proving gauge invariance is a lot easier if you use the finite gauge transformation $F_{\mu\nu} \to U^{-1}F_{\mu\nu}U$ and the trace property $\mbox{Tr} (AB) = \mbox{Tr} (BA)$: $$\mbox{Tr} \left( U^{-1}F_{\mu\nu}UU^{-1}F^{\mu\nu}U\right) = \mbox{Tr}\left( U^{-1} (F_{\mu\nu}F^{\mu\nu}U)\right) = \mbox{Tr} \left( F_{\mu\nu}F^{\mu\nu}\right) .$$

#### Geonaut

That makes perfect sense, that is exactly the information that I was looking for! I was mislead a little bit by other sources, and you just completely cleared everything up for me, thank you!

"Proof - gauge transformation of yang mills field strength"

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