# I How to show that Electrodynamics is conformally invariant?

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1. Jul 27, 2016

### physicality

[Moderator's note: changed thread title to be more descriptive of the actual question.]

Consider Maxwell's action $S=\int L$ over Minkovski space, where the Lagrangian density is $L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$, and the Electromagnetic tensor is given by $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ for some potential $A$. How can we see that this action is invariant under dilations and special conformal transformations?

A conformal transformation of space takes the form $$\xi^\mu(x)=a^\mu+\lambda^{\mu\nu}x_\nu+\lambda_Dx^\mu+x^2\lambda^\mu_K-2x^\mu x\cdot \lambda_K$$

The potential $A$ transforms as $$\delta A(x) = \xi^\mu(x)\partial_\mu A(x) - \Lambda^{\mu\nu}_M(x) \eta_{\mu[\nu}\delta^i_{j]} A^j+\Lambda_D(x) A(x)$$

where $\eta$ is the metric, $\delta_i^j$ is Kronekers' symbol, $${\Lambda_M}_{\mu\nu}(x)=\partial_{[\nu}\xi_{\mu]}=\lambda_{\mu\nu}-4x_{[\mu}\lambda_{K\nu]}$$ $$\Lambda_D(x)=\frac{1}{4}\partial_\rho \xi^\rho=\lambda_D-2x\cdot \lambda_K$$

Above, the extra terms are due to intrinsic parts of conformal transformations that act on the representation indices.

We would know that the electromagnetic tensor has conformal weight $2$, if we could show that $$\delta F_{\mu\nu}=\xi^\rho\partial_\rho F_{\mu\nu}+2\Lambda_{M[\mu}^\rho F_{\nu]\rho}+2\Lambda_D F_{\mu\nu}$$

Who can we show it and then use it to see that the action is invariant? Do we need the value of $\Lambda_D$ in terms of $\xi^\rho$ to this end?

Last edited by a moderator: Jul 27, 2016
2. Jul 28, 2016

### vanhees71

Well, you have to define, how the transformation acts on the fields (i.e., $A^{\mu}$) to fully define the symmetry transformation. Then you can use the complete definition of your transformation to check whether the equations of motion are form invariant, i.e., if they are a symmetry by plugging in the transformed quantities in the Lagrangian and show that it differs from the original Lagrangian only by a complete four-divergence.

3. Jul 28, 2016

### physicality

Yes, I know this, Vanhees. I did define the transformation. It is in explicit form. I'm asking how we can see that the action is invariant for this particular definition. I expect that the Lagrangian will not differ at all but I cannot go through with the computation.

4. Jul 28, 2016

### haushofer

See Zee's GR book, page 621. He does it explicitly.

5. Jul 29, 2016

### samalkhaiat

The invariance (of the free Maxwell theory) under the Poincare transformations (in n dimensions) is obvious. So, you only need to examine the invariance under dilatation and special conformal transformations.
1) Scale (Dilatation) Transformation:
Under dilatation, $\delta_{D}$, the vector potential transforms according to
$$\delta_{D}A_{\mu}(x) = \left( x^{\sigma}\partial_{\sigma} + d \right) A_{\mu}(x) , \ \ \ \ (1)$$ where $d$ is the scaling dimension of $A_{\mu}$ in n-dimensional space-time ($n = \partial_{\mu}x^{\mu}$). For the field tensor $F_{\mu\nu}$, equation (1) implies the following scale transformation
$$\delta_{D}F_{\mu \nu} = \left( x^{\sigma}\partial_{\sigma} + (d +1) \right) F_{\mu \nu} . \ \ \ \ (2)$$ Applying $\delta_{D}$ to the free Maxwell Lagrangian, we get
$$\delta_{D}\mathcal{L} = -\frac{1}{4}\eta^{\alpha \beta}\eta^{\mu\nu} \left( F_{\alpha \mu} \delta_{D}F_{\beta \nu} + (\delta_{D}F_{\alpha \mu}) F_{\beta \nu} \right) . \ \ \ (3)$$ Now substitute (2) in (3), then if you add and subtract the quantity $$F_{\alpha \mu}F_{\beta \nu} \ \partial_{\sigma}x^{\sigma} = n \ F_{\alpha \mu}F_{\beta \nu} ,$$ you obtain
$$\delta_{D}\mathcal{L} = \partial_{\sigma} \left( x^{\sigma} \mathcal{L} \right) + \left( 2(d+1) - n \right) \mathcal{L} .$$
Thus, the free Maxwell theory is scale invariant in n-dimensional space-time provided that $$d = \frac{n-2}{2} .$$
2) Special Conformal Transformation:
$$\delta_{C}^{\mu}x^{\nu} = 2x^{\mu}x^{\nu} - \eta^{\mu\nu}x^{2} . \ \ \ (4)$$
For a generic field, $\varphi (x)$, having scale dimension $d$ and transforming irreducibly by finite-dimensional (matrix) representation $\Sigma^{\mu\nu}$ of the Lorentz group $SO(1, n-1)$, we can show that $\delta_{C}^{\mu}\varphi (x)$ has the following form
$$\delta_{C}^{\mu}\varphi (x) = \left( \delta_{C}^{\mu}x^{\nu} \right) \partial_{\nu}\varphi (x) + 2 \left( x^{\mu} d + x_{\nu}\Sigma^{\mu\nu} \right) \varphi (x) . \ \ \ \ \ (5)$$
Fields that transform according to (5), under special conformal transformation, are called primary fields. For Maxwell theory in n-dimensional space-time, if we take the vector potential $A_{\alpha}$ to be a primary field having
$$d = \frac{n}{2} - 1, \ \ \mbox{and} \ \ \left( \Sigma^{\mu \nu}\right)_{\alpha}{}^{\rho}A_{\rho} = \delta^{\mu}_{\alpha}A^{\nu}- \delta^{\nu}_{\alpha}A^{\mu} ,$$ then (5) becomes
$$\delta_{C}^{\mu}A_{\alpha} = \left( \delta_{C}^{\mu}x^{\nu}\right) \partial_{\nu}A_{\alpha} + (n-2) x^{\mu}A_{\alpha} - 2 x_{\alpha}A^{\mu} + 2 \delta^{\mu}_{\alpha} x^{\nu}A_{\nu} . \ \ (6)$$
To obtain the action of $\delta_{C}^{\mu}$ on the field tensor $F_{\alpha \beta}$, we differentiate (6) with respect to $x^{\beta}$ and form the following anti-symmetric combination
$$\partial_{\alpha}\left(\delta_{C}^{\mu}A_{\beta}\right) - \partial_{\beta}\left(\delta_{C}^{\mu}A_{\alpha}\right) = \delta_{C}^{\mu} \left( \partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}\right) \equiv \delta_{C}^{\mu}F_{\alpha \beta} .$$
Doing the easy but boring algebra, you get
$$\delta_{C}^{\mu}F_{\alpha \beta} (x) = \Gamma^{\mu}_{\alpha \beta} (F) + 2(n-4) \delta^{\mu}{}_{[ \alpha}A_{\beta ]} , \ \ \ \ \ (7)$$ where
$$\Gamma^{\mu}_{\alpha \beta} (F) = \left( \delta_{C}^{\mu} x^{\nu}\right) \partial_{\nu}F_{\alpha \beta} + nx^{\mu}F_{\alpha \beta} + F^{\mu}{}_{[ \alpha}x_{\beta ]} - x^{\nu} \delta^{\mu}{}_{[ \alpha}F_{\beta ] \nu} . \ \ (8)$$
The presence of the last term in (7) means that $F_{\mu\nu}$ is a non-primary field when $n \neq 4$. The next boring step (which I let you do) consists of applying the special conformal variation-operator $\delta_{C}^{\mu}$ to the Maxwell Lagrangian (remember to use Leibniz rule). This will give you
$$\delta_{C}^{\mu}\mathcal{L} = \partial_{\nu}\left( (\delta_{C}^{\mu}x^{\nu} ) \mathcal{L} \right) + (4 - n) F^{\mu\nu}A_{\nu} . \ \ \ \ \ \ (9)$$
Since, in 4-dimensional space-time (n = 4), the change in Lagrangian is given by a total divergence, the action remains invariant. Thus, the full conformal group is the maximal space-time symmetry of the free Maxwell theory in $n = 4$ dimensions. However, when $n \neq 4$ the full conformal group is not an invariance group of the free Maxwell action. This follows from the fact that the term $F^{\mu\nu}A_{\nu}$ in (9) is not a total divergence of some object.
The same conclusions, about the invariance/non-invariance of Maxwell action, can be reached without calculating the variation $\delta_{C}^{\mu}\mathcal{L}$:
We already established that Maxwell’s theory is invariant under dilatation and Poincare transformations in n-dimensional space-time. So, to establish the invariance/non-invariance under the full conformal group, we only need to construct symmetric and conserved energy-momentum tensor, $T^{\mu\nu} = T^{\nu\mu}, \ \ \partial_{\mu}T^{\mu\nu} = 0$, examine its trace $\eta_{\mu\nu}T^{\mu\nu} = T$, and see whether or not $T = 0$. Now, in n-dimensional space-time, Maxwell theory allows us to construct such tensor
$$T^{\mu\nu} = -F^{\mu\rho}F^{\nu}{}_{\rho} - \eta^{\mu\nu} \mathcal{L} .$$ Taking the trace, we get
$$T = - F^{2} - n \mathcal{L} = (4 - n) \mathcal{L} .$$
Clearly, the trace vanishes only at $n = 4$.

6. Jul 30, 2016

### haushofer

Sam's post should help you through the calculation. If you still want me to scan Zee's analysis, let me know and I'll try. It's basically the variation of the Lagrangian, some partial integration and using the conformal Killing equation. The result is Sam's eqn.(9).

7. Jul 30, 2016

### ShayanJ

How can you check whether this condition is met or not? Or better...how do you determine d?

8. Jul 30, 2016

### samalkhaiat

There are many ways to determine the value of the scale dimension, beside the one that I have already described (i.e., the invariance of the action integral under dilatation).
A) Field Theoretical Methods:
1) Free fields: The scale dimension of a bosonic/fermionic field is determined from the invariance of the canonical equal-time commutation/anti-commutation relations under the scale transformation
$$x^{\mu} \to \bar{x}^{\mu} = e^{- \lambda} x^{\mu} . \ \ \ \ \ (1)$$
If the coordinates are scaled (down) as in (1), a generic local (boson/fermion) field $\varphi (x)$ will be subject to the following unitary transformation
$$\bar{\varphi}(\bar{x}) = U(\lambda) \varphi (\bar{x}) U^{\dagger}(\lambda) = e^{\lambda d} \varphi (x) , \ \ \ \ \ (2)$$ where $U = \exp (i\lambda D)$, $D$ is the hermitian generator (i.e., the Noether symmetry charge), $T(\lambda) = \exp (\lambda d)$ is a finite-dimensional (fully reducible) representation of the (non-compact) abelian group of dilatations $SO(1,1)$, and the real number $d$ is called the scale dimension of the field $\varphi (x)$. The infinitesimal version of (2) reads (the parameter $\lambda$ being factored out)
$$\delta_{D}\varphi (x) = [iD , \varphi (x)] = (d + x^{\mu} \partial_{\mu}) \varphi (x) . \ \ \ \ (3)$$
Case One: $\varphi (x)$ is a free bosonic field.
In this case, the conjugate momentum is given by $\pi (x) = \dot{\varphi}(x)$, and the canonical bracket by the fundamental equal-time commutator $[ \varphi (x , t) , \pi (y , t)] = i\delta^{n-1}(x-y)$.
To see how $\pi (x)$ transforms under the scale transformation (1), we differentiate (2) or (3) with respect to time:
From (2) and (1), we obtain (using the chain rule and the fact that $dD/dt = 0$)
$$\bar{\pi} (\bar{x}) = U \pi (\bar{x}) U^{\dagger}= e^{\lambda (d + 1)} \pi (x) . \ \ \ \ \ (4)$$
Infinitesimally, this becomes
$$\delta_{D}\pi (x) = [iD , \pi (x)] = (d + 1 + x^{\mu}\partial_{\mu}) \pi (x) \ \ \ \ (5)$$
Now write the commutator $$[ \bar{\varphi} (\bar{x} , \bar{t}) , \bar{\pi} (\bar{y} , \bar{t})] = i\delta^{n-1}(\bar{x} - \bar{y}) , \ \ \ \ (6)$$ substitute (1) into the RHS, and (2) & (4) into the LHS of (6) and obtain
$$[ \varphi (x , t) , \pi (y , t)] \ e^{\lambda (2d + 1)} = i e^{\lambda (n - 1)} \ \delta^{n-1}(x-y) .$$
Thus, the scale dimension of the free bosonic field is given by $$d = \frac{n - 2}{2} .$$
Case Two: $\varphi (x)$ is a free Fermi field. This case is easy because the conjugate momentum is given by $\pi (x) = i \varphi^{\dagger}(x)$, and it has the same transformation law as that of $\varphi (x)$. So, in this case, the equal-time anti-commutation relation $$\{ \bar{\varphi} (\bar{x} , \bar{t}) , \bar{\pi} (\bar{y} , \bar{t}) \} = i \delta^{n-1}(\bar{x} - \bar{y}) ,$$ leads to $$d( \mbox{fermion} ) = \frac{n - 1}{2} .$$
2) Consistency with the power-law behaviour of the correlation function:
$$\langle \phi_{1}(x) \phi_{2}(0) \rangle \sim \int d^{n}k \frac{e^{ikx}}{k^{2}} \sim x^{2-n} . \ \ \ \ (7)$$
Let $d(\phi_{1}) = d(\phi_{2}) \equiv d$. In fact, it is easy to show (using the conformal algebra) that the 2-point functions of fields of different scale dimensions vanish. Now, consider the identity
$$[iD , \phi_{1}(x)\phi_{2}(0)] = [iD , \phi_{1}(x)] \phi_{2}(0) + \phi_{1}(x) [iD , \phi_{2}(0)] . \ \ \ (8)$$
Exact dilatation symmetry implies that $D|0 \rangle = \langle 0 | D = 0$. Thus, Eq(8) leads to
$$\langle 0 | [ iD , \phi_{1}(x)] \phi_{2}(0) | 0 \rangle + \langle 0 | \phi_{1}(x) [ iD , \phi_{2}(0) ] | 0 \rangle = 0 . \ \ \ (9)$$
Applying Eq(3) to our fields, we get
$$[iD , \phi_{1}(x) ] = (d + x \cdot \partial ) \phi_{1}(x) ,$$ $$[iD , \phi_{2}(0) ] = d \ \phi_{2}(0) .$$
Substituting these equations into Eq(9), we find
$$( x \cdot \partial + d ) \langle \phi_{1}(x) \phi_{2}(0) \rangle + d \ \langle \phi_{1}(x) \phi_{2}(0) \rangle = 0 . \ \ \ (10)$$
Now, substitute the power-law behaviour (7) into equation (10), and do the differentiation:
$$(2 - n) x^{2-n} + 2d x^{2-n} = 0 , \ \ \Rightarrow \ \ d = \frac{n-2}{2} .$$
B) Group Theoretical Method:
This is too complicated and lengthy business to do on PF. Basically, the representations of the conformal group $SO(2 , n)$, corresponding to primary fields, are classified by the finite-dimensional irreducible representations, $(j_{1}, j_{2})$, of the Lorentz group and the scale dimension $d$. These determine the invariant Casimir operators of the conformal group. The representations of $SO(2,n)$ can also be classified in terms of its maximal compact subgroup $SO(2) \times SO(n)$. In all methods, unitarity imposes the following lower bounds on the scale dimension of the fields: mass is denoted by $m$, and spin & helicity by $s/h$
$$\mbox{For} \ \ j_{1}j_{2} = 0, \ \ \ \ d \geq \frac{n-2}{2} + j_{1} + j_{2} .$$ This contains $m \geq 0$ and $s / h = j_{1} + j_{2} / j_{1} - j_{2}$,
$$\mbox{For} \ \ j_{1}j_{2} \neq 0, \ \ \ \ d \geq (n-2) + j_{1} + j_{2} .$$ This contains $m > 0$ and spins $s = |j_{1} - j_{2}|$ all the way to $j_{1} + j_{2}$ in integer steps.
Notice, as in the case of the Lorentz group, that the vector potential $A_{\mu}$ has no room in the representation theory of the conformal group.

9. Jul 31, 2016

### ShayanJ

Wow...thanks Sam!
But I thought you derived the condition for Maxwell's equations to be invariant under dilatations. So I thought we should determine d with some procedure and then see whether it satisfies that condition. But if you call that "determining d", then it seems you want Maxwell's equation to be invariant under dilatations and then choose d somehow that they are.

10. Aug 1, 2016

### vanhees71

Sure, that's the way you find symmetries. The point is that you must find a way to define how the symmetry transformations act on all elements of the (variation of the) action. If you find a way to define such transformation laws under a certain group, which leave the variation of the action invariant, you have defined a symmetry and also found corresponding conserverd quantities. Noether's theorem also works in the other direction, i.e., if you have found some conserved quantity it defines a one-parameter symmetry group of the theory. It's very elegant in the Hamilton version of the action principle: The conserved quantities are the generators of the symmetry transformations in the sense of the corresponding Lie algebra defined by the Poisson brackets (in field theory of course defined in terms of functional derivatives; otherwise it's very similar to what's done in point mechanics).

11. Aug 1, 2016

### physicality

12. Aug 9, 2016

### physicality

I'm looking at your original answer, samalkhaiat, and at the section treating special conformal transformations in particular.
You say that a special conformal transformation take the form
Transforming a single coordinate component of a point is meaningless. A special conformal transformation of a point is generally $\delta_{C}^{\mu}x = 2 x\cdot C x^{\mu} - x^{2} C^\mu$. Still, trying to assign meaning to a special conformal transformation of a single component of a space-time point, in the same way that you do it, while following my original post, I arrive at
$$\delta_{C}^{\mu}x^{\nu} = 2x^{\mu}x^{\nu}C_{\nu} - \eta^{\mu\nu}C^{\mu}x^{2} . \ \ \ (4')$$

However, the potential A does not transform like a primary field. In four dimensions, the scale dimension of A is $d=K_D=1$. Now, following my original post, and taking a special conformal transformation, i.e. the only non-zero parameter is $C=\lambda_K,$ we see that A transforms by $$\delta_{C} A_i(x) = \delta_{C}^{\mu}(x) \partial_{\mu}A_i(x) - \frac{1}{2} \eta^{\mu\gamma} \eta^{\nu\delta} ( -4x_{[\gamma} C_{\delta]} ) ( 2 ( \delta^\mu_i A^\nu(x) - \delta^\nu_i A^\mu(x) ) ) - (2x \cdot C)A_i(x)$$

How can we reconcile these apparent differences? How should we modify the rest of your argument?

13. Aug 9, 2016

### samalkhaiat

Who did that? The $x^{\nu}$’s are coordinates of an arbitrary point in space-time. And, why is it “meaningless” to look at how, say, the $x^{1}$ transforms?

What is the meaning of your symbol $\delta^{\mu}_{C}$ ?

Clearly, this equation is wrong. Notice the index structure of both sides. Let me explain to you what I did.
For an arbitrary infinitesimal coordinate transformations, we usually write $$x^{\nu} \to \bar{x}^{\nu} = x^{\nu} + \delta x^{\nu} , \ \ \mbox{for all} \ \ \nu = 0 , 1 , 2 , \cdots ,$$ where $\delta x^{\nu}$ represents some function $f^{\nu}(x ; C)$ of the coordinates and the infinitesimal parameters which define the transformations. Mathematically speaking, the above coordinate transformations can be defined once the vector field $\delta x^{\nu} \partial_{\nu} = f^{\mu}\partial_{\mu}$ is given. So, the infinitesimal special conformal transformations are generated by the following vector field (components)
$$f^{\nu} (x;C) = \delta x^{\nu} \equiv 2 (C \cdot x) x^{\nu} - x^{2}C^{\nu} .$$
This can be rewritten as
$$\delta x^{\nu} \equiv \left( 2 x^{\mu} x^{\nu} - x^{2} \eta^{\mu\nu} \right) C_{\mu} . \ \ \ \ \ (1)$$
Now, I don’t like the parameter $C_{\mu}$ to appear in every equation I write. And since Eq(1) is a definition, I can factor out $C_{\mu}$ by the following definition $$\delta x^{\nu} \equiv C_{\mu} \delta^{\mu}_{C} \ x^{\nu} . \ \ \ \ \ \ \ \ \ (2)$$ Clearly, (1) and (2) allow you to write the infinitesimal SCT’s in the form
$$\delta^{\mu}_{C}x^{\nu} = 2 x^{\mu} x^{\nu} - \eta^{\mu\nu} x^{2} . \ \ \ \ (4)$$

Who told you that? Can you write a non-trivial $A^{\mu}$ as a derivative of another field? To be precise, the vector potential is a primary field up to gauge transformation. In the same way, $A^{\mu}$ is a Lorentz vector up to gauge transformation.
A local field (operator) $\varphi_{r}(x)$ is called primary if it is "annihilated" by the special conformal generator $K^{\mu}$ at $x = 0$, i.e.,
$$\delta^{\mu}_{C} \varphi_{r}(0) = [i K^{\mu} , \varphi_{r}(0)] = 0 .$$
I cannot check this equation unless you position your indices properly.

Last edited: Aug 9, 2016
14. Aug 14, 2016

### physicality

Dear samalkhaiat,

Thank you for your help. I will address each issue in the order you raise it.

The first misunderstanding stems from the fact that I take symbols like $x,f$ to mean a point and a field and I take $x^\mu,f^\mu$ to mean their $\mu$-th component. Nevermind this piece of semantics.

Secondly, the meaning of $\delta_C^\mu$ is what you see on the left hand side of (4'). This is a meaningful expression. I have two conciding fixed indices on both sides. Nevermind this naive attempt either. The right equation is equation (1) from your last response.

Now, no one told me that the electromagnetic potential does not transform as a primary field. I just don't see it from the information that I have. I mean the equations in my original post. And so following the notation from my original post, I took $\lambda_D=0,a^\mu=0,\lambda_K=C,\lambda^{\mu\nu}=0$, i.e. I took a special conformal transformation determined by the vector $C$. Considering that $(\Sigma^{\mu\nu})_\alpha^\rho A_\rho = \eta^{\mu[\nu}\delta_\alpha^{\rho]} A_\rho=\delta^\mu_\alpha A^\nu - \delta^\nu_\alpha A^\mu$, I arrive at
$$\delta A_i =\delta^{\mu}(x) \partial_{\mu}A_i(x) - \frac{1}{2} ( -4x_{[\mu} C_{\nu]} ) ( 2 ( \delta^\mu_i A^\nu(x) - \delta^\nu_i A^\mu(x) ) ) - (2x \cdot C)A_i(x)$$

I hope that I have substituted right this time.

15. Aug 14, 2016

### vanhees71

Good! I recommend

M. Schwartz, Quantum field theory and the standard model, Cambridge University Press

instead. For the very fine details and the most general cases (like particles with spins $\geq 3/2$ and the like) see

S. Weinberg, Quantum Theory of Fields, Cambridge University Press (2 vols. + 1 vol. SUSY QFT).

16. Aug 16, 2016

### samalkhaiat

1) Piece of advice: To avoid confusing, never use the variation symbol $\delta$ to mean an ordinary function. If you are dealing with group of transformations, $T(\vec{\alpha})$, with $\alpha_{a}$ being the corresponding parameters, then you can think of $\delta = \alpha_{a}\delta_{(T)}^{a}$ as an operator.
2) I cannot comment on the textbook you mentioned in your PM to me, because I don’t know their notations for Lorentz group element. Some people write $\exp (\omega_{\mu\nu}\Sigma^{\mu\nu})$, which leads to a factor of 2 appearing in the transformation law for the fields and also in the algebra. I use the most frequently used convention, i.e., $\exp(\frac{1}{2}\omega_{\mu\nu}\Sigma^{\mu\nu})$.
3) The information you have still say that $A_{\mu}$ is a primary: Regardless of the messy-looking transformation law that you wrote, you still have $\delta_{SCT}A_{\mu}(0) \equiv \bar{A}_{\mu}(x) - A_{\mu}(x)|_{x = 0} = 0$, and that is the definition of primary field.
4) You need to understand why the fields transform the way they do under the special conformal transformation (SCT.). Consider the following infinitesimal special conformal diffeomorphism on flat n-dimensional space-time
$$x^{\mu} \to \bar{x}^{\mu} = x^{\mu} - \xi^{\mu}(x; C) , \ \ \ \ |C^{\mu}| \ll 1 ,$$ $$\xi^{\mu}(x;C) \equiv - \delta x^{\mu} = 2 (C \cdot x) x^{\mu} - x^{2} C^{\mu} ,$$ where $C^{\mu}$, $\mu = 0,1, \cdots , n-1$ are the constant parameters of SCT. To understand the structure of the SCT, we need the Jacobian of the transformations:
$$\frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} = \delta^{\mu}_{\nu} - 2 (C \cdot x) \delta^{\mu}_{\nu} + 2 ( C^{\mu}x_{\nu} - C_{\nu} x^{\mu} ) .$$
This says that (at each space-time point) an infinitesimal SCT is a local scaling by $\alpha (x) = -2 C \cdot x$ followed by local Lorentz transformation by $\omega_{\mu\nu}(x) = 2 (C_{\mu} x_{\nu} - C_{\nu} x_{\mu})$. Therefore, the field, $\varphi_{r}(x)$, which transforms irreducibly under the Lorentz group (i.e., having intrinsic spin matrix $\Sigma^{\mu\nu}$) and, therefore, having intrinsic scale matrix $(\Delta)_{r}{}^{s} = d \ \delta^{s}_{r}$, should transform as follow (under the SCT)
$$\bar{\varphi}_{r}(\bar{x}) = \left( e^{\frac{1}{2}\Sigma^{\mu\nu} \omega_{\mu\nu}(x) - \Delta \alpha (x)} \right)_{r}{}^{s} \ \varphi_{s}(x) .$$
Expanding both sides to first order in $C^{\mu}$ and using the identity $\Sigma^{\mu\nu}\omega_{\mu\nu}(x) = 4 C_{\mu}x_{\nu}\Sigma^{\mu\nu}$, we obtain
$$\bar{\varphi}_{r}(x) - \xi^{\mu}(x;C) \partial_{\mu}\varphi_{r}(x) = \left( 1 + 2 C_{\mu}x_{\nu}\Sigma^{\mu\nu} + 2 (C \cdot x) \Delta \right)_{r}{}^{s} \ \varphi_{s}(x) .$$ Substituting $\Delta_{r}{}^{s} = d \ \delta^{s}_{r}$ and $\delta_{SCT}\varphi_{r}(x) \equiv \bar{\varphi}_{r}(x) - \varphi_{r}(x)$, we get
$$\delta_{SCT}\varphi_{r}(x) = \xi^{\mu}(x;C) \partial_{\mu}\varphi_{r}(x) + 2 C_{\mu}x_{\nu} \left( \Sigma^{\mu\nu}\right)_{r}{}^{s}\varphi_{s}(x) + 2d \left( C \cdot x \right) \varphi_{r}(x) .$$
To specialize this equation to the vector potential $A_{\sigma}$, let $r \to \sigma$, $s \to \rho$ and $\varphi (x) \to A(x)$, then use
$(\Sigma^{\mu\nu})_{\sigma}{}^{\rho}A_{\rho} = \delta^{\mu}_{\sigma}A^{\nu} - \delta^{\nu}_{\sigma}A^{\mu}$ and $2d = (n - 2)$ (if you work in 4 dimensions, then $d = 1$). This way you will have no difficulty in obtaining
$$\delta_{SCT}A_{\sigma}(x) = \xi^{\mu}(x;C) \partial_{\mu}A_{\sigma} + 2 (C_{\sigma}x_{\nu} - C_{\nu}x_{\sigma}) A^{\nu} + (n - 2) (C \cdot x ) A_{\sigma} .$$
This transformation law contains the special conformal parameter $C_{\mu}$. To factor out the parameter and obtain the equation I wrote in previous post, write
$$\delta_{SCT} \equiv C_{\tau} \delta_{C}^{\tau} , \ \ \mbox{and} \ \ \xi^{\mu}(x;C) \equiv C_{\tau} \xi^{\mu \tau}(x) \equiv C_{\tau} \delta_{C}^{\tau}x^{\mu} .$$
For more details, you can see the following old thread