QFTs & Epsilons: Unraveling Beta Functions in Dim Reg

[lonelyphysicist]

Question about epsilons that occur in calculating beta functions in QFTs w dim reg

In deriving the beta function of, say, QED using dimensional regularization we get the relation (up to 1 loop)

\beta[e] = - \frac{\epsilon}{2} e - e \frac{d ln[Z_{e}]}{d ln[\mu]} \quad (1)

and

Z_{e} = 1 + \frac{e^{2} A}{\epsilon}

where e is the coupling, Z_{e} is the renormalization of the coupling, \mu is the arbitrary scale introduced to make the dimension of the coupling the same as if \epsilon was zero, and A is some constant that does not depend on \mu or \epsilon.

Now if I do the following math

\frac{d ln[Z_{e}]}{d ln[\mu]}
= \frac{1}{1 + \frac{e^{2} A}{\epsilon} } \frac{d}{d ln[\mu]} \left( 1 + \frac{e^{2} A}{\epsilon} \right)
= \frac{2 e A}{\epsilon + e^{2} A} \beta[e]

Is this correct? It seems extremely elementary, but if I trust my results

\beta[e] \left(1 + \frac{2 e^2 A}{\epsilon + e^{2} A} \right) = - \frac{\epsilon}{2}e

Taking the limit \epsilon going to zero I get that the beta function for QED is zero up to one loop!

What seems to be usually done is we taylor expand \frac{1}{1 + \frac{e^{2} A}{\epsilon} } \approx 1 - \frac{e^2 A}{\epsilon} and so

\frac{d ln[Z_{e}]}{d ln[\mu]}
= \frac{2 A e \beta[e]}{\epsilon} + \mathcal{O}[e^{4}] \quad (2)

and next we put (2) into (1) and iterate

\beta[e]
= - \frac{\epsilon}{2}e - \frac{2 A e^{2} \beta[e]}{\epsilon} + \mathcal{O}[e^{4}]
= - \frac{\epsilon}{2}e - 2 A e^{2} \left( - \frac{\epsilon}{2}e - \frac{2 A e^{2} \beta[e]}{\epsilon} + \mathcal{O}[e^{4}] \right) \frac{1}{\epsilon} + \mathcal{O}[e^{4}]
= - \frac{\epsilon}{2} e + A e^{3} + \frac{(2 A e^{2})^{2} \beta[e]}{\epsilon^{2}} + . . .

and we say as epsilon goes to zero it's approximately equal to

A e^{3}

(For instance A = 1/12 \pi^{3} for pure QED.)

How can we taylor expand in powers of e when \epsilon is supposed to be tiny in dimensional regularization? Even if we do taylor expand and iterate what about the 1/\epsilon^{2} and possibly even more infinite quantities?

This issue really bothers me a lot because I'm sure there's something I'm not understanding here, since the QED coupling does indeed run as calculated, right? Any clarification would be deeply appreciated.

 

[R0man]

The use of dimensional regularization in calculating beta functions in QFTs can be quite confusing, especially when it comes to the role of epsilons. Your calculations are correct, but it is important to understand the concept of "renormalization group flow" in order to make sense of the results.

In dimensional regularization, we introduce an arbitrary scale \mu to make the dimension of the coupling the same as if \epsilon was zero. This allows us to avoid infinities and perform calculations in a more manageable way. However, the presence of \epsilon in the equations indicates that we are working in a theory with a non-integer number of dimensions. This is not physical, and the goal is to take the limit \epsilon \rightarrow 0 at the end of the calculations.

In the expansion of (1/\epsilon), we are essentially taking into account the effects of higher dimensional operators. This is why we are able to get a non-zero beta function even though the one-loop result seems to indicate that the coupling does not run. However, in order to get a physical result, we must take the limit \epsilon \rightarrow 0.

The taylor expansion in powers of e is a way of approximating the results in terms of the coupling constant, which is a small parameter. This is a common technique in perturbative calculations. The higher order terms involving 1/\epsilon^{2} and higher powers of e are actually finite and do not cause any issues. In fact, they are necessary in order to get a physical result when taking the limit \epsilon \rightarrow 0.

So in summary, the use of epsilons in dimensional regularization is a mathematical tool that allows us to perform calculations without dealing with infinities. It is important to remember that the ultimate goal is to take the limit \epsilon \rightarrow 0 in order to get a physical result. The taylor expansion in powers of e is a useful approximation technique, and the higher order terms involving 1/\epsilon^{2} and higher powers of e are necessary for obtaining a physical result.