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I am recently trying to understand dimensional regularization in the context of quantum field theory. So to solve an integral $$ \int_{\mathbb R^d} \frac{\text d ^d p}{(2 \pi)^d} \frac{1}{(p^2 + m^2)^2} $$ which is divergent for $d = 4$ it is often said that if one assumes that $d \in \mathbb C$ with sufficiently small $\text{Re}(d)$ it is possible to do some kind of analytic continuation. But in examples it is often shown that one uses $d = 4 - \varepsilon$ and then transforms to hyper spherical coordinates to get an expression like: $$ \int_{\mathbb R^+}\frac{\text d p}{(2\pi)^{4-\varepsilon}} \frac{2\pi^{(4-\varepsilon)/2}}{\Gamma\left(\frac{4 - \varepsilon}{2}\right)} \frac{p^{3-\varepsilon}}{(p^2 + m^2)^2} $$ which has a solution. We then expand for small values of $\varepsilon$.

Question:

Why is it necessary for $d$ to be complex?$\quad\rightarrow$ is there a possibility to stay real ?

Why is it legal to transform in hyper spherical coordinates when complex dimension is assumed ?

Do i get right that we expand the solution to the second integral in a laurent series for small $\varepsilon$ to extract terms which do not diverge at $\varepsilon=0$ and then when adding amplitudes the divergent terms cancel out ( dependent on the process of course ) ?

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  • $\begingroup$ FWIW, analytic continuation into the complex plane comes for free. Do you want to stay on the real line? $\endgroup$
    – Qmechanic
    May 2 '20 at 14:00
  • $\begingroup$ @Qmechanic i reread my question and commit that it was kind of unclear what i am asking. So i edited it. I know it is not welcome to ask multiple questions in one thread but since these are kind of related to the same problem i hope it is excused in this case. $\endgroup$ May 2 '20 at 14:40
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1) As much as I understand it, the value of d which you choose doesn't necessary need to be complex to solve the integral. The point is that you have the freedom to choose it to be a complex number but you are not forced to do it. Usually you choose a value of d for which the resulting integral is convergent and then you can analytically continue your result. But the analytical continuation always takes place on an open subset of the complex plane which gives you the "right" to also choose complex d.

2)This is a matter of definition. When you dimensionally regularize, you define the integral as an function of the integrand and the parameter d to the complex numbers with the special property(and some others) that for integer d it takes the same values as a usual d-dimensional integral. This condition allows you to write it as a hyperspherical integral, because the resulting formula is meromorphic and satisfies all the conditions imposed (e.g. linearity). In his book on renormalization collins proves (in a physics manner) that this conditions fixes the function uniquely and since the hyperspherical integral satisfies all the conditions, it is the right way to go.

3) Well the expansion in $\epsilon$ is right, you expand in a laurent series in $\epsilon$ (This uniquely defines your analytic continuation btw). You don't just add amplitudes that you got from your previous lagrangian. There are several ways to formulate the process of Renormalization but what I find the most convenient is via counterterms. You include additional, divergent terms in your lagrangian by redefinition of your masses, fields and coupling constants. These new terms induce new Feynman diagrams which you have to include in your amplitude calculation and the divergencies from your previous loop integrals cancel with these new feynman diagrams (at least for uv divergencies as far as I know, I'm still learning so I haven't gotten to infrared divergencies yet)

I hope this helps. Best

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