Dimensional regularization

Example: potential of an infinite charged line

Consider an infinite charged line with charge density s, and we calculate the potential of a point distance x away from the line. The integral diverges:V(x) = A \int_{-\infty}^\infty \frac{dy}{\sqrt{x^2 + y^2}}where A = s/(4\pi\epsilon_0).

Since the charged line has 1-dimensional "spherical symmetry" (which in 1-dimension is just mirror symmetry), we can rewrite the integral to exploit the spherical symmetry:\int_{-\infty}^\infty \frac{dy}{\sqrt{x^2 + y^2}} = \int_{-\infty}^\infty \frac{dt}{\sqrt{(x/x_0)^2 + t^2}} = \int_{0}^\infty \frac{\mathrm{vol}(S^1) dr}{\sqrt{(x/x_0)^2 + r^2}}where we first removed the dependence on length by dividing with a unit-length x_0, then converted the integral over \R^1 into an integral over the 1-sphere S^1, followed by an integral over all radii of the 1-sphere.

Now we generalize this into dimension d. The volume of a d-sphere is \frac{2\pi^{d/2}}{\Gamma(d/2)}, where \Gamma is the gamma function. Now the integral becomes \frac{2\pi^{d/2}}{\Gamma(d/2)}\int_{0}^\infty \frac{r^{d-1}dr}{\sqrt{(x/x_0)^2 + r^2}}When d = 1-\epsilon, the integral is dominated by its tail, that is, \int_{0}^\infty \frac{r^{d-1}dr}{\sqrt{(x/x_0)^2 + r^2}} \sim \int_c^\infty r^{d-2}dr = \frac{1}{d-1}c^{d-1} = \epsilon^{-1} c^{-\epsilon}, where c = \Theta(x/x_0) (in big Theta notation). Thus V(x)\sim (x_0/x)^\epsilon/\epsilon , and so the electric field is V'(x) \sim x^{-1}, as it should.

Example

Suppose one wishes to dimensionally regularize a loop integral which is logarithmically divergent in four dimensions, like

:I = \int\frac{d^4p}{(2\pi)^4}\frac{1}{\left(p^2+m^2\right)^2}.

First, write the integral in a general non-integer number of dimensions d = 4 - \varepsilon, where \varepsilon will later be taken to be small,I = \int\frac{d^dp}{(2\pi)^d}\frac{1}{\left(p^2+m^2\right)^2}. If the integrand only depends on p^2, we can apply the formula\int d^dp , f(p^2) = \frac{2 \pi^{d/2}}{\Gamma(d/2)} \int_0^\infty dp , p^{d-1} f(p^2). For integer dimensions like d = 3, this formula reduces to familiar integrals over thin shells like \int_0^\infty dp , 4 \pi p^2 f(p^2). For non-integer dimensions, we define the value of the integral in this way by analytic continuation. This givesI = \int_0^\infty \frac{dp}{(2\pi)^{4-\varepsilon}} \frac{2\pi^{(4-\varepsilon)/2}}{\Gamma\left(\frac{4-\varepsilon}{2}\right)} \frac{p^{3-\varepsilon}}{\left(p^2+m^2\right)^2} = \frac{2^{\varepsilon -4}\pi^{\frac{\varepsilon}{2}-1}}{\sin \left(\frac{\pi\varepsilon}{2}\right) \Gamma \left(1-\frac{\varepsilon}{2}\right)}m^{-\varepsilon} = \frac{1}{8\pi^2\varepsilon}-\frac{1}{16\pi^2}\left(\ln \frac{m^2}{4\pi}+\gamma\right)+ \mathcal{O}(\varepsilon). Note that the integral again diverges as \varepsilon \rightarrow 0, but is finite for arbitrary small values \varepsilon \neq 0.

References

References

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