dimensional regularization

Last updated Dec 10, 2022

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A function that goes through the points $(x,y)$ with $y = (x-1)!$

Definition: Gamma function For $n\in\N$: $\Gamma(n)=(n-1)!$ Extension to complex numbers $z\in\cnums$ $$\Gamma(z) = \int^\infty_0\text{d}x x^{z-1} e^{-x}$$

Note:

• $\Gamma(1) = 1$
• For $n\in\N$ we have $\Gamma(n+1)=n\Gamma(n)$
• Also for $z\in\cnums$ we have $\Gamma(z+1)=z\Gamma(z)$: proove: by partial integration $$\int^\infty_0\text{d}x x^z e^{-x} = x^{z-1}e^{-x}\big|^\infty_0 + \int^\infty_0\text{d}x x^{z-1}e^{-x}$$
• Useful values: $\Gamma(\frac{1}{2})=\sqrt{\pi}$, $\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}$

The Gamma function in this form does not converge for negative values. One can define an analytic continuation through the recurrence formula $$\Gamma(z) = \frac{\Gamma(z + n + 1)}{z(z+1)\cdots (z+n)}.$$ Still this continuation is singular for negative integers.

Through a different form of the Gamma function we can arrive at $$\Gamma(-n + \epsilon) = \frac{(-1)^n}{n!}(\frac{1}{\epsilon} + \sum_i \frac{1}{i} - \gamma_E + \mathcal{O}(\epsilon)),$$ where $\gamma_E = 0.5772$ is the Euler-constant.

Example:

• $\Gammae(\epsilon) = \frac{1}{\epsilon} - \gamma_E + \mathcal{O}(\epsilon)$
• $\Gammae(-1+\epsilon) = -\frac{1}{\epsilon} - 1 + \gamma_E + \mathcal{O}(\epsilon)$

• We define a $d$-dimensional integral as $$\int \text{d}^dke^{-k^2x} = \left(\int^\infty_{-infty}\text{d}xe^{-k^2x}\right)^d = (\frac{\pi}{x})^{d/2}$$ I only see this work for functions $f(k^2)$ with $f(x+y) = f(x)f(y)$

We can split the angular part $\text{d}\Omega_d$ in general spherical coordinates $$\int\text{d}^dk f(k^2) = \int \text{d}\Omega_d \int^\infty_0 \text{d}k k^{d-1} f(k^2)$$

We can apply this to calculate high dimensional integrals of $e^{-k^2x}$ $$\int \text{d}^dk e^{-k^2x} = \int \text{d}\Omega_d \int^\infty_0\text{d}k k^d-1e^{-k^2x}=\frac{1}{2}\int\text{d}\Omega_d\int^\infty_0\text{d}k^2 (k^2)^{\frac{d-2}{2}}e^{-k^2x} = \frac{1}{2}\int\text{d}\Omega_d \frac{\Gamma(\frac{d}{2})}{x^{\frac{d}{2}}}$$

comparing the two computations of the integral we find the general formular for the are of a $d$-dimensional sphere

$$\int \text{d}\Omega_d = 2\frac{\pi^\frac{d}{2}}{\Gamma(\frac{d}{2})}$$

We can extend this definition to spheres with dimension $d\in\Reals$

General formula $$\int \text{d}^d k f(k^2) = 2 \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}\int^\infty_0\text{d}k k^{d-1}f(k^2)$$

A useful identity: $$\alpha^-S\Gamma(s) = \int^\infty_0 \text{d}x x^{S-1} e^{-x\alpha}$$

Calculating some integrals (using the identity above):

1. $\int \text{d}^d k \frac{1}{(k^2+a^2)^n = \int \text{d}^dk \frac{1}{\Gamma(n)} \int^\infty_0 \text{d}x x^{n-1} e^{-(k^2 + a^2)x} = \frac{\pi^{\frac{d}{2}}}{\Gamma(n)}\Gamma(n-\frac{d}{2}) a^{d-2n}$

2. $\int \text{d}^d k \frac{k^2}{(k^2+a^2)^n$ Partial fraction decomposition

3. $\int \text{d}^d k \frac{k^\mu k^\nu}{(k^2+a^2)^n} = \frac{\delta^{\mu\nu}}{d}\int\text{d}^dk \frac{k^2}{(k^2+a^2}^n}$ by inserting $\delta^{\mu\nu}\detla_{\mu\nu}$

$$\int \frac{\text{d}^dk}{(k^2)^\alpha} = 0$$

Setting the dimension $d=4-2\epsilon$ consider

$$\mu^{2\epsilon}\int\frac{\text{d}^{4-2\epsilon}k}{(2\pi)^{4-2\epsilon}}f(k)$$

where $\mu$ is some scale of mass dimension one so that $[\mu^{2\epsilon}d^{4-2\epsilon}k]=4$. Using the integrals with the Gamma function one gets

$$\mu^{2\epsilon}\int\frac{\text{d}^{4-2\epsion}k}{(2\pi)^{4-2\epsilon}}\frac{1}{k^2+m^2} = \frac{m^2}{16\pi^2}\left(-\frac{1}{\epsilon}-1+\gamma_E-\log 4\pi+\log\frac{m^2}{\mu^2}+\mathcal{O}(\epsilon)\right)$$ and the squared version $$\mu^{2\epsilon}\int\frac{\text{d}^{4-2\epsion}k}{(2\pi)^{4-2\epsilon}}\frac{1}{(k^2+m^2)^2} = \frac{1}{16\pi^2}\left(\frac{1}{\epsilon}-\gamma_E+\log 4\pi-\log\frac{m^2}{\mu^2}+\mathcal{O}(\epsilon)\right)$$

General formula $$\mu^{2\epsilon}\int\frac{\text{d}^{4-2\epsilon}k}{(2\pi)^{4-2\epsion}}\frac{k^l}{(k^2+m^2)^n} = \frac{\mu^{2\epsilon}4^\epsilon}{16\pi^{2-\epsilon}}m^{4-2n+l-2\epsilon}\frac{\Gamma(2+\frac{l}{2}-\epsilon)\Gamma(n-2-\frac{l}{2}+\epsilon)}{\Gamma(n)\Gamma(2-\epsilon)}$$

After dimensional regularization the integral for the action is $d=4-2\epsilon$-dimensional

$$S = \int\text{d}^{4-2\epsilon}x \mathcal{L}$$

Therefore fields and couplings get a dimension different from the $4$-dim case

• $[\mathcal{L}]=d$
• $[\psi]=\frac{d-1}{2}$
• Coupling $g$: $[g]=\epsilon$ … etc