Feynman Parameters¶

When integrating a denominator like ${1\over AB}$ , the idea is to introduce auxiliary parameters in order to make the denominator simpler. We start with the identity:

(1) ${1\over AB} = \int_0^1 \d x {1\over \left(xA + (1-x)B\right)^2} = \int_0^1 \d x \int_0^1 \d y {\delta(x+y-1) \over \left(xA + yB\right)^2}$

which can be proven easily:

>>> var("A B")
(A, B)
>>> integrate(1/(x*A + (1-x)*B)**2, (x, 0, 1))
1/(A*B - A**2) - 1/(-A*B + B**2)
>>> simplify(_)
1/(A*B)

By repeatedly differentiating with respect to B:

(2) ${1\over AB^2} = \int_0^1 \d x \int_0^1 \d y {2 y \delta(x+y-1) \over \left(xA + yB\right)^3} {1\over AB^3} = \int_0^1 \d x \int_0^1 \d y {3 y^2 \delta(x+y-1) \over \left(xA + yB\right)^4} {1\over AB^n} = \int_0^1 \d x \int_0^1 \d y {n y^{n-1} \delta(x+y-1) \over \left(xA + yB\right)^{n+1}}$

Then we prove:

(3) ${1\over A_1A_2\cdots A_n} = \int_0^1 \d x_1\cdots\int_0^1 \d x_n {(n-1)!\, \delta(x_1 + \cdots + x_n-1) \over \left(x_1A_1 + \cdots + x_nA_n\right)^n}$

For $n=2$ we get (1) and if it holds for $n$ it also holds for $n+1$ , because we multiply (3) by ${1\over A_{n+1}}$ and get:

${1\over A_1A_2\cdots A_n}{1\over A_{n+1}} = = \int_0^1 \d x_1\cdots\int_0^1 \d x_n (n-1)!\,\delta(x_1 + \cdots +x_n-1) {1 \over \left(x_1A_1 + \cdots + x_nA_n\right)^n A_{n+1}} = = \int_0^1 \d x_1\cdots\int_0^1 \d x_n (n-1)!\,\delta(x_1 + \cdots +x_n-1) \int_0^1 \d x \int_0^1 \d y {n y^{n-1} \delta(x+y-1) \over \left(xA_{n+1} + y\left(x_1A_1 + \cdots + x_nA_n\right)\right)^{n+1}} = \int_0^1 \d x_1\cdots\int_0^1 \d x_n \int_0^1 \d y {n!\,\delta(x_1 + \cdots +x_n-1) y^{n-1} \over \left((1-y)A_{n+1} + y\left(x_1A_1 + \cdots + x_nA_n\right)\right)^{n+1}}= = \int_0^{1\over y} \d x_1\cdots\int_0^{1\over y} \d x_n \int_0^1 \d y {n!\,\delta(yx_1 + \cdots +yx_n-y) y^n \over \left((1-y)A_{n+1} + y\left(x_1A_1 + \cdots + x_nA_n\right)\right)^{n+1}}= = \int_0^{1\over y} y\d x_1\cdots\int_0^{1\over y} y\d x_n \int_0^1 \d y {n!\,\delta(yx_1 + \cdots +yx_n-y) \over \left((1-y)A_{n+1} + \left(yx_1A_1 + \cdots + yx_nA_n\right)\right)^{n+1}}= = \int_0^1 \d z_1\cdots\int_0^1 \d z_n \int_0^1 \d y {n!\,\delta(z_1 + \cdots +z_n-y) \over \left((1-y)A_{n+1} + \left(z_1A_1 + \cdots + z_nA_n\right)\right)^{n+1}}= = -\int_0^1 \d z_1\cdots\int_0^1 \d z_n \int_1^0 \d y' {n!\,\delta(z_1 + \cdots +z_n+y'-1) \over \left(y'A_{n+1} + \left(z_1A_1 + \cdots + z_nA_n\right)\right)^{n+1}}= = \int_0^1 \d x_1\cdots\int_0^1 \d x_{n+1} {n!\, \delta(x_1 + \cdots + x_{n+1}-1) \over \left(x_1A_1 + \cdots + x_{n+1}A_{n+1}\right)^{n+1}}$

Where we used (2) and the fact, that $\delta(x_1 + \cdots +x_n-1)=y\,\delta(yx_1 + \cdots +yx_n-y)$ , after the substituation we also restricted the limits of integration from 1 to $1\over y$ , since $x_1$ , $x_2$ , ... are all positive.

Example 1¶

$\int {\d^4 k\over (k-p)^2 (k^2 - m^2)} = \int d^4 k \int_0^1\d x \d y {\delta(x+y-1)\over D^2}$

where

$D = x (k-p)^2 + y(k^2 - m^2) = (x+y)k^2 - 2xk\cdot p +xp^2-ym^2 = k^2 - 2xk\cdot p +xp^2-ym^2$

In the last part we used $x+y=1$ . We now shift $k$ by introducing:

$l = k - xp \d^4k = \d^4 l$

and we get:

$D = k^2 - 2xk\cdot p +xp^2-ym^2 = l^2 - x^2p^2 + xp^2 - ym^2$

thus:

$\int d^4 k \int_0^1\d x \d y {\delta(x+y-1)\over D^2} = = \int l^4 k \int_0^1\d x \d y {\delta(x+y-1)\over (l^2 - x^2p^2 + xp^2 - ym^2)^2}$

Example 2¶

$\int {\d^4 k\over (k^2-m^2+i\epsilon)((k+p)^2 - m^2 + i\epsilon) ((k-p)^2+i\epsilon)} = \int d^4 k \int_0^1\d x \d y\d z {2\delta(x+y+z-1)\over D^3}$

where

$D = x (k^2-m^2+i\epsilon) + y((k+p)^2 - m^2 + i\epsilon) + z((k-p)^2+i\epsilon) = = (x+y+z)k^2 + 2k\cdot(yq-zp) + yq^2 + zp^2 - (x+y)m^2 + (x+y+z)i\epsilon = = k^2 + 2k\cdot(yq-zp) + yq^2 + zp^2 - (x+y)m^2 + i\epsilon$

In the last part we used $x+y+z=1$ . We now shift $k$ by introducing:

$l = k + yq - zp \d^4k = \d^4 l$

and we get:

$D = k^2 + 2k\cdot(yq-zp) + yq^2 + zp^2 - (x+y)m^2 + i\epsilon = = l^2 - \Delta + i\epsilon$

where

$\Delta = -xyq^2 + (1-z)^2m^2$

thus:

$\int d^4 k \int_0^1\d x \d y\d z {2\delta(x+y+z-1)\over D^3} = = \int \d^4 l \int_0^1\d x \d y\d z {2\delta(x+y+z-1)\over (l^2 - \Delta + i\epsilon)^3 } = = (-i)\int \d^4 l_E \int_0^1\d x \d y\d z {2\delta(x+y+z-1)\over (l_E^2 + \Delta)^3 } = = (-i)\int \d\Omega_4 \int_0^\infty \d l_E \int_0^1\d x \d y\d z {2\delta(x+y+z-1)l_E^3\over (l_E^2 + \Delta)^3 } = = (-i4\pi^2)\int_0^1\d x \d y\d z \delta(x+y+z-1) \int_0^\infty \d l_E {l_E^3\over (l_E^2 + \Delta)^3 } = = (-i4\pi^2)\int_0^1\d x \d y\d z \delta(x+y+z-1) \int_\Delta^\infty \d h {h-\Delta \over 2 h^3 } = = (-i4\pi^2)\int_0^1\d x \d y\d z \delta(x+y+z-1) {1\over 4\Delta} = = (-i\pi^2)\int_0^1\d x \d y\d z {\delta(x+y+z-1) \over \Delta} = = (-i\pi^2)\int_0^1\d x \d y\d z {\delta(x+y+z-1) \over (1-z)^2m^2 - xyq^2}$

This integral has an infrared divergence. We can cure this by pretending that the photon has a small nonzero mass $\mu$ , then in the denominator of the photon propagator we need to change:

$(k-p)^2 \to (k-p)^2 - \mu^2$

This denominator is multiplied by $z$ later on, so at the end we need to do the change:

$\Delta \to \Delta + z\mu^2$

and we get:

$(-i\pi^2)\int_0^1\d x \d y\d z {\delta(x+y+z-1) \over (1-z)^2m^2 - xyq^2} \to \to (-i\pi^2)\int_0^1\d x \d y\d z {\delta(x+y+z-1) \over (1-z)^2m^2 - xyq^2 + z\mu^2}$

for $q^2=0$ we get:

$(-i\pi^2)\int_0^1\d x \d y\d z {\delta(x+y+z-1) \over (1-z)^2m^2 - xyq^2 + z\mu^2} \to \to (-i\pi^2)\int_0^1\d x \d y\d z {\delta(x+y+z-1) \over (1-z)^2m^2 + z\mu^2} = = (-i\pi^2)\int_0^1\d z\int_0^{1-z} \d y{1 \over (1-z)^2m^2 + z\mu^2} = = (-i\pi^2)\int_0^1\d z {1-z \over (1-z)^2m^2 + z\mu^2}$

We can use the following integral:

$\int_{0}^{1} \frac{1 - z}{1 - 2 z + z^{2} + z \mu^{2}}\,dz = \frac{1}{2} \operatorname{log}\left(\mu^{-2}\right) + \frac{\operatorname{atan}\left(\frac{1}{\sqrt{-1 + \frac{4}{\mu^{2}}}}\right)}{\sqrt{-1 + \frac{4}{\mu^{2}}}} - \frac{\operatorname{atan}\left(\frac{1 - \frac{2}{\mu^{2}}}{\sqrt{-1 + \frac{4}{\mu^{2}}}}\right)}{\sqrt{-1 + \frac{4}{\mu^{2}}}}$

that is equal to $\half\log({1\over\mu^2})$ in the limit $\mu\to0$ .

here are a few special cases for $\mu = 1$ , $\mu=1/2$ and $\mu=1/3$ :

$\int_{0}^{1} \frac{1 - z}{1 - z + z^{2}}\,dz = \frac{1}{9} \pi \sqrt{3} \int_{0}^{1} \frac{1 - z}{1 - \frac{7}{4} z + z^{2}}\,dz = \frac{1}{2} \operatorname{log}\left(4\right) + \frac{1}{15} \sqrt{15} \operatorname{atan}\left(\frac{1}{15} \sqrt{15}\right) + \frac{1}{15} \sqrt{15} \operatorname{atan}\left(\frac{7}{15} \sqrt{15}\right) \int_{0}^{1} \frac{1 - z}{1 - \frac{17}{9} z + z^{2}}\,dz = \frac{1}{2} \operatorname{log}\left(9\right) + \frac{1}{35} \sqrt{35} \operatorname{atan}\left(\frac{1}{35} \sqrt{35}\right) + \frac{1}{35} \sqrt{35} \operatorname{atan}\left(\frac{17}{35} \sqrt{35}\right)$

Code:

>>> from sympy import log, atan, var, sqrt, Eq, Integral, S
>>> var("z m mu")
>>> F = -log(z*(1 - 2/m) + 1/m + z**2/m)/2 + \
        atan((1 - 2/m + 2*z/m)/sqrt(-1 + 4/m))/sqrt(-1 + 4/m)
>>> f = F.diff(z).simplify()
>>> print f
(1 - z)/(1 - 2*z + m*z + z**2)
>>> integ_f_0_1 = F.subs(z, 1) - F.subs(z, 0)
>>> e = Eq(Integral(f.subs(m, mu**2), (z, 0, 1)), integ_f_0_1.subs(m, mu**2))
>>> print e
Integral((1 - z)/(1 - 2*z + z**2 + z*mu**2), (z, 0, 1)) == log(mu**(-2))/2 + atan((-1 + 4/mu**2)**(-1/2))/(-1 + 4/mu**2)**(1/2) - atan((1 - 2/mu**2)/(-1 + 4/mu**2)**(1/2))/(-1 + 4/mu**2)**(1/2)
>>> print e.subs(mu, 1)
Integral((1 - z)/(1 - z + z**2), (z, 0, 1)) == pi*3**(1/2)/9
>>> print e.subs(mu, S(1)/2)
Integral((1 - z)/(1 - 7*z/4 + z**2), (z, 0, 1)) == log(4)/2 + 15**(1/2)*atan(15**(1/2)/15)/15 + 15**(1/2)*atan(7*15**(1/2)/15)/15
>>> print e.subs(mu, S(1)/3)
Integral((1 - z)/(1 - 17*z/9 + z**2), (z, 0, 1)) == log(9)/2 + 35**(1/2)*atan(35**(1/2)/35)/35 + 35**(1/2)*atan(17*35**(1/2)/35)/35

Then for $m=1$ and small $\mu$ we get:

$(-i\pi^2)\int_0^1\d z {1-z \over (1-z)^2m^2 + z\mu^2} = = (-i\pi^2)\half\log{1\over\mu^2}$

Feynman Parameters¶

Example 1¶

Example 2¶

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Feynman Parameters¶

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Example 2¶

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