# Some Useful Integrals In Quantum Field Theory

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In my paper on renormalisation I mentioned what most who have studied calculations in Quantum Field Theory find, its rather complicated and mind numbing. I am not sure doing these is that illuminating of anything, but every now and then you gotta to do, what you gotta to do, so I would like to go through one of those, Pauli–Villars regularization, as an example of the general theory in my second paper.

This makes use of a number of multidimensional integrals which is the subject of this paper.

The Gaussian Integral

The most basic of the integrals considered here is the good old Gaussian integral which you likely have seen before, but will do it anyway.  It’s:

$I = \int_{-∞}^{∞} e^{-x^2} dx$

Here is the trick:

$I^2 = \int_{-∞}^{∞} e^{-x^2} dx \int_{-∞}^{∞} e^{-y^2} dy = \int_{-∞}^{∞} \int_{-∞}^{∞} e^{-(x^2+y^2)} dx dy$

Going to polar coordinates:

$I^2 =\int_{0}^{∞} \int_{0}^{2\pi} e^{-r^2} r drdθ = 2\pi \int_{0}^{∞} e^{-r^2} r dr$

Substituting $s = r^2$

$I^2 = \pi \int_{0}^{∞} e^{-s} ds = \pi$

Hence $I = \int_{-∞}^{∞} e^{-x^2} dx =\sqrt {\pi}$

The Gamma Function

Another function that tends to crop up is the so called Gamma Function $\Gamma(x)$ defined by:

$\Gamma(x) = \int_{0}^{∞} t^{x-1} e^{-t} dt$

It has two rather interesting properties:

Using integration by parts $\Gamma(x+1) = x \Gamma(x)$.  But since  $\Gamma(1) = 1$ we have $\Gamma(n+1) = n!$.  The Gamma function generalises the factorial function.

$\Gamma ( \frac{1}{2} ) = \int_{0}^{∞} t^{-\frac{1}{2}} e^{-t} dt$.  Substitute $t = u^2$.  $\Gamma ( \frac{1}{2} ) = 2 \int_{0}^{∞} e^{-u^2} du = \int_{-∞}^{∞} e^{-u^2} du = \sqrt {\pi}$

Generalisations To Arbitrary Dimensions

One interesting thing about the Gaussian integral is it easily generalises to arbitrary dimensions because:

$e^{-k^2} = e^{-k_1^2} e^{-k_2^2}……..e^{-k_d^2}$

so:

$J = \int_{-∞}^{∞} d^dk e^{-k^2} = {\pi}^\frac{d}{2}$

Now lets see how this helps in a more general case where $F$ is any reasonable function of $k^2$:

$I = \int_{-∞}^{∞} d^dk F({k^2})$

We could go and look it up in some tome, or feed it into a math package, but I think most into math dread the – and it can be shown – and would like to know why.  With that in mind lets see how far we can get with just the above. As you probably have guessed there is a trick involved.

To do the integral we, similar to polar coordinates, go over to generalised spherical coordinates in arbitrary dimensions. First have a look on Wikipedia at what generalised spherical coordinates look like:

N Sphere

The associated volume element is:

$d^dV = k^{d-1} sin^{d-2}(Θ_1) sin^{d-3}(Θ_2) …… sin (Θ_{d-2}) dk dΘ_1 dΘ_2 ….. dΘ_{d-1}$

Plugging this into the the integral, $I$, we have:

$I = C(d) \int_{0}^{∞} dk k^{d-1} F({k^2})$

where $C(d)$ is some constant depending only on the dimension.

You probably have guessed the trick – since we know $J$, we can determine $C(d)$:

$J = {\pi}^\frac{d}{2} = C(d) \int_{0}^{∞} dk k^{d-1} e^{-k^2} = \frac{C(d)}{2} \int_{0}^{∞} dt t^{\frac{d}{2}-1} e^{-t} = \frac{C(d)}{2} \Gamma (\frac{d}{2})$

A substiution $t = k^2$ was used.  Hence:

$C(d) = \frac {2 \Pi ^{ \frac{d}{2}}}{\Gamma (\frac{d}{2})}$

And finally we have:

$I = \frac {2 \Pi ^{ \frac{d}{2}}}{\Gamma (\frac{d}{2})} \int_{0}^{∞} dk k^{d-1} F({k^2})$

The above formula is useful for what’s called dimensional regularisation, but most of the time we are interested in the the case when $d=4$:

$I = 2 \Pi ^2 \int_{0}^{∞} dk k^3 F({k^2}) = \Pi ^2 \int_{0}^{∞} dk^2 k^2 F({k^2})$    $(1)$

Wick Rotation

The above is interesting, but unfortunately as it stands isn’t much use for calculating integrals that occur in Quantum Field Theory. That’s because for those integrals, the $k^2$ in $F({k^2})$ is not the Euclidean norm:

$k^2 = k_1^2 + k_2^2 + k_3^2 + k_4^2$

It’s the relativistic one:

$k^2 = k_0^2 – k_1^2 – k_2^2 – k_3^2$

This is where we need another trick.  Suppose we have $F(k^2) = F( k_0^2 – k_1^2 – k_2^2 – k_2^2)$.  Then:

$\int_{-∞}^{∞} d^4k F({k^2}) = \int_{-∞}^{∞} d^3k \int_{-∞}^{∞} dk_0 F( k_0^2)$

The integral we are interested in is $\int_{-∞}^{∞} dk_0 F( k_0^2)$.

Many integrals are easier when done in the complex plane, so first we consider it a complex function and integrate along the real axis.  We suppose that at infinity $F$ is zero and any poles occur in the 2nd and 4th quadrant.   This is usual for the functions we work with.

First consider the following contour integral in the first quadrant, from zero along the real axis to a large positive value, then a circular arc to the positive imaginary axis, then, from a large positive value on the imaginary axis, along the imaginary axis, back to zero. Since there are no poles the Residue Theorem says this is zero.  Along the arc $F$ is zero, so the integral from zero to infinity along the real axis is the same as the integral along the imaginary axis from zero to infinity.   Actually we would need to show as the radius of the arc goes to infinity the integral along the arc goes to zero, but I wont be that rigorous.  It’s an interesting exercise showing this in the examples later, and the interested reader may like to chug through it.  Doing a similar thing in the 3rd quadrant, going from zero to a large negative value along the real axis, along a circular arc to the negative imaginary axis, then from a large negative value on the imaginary axis to zero.  Again the integral along the arc is zero, so the integral from minus infinity along the real axis is the same as along the imaginary axis from minus infinity to zero.

$\int_{-∞}^{∞} dk_0 F( k_0^2) = \int_{-i∞}^{i∞} dk_0 F( k_0^2) = i \int_{-∞}^{∞} dk_4 F((i k_4)^2)$

Where the substitution $k_0 = i k_4$ has been made.  This has now put the integral in the form with the usual Euclidean metric because $( i k_4) ^2 = – k_4^2$ so:

$\int_{-∞}^{∞} d^3k \int_{-∞}^{∞} dk_0 F( k_0^2) = i \int_{-∞}^{∞} d^3k \int_{-∞}^{∞} dk_0 F( – k_4^2) = i \int_{-∞}^{∞} d^4k’ F(-k’^2) = i \Pi ^2 \int_{0}^{∞} dk’^2 k’^2 F(-{k’^2})$  $(2)$

Where, because $k’$ has the ordinary Euclidean metric $k’^2 = k_1^2 + k_2^2 + k_3^2 + k_4^2$, we have used equation $(1)$.

This trick of changing the integration variable of $k_0$ to $ik_4$ is called Wick Rotation.

Two Examples

An integral that pops up in the large momentum approximation of the meson $Φ^4$ theory is actually simple and illustrative.  It’s:

$\int_{-∞}^{∞} d^4k \frac{1}{k^4}$

First we need to show it obeys the assumptions in deriving formula $(2)$ ie $F (k_0^2)$ is zero at infinity and has no poles in the first or third quadrant.

$F (k_0^2) = \frac{1}{(k_0^2 – k_1^2 – k_2^2 – k_2^2)^2} = \frac{1}{(k_0^2 – k^2)^2}$

This goes to zero at infinity, but has poles on the real axis at $± k$.  To get around this we modify it:

$\frac{1}{(k_0^2 – k^2 + iε)^2}$

Here $ε$ is a positive infinitesimal.  It shifts the poles to the 2nd and 4th quadrants.  In equation $(2)$ we take $ε→0$.

$\int_{-∞}^{∞} d^4k \frac{1}{k^4} = i \Pi ^2 \int_{0}^{∞} dk’^2 \frac{1}{k’^2} =$ $Λ^2 → ∞$ $i \Pi ^2 \ln (Λ^2)$

This is the origin of the equation requiring renormalisation in the first paper.

The previous integral was an approximation.  Removing that requires some tricky math, which will be the subject of another paper where Pauli–Villars regularization will be used, and the following integral will be found useful:

$\int_{-∞}^{∞} d^4k \frac{1}{(k^2 – c^2)^3}$

Checking if it obeys our assumptions:

$F (k_0^2) = \frac{1}{(k_0^2 – k^2 – c^2)^3}$

It goes to zero at infinity, but again it has poles on the real line.  We use the same trick to shift them to the 2nd and 4th quadrant, then take $ε→0$.

$\int_{-∞}^{∞} d^4k \frac{1}{(k^2 – c^2)^3} = -i \Pi ^2 \int_{0}^{∞} dk’^2 \frac{k’^2}{(k’^2 + c^2)}$

$\int_{}^{} dx \frac{x}{(x^2 + c^2)} = – \frac{c^2 + 2x}{2 (c^2 + x^2)}$

$-i \Pi ^2 \int_{0}^{∞} dk’^2 \frac{k’^2}{(k’^2 + c^2)} = i \Pi^2 \frac{c^2 + 2k’^2}{2 (c^2 + k’^4)^2} \Big|_0^∞ = \frac{ -i \Pi^2 }{2 c^2}$

Conclusion

I wanted this to be about some of tricks used in evaluating the integrals in Quantum Field Theory and renormalisation, and I think I have done that.  But while doing it I noticed something and maybe you did to.  We had all these things appearing you wouldn’t think had anything to do with it.

The gernralisation of the factorial function, the gamma Fumction.

π appeared all over the place

An apparently real integral turns out secretly to involve i, simply by using a slightly different metric.

What going on, its all very strange.  Well Wigner wrote an essay about it:

The Unreasonable Effectiveness of Mathematics in the Natural Sciences

Thought provoking.

My favourite interest is exactly how can we view the world so what science tells us is intuitive.

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