One loop integration evaluation

[Sekonda]

Hey guys,

I have a loop integration of the form below:

$$-i\lambda\int_{-\infty}^{\infty}\frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^{2}}$$

Where 'k' is the four vector:

$$k=(E,\mathbf{p})$$

And so this integration could also be represented by expanding out the four vector:

$$-i\lambda\int_{-\infty}^{\infty}dE\int_{-\infty}^{\infty}\frac{d^3p}{(2\pi)^4}\frac{i}{E^2-\mathbf{p}^2-m^2}$$

I have to evaluate this integration and was introduced to the Cauchy-Rienmann equations, Contour integrals, Calculus of residues, Green's functions, Regularization & Wick's Rotations. So can anyone give me a starting point on how I'd go about evaluating the above integration, I know there are poles where:

$$E^{2}-\mathbf{p}^2=m^2$$

and so I think the idea of evaluating the integration is to somehow exclude these poles from the integration, but I'm not sure where to start and if this is the correct idea.

Thanks,
SK

 

[strangerep]

Do the E integration first, using contour integration and calculus of residues. (If you don't have a thorough knowledge of these techniques, then you're going to need a complex variables textbook and some time to master those techniques.)

 

[andrien]

without giving definitive meaning to poles,it is not possible to do it(poles on real axis).the easy way is to give mass a negative small imaginary part(retarded) and then it will be possible to do it.But it seems there is an exponential factor missing,can you fix it.

 

[Sekonda]

So I can do the E integration but do I still find that when I come to the p-integration, despite manipulation by Wick's Rotation, the integral still blows up and goes to infinity - which is incorrect and so we introduce some 'cut-off' value for the momentum p?

 

[Sekonda]

Also I have seen the manipulation of the propagator term from:

$$\frac{i}{E^2+\mathbf{p}^2-m^2}$$

to adding this term in the denominator:

$$\frac{i}{E^2+\mathbf{p}^2-m^2+i\epsilon}$$

What is this term? Why can we add it without changing the integral and how do we realize we need this term?

 

[Mute]

Also I have seen the manipulation of the propagator term from:

$$\frac{i}{E^2+\mathbf{p}^2-m^2}$$

to adding this term in the denominator:

$$\frac{i}{E^2+\mathbf{p}^2-m^2+i\epsilon}$$

What is this term? Why can we add it without changing the integral and how do we realize we need this term?

When you perform the contour integration to do the E integral, the pole will lie along your contour. This has to be treated carefully. One way to treat it is to have your contour divert around the pole in a small semi-circle of radius $\epsilon$, which is taken to zero at the end of the calculation. The typical thing to do in field theory calculations is similar and acheives the same effect: move the pole off the contour by giving the pole a small imaginary component to move it off the contour and inside the contour. The sign of the shift determines whether you move the pole up or down.

 

[Sekonda]

Okay, so we aren't really moving the pole positions when we take the limit where ε goes to zero, though we do move them for the purposes of manipulating the contour integral into something solvable - though we still take the limit where ε goes to zero but still excluding the pole - I'm guessing we can do this as the pole is at some point and so we can go infinitesimally close to it.

We exclude the poles so we can get a finite integration result, but in terms of positive & negative energies - we exclude the positive pole when considering negative energies and include the negative pole, and vice versa when considering the negative pole. Though surely inclusion of a pole in the contour integral gives an infitity? Or is this not so because we are shifting it and then taking the limit?

Thanks Mute, Andrien & Strangerep.

 

[andrien]

there is still problem,I think.regardless of all efforts of shifting the poles,the integral is divergent in high energy limit.I mean that in high energy limit, one gets a quadratic divergence!

 

[Sekonda]

Indeed, is this where we require renormalization? Which may involve some cut-off energy approximated for the UV region? I'm not really quite sure if this would be the correct progression from my previous point.

 

[andrien]

I don't think that it will be right to introduce any cut-off for it.Integrals of these type do appear as feynman integrals (not path integrals),and can be solved without any cutoff.Since the integral is very badly divergent,no renormalization can put it in a proper way.

 

[Sekonda]

Ok, but I presume that for some feynman diagrams "vaguely" similar to my one aren't as 'divergent' and thus can be renormalized, when and why do we introduce to a cut-off in the UV region?

 

[andrien]

Ok, but I presume that for some feynman diagrams "vaguely" similar to my one aren't as 'divergent' and thus can be renormalized, when and why do we introduce to a cut-off in the UV region?

yes,there are diagrams which are not as divergent and are used.In self energy diagram of electron,the virtual photon can have a very high momenta because of the off-shell condition.one integrates over the four momenta of it as usual feynman rules,but it is divergent in high energy limit(upper limit) so one modifies the photon propagator by a factor C which is dependent on the cut-off energy used in high energy region.Integration is carried out with upper limit set to cut-off,integrand is found to vary logarithmically with cut-off which is really much better than the previous quadratic divergence.You can see the relevant calculation here,from page 327 to 329
http://books.google.co.in/books?id=lvmSZkzDFt0C&pg=PA327&dq=sakurai+feynman+integrals&hl=en#v=onepage&q=sakurai%20feynman%20integrals&f=false

 

[strangerep]

Sekonda,

The propagators for the free particle should not be regarded as ordinary functions, but rather as generalized functions (aka distributions).
http://en.wikipedia.org/wiki/Distribution_(mathematics)

This means they only make sense when integrated against a test function (for which it and all its derivatives fall off fast at infinity). All this is mathematically fine in the free case: even though that integral looks divergent, it's ok when the result is integrated against a test function, which is all that matters here.

In the interacting case, one encounters products of these distributions but such products are ill-defined in general. Distributions are not like ordinary functions on which you can perform pointwise multiplication with no problems.

Regularization + renormalization are (partly) about dealing with this problem. (Scharf has a whole textbook on finite quantum electrodynamics devoted to the details of how we deal with these inconvenient properties of distributions.)

Regularization (by some form of cutoff, or other means) just isolates the divergent bits into terms which can be absorbed by a redefinition of the physical constants of the theory. If such absorption is possible at all orders of perturbation theory with only a finite number of such constants, the theory is said to be "renormalizable".

I always recommend that QFT students (re-)study a much simpler version of this process, known as the Poincare-Lindstedt method in classical dynamics.
http://en.wikipedia.org/wiki/Poincaré–Lindstedt_method
In the textbook by Jose & Saletan, you'll find a worked example of how this is applied to the case of a quartic anharmonic oscillator. After one understands the process of perturbatively adjusting the so-called constants of a theory, renormalization in QFT becomes less puzzling, (or so I found).