Analytically continuing 2,3,4pt integrals

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In summary, the analytic continuation of a result computed in a kinematic region allows for the ##i\epsilon## to be dropped and then one can analytically continue results via the prescription $$p_i^2 \rightarrow p_i^2 + i\varepsilon, s_{ij} \rightarrow s_{ij} + i\varepsilon, m_i \rightarrow m_i - i\varepsilon.$$
  • #1
CAF123
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I was reading a paper that gives a nice collection of all scalar integrals that crop up in QCD loop calculations. Such integrals are computed in some kinematic region and then the authors say the results may be analytically continued if so desired. I just wonder how is this analytic continuation done in practice? It's a relatively short paper and I've pasted it here https://arxiv.org/pdf/0712.1851.pdf

Thanks!
 
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  • #3
I was wondering how it applies to the loop integrals with the feynman +iepsilon prescription, e.g

The authors state that a particular kinematic region allows for the ##i\epsilon## to be dropped and then one can analytically continue results via the prescription $$p_i^2 \rightarrow p_i^2 + i\varepsilon, s_{ij} \rightarrow s_{ij} + i\varepsilon, m_i \rightarrow m_i - i\varepsilon.$$ I just wonder why this is the case and if the sign choices here are significant?

As a simple example, the analytic continuation of the massive tadpole is given as $$I_1^D(m^2) = -\mu^{2\epsilon} \Gamma(\epsilon-1) (m^2-i\varepsilon)^{1-\epsilon}$$ but what should I do with this result as it contains an explicit ##\varepsilon## in ##m^2-i\varepsilon##?
 
  • #4
Since you are interested in the limit ##\epsilon \rightarrow 0## the ##\mathrm{i} \epsilon## is unproblematic.
 
  • #5
vanhees71 said:
Since you are interested in the limit ##\epsilon \rightarrow 0## the ##\mathrm{i} \epsilon## is unproblematic.
Could you elaborate? There is an ##\epsilon## (epsilon) from dim reg and there is a ##i\varepsilon## (varepsilon) from the feynman prescription.
 
  • #6
Argh, yes. I meant the ##\mathrm{i} \varepsilon## of course. Don't use ##\epsilon## for both quantities :-).
 
  • #7
Ok. I'm just wondering though if I take the ##\varepsilon \rightarrow 0## limit don't I just get the result before analytic continuation? E.g for the massive tadpole given earlier, I was presuming the result before analytic continuation is with ##m^2-i\varepsilon## replaced with ##m^2##?
 
  • #8
Let's write your expression as
$$I_1^D(m^2) = -\mu^{2\epsilon} \Gamma(\epsilon-1) (m^2-i\eta)^{1-\epsilon}.$$
What I meant is that you are interested in the limit ##\epsilon \rightarrow 0##. Then the ##\mathrm{i} \eta## is irrelevant for ##\eta \rightarrow 0^+##.

The Laurent expansion is
$$I_1^D=\frac{m^2 - \mathrm{i} \eta}{\epsilon} - (m^2 - \mathrm{i} \eta) \left [ -1 +\gamma_{\text{E}} + \ln \left (\frac{m^2-\mathrm{i} \eta}{\mu^2} \right) \right ] + \mathcal{O}(\epsilon).$$
As you see, you can just take ##\eta=0## without problems.
 
  • #9
I see thanks, a few follow up questions

1)I guess the idea of analytic cont is the process of taking a result computed in some kinematic region and looking for the result in some other kinematic region. So for the tadpole example, is the analytic cont a means of finding the result for ##m^2<0## if the original result was computed in ##m^2>0##?

2) What does the fact that one can take ##\eta \rightarrow 0## without problems mean? I suppose the original result before analytic cont. was computed in the region ##m^2>0## but if I take ##m^2<0## and send ##\eta \rightarrow 0## the log has a negative argument. Perhaps I'm thinking about this in the wrong way.

2) why is it necessary that the correct analytic cont for ##m^2## is ##m^2-i\eta## and not, say, ##m^2+i\eta##? I suppose this is why you made ##\eta \rightarrow 0^+##.

Thanks!
 
  • #10
The ##\mathrm{i} \eta## comes into the game, because in vacuum-QFT pertubation theory you need the time-ordered propagator, i.e., you have a pole structure (in Minkowski space with the (+---) signature)
$$G(p) \propto \frac{1}{p^2-m^2+\mathrm{i} \eta}.$$
Alternatively you can use this pole structure of the free propagators to Wick rotate the energy component ##p^0## to make it purely imaginary. Then you deal with Euclidean QFT since the Minkowski product becomes (up to a cumbersome sign) the positive definite Euclidean scalar product. The prize to pay is that you have to Wick rotate back after you've done your integrals, which is not as simple as it looks at the first glance, particularly if you have vertex functions with ##\geq 3## legs, which depend on more than one independent four-momentum.

In the case of the tadpole loop you get of course a function that does not depend on any external momentum, because it's effectively a one-point function and the vacuum is translation invariant. With ##m^2>0## the tadpole loop with a scalar propagator has superficial (and since there are no subdivergences also true) degree of UV divergence 2 and no infra-red/collinear singularities. The UV divergence has to be regularized, and in your case this has been done using dimensional regularization. As long as ##m^2>0## and because the degree of divergence is positive, there are no poles or other singularities, and thus the ##\mathrm{i} \eta## has no function anymore, and you can make ##\eta \rightarrow 0## without trouble.

Since the superficial degree of divergence is positive, there's also no problem in making ##m^2 \rightarrow 0^+##. This is because there are no IR problems with the diagram and the fact that dimensional regularization is a socalled mass-independent, when the (modified) minimal-subatraction scheme is applied, i.e., you don't subtract any log terms containing the mass. Thus both the regularized as well as the MS renormalized tadpole diagram vanishes for ##m^2 \rightarrow 0^+##.

For more on renormalization, see my QFT manuscript:

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
 
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  • #11
If I put ##m^2## to zero in the integral representation for the tadpole then there is now an IR singularity ( to be canceled with the UV divergence to render the result zero in dim reg because integral is scaleless). How does this agree with what you said about being able to take ##m^2 \rightarrow 0^+## ?

Also , regarding the analytical continuation , in that paper I referenced they mention one can correctly analytically continue expressions using some relations between logs and dilogs which crop up in massive triangles. Can I ask if you understand what is being said in this section of the paper (sect. 2.2) so I can ask a few things ?

Thanks!
 

Related to Analytically continuing 2,3,4pt integrals

1. What is the purpose of analytically continuing 2,3,4pt integrals?

Analytically continuing 2,3,4pt integrals is a technique used in mathematical analysis to extend the domain of a function beyond its original definition. This allows for the evaluation of integrals in complex domains that were previously not possible.

2. How does analytically continuing 2,3,4pt integrals work?

The process of analytically continuing 2,3,4pt integrals involves applying a series of mathematical operations to a given function in order to extend its domain. This typically involves using complex numbers and conformal mappings to transform the function into a new domain where the integral can be evaluated.

3. What are some real-world applications of analytically continuing 2,3,4pt integrals?

Analytically continuing 2,3,4pt integrals have many applications in physics, engineering, and finance. For example, they are used in quantum field theory to calculate scattering amplitudes, in signal processing to analyze complex signals, and in option pricing models in finance.

4. Are there any limitations to analytically continuing 2,3,4pt integrals?

While analytically continuing 2,3,4pt integrals can greatly expand the domain of a function, there are some limitations to this technique. It may not work for functions with essential singularities or for functions that do not have a well-defined analytic continuation.

5. Is there a specific method for analytically continuing 2,3,4pt integrals?

There are various methods for analytically continuing 2,3,4pt integrals, such as the Cauchy integral formula, the residue theorem, and the method of steepest descent. The choice of method depends on the specific function and the desired domain of evaluation.

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