Renormalisation Question

  • #1
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I'm trying question 1 in this past paper:
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2009/Paper48.pdf [Broken]

I'm on the bit "Show that the divergent part of this diagram can be written in the form [itex]Ap^2+B[/itex] where [itex]p=p_1=-p_2[/itex] and A and B are momentum independent.

Now, as far as I can tell, this calculation is going to be exactly the same as the one on p48 of these notes:
http://www.damtp.cam.ac.uk/user/ho/Notes.pdf [Broken]
(the final result being given at the top of p49) with the only difference being that when we use dimensional regularisation to analytically continue to [itex]d \in \mathbb{C}[/itex], we should set [itex]d=6-\epsilon[/itex] instead of [itex]d=4-\epsilon[/itex]. However, this will change the powers of some factors in the final answer, it won't change the pole structure, will it? I certainly don't see how we're going to need an [itex]Ap^2[/itex] counter term as there is no divergent piece multiplying a [itex]p^2[/itex] that I can see...

Can anyone offer some help please...
 
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  • #2
fzero
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I'm trying question 1 in this past paper:
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2009/Paper48.pdf [Broken]

I'm on the bit "Show that the divergent part of this diagram can be written in the form [itex]Ap^2+B[/itex] where [itex]p=p_1=-p_2[/itex] and A and B are momentum independent.

Now, as far as I can tell, this calculation is going to be exactly the same as the one on p48 of these notes:
http://www.damtp.cam.ac.uk/user/ho/Notes.pdf [Broken]
(the final result being given at the top of p49) with the only difference being that when we use dimensional regularisation to analytically continue to [itex]d \in \mathbb{C}[/itex], we should set [itex]d=6-\epsilon[/itex] instead of [itex]d=4-\epsilon[/itex]. However, this will change the powers of some factors in the final answer, it won't change the pole structure, will it? I certainly don't see how we're going to need an [itex]Ap^2[/itex] counter term as there is no divergent piece multiplying a [itex]p^2[/itex] that I can see...

Can anyone offer some help please...
For [tex]d=4-\epsilon[/tex] the divergence is in [tex]\Gamma(0)[/tex]. For [tex]d=6-\epsilon[/tex] it's in [tex]\Gamma(-1)[/tex].
 
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  • #3
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For [tex]d=4-\epsilon[/tex] the divergence is in [tex]\Gamma(0)[/tex]. For [tex]d=6-\epsilon[/tex] it's in [tex]\Gamma(-1)[/tex].
So I find that [itex]\hat{\tau}_2^{(1)}(p,-p)=\frac{1}{2}g^2 \frac{1}{(4 \pi)^3} (4 \pi)^\frac{\epsilon}{2} \Gamma(-1+\frac{\epsilon}{2}) \int_0^1 d\alpha AA^{-\frac{\epsilon}{2}}[/itex]
[itex]=\frac{1}{2}g^2 \frac{1}{(4 \pi)^3} (4 \pi)^\frac{\epsilon}{2} \Gamma(\frac{\epsilon}{2})\frac{1}{\frac{\epsilon}{2}-1} \int_0^1 d\alpha Ae^{-\frac{\epsilon}{2} \log{A}}[/itex]
[itex]=\frac{1}{2}g^2 \frac{1}{(4 \pi)^3} (4 \pi)^\frac{\epsilon}{2} \Gamma(\frac{\epsilon}{2}+1)\frac{1}{(\frac{\epsilon}{2}-1)\frac{\epsilon}{2}} \int_0^1 d\alpha Ae^{-\frac{\epsilon}{2} \log{A}}[/itex]
[itex]=\frac{1}{2}g^2 \frac{1}{(4 \pi)^3} (4 \pi)^\frac{\epsilon}{2} e^{-\gamma_E \frac{\epsilon}{2}} \frac{1}{(\frac{\epsilon}{2}-1)\frac{\epsilon}{2}} \int_0^1 d\alpha Ae^{-\frac{\epsilon}{2} \log{A}}[/itex]
[itex]=\frac{1}{2}g^2 \frac{1}{(4 \pi)^3} ((4 \pi)e^{-\gamma_E})^\frac{\epsilon}{2} (-\frac{2}{\epsilon} ( 1 + \frac{\epsilon}{2} + \mathcal{O}(\epsilon^2)) \int_0^1 d\alpha A (1-\frac{\epsilon}{2} \log{A})[/itex]

So this I think is my divergent piece. The p^2 term is contained in [itex]A=\alpha(1-\alpha)p^2+m^2[/itex].
What would you recommend now? Should I go through and find all pieces that are of [itex]\mathcal{O}(\frac{1}{\epsilon})[/itex] and then I should be able to read of A and B in theory, right?
 
  • #4
fzero
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Should I go through and find all pieces that are of [itex]\mathcal{O}(\frac{1}{\epsilon})[/itex] and then I should be able to read of A and B in theory, right?
Yes.
 
  • #5
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Yes.
There is a bit of a complication though because, when I multiply this out I get

[itex]=-\frac{g^2}{2} \frac{1}{(4 \pi)^3} ( 4 \pi e^{-\gamma_E )^{\epsilon}{2} \int_0^1 A ( \frac{2}{\epsilon}-1+1-\frac{\epsilon}{2}) \log{A} d \alpha[/itex]

Now this means I get a piece that goes as [itex]\frac{2}{\epsilon}[/itex] and a piece that goes as [itex]{\epsilon}{2}[/itex] (which I'm assuming we can neglect as its not divergent in the regulator limit)

So the divergent bit is [itex]=-\frac{g^2}{2} \frac{1}{(4 \pi)^3} ( 4 \pi e^{-\gamma_E )^{\epsilon}{2} \int_0^1 A \frac{2}{\epsilon} \log{A} d \alpha[/itex]

But [itex]A=\alpha(1-\alpha)p^2_m^2[/itex] so I don't see how I can isolate a coefficient A of p^2 as there is a p^2 in the log as well, isn't there?
 
  • #6
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It's clear from the last line in post #3 that the [tex]\epsilon[/tex] in front of the log term means that it doesn't contribute to the divergent part of the expression, so check your algebra.
 
  • #7
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It's clear from the last line in post #3 that the [tex]\epsilon[/tex] in front of the log term means that it doesn't contribute to the divergent part of the expression, so check your algebra.
Yeah. I made a mistake half way through and didn't correct it properly throughout.
So anyway I find [itex]A=-\frac{g^2}{2} \frac{1}{(4 \pi)^3} \frac{1}{\epsilon}[/itex] and [itex]B=-g^2 \frac{m^2}{(4 \pi)^3} \frac{1}{\epsilon}[/itex]

Now for the final part he asks me to calculate B using a UV cut-off. I don't understand how he wants me to do that. There is no integral in B and we need an integral to be present to use UV cut-off regularisation, don't we?
 
  • #8
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Yeah. I made a mistake half way through and didn't correct it properly throughout.
So anyway I find [itex]A=-\frac{g^2}{2} \frac{1}{(4 \pi)^3} \frac{1}{\epsilon}[/itex] and [itex]B=-g^2 \frac{m^2}{(4 \pi)^3} \frac{1}{\epsilon}[/itex]

Now for the final part he asks me to calculate B using a UV cut-off. I don't understand how he wants me to do that. There is no integral in B and we need an integral to be present to use UV cut-off regularisation, don't we?
B is (part of) the divergent part of an integral. Use that integral.
 
  • #9
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B is (part of) the divergent part of an integral. Use that integral.
Well certainly B came from an integral over the loop momenta i.e. we had [itex]\int \frac{d^dk}{(2 \pi)^d} \dots[/itex]

However, I used that integral when I introduced a gamma function during the dimensional regularisation procedure - there don't appear to be any integral signs left over i.e. the whole integral has been evaluated and has produced [itex]Ap^2+B[/itex] as an answer...
 
  • #10
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Well certainly B came from an integral over the loop momenta i.e. we had [itex]\int \frac{d^dk}{(2 \pi)^d} \dots[/itex]

However, I used that integral when I introduced a gamma function during the dimensional regularisation procedure - there don't appear to be any integral signs left over i.e. the whole integral has been evaluated and has produced [itex]Ap^2+B[/itex] as an answer...
The problem is asking you to evaluate B using primitive cutoff regularization, not dimensional regularization.
 
  • #11
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The problem is asking you to evaluate B using primitive cutoff regularization, not dimensional regularization.
Yes but B is the result of the dimensional regularisation is it not? Perhaps if it asked me to evaluate the original integral using cutoff regularisation i could try that? But B as it stands doesn't have an integral in it so it cant be evaluated any further - by any means.

So I still don't really get what I'm supposed to do because I don't know what the integral im meant to be evaluating is?
 
  • #12
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Yes but B is the result of the dimensional regularisation is it not? Perhaps if it asked me to evaluate the original integral using cutoff regularisation i could try that? But B as it stands doesn't have an integral in it so it cant be evaluated any further - by any means.

So I still don't really get what I'm supposed to do because I don't know what the integral im meant to be evaluating is?
The problem didn't specify how you were supposed to regulate the integral. Dimensional regularization was convenient for deriving the form of the divergence, but wasn't necessarily the only way to do it. Go back to the original integral and see if there's a limit in which you can extract B using a cutoff.
 
  • #13
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The problem didn't specify how you were supposed to regulate the integral. Dimensional regularization was convenient for deriving the form of the divergence, but wasn't necessarily the only way to do it. Go back to the original integral and see if there's a limit in which you can extract B using a cutoff.
[itex]\hat{F}_2(p_1,p_2)= \frac{g^2}{2} \int_{\mathbb{R}^6} \frac{d^6k}{(2 \pi)^6} \frac{1}{(k^2+m^2)((p-k)^2+m^2)}[/itex]
[itex]= \frac{g^2}{2} \frac{1}{(2 \pi)^6} \int_{S^5} d \Omega_5 \int \frac{k^5 dk}{(k^2+m^2)((p-k)^2+m^2)}[/itex]
[itex]= \frac{g^2}{2^7 \pi^3} \int \frac{k^5 dk}{(k^2+m^2)((p-k)^2+m^2)}[/itex] as [itex]\text{Vol}(S^5)=\pi^3[/itex]
[itex]\rightarrow \frac{g^2}{2^7 \pi^3} \int_{k \leq \Lambda} \frac{k^5 dk}{(k^2+m^2)((p-k)^2+m^2)}[/itex] by imposing a UV cutoff after wick rotation

Now I am wondering if there is some trick to evaluating this. the only way i can think of is introducing integrals over [itex]\alpha[/itex] so that we can right [itex]\frac{1}{k^2+m^2} = \int d \alpha e^{-\alpha (k^2+m^2)}[/itex] but this was from dimensional regularisation so not sure if it will be any use here!

What would you advise for evaluating this?
 
  • #14
fzero
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You don't need to compute the integral, you only need to obtain the divergent part. So you can consider the range where that comes from. Furthermore you don't have to compute the whole divergent part, only the part that is independent of [tex]p[/tex].
 
  • #15
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You don't need to compute the integral, you only need to obtain the divergent part. So you can consider the range where that comes from. Furthermore you don't have to compute the whole divergent part, only the part that is independent of [tex]p[/tex].
I don't understand. I thought when we used primitive cutoff regularisation we explicitly did the integral but over a range [itex]0 \leq k \leq \Lambda[/itex] rather than [itex]0 \leq k \leq \infty[/itex]?

What should I do here then?
 
  • #16
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I don't understand. I thought when we used primitive cutoff regularisation we explicitly did the integral but over a range [itex]0 \leq k \leq \Lambda[/itex] rather than [itex]0 \leq k \leq \infty[/itex]?

What should I do here then?
Suppose that we have

[tex]I = \int \frac{d^4 k}{k^2 (k^2+m^2)}.[/tex]

The divergence is in the UV, so we can introduce an intermediate scale [tex]M\gg m[/tex] and write

[tex] I = \int_0^M \frac{dk k^3}{ k^2 (k^2+m^2)} + \int_M^\Lambda \frac{dk k^3}{ k^2 (k^2+m^2)} .[/tex]

The divergence is completely contained in the 2nd term, which we can write as

[tex] \int_M^\Lambda \frac{dk k^3}{ k^2 (k^2+m^2)} = \int_M^\Lambda dk \frac{dk}{ k} \left( 1 - \frac{m^2}{k^2} + \cdots \right).[/tex]

Now the divergence is completely contained in the leading term, since the higher order terms are finite.
 
  • #17
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Suppose that we have

[tex]I = \int \frac{d^4 k}{k^2 (k^2+m^2)}.[/tex]

The divergence is in the UV, so we can introduce an intermediate scale [tex]M\gg m[/tex] and write

[tex] I = \int_0^M \frac{dk k^3}{ k^2 (k^2+m^2)} + \int_M^\Lambda \frac{dk k^3}{ k^2 (k^2+m^2)} .[/tex]

The divergence is completely contained in the 2nd term, which we can write as

[tex] \int_M^\Lambda \frac{dk k^3}{ k^2 (k^2+m^2)} = \int_M^\Lambda dk \frac{dk}{ k} \left( 1 - \frac{m^2}{k^2} + \cdots \right).[/tex]

Now the divergence is completely contained in the leading term, since the higher order terms are finite.
Why did you need to split the integral at all?

I can't see why the higher order terms should be finite?

Lastly, should I use binomial on both the terms I have in the denominator? That [itex]((p-k)^2+m^2)[/itex] term always confuses me on how to deal with it

Thanks.
 
  • #18
fzero
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Why did you need to split the integral at all?
You're right that you don't if you use the steps above. I was originally going to say that in the 2nd term we could just set [tex]m=0[/tex], but it's clearer to expand the way I ended up doing.

I can't see why the higher order terms should be finite?
Try to do the integral of the subleading term.

Lastly, should I use binomial on both the terms I have in the denominator? That [itex]((p-k)^2+m^2)[/itex] term always confuses me on how to deal with it

Thanks.
You have to be a bit careful, since [tex](p-k)^2[/tex] depends on the angles in k space. If you had to compute the whole integral, you'd introduce Feynman parameters. However for your problem, you only have to compute the part of the divergence that is independent of [tex]p[/tex], so you can simplify things a bit before manipulating the integral.
 
  • #19
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You're right that you don't if you use the steps above. I was originally going to say that in the 2nd term we could just set [tex]m=0[/tex], but it's clearer to expand the way I ended up doing.




Try to do the integral of the subleading term.
Yeah because it only becomes divergent for powers -1 or more and subleading term is -3 therefore finite.

You have to be a bit careful, since [tex](p-k)^2[/tex] depends on the angles in k space. If you had to compute the whole integral, you'd introduce Feynman parameters. However for your problem, you only have to compute the part of the divergence that is independent of [tex]p[/tex], so you can simplify things a bit before manipulating the integral.

Well to isolate the p terms, i tried to complete the square on the denominator?

Is that right? It's not going very well:

[itex](k^2+m^2)(p^2-2pk+k^2+m^2) = p^2k^2-2pk^3+k^4+k^2m^2 +m^2p^2-2pkm^2+k^2m^2+m^4[/itex]
There's too many terms to complete the square here so I don't really know how to simplify it to get rid of the p terms?
 
  • #20
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Well to isolate the p terms, i tried to complete the square on the denominator?

Is that right? It's not going very well:

[itex](k^2+m^2)(p^2-2pk+k^2+m^2) = p^2k^2-2pk^3+k^4+k^2m^2 +m^2p^2-2pkm^2+k^2m^2+m^4[/itex]
There's too many terms to complete the square here so I don't really know how to simplify it to get rid of the p terms?
Independent of [tex]p[/tex] means that you can set [tex]p=0[/tex] to evaluate the part that you want.
 
  • #21
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Independent of [tex]p[/tex] means that you can set [tex]p=0[/tex] to evaluate the part that you want.
OK I think I have it. So we get

[itex]\int_0^\Lambda \frac{k^5 dk}{(k^2+m^2)^2} = \int_0^\Lambda \frac{k^5 dk}{k^4(1+\frac{m^2}{k^2})^2} = \int_0^\Lambda dk k ( 1 - \frac{2m^2}{k^2} + \dots )[/itex] where all other pieces are finite

So the divergent pieces are equal to [itex]\int_0^\Lambda dk k - \frac{2m^2}{k} = \lambda - 2m^2 ( \log{\Lambda} - \log{0})[/itex]

Now I'm assuming this is why you wanted me to split the integral earlier? To avoid the log(0)? Why is the log(0) appearing at all though? Isn't that a result of us having cut a corner somewhere?

In which case we get a quadratic and a logarithmic divergence of the form [itex]\frac{\Lambda^2}{2} - 2m^2 \log{\Lambda}[/itex]

This doesn't quite seem to match up with the B we found earlier in post 7 though?
 
  • #22
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OK I think I have it. So we get

[itex]\int_0^\Lambda \frac{k^5 dk}{(k^2+m^2)^2} = \int_0^\Lambda \frac{k^5 dk}{k^4(1+\frac{m^2}{k^2})^2} = \int_0^\Lambda dk k ( 1 - \frac{2m^2}{k^2} + \dots )[/itex] where all other pieces are finite

So the divergent pieces are equal to [itex]\int_0^\Lambda dk k - \frac{2m^2}{k} = \lambda - 2m^2 ( \log{\Lambda} - \log{0})[/itex]

Now I'm assuming this is why you wanted me to split the integral earlier? To avoid the log(0)? Why is the log(0) appearing at all though? Isn't that a result of us having cut a corner somewhere?
That divergence is fictitious. It arises from the [tex]k\rightarrow 0[/tex] part of the integral, but that's precisely where the expansion in [tex]m/k[/tex] is bad. For small k, we should expand in [tex]k/m[/tex] and we find no divergence at [tex]k=0[/tex].

In which case we get a quadratic and a logarithmic divergence of the form [itex]\frac{\Lambda^2}{2} - 2m^2 \log{\Lambda}[/itex]

This doesn't quite seem to match up with the B we found earlier in post 7 though?
That's because you're comparing results in two different regularization schemes. In general, physical quantities should agree after renormalization, but the divergent parts are regulator dependent and won't agree.
 
  • #23
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That divergence is fictitious. It arises from the [tex]k\rightarrow 0[/tex] part of the integral, but that's precisely where the expansion in [tex]m/k[/tex] is bad. For small k, we should expand in [tex]k/m[/tex] and we find no divergence at [tex]k=0[/tex].



That's because you're comparing results in two different regularization schemes. In general, physical quantities should agree after renormalization, but the divergent parts are regulator dependent and won't agree.
These are exactly the same poles we found in 4D i.e. we got a 1/epsilon (or a quadratic+logarithmic when we used UV cutoff). So what was the effect of higher dimensions? To change the A and B coefficients?
 
  • #24
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These are exactly the same poles we found in 4D i.e. we got a 1/epsilon (or a quadratic+logarithmic when we used UV cutoff). So what was the effect of higher dimensions? To change the A and B coefficients?
It's not the same pole, as I pointed out in post #2. In 4d there was only a log divergence, here in 6d we find a quadratic divergence.
 
  • #25
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It's not the same pole, as I pointed out in post #2. In 4d there was only a log divergence, here in 6d we find a quadratic divergence.
Is the way to see that 4D only gives logarithmic because we produced a [itex]\frac{1}{\epsilon}[/itex] pole by dim reg and since a [itex]\frac{1}{\epsilon}[/itex] in dim reg corresponds to a [itex]\log{\frac{\Lambda}{m}}[/itex] in UV cutoff we only have a log. Is that correct?
 

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