• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

156 submissions , 130 unreviewed
4,065 questions , 1,471 unanswered
4,951 answers , 21,107 comments
1,470 users with positive rep
554 active unimported users
More ...

  One Loop Higgs Mass Correction

+ 5 like - 0 dislike

I am attempting to compute the one loop correction to the Higgs mass, which requires the evaluation of a scattering amplitude, namely

$$\require{cancel} \mathcal{M} = (-)N_f \int \frac{\mathrm{d}^4 k}{(2\pi)^4} \mathrm{Tr} \, \left[ \left( \frac{i\lambda_f}{\sqrt{2}}\right) \frac{i}{\cancel{k}-m_f} \left( \frac{i\lambda_f}{\sqrt{2}} \right) \frac{i}{\cancel{k} + \cancel{p}-m_f}\right]$$

which corresponds to the Feynman diagram:

enter image description here

After combining constants, and rationalizing the denominators, I obtain,

$$-\frac{N_f \lambda_f^2}{2} \int \frac{\mathrm{d}^4 k}{(2\pi)^4} \frac{\mathrm{Tr}\left[ \cancel{k}\cancel{k} + \cancel{k}\cancel{p} +2m_f \cancel{k} + m_f \cancel{p} + m_f^2\right]}{\left(k^2-m_f^2\right)\left((k+p)^2 -m_f^2 \right)}$$

Computing traces, via the relation $\mathrm{Tr}[\cancel{a}\cancel{b}] = 4(a\cdot b)$ yields,

$$-2N_f \lambda_f^2 \int \frac{\mathrm{d}^4 k}{(2\pi)^4} \frac{k^2 +k\cdot p + m_f^2}{\left(k^2-m_f^2\right)\left((k+p)^2 -m_f^2 \right)}$$

At this point, I employed dimensional regularization, followed by Feynman reparametrization to combine the denominators, and then completed the square, yielding

$$-\frac{2^{2-d}\pi^{-d/2}}{\Gamma (d/2)}N_f \lambda_f^2 \int_{0}^1 \mathrm{d}x \int_0^\infty \mathrm{d}k \frac{k^{d-1}(k^2 +kp + m_f^2)}{\left[ \left(k-(x-1)p\right)^2 +p^2(x-x^2 -1)\right]^2}$$

Additional Calculations (Edit)

I attempted to further simplify the integrand using a substitution in only the first integral, namely $\ell = k-(1-x)p$ which implies $\mathrm{d}\ell = \mathrm{d}k$, yielding (after several manipulations),

$$-\frac{2^{2-d}\pi^{-d/2}}{\Gamma(d/2)}N_f \lambda_f^2 \int_0^1 \mathrm{d}x \, \int_{(x-1)p}^{\infty} \mathrm{d}\ell \frac{(\ell + (1-x)p)^{d-1}[(\ell + \frac{1}{2}p(3-2x))^2 - \frac{1}{4}p^2 + m_f^2]}{[\ell^2 + p^2(x-x^2-1)]^2}$$

N.B. Mathematica evaluated the original integral over $k$, and outputted a combination of the first Appell hypergeometric series, which possess the integral representation,

$$F_1(a,b_1,b_2,c;x,y) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)} \int_0^1 \mathrm{d}t \, t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b_1}(1-yt)^{-b_2}$$

with $\Re c >\Re a >0$, which has a structure similar to the beta function. If I can express the loop integral in a similar form, I may be able to express it in terms of these functions. At the end of the calculation, I will take $d \to 4-\epsilon$ to obtain poles in $\epsilon$, using the usual expansion

$$\Gamma(x) = \frac{1}{x} -\gamma + \mathcal{O}(x)$$

and a similar expansion should the final answer indeed contain the Appell hypergeometric series.

Passarino-Veltmann Reduction (Edit):

Based on my understanding of Veltmann-Passarino reduction, it is not applicable as the numerator contains an arbitrary power of loop momentum. I could plug in $d=4$, and impose a high momentum cut off, but this has already been done in many texts. As aforementioned, I would like a dimensionally regularized amplitude.

I am stuck at this point, can anyone give some details as to how to proceed? In addition, I have a query regarding the hierarchy problem. If using a simple cut-off regularization, the one loop correction can be shown to be quadratically divergent. But why is this an issue that needs to be remedied, by for example, the minimally supersymmetric standard model? Can't the divergence be eliminated by a regular renormalization procedure?

This post imported from StackExchange Physics at 2014-05-04 11:44 (UCT), posted by SE-user JamalS
asked Mar 30, 2014 in Theoretical Physics by JamalS (885 points) [ no revision ]

2 Answers

+ 4 like - 0 dislike

I go through the calculation below. However, I won't calculate the integral myself since its very impractical and not what you want to do in practice. You need a quick formula to simplify your integrals. Thanksfully, such a formula is provided in any standard textbook in QFT. You should derive this formula once and then move on. I will do the calculation using this formula and if you would like to see the derivation its done in Peskin and Schroeder, when they introduce dim-reg.

I dropped the $N_f$ factor because its not quite right due to the sum over the masses of flavor states. As you mentioned the diagram is given by (I kept your other conventions for the couplings, I presume they are correct) \begin{equation} {\cal M} = - \int \frac{ d ^4 k }{ (2\pi)^4 } \left( \frac{ i \lambda _f }{ \sqrt{ 2}} \right) ^2 ( i ) ^2 \mbox{Tr} \left[ \frac{ \cancel{k} + m _f }{ k ^2 - m ^2 _f } \frac{ \cancel{k} +\cancel{p} + m _f }{ (k+p) ^2 - m ^2 _f } \right] \end{equation} You can combine the denomenators using Feynman parameters (this is the first of two formulas you may want to write down and refer to in the future, but I'll do it explicitly here): \begin{align} \frac{1}{ D} & = \frac{1}{ ( k ^2 - m ^2 ) \left( ( k + p ) ^2 - m ^2 \right) } \\ & = \int d x \frac{1}{ \left[ x ( ( k + p ) ^2 - m ^2 ) + ( 1 - x ) ( k ^2 - m ^2 ) \right] ^2 } \\ & = \int d x \frac{1}{ \left[ k ^2 + 2 k p x + p ^2 x ^2 - p ^2 x ^2 + p ^2 x - m ^2 x - m ^2 + x m ^2 \right] ^2 } \\ & = \int d x \frac{1}{ \left[ ( k + p x ) ^2 - ( p ^2 x ^2 - p ^2 x + m ^2 ) \right] ^2 } \\ & = \int d x \frac{1}{ \left[ ( k + p x ) ^2 - \Delta \right] ^2 } \end{align} where $ \Delta \equiv p ^2 x ^2 - p ^2 x + m ^2 $.

To get rid of the $ k + p x $ factor we shift $ k: k \rightarrow k - p x $. Then the denomenator is even in $k$. The trace is given by: \begin{align} \mbox{Tr} \left[ ... \right] & \rightarrow \mbox{Tr} \left[ ( \cancel{k}-\cancel{p}x + m _f ) ( \cancel{k} + \cancel{p} ( 1-x ) + m _f ) \right] \\ & = 4 \left[ ( k - p x ) ( k + p ( 1-x ) ) + m ^2 _f \right] \end{align} All linear terms are zero since the denominator is even. Thus the trace becomes: \begin{equation} \mbox{Tr} \left[ ... \right] \rightarrow 4 \left[ k ^2 - p ^2 x ( 1 - x ) + m ^2 _f \right] \end{equation}

The amplitude now takes the form, \begin{equation} - \left( 2\lambda _f ^2 \right) \mu ^\epsilon \int \,dx \frac{ \,d^dk }{ (2\pi)^4 }\frac{ k ^2 - p ^2 x ( 1 - x ) + m _f ^2 }{\left[ k ^2 - \Delta \right] ^2 } \end{equation} where I moved to $ d $ dimensions and introduce a renormalization scale, $ \mu $, to keep the coupling dimensionless.

I now use two formula out of Peskin and Schroeder, Eq A.44 and A.46, and simplify the final result, \begin{align} & \int \frac{ \,d^4k }{ (2\pi)^4 } \frac{ k ^2 }{ ( k ^2 - \Delta ) ^2 } = \frac{ i \Delta }{ 16 \pi ^2 } \left( \frac{ 2 }{ \epsilon } + \log \frac{ \mu ^2 }{ \Delta } + \log 4\pi + 2 \gamma + 1 \right) \\ & \int \frac{ \,d^4k }{ (2\pi)^4 } \frac{ 1 }{ ( k ^2 - \Delta ) } = \frac{ i }{ 16 \pi ^2 } \left( \frac{ 2 }{ \epsilon } + \log \frac{ \mu ^2 }{ \Delta } + \log 4\pi - \gamma \right) \end{align}
where I used $ d = 4 - \epsilon $.

For simplicity lets only focus on the most divergent part (of course to calculate the physical cross-sections you'll need the full amplitude). Its easy, but more cumbersome, to include all the finite pieces. In that case we have, \begin{align} {\cal M} &= - \frac{ 2 i \lambda _f ^2 }{ 16 \pi ^2 \epsilon } \int d x \left[ \Delta - p ^2 x ( 1 - x ) + m ^2 _f \right] \\ & = - \frac{ 2 i \lambda _f ^2 }{ 16 \pi ^2 \epsilon } \left[ -\frac{ p ^2}{3} + 2m ^2 _f \right] \end{align}

Now with regards to your question about the hierarchy problem. Yes, the divergence can and is cancelled by a counterterm. But, the modern view of QFT says that renormalization is not an artificial procedure, but instead a physical consequence of quantum corrections. That being said, if the Higgs mass is at the TeV scale but the amplitude is at the Planck scale, the counterterms must be huge. This means that while the physical mass is still at the TeV scale very precise cancellation need to occur for this to happen which is very unnatural. Such cancellation don't happen anywhere else in Nature!

This post imported from StackExchange Physics at 2014-05-04 11:44 (UCT), posted by SE-user JeffDror
answered Apr 4, 2014 by JeffDror (650 points) [ no revision ]
+ 2 like - 0 dislike

These type of integrals are very common in 1-loop (and higher order) calculations, so they were categorized according to the number of legs attached to the loop. The convention is to label them starting with $A$, so one-point function is $A(m)$, two point function is $B(p,m_1,m_2)$, etc. Depending on the numerator of the integrand they might also have tensor components. I believe the first complete treatment of these were given by Passarino and Veltman in their "One loop corrections for $e^+ e^- \rightarrow \mu^+ \mu^-$" paper. You can check appendix D. Both $A$ and $B$ are divergent, and closed form expressions are given in $n=4-\epsilon$ dimensions, so you can directly see the $1/\epsilon$ term.

This post imported from StackExchange Physics at 2014-05-04 11:44 (UCT), posted by SE-user henry mcfly
answered Apr 2, 2014 by henry mcfly (35 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights