# One Loop Higgs Mass Correction

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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:

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

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

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

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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) $${\cal M} = - \int \frac{ d ^4 k }{ (2\pi)^4 } \left( \frac{ i \lambda _f }{ \sqrt{ 2}} \right) ^2 ( i ) ^2 \mbox{Tr} \left[ \frac{ \cancel{k} + m _f }{ k ^2 - m ^2 _f } \frac{ \cancel{k} +\cancel{p} + m _f }{ (k+p) ^2 - m ^2 _f } \right]$$ 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: $$\mbox{Tr} \left[ ... \right] \rightarrow 4 \left[ k ^2 - p ^2 x ( 1 - x ) + m ^2 _f \right]$$

The amplitude now takes the form, $$- \left( 2\lambda _f ^2 \right) \mu ^\epsilon \int \,dx \frac{ \,d^dk }{ (2\pi)^4 }\frac{ k ^2 - p ^2 x ( 1 - x ) + m _f ^2 }{\left[ k ^2 - \Delta \right] ^2 }$$ 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 (650 points)
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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 (35 points)

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