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Renormalised Mass and the Higgs Boson 
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#1
Dec2612, 12:05 PM

P: 209

Hey,
I have the following equation for the renormalised mass of some scalar particle (i.e. Higgs boson) [tex]m_{r}^{2}=m_{0}^{2}+\frac{\lambda}{32\pi^{2}}( \Lambda ^{2}m_{0}^{2}ln(1+\frac{\Lambda^{2}}{\mu^{2}}))[/tex] Where I have the first order correction to the mass of the loop in a two point function as [tex]\delta m^{2}=\frac{\lambda}{32\pi^{2}}(\Lambda^{2}m^{2}ln(1+\frac{\Lambda^{2}}{m^{2}}))[/tex] Now I'm a bit confused with all these different terms and wanted to know what μ means in the top equation as I'd of thought it was just our mass squared in the second equation? I suppose, more importantly, what I want to ask is the 'Renormalised Mass' the mass we measure (so for the Higgs Boson 125GeV) and the capital Lambda is our UVlimit (on planck scale 10^(19)GeV). If this is so then our bare mass is negative?? Also is it safe to assume the selfcoupling constant of a Higgs boson is 1/8? I'm sure we measure the mass to be about 125GeV however quantum corrections due to the 'divergences' (from virtual particles) want to push this mass up by about an order of 10^(17) which is not observed... so there 'must' be something else going on... Is this correct? Thanks, SK 


#2
Dec2612, 04:08 PM

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P: 11,589

Supersymmetric models replace the linear term by a logarithm, which requires less finetuning. 


#3
Dec2612, 05:07 PM

P: 209

Well I think I have understood this correctly to minute detail! I want to know how I show with my given equation the order of the 'quantum corrections'.... I'll keep looking at it but I'm getting the m_0 as negative  which is wrong!
Thanks for getting back to me, SK 


#4
Dec2612, 06:07 PM

Mentor
P: 11,589

Renormalised Mass and the Higgs Boson
Maybe a sign error for the quantum correction?
I think I saw a negative m_0^2 somewhere. 


#5
Dec2712, 07:46 AM

P: 209

Would I be correct in saying that the renormalised mass m_(r) is the 125GeV and the capital λ is the 10^(19)GeV?



#6
Dec2812, 09:19 AM

P: 209

It seems to make sense that we say
[tex]m_r^{2}=m_{0}^{2}+\delta m^{2}[/tex] Where I have [tex]\delta m^{2}=\frac{\lambda}{32\pi^{2}}(\Lambda^{2}m^{2}ln(1+\frac{\Lambda^{2}}{m^{2}}))[/tex] I believe I use values for: [tex]m_r=125GeV\: ,\: \Lambda=10^{19}GeV[/tex] However I'm not sure how I can use these to show what value the bare mass is because I keep getting negative masses for m_0. Though saying this, I'm not sure what mass we use in the correction δm^2 ; I assumed it was just the bare mass  but I'm not sure. 


#7
Dec2812, 11:41 PM

P: 1,020

bare mass can not be measured because for that you will have to take the self interaction away which is not the case.In some case an infinite negative bare mass is used to cancel other positive infinity to get sensible terms.This is how it works. the statement here is due to freeman dyson.



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