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[itex]K^{0}\bar{K}^{0}[/itex] mixing SQCD 
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#1
Jun114, 01:25 PM

P: 739

I am trying to understand what the author means by the attachment (the underlined phrase).
In my understanding if the masses were equal I should have: [itex] \frac{1}{(p^{2}m^{2}+i \epsilon)^{2}} \sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js}[/itex] why is the unitary supposed to make it vanish? (the attachment is from page 202 of Theory and Phenomenology of Sparticles (2004) M.Drees, R.Godbole,P.Roy) 


#2
Jun114, 02:59 PM

Sci Advisor
Thanks
P: 4,160

I think because then the numerators become UU^{† }= I, and I is diagonal and would not couple d to s, which is offdiagonal.



#3
Jun214, 02:33 AM

P: 739

Thanks, I just realized that [itex]d,s[/itex] where not free indices, but they denoted the corresponding quarks coming in and out...



#4
Jun214, 03:27 AM

P: 739

[itex]K^{0}\bar{K}^{0}[/itex] mixing SQCD
Also the last equation there is in the attachment should be some kind of Taylor expansion of the above factor, around [itex] m_{i}^{2} = m_{d}^{2} + Δm_{i}^{2}[/itex]
I am also having 1 question... [itex] \sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js}\frac{1}{(p^{2}m_{d}^{2}Δm_{i}^{2}+i \epsilon)(p^{2}m_{d}^{2}Δm_{j}^{2}+i \epsilon)}[/itex] [itex]\frac{1}{p^{2}m_{d}^{2}Δm_{i}^{2}+i \epsilon}= \frac{1}{p^{2}m_{d}^{2}+i \epsilon} + \frac{Δm_{i}^{2}}{(p^{2}m_{d}^{2}+i \epsilon)^{2}}+O(\frac{1}{(p^{2}m_{d}^{2})^{3}})[/itex] So for j... [itex]\frac{1}{p^{2}m_{d}^{2}Δm_{j}^{2}+i \epsilon}= \frac{1}{p^{2}m_{d}^{2}+i \epsilon} + \frac{Δm_{j}^{2}}{(p^{2}m_{d}^{2}+i \epsilon)^{2}}+O(\frac{1}{(p^{2}m_{d}^{2})^{3}})[/itex] And these I multiply... the 1st terms will give zero because of the unitarity of [itex]U[/itex]'s as before... the 12 and 21 terms will also give zero because of the unitarity of one of the [itex]U[/itex] each time ... So the only remaining term is the 22: [itex] \frac{1}{(p^{2}m_{d}^{2}+i \epsilon)^{4}}\sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js} Δm_{i}^{2}Δm_{j}^{2}+O(\frac{1}{(p^{2}m_{d}^{2})^{5}})[/itex] Which I think is equivalent to the last expression... However isn't it also zero? because "i" can't be "s" and "d" at the same time... Shouldn't [itex]Δm_{i}[/itex] be between the [itex]U[/itex]'s? if it could be regarded as a matrix to save the day... 


#5
Jun214, 03:42 AM

P: 739

Oh I'm stupid... I just saw what's going on....
[itex]Δm_{i}[/itex] is going to change the summation way, so it won't give the δ_{ds} anymore... 


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