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[itex]K^{0}-\bar{K}^{0}[/itex] mixing SQCD

by ChrisVer
Tags: k0bark0, mixing, sqcd
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ChrisVer
#1
Jun1-14, 01:25 PM
P: 754
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)
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Bill_K
#2
Jun1-14, 02:59 PM
Sci Advisor
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I think because then the numerators become UU = I, and I is diagonal and would not couple d to s, which is off-diagonal.
ChrisVer
#3
Jun2-14, 02:33 AM
P: 754
Thanks, I just realized that [itex]d,s[/itex] where not free indices, but they denoted the corresponding quarks coming in and out...

ChrisVer
#4
Jun2-14, 03:27 AM
P: 754
[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 1-2 and 2-1 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 2-2:

[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...
ChrisVer
#5
Jun2-14, 03:42 AM
P: 754
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 δds anymore...


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