# $K^{0}-\bar{K}^{0}$ mixing SQCD

by ChrisVer
Tags: k0bark0, mixing, sqcd
 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: $\frac{1}{(p^{2}-m^{2}+i \epsilon)^{2}} \sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js}$ 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) Attached Thumbnails
 P: 754 Thanks, I just realized that $d,s$ where not free indices, but they denoted the corresponding quarks coming in and out...
 P: 754 $K^{0}-\bar{K}^{0}$ mixing SQCD Also the last equation there is in the attachment should be some kind of Taylor expansion of the above factor, around $m_{i}^{2} = m_{d}^{2} + Δm_{i}^{2}$ I am also having 1 question... $\sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js}\frac{1}{(p^{2}-m_{d}^{2}-Δm_{i}^{2}+i \epsilon)(p^{2}-m_{d}^{2}-Δm_{j}^{2}+i \epsilon)}$ $\frac{1}{p^{2}-m_{d}^{2}-Δm_{i}^{2}+i \epsilon}= \frac{1}{p^{2}-m_{d}^{2}+i \epsilon} + \frac{Δm_{i}^{2}}{(p^{2}-m_{d}^{2}+i \epsilon)^{2}}+O(\frac{1}{(p^{2}-m_{d}^{2})^{3}})$ So for j... $\frac{1}{p^{2}-m_{d}^{2}-Δm_{j}^{2}+i \epsilon}= \frac{1}{p^{2}-m_{d}^{2}+i \epsilon} + \frac{Δm_{j}^{2}}{(p^{2}-m_{d}^{2}+i \epsilon)^{2}}+O(\frac{1}{(p^{2}-m_{d}^{2})^{3}})$ And these I multiply... the 1st terms will give zero because of the unitarity of $U$'s as before... the 1-2 and 2-1 terms will also give zero because of the unitarity of one of the $U$ each time ... So the only remaining term is the 2-2: $\frac{1}{(p^{2}-m_{d}^{2}+i \epsilon)^{4}}\sum_{ij} U^{d_{L}}_{di} U^{d_{L} \dagger}_{is}U^{d_{L}}_{dj} U^{d_{L} \dagger}_{js} Δm_{i}^{2}Δm_{j}^{2}+O(\frac{1}{(p^{2}-m_{d}^{2})^{5}})$ 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 $Δm_{i}$ be between the $U$'s? if it could be regarded as a matrix to save the day...
 P: 754 Oh I'm stupid... I just saw what's going on.... $Δm_{i}$ is going to change the summation way, so it won't give the δds anymore...