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

  1. ChrisVer

    ChrisVer 2,297
    Gold Member

    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)
     

    Attached Files:

  2. jcsd
  3. Bill_K

    Bill_K 4,159
    Science Advisor

    I think because then the numerators become UU= I, and I is diagonal and would not couple d to s, which is off-diagonal.
     
    1 person likes this.
  4. ChrisVer

    ChrisVer 2,297
    Gold Member

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

    ChrisVer 2,297
    Gold Member

    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...
     
    Last edited: Jun 2, 2014
  6. ChrisVer

    ChrisVer 2,297
    Gold Member

    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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