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

  1. Jun 1, 2014 #1

    ChrisVer

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    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. Jun 1, 2014 #2

    Bill_K

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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.
     
  4. Jun 2, 2014 #3

    ChrisVer

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    Thanks, I just realized that [itex]d,s[/itex] where not free indices, but they denoted the corresponding quarks coming in and out...
     
  5. Jun 2, 2014 #4

    ChrisVer

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    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. Jun 2, 2014 #5

    ChrisVer

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    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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