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Electro weak vertices

  1. Mar 29, 2012 #1
    I have a question about Feynman Diagrams:

    let's say we have a process: up antidown -> W+ -> up antidown...

    the first vertex is like V_CKM G PL ( mixing, gamma, projector)
    the second is the same..only with the complex conjugate CKM matrix...
    but why?...

    If I compute the M* I have to bar the vertices..and there I got the same vertex..with the same flow..but there I would change PL to PR and interchange PL and Gamma..why is that the case?
     
  2. jcsd
  3. Mar 30, 2012 #2
    Hello, in my opinion the answer is the following: the Feynman diagram you are considering is composed of two vertices: in the first an up is destroyed, an antidown is destroyed and a W+ is created; in the second vertex an up is created, an antidown is created and a W+ is destroyed; so, roughly speaking, the first is associated with a term in the lagrangian like (u dbar W-), while the second with (ubar d W+), that is its hermitian conjugate (of course I have forgot all the contraction matrices...); this is the origin of the conjugation of the CKM matrix paramters (and, of course, one should be careful with the imaginary units!)
    Best,
    Francesco
     
    Last edited: Mar 30, 2012
  4. Mar 30, 2012 #3
    I think another way of looking at it is like this.

    That Feynman diagram also describes the processes

    u → W+ + d
    W+ + d → u​

    as all I have done here is replace the incoming anti-d with an outgoing d, and the outgoing anti-d with an incoming d.

    The CKM matrix, as defined, is the factor for 'converting' down-type quarks to up-type, eg

    |u> = Vud |d>​

    Provided only the three known generations of quarks exist, the CKM matrix must be unitary, and hence

    V-1 = V

    so

    |d> = V*ud |u>​
     
    Last edited: Mar 30, 2012
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