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Calculation of leptonic decay widths

  1. Dec 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Halzen & Martin, problem 2.25


    2. Relevant equations

    The ρ and ω wavefunctions are u[itex]\overline{u}[/itex]-d[itex]\overline{d}[/itex] and u[itex]\overline{u}[/itex]+d[itex]\overline{d}[/itex] except for a normalization factor.

    3. The attempt at a solution

    In this problem one has to evaluate the expectation value of the charge operator for each of the mesons listed using their quarks wavefunctions and then square them but I get the same value of that expectation value for both ρ and ω so their squares will never be in the ratio 9:1.

    Any ideas of what I'm doing wrong?

    Thanks!
     
  2. jcsd
  3. Dec 18, 2012 #2

    mfb

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    You have to add the individual contributions and square the sum, not the other way round. This should give a ratio of 3:1 for the sum (as 2/3+1/3 != 2/3-1/3) and 9:1 for its square.
     
  4. Dec 18, 2012 #3
    Thank you for your reply, mfb.

    I appreciate the point that one has to square afterwards.

    I think that my problem lies in the evaluation of the expectation value of the charge operator for each meson. What I'm doing for the ρ meson, for instance, is

    [itex] \langle u \overline{u}-d \overline{d}|e_1+e_2|u \overline{u}-d \overline{d} \rangle=\frac{2}{3}-\frac{1}{3}-\frac{2}{3}+\frac{1}{3}=0[/itex],

    where [itex]e_1[/itex] and [itex]e_2[/itex] represent the charge operators for each of the quark/antiquarks.

    Obviously conceptually there is something wrong but I'm not sure what.

    Thank you for your attention.
     
  5. Dec 19, 2012 #4

    mfb

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    u can go to d + positive lepton only, u-bar can go to d-bar + negative lepton only - they are two different processes and do not add.
     
  6. Dec 19, 2012 #5
    I do not quite understand your reply.

    So, what I'm doing for calculating the expectation value of [itex]e_1[/itex], for the sake of the argument, is

    [itex]\langle u\overline{u}-d\overline{d} | e_1|u\overline{u}-d\overline{d}\rangle=\langle u\overline{u}|e_1|u\overline{u}\rangle + \langle d\overline{d}|e_1|d\overline{d}\rangle = \frac{2}{3}-\frac{1}{3}=\frac{1}{3}[/itex]

    and analogously for [itex] e_2[/itex].
     
  7. Dec 19, 2012 #6

    mfb

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    Calculate the decay widths for positive and negative leptons separately - they are two different processes, the amplitudes do not add.
     
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