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Selection Rule: delta_I = 1/2 in the strangeness-changing weak currents

  1. Oct 1, 2012 #1
    I read in a paper http://nngroup.physics.sunysb.edu/~nngroup/misc/Documents/NeutrinoReactionsAtAcceleratorEnergies.pdf that delta_I = 1/2 rule of the strangeness-changing weak current implies the ratio of cross section of proton->Ʃ0 process to neutron->Ʃ- process to be 1/2. The delta_I = 1/2 rule is mentioned just before equation (3.40) on page 317.

    I do not understand what is this selection rule, and how does it effect the cross section ratio of the 2 processes? Is delta_I = 1/2 rule is that the change in the isospin of initial and final particle which has to be 1/2 for these processes?

    These are the cabbibo suppressed 'charged current processes', where when a neutrino interacts with a nucleon, converts to a lepton producing a W-, which converts a u-quark in the nucleon to a s-quark. So this is a strangness changing process (delta_S = 1) where a 'proton converts to a Ʃ0' or a 'neutron converts to a Ʃ-', both processes are written in equation (3.39) and their cross section ratio is written to be 1/2 accodording to delta_I = 1/2 rule in equation (3.40) which I want to understand.

    Thanks much for the help!
  2. jcsd
  3. Oct 1, 2012 #2


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    Ok, what he is calling a ΔI = 1/2 rule is not really a "rule", but an empirical observation. You're comparing the cross-sections for two semi-leptonic interactions. Rewrite them as

    1) μ+ν → ∑0p
    2) μ+ν → ∑+n

    The right-hand sides couple an isospin 1 particle (∑) with an isospin 1/2 particle (p or n). In general these can form an I = 1/2 state or an I = 3/2 state. Copying the Clebsch-Gordan coefficients from Wikipedia,

    1) (for ∑0p) |1, 0> ⊗ |1/2, 1/2> = (√2/3) |3/2, 1/2> ⊕ (-√1/3)|1/2, 1/2>
    2) (for ∑+n) |1, 1> ⊗ |1/2, -1/2> = (√1/3) |3/2, 1/2> ⊕ (√2/3)|1/2, 1/2>

    The ΔI = 1/2 rule says the reaction produces only I = 1/2. Comparing the coefficients, (-√1/3) for ∑0p and (√2/3) for ∑+n, we find the amplitude for the second reaction is greater by a factor of √2. Therefore the cross-section will be greater by the square of this, namely 2.
  4. Oct 1, 2012 #3
    Great! I understand now the equations you wrote and such a neat proof of cross section ratio to be 2. However, why did u rewrite the reactions as

    1) μ+ν → ∑0p
    2) μ+ν → ∑+n

    What is the logic behind rewriting the processes as above instead of

    1') vp → μ+0
    2') vn → μ+-

    This actually changes the Clebsch-Gordan coefficients, as then for (2') I will have

    (for ∑-n) |1, -1> ⊗ |1/2, -1/2> = (1) |3/2, -3/2>
    and all other terms will be zero obeying m1+m2 = m rule.

    I am sorry I might be missing some very basic and abvious point here. Thanks for the help again.
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