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Coupling the gauge bosons to the Higgs scalars

  1. Jun 14, 2008 #1
    At page 52 of 4th chapter in "An Introduction to gauge theories & modern particle physics" by Leader & Predazzi one can find such statement:

    "We must therefore rearrange (4.2.4) so that we can identify the field that multiplies[tex]\frac{1}{2}\left(1+\tau_{3}\right)[/tex] as gauge boson that remains massless i.e. as photon"

    But I still don't understand why such field is needed and what it mean that "field multiplies (transformation mentioned above) as gauge massles boson". For me this statement sounds "slangish".

    Next question: unbroken generator - what's that? And for what purpose one impose condition(4.2.8)

    I will appreciate any kind of advice.
    QuantumDevil.
     
  2. jcsd
  3. Jun 19, 2008 #2
    I'm not familiar with this text, but perhaps the following will help:

    It's not really about 'why such a field is needed'; the gauge fields follow from assuming some gauge symmetry in the Lagrangian, and in this case we're considering the Higgs scalar (2 complex components), which naturally admits SU(2)xU(1) symmetry. [tex]\frac{1}{2}\left(1+\tau_{3}\right)[/tex] is a 2-by-2 Hermitian matrix with complex entries, and if you exponentiate it you get a gauge transformation from SU(2)xU(1).... so it's called a generator. It's an unbroken generator, because the gauge transformations it generates do not change the Higgs' vacuum state. In more detail, we can write the Higgs vacuum state as
    [tex]
    v \begin{bmatrix}
    0 \\ 1
    \end{bmatrix}
    [/tex]
    with v a real number (the vacuum expectation value). Note that
    [tex]\frac{1}{2}\left(1+\tau_{3}\right) = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}[/tex]
    and hopefully it's clear that why the Higgs' vacuum state is unaffected under these gauge transformations. However, the other generators of SU(2)xU(1) do change the Higgs vacuum state.... so these are called broken generators. The broken generators correspond to massive gauge bosons, while the unbroken generators are massless - this comes from the covariant derivative in the Higgs part of the Lagrangian, where only broken generators will couple to the Higgs' nonzero vacuum state.

    Best wishes

    Dave
     
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