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A Electric charge as ##Q = T_{3} + Y##

  1. Feb 4, 2017 #1
    My question is about the formula ##Q = T_{3} + Y##.

    Let us say that there is some complex scalar field that transforms as a triplet of ##SU_{L}(2)##; i.e.

    ##\psi = \begin{pmatrix} \psi_{1}\\ \psi_{2} \\ \psi_{3} \end{pmatrix}##

    and

    ##\delta_{2}\psi = i\omega_{2}^{a}t_{a}\psi##

    with

    ##t_{1} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \qquad
    t_{2} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix}, \qquad
    t_{3} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}##

    Let us suppose also that the hypercharge, ##Y##, of the field ##\psi## is zero.

    How do we now obtain the electric charges of the component fields ##\psi_{1}##, ##\psi_{2}##, and ##\psi_{3}##?

    Is it ##1/\sqrt{2}##, ##0## and ##-1/\sqrt{2}##, because these are the eigenvalues of the eigenstates ##\psi_{1}##, ##\psi_{2}## and ##\psi_{3}## of ##\psi##?
     
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  3. Feb 4, 2017 #2

    Orodruin

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    You have gotten the normalisation of ##t_3## wrong. The eigenvalues should be -1, 0, and +1.
     
  4. Feb 4, 2017 #3
    Ah! I see!

    So, now if you wanted to transform to unitary gauge, how would you do so?

    I have difficulty understanding what the unitary gauge transformation explicitly looks like.
     
  5. Feb 12, 2017 #4
    Yes! (EDIT: sorry, it seems according to Orodruin's reply to this post that I haven't made my yes clear enough: let's suppose you had the normalisation right, it would indeed be the way you would calculate the charges: the diagonal values give you the ##T_3## charge of each component, to which you just have to add the hypercharge)

    Unitary gauge is usually defined when you have Goldstone bosons and want to get rid of them. For instance, for the Higgs field, which you can write ##h=e^{(-i\eta_at^a)}(0, h+v)##, then the change to unitary gauge would be one of parameters the ##\eta##'s to make them disappear. But it does not change anything to the generators of your group.
     
    Last edited: Feb 12, 2017
  6. Feb 12, 2017 #5

    Orodruin

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    No. Why would you bother to give a wrong answer to a question that has already been answered? The normalisation of ##t_3## should not involve the ##1/\sqrt{2}## and the third isospin component of the same multiplet differ by one.
     
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