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W-Z boson difference mass

  1. Nov 18, 2013 #1

    ChrisVer

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    I would like to ask you 2 completely separate questions... Since both concern the W,Zs I don't want to create more than one post (that could be spamming)...

    1. what are the consequences (theoretically and experimentally) ,if there are, of the fact that the masses of W and Z bosons don't match?
    For example experimentally the only thing I can think about, is that the Z and Ws couple differently because of the propagator's dependence on mass (~/M)
    Theoretically there is the Weinberg's angle θW that is given by the ratio of that mass... Where does that angle appear in the theoretical scheme?

    2. When we create the SU(2)LxU(1)Y theory we have 4 massless boson fields (W+, W-,W0 and B0) which after the higgs mechanism mass-giving we end up with the massive W,Z and the photon. My question is about Z, why do we write it as a combination of W0 and B0 alone? Couldn't it be a combination of W- and W+ as well ? (since the total charge will remain 0). Except for that it'd give an instinct polarization to the Z0 boson, is there any more serious problem (that I'm missing) in that approach?

    by the way, hello PF!
     
    Last edited: Nov 18, 2013
  2. jcsd
  3. Nov 19, 2013 #2
    Question 2) Mixing the W+ and W- into the Z would lead to electric charge conservation violation.
     
  4. Nov 19, 2013 #3

    ChrisVer

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    Hey, thanks for the answer. But I don't really get it. A W+ and W- would give charge 0 to the boson...
    But I think I have to relook in the procedure of Higgs SSB for SU2 U1, since it comes naturally in it, that Z is a mixture of uncharged fields
     
  5. Nov 19, 2013 #4

    Bill_K

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    You're misinterpreting the word "combination". Sounds like you're thinking of the Z as a composite particle containing both a W+ and W- side by side, so their charges add to zero. What they really mean is that the Z state is a mixture, a superposition or linear combination,

    |Z> = cos θW |W0> - sin θW |B>

    You can only superpose states that both have the same charge, as the W0 and B do.
     
  6. Nov 19, 2013 #5
    You might want to read about superselection rules which explain why such mixtures are forbidden.
     
  7. Nov 19, 2013 #6
    In any case, even if you could superpose a state with charge +1 and a state with charge -1, you wouldn't end up with a state with 0 charge. You'd end up with a state of indefinite charge. It's the same as if you superposed two plane waves, each with definite momentum:

    ##e^{ip_1 x / \hbar} + e^{i p_2 x /\hbar}##

    The resulting superposition does not have momentum ##(p_1 + p_2)/2##, but rather is a state of indefinite momentum.
     
    Last edited: Nov 19, 2013
  8. Nov 19, 2013 #7

    fzero

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    The consequences of the mass difference between the W and Z is less important than the fact that the W is electrically charged and corresponds to the off-diagonal components of the SU(2) gauge boson, while the Z is neutral and corresponds to the diagonal component of the SU(2) gauge boson. This means that the structure of the interaction of these particles with fermions is very different. In particular, the W boson couples up-type fermions to down-type fermions (whose charges differ by ##\pm 1##), while the Z boson interaction couples the chiral components of the same fermion to one another. In terms of Dirac spinors, the Z-vertex has the form

    $$ Z_\mu \bar{f} \gamma^\mu (1-\gamma^5)f. $$

    There are processes like ##e^+ e^- \rightarrow \nu_e \bar{\nu}_e ## that can proceed by either W or Z exchange at tree-level. The corresponding amplitudes will in fact differ due to the W-Z mass difference, however they will also have a different dependence on the momenta because of the structure of the vertices. The Z process can be said to proceed via the s-channel, while the W process is via the t-channel.

    The Weinberg angle is introduced as a mixing angle in defining the linear combinations of ##SU(2)\times U(1)## bosons that become the photon and Z after electroweak symmetry breaking, as Bill_K illustrated. The additional parameters involved are the couplings ##g,g'## for the different factors in the gauge interaction and the Higgs vacuum expectation value ##v##. Once the charges of all of the particles in the electroweak theory are assigned, formulas for the masses and Weinberg angle can be derived in terms of ##g,g',v##. The ratio of the W mass to the Z mass turns out to be the same combination of coupling constants that appears in the formula for ##\cos\theta_W##, hence the formula you refer to. The details are worked out in this review.
     
  9. Nov 19, 2013 #8

    Bill_K

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    I believe the Z vertex is not pure V - A. (See the additional sin2θW term at the end of Eq (49) in the paper you cited.)
     
  10. Nov 19, 2013 #9

    fzero

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    Yes, you're correct. The term I left out follows from the ##B## component of the ##Z##, which has the photon-like pure vector couplings. Thanks for catching that.
     
  11. Nov 19, 2013 #10

    Vanadium 50

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    You are getting confused because before - or rather, without - the symmetry breaking there is no such thing as a W+ or W-. There are w1, w2, w3 and B. What we call "electric charge" is one of an infinitely many conserved quantities. It's only after the symmetry breaking where a particular combination of isospin and hypercharge combines to produce electric charge does it make sense to talk about W+.
     
  12. Nov 20, 2013 #11

    ChrisVer

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    I think you are wrong here. W + W- appear as combinations of W1 W2 and represent creation and annihilations of e and neutrinos.
     
  13. Nov 20, 2013 #12

    Vanadium 50

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    Yes, post-SSB the W+ is a combination of w1 and w2. But until you break the symmetry, you can redefine the w1 and w2 any way you want.
     
  14. Nov 20, 2013 #13

    ChrisVer

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    One more question then about that,
    what are the goldstone bosons we get after the SSB?
    For what I know is that when you sent at first the Higgs field under an SU2 transformation to the form Φ=[0; v+n] you had an exponential in front with a field ξ on it (i.e. exp[i ξ^α τ^α/2] )... That field couplings don't appear though in the final SM lagrangian...
     
  15. Nov 20, 2013 #14

    Vanadium 50

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    The goldstones become the longitudinal degrees of freedom of the W's, Z's.
     
  16. Nov 20, 2013 #15
    You only get Goldstone bosons if the symmetry that is being broken is a global symmetry. If the symmetry being broken is a local symmetry, the Goldstone boson doesn't appear and its degree of freedom becomes the longitudinal component of the (now massive) gauge boson. That's sometimes jokingly described as "The gauge boson eats the Goldstone boson and gets fat"
     
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