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I Electroweak symmetry and unbroken generators

  1. Mar 17, 2016 #1

    CAF123

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    I realise there has been quite a few threads on this topic lately but I wanted to ask some questions that I have here. The following are statements from my lecture notes and I have write my questions after each statement.

    In the gauge sector of the electroweak theory, we can write down ##\mathcal L_{\text{gauge}}## and for this to be invariant under local transformations,we introduce a covariant derivative $$D_{\mu} = \partial_{\mu} + ig \frac{1}{2} \mathbf{\sigma^j} \mathbf{W^j_{\mu}} + ig' Y B_{\mu}$$ Is this the analogous covariant derivative from what we define in, say, QCD with ##D_{\mu} = \partial_{\mu} + igA_{\mu}^a t^a## but this time ##j## is a flavour index?

    To each broken generator we can find, i.e ##T^a \langle \phi_{\text{VEV}}\rangle \neq 0,## where ##\langle \phi_{\text{VEV}}\rangle## is the VEV of the Higgs field, we have a massless Goldstone boson. Choosing ##\langle \phi_{\text{VEV}}\rangle \propto (0,v)## we find that ##\sigma_1, \sigma_2## and ##\sigma^3-Y## are broken generators and thus give rise to three massless Goldstone bosons by Goldstone's theorem. But can't I say the same about ##\sigma^3## alone and Y = 1/2 alone? Or is it because they are not linearly independent from ##\sigma^3-Y## so should not also be considered in addition?

    We find that ##\sigma^3/2+Y = Q## is a broken generator and thus this implies there exists a linear combination of ##W_{\mu}## and ##B_{\mu}## such that $$\exp \left( i \left( \frac{1}{2}\sigma^3 W^3_{\mu} + B_{\mu} \text{Id}\right)\right)\langle \phi_{\text{VEV}}\rangle = 0 $$ Why is this so? It looks like some group element of ##SU(2) \times U(1)## which upon acting on the VEV leaves it invariant. But then if the generic structure of a group element is ##e^{i \lambda_a t^a}## then this seems to imply that ##W_{\mu}## and ##B_{\mu}## are to be treated as group parameters rather than massless gauge fields.

    Thanks!
     
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  3. Mar 17, 2016 #2

    ChrisVer

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    The Covariant derivatives are the "same"... they all introduce a vector field... now what that field is and how many there are depends on your symmetry group under consideration.
    The one that you wrote down has the [itex]W^{1,2,3},B[/itex] vector fields/gauge bosons of the [itex]SU(2) \times U(1)[/itex]... The covariant derivative of the Standard Model would be:
    [itex] D^{SM}_{\mu} = \partial_\mu + ig_1 Y B_\mu + i g_2 \frac{\tau^a}{2} W^a_\mu + i g_3 \frac{\lambda^b}{2} G^b_\mu[/itex]
    with [itex]a=1,2,3 ~~,b=1,2,...,8[/itex], the G is the gluon field. An SU(5) would have a covariant derivative:
    [itex]D^{SU(5)}_\mu = \partial_\mu + ig_5 T^a X^a_\mu~~,a=1,2,...,24[/itex]
     
  4. Mar 18, 2016 #3

    CAF123

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    Hi Chrisver, I see, many thanks - in this case ##b## would correspond to a colour index and ##a## the weak isospin index I guess? Do you have any comments regarding the other questions I wrote down?

    Thanks!
     
  5. Mar 18, 2016 #4

    ChrisVer

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    "yes". I am not sure if "color" is the right word to use for that, since colors exist in the fundamental representation of the SU(3) , i.e. [itex]\textbf{3}, \bar{\textbf{3}}[/itex]. If I recall well we say R,G,B and they are transformed like a triplet? However b runs through the adjoint representation dimension which for the SU(n) groups is the [itex]n^2-1[/itex]... In QCD I would say it is the number of the Gellman matrices, or a combination of color-anticolor (?? some intro books like to say that gluons don't have a single color but a pair of them).
    Similarily for the isospin index.


    For the other questions, I didn't really understand the 2nd. I think that trying [itex]\sigma^3[/itex] or [itex]Y[/itex] alone wouldn't help much... at the end of the day you don't want to completely break the symmetry but reach [itex]SU(2) \times U(1) \rightarrow U(1)[/itex]. Of course neither the hypercharge ##U(1)## nor the ##SU(2)## symmetry remain (so neither ##sigma3## nor ##Y## should be a symmetry, or neither the ##W^3## nor the ##B## are the photon) however a mixture of the 2 happens to remain.... I know this might not be satisfactory to you because I cannot really "explain" in this context how this 'happens to do' apart from seeing it right away by doing the calculation [acting with ##Q##].
    One thing you can try is to try and break just an [itex]SU(2)[/itex], with the higgs transforming under the adjoint representation 3 ? That'd be the Weinberg Georgi model...and supposedely you have to obtain 2 massive and 1 massless vector bosons and one massive scalar.

    The 3rd question I think is somewhat connected to the 2nd??? if not, then I am not unerstanding it.

    IMO it's easier for someone to start from the general [itex]\Phi= \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix}[/itex] Higgs field at the point of the minimum energy and make an SU(2) transformation and reach the point where [itex] \Phi = \begin{pmatrix} 0 \\ v \end{pmatrix}[/itex].
    now if you expand around the minimum of the potential, you can do it along the flat direction (around the mexican hat) or try to rise higher in the potential.
    So at the excitation the Higgs field around the minimum is:
    [itex] \Phi (x) = \exp \Big( \frac{i}{u} T^a g^a(x) \Big) \begin{pmatrix} 0 \\ \frac{1}{\sqrt{2}} \big( u + h(x) \big) \end{pmatrix} [/itex]
    Again by making an SU(2) rotation again to go to the so called 'unitary gauge', can destroy those massless degrees of freedoms in the exponential (which has the 3 goldstone bosons [itex]g^a(x)[/itex]) ... and be left with the 1 physical higgs...
     
  6. Mar 24, 2016 #5
    Mostly to further my own understanding, I will explain my view of it and hope it helps.

    By construction, the scalar field ##\phi## transforms under the gauge symmetry group ##SU_c(3) \times SU_L(2) \times U_Y(1)## as (1,2,1/2). That is, ##\phi## is a color singlet, ##SU_L(2)## doublet and has a weak hypercharge eigenvalue of ##y = \frac{1}{2}##. Hence, under a gauge transformation, ##\phi## changes by,
    $$\delta\phi=(\frac{i}{2}\omega_1(x)+\frac{i}{2}\omega_2^a \tau_a)\phi ,$$
    where ##\omega_1,\omega_2## are the infinitesimal angles and ##\tau_a## are the usual Pauli-matrices (##\sigma## in your post). With this, it is straightforward to find how ##\phi= \begin{pmatrix} 0 \\ v \end{pmatrix}## transforms:
    $$\delta \begin{pmatrix} 0 \\ v \end{pmatrix}=(\frac{i}{2}\omega_1(x)+\frac{i}{2}\omega_2^a \tau_a)\begin{pmatrix} 0 \\ v \end{pmatrix}=\frac{i}{2}\begin{pmatrix} [\omega_2^1-i\omega_2^2]v \\ [\omega_1-\omega_2^3]v \end{pmatrix}$$
    And so, ##\omega_2^1=\omega_2^2 \ \text{and} \ \omega_2^3=\omega_1=\omega## leave the ground state unchanged. The latter condition then implies that ##Q \equiv T_3+Y## is the generator of the unbroken symmetry, which mixes ##W_1## and ##W_2##. Hopefully, that sheds light on why it's the combination of that is the unbroken symmetry. The ##SU_L(2) \times U_Y(1)## is broken to ##U_{em}(1)##.

    Now, I am not sure what you mean by "... this seems to imply that ... are to be treated as group parameters rather than massless gauge fields. " so hopefully, I'm not talking about something unrelated to your question.
     
  7. Apr 3, 2016 #6

    CAF123

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    Thanks ChrisVer and DelcrossA, apologies again for late reply
    I am not quite seeing why ##\omega_1 = \omega_2^3 ## implies that ##Q=T_3+Y## is a generator of unbroken symmetry. Could you explain this part?

    I wrote $$\exp \left(i\left((B_{\mu} + \frac{1}{2} W_{\mu}^3 \sigma^3\right)\right) \begin{pmatrix} 0 \\ v \end{pmatrix} = 0.$$ This is similar to what you wrote in your exponential on the lhs with now the ##\tau^a##'s being the ##\sigma## and the ##\omega##'s are the ##W_{\mu}^3 ## and ##B_{\mu}## no? That's why I thought the B and W field were interpreted as a parameter in this set up but didn't know where such an equation came from in the first place.

    Thanks!
     
  8. Apr 3, 2016 #7

    samalkhaiat

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  9. Apr 4, 2016 #8

    CAF123

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  10. Apr 4, 2016 #9

    samalkhaiat

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    1) Did you understand it?
    2) Can you tell me what is the meaning of exponentiating vector bosons, i.e., what is the meaning of writing [itex]e^{iB_{\mu} + ...}[/itex]?
    3) As for why [itex]Q = (1/2)(Y-T_{3})[/itex] is the generator of [itex]U_{em}(1)[/itex]: read the paragraph after eq(30) and eqs(31-33).
     
  11. Apr 4, 2016 #10

    CAF123

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    For the most part, yes.

    1) In using a particular choice of ##\theta##, the ##A_{\mu}## field (involved in the redefinition of ##W^3_{\mu}## and ##Z_{\mu}##) no longer appears in the covariant derivative. I was just wondering if this choice of theta is equivalent to the unitary gauge?

    2) When you mention the charged current interaction and neutral current interactions are these referred to ##W^+/W^-## couplings with quark lines for example in the first case and ##Z^0## couplings in the latter case?
    I am not sure - I didn't understand it so was trying to interpret such an exponentation as a group element of ##SU(2) \times U(1)## acting on the VEV. My notes say that ##T^3+Y## is an unbroken generator (ie. this acting on the VEV gives zero) and then says that this implies that there exists a linear combination of ##B## and ##W_3^{\mu}## which creates rotations $$\exp \left(i (T^3 W_3^{\mu} + B_{\mu})\right) \langle \phi_o \rangle = 0$$ But I couldn't understand what this meant and why it was true.

    Yup it's clear from what you write but I was interested in seeing how one obtains the same conclusion from what DelcrossA wrote.

    Thanks!
     
  12. Apr 5, 2016 #11

    samalkhaiat

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    No. Unitary gauge allows you to get rid of “the massless Goldstone bosons” but leaves residual [itex]U(1)[/itex] invariance. After redefining the fields, the choice [itex]g \sin \theta = \bar{g} \cos \theta = e[/itex] identifies that residual symmetry with electromagnetic gauge symmetry [itex]U_{em}(1)[/itex]. See below.

    Yes. For example the lowest order elastic [itex]\nu_{e}e[/itex] scattering consists of two diagrams: the charged-current contribution via [itex]W-[/itex]exchange and the neutral-current contribution via [itex]Z-[/itex]exchange.

    Mathematically, the argument of any function must be a scalar field. So, the expression [itex]e^{i B^{\mu} + …}[/itex] has no meaning. The equation you are after is the following one
    [tex]e^{i\theta Q}\langle \phi \rangle \equiv e^{i\theta (T_{3} + \frac{1}{2}Y I_{2})} \begin{pmatrix} 0 \\ v \end{pmatrix} = \begin{pmatrix} 0 \\ v \end{pmatrix} .[/tex]
    Okay, I will try to explain things with minimum use of group theory and other fancy mathematics. Our gauge group is [itex]SU(2) \times U_{y}(1)[/itex]. For general group element, we will use
    [tex]U(x) = \exp \Big \{ - i \left( \frac{g}{2} \vec{\alpha}(x) \cdot \vec{\tau} + \frac{\bar{g}}{2}y \theta (x) I_{2} \right) \Big \} . \ \ \ \ (1)[/tex]
    Our complex iso-doublet is
    [tex]\phi(x) = \begin{pmatrix} \phi^{+}(x) \\ \phi^{0}(x) \end{pmatrix} , \ \ \ y(\phi) = 1.[/tex]
    The assignment [itex]y(\phi) = 1[/itex] means that [itex]\phi^{+}[/itex] is positively charged, and [itex]\phi^{0}[/itex] is neutral. Under [itex]SU(2) \times U_{y}(1)[/itex], the doublet transforms as
    [tex]\phi (x) \to \phi^{'}(x) = U(x) \phi (x) , \ \ \ \ \ \ \ \ \ (2)[/tex]
    and, infinitesimally, the gauge fields transform according to
    [tex]W_{i}^{\mu} \to W_{i}^{\mu} + g \epsilon_{ijk} \alpha_{j} W_{k}^{\mu} + \partial^{\mu}\alpha_{i} , \ \ (3a)[/tex] [tex]B^{\mu} \to B^{\mu} + \partial^{\mu}\theta . \ \ \ \ \ \ \ \ \ \ \ (3b)[/tex]
    The minimum of the potential [tex]V(\phi) = \mu^{2}\phi^{\dagger}\phi + \lambda (\phi^{\dagger}\phi)^{2} ,[/tex] is attained where [tex]\phi^{\dagger}\phi = - \mu^{2}/ 2 \lambda \equiv v^{2}/2 > 0 . \ \ \ \ \ \ \ \ (4)[/tex] Since this manifold is gauge invariant, infinitely many values of [itex]\phi[/itex] will satisfy (4). Let us choose [tex]\langle \phi \rangle = \begin{pmatrix} 0 \\ \frac{v}{\sqrt{2}}\end{pmatrix}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)[/tex] as our vacuum. This choice spontaneously breaks the global symmetry, since it selects a particular direction in the internal charge space. Other choices of [itex]\langle \phi \rangle[/itex] can be obtained from (5) by a global phase transformation. Note that (5) leads to spontaneous symmetry breaking for the neutral component only. Thus, the vacuum is still invariant under the electromagnetic [itex]U_{em}(1)[/itex] gauge transformation of the scalar field, i.e., [itex]e^{i \beta Q}\langle \phi \rangle = \langle \phi \rangle[/itex].
    Let us now parameterize the scalar doublet in terms of 4 real fields, all with zero vacuum expectation values, such that
    [tex]\phi(x) = \frac{1}{\sqrt{2}} \begin{pmatrix} \rho_{2}(x) + i \rho_{1}(x) \\
    v + h(x) + i \rho_{3}(x) \end{pmatrix} \ \ \ \ \ \ \ (6)[/tex]
    Substituting this in the Lagrangian
    [tex]\mathcal{L}(\phi) = (D_{\mu}\phi)^{\dagger} D^{\mu}\phi - V(\phi) ,[/tex]
    we obtain 3 massless Goldstone bosons [itex](\rho_{1}, \rho_{2}, \rho_{3})[/itex], together with a massive scalar field [itex]h(x)[/itex], with mass [itex]m_{h} = v \sqrt{2\lambda}[/itex].
    Next, we gauge away the 3 unphysical Goldstone bosons, so that the resulting Lagrangian contains physical fields with correct number of degrees of freedom. This is done by finding a transformation [itex]U(x) \in SU(2) \times U_{y}(1)[/itex] such that
    [tex]\phi^{'}(x) = U(x) \phi(x) = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ v + h(x) \end{pmatrix} [/tex]
    It is not that hard to see that the following choices of the gauge functions do the job for us
    [tex]\alpha_{1}(x) = \frac{2}{gv}\rho_{1}(x) , \ \ \alpha_{2}(x) = \frac{2}{gv}\rho_{2}(x) , \ \ \alpha_{3}(x) = - \frac{1}{gv}\rho_{3}(x) , \ \ \theta(x) = \frac{1}{y\bar{g}v}\rho_{3}(x) .[/tex]
    That is
    [tex]U(x) = \exp \left[ -\frac{i}{v}\left( \rho_{1}\tau^{1} + \rho_{2}\tau^{2} + \frac{1}{2}( I_{2} - \tau^{3}) \rho_{3} \right) \right] . \ \ \ \ (7)[/tex]
    Indeed, retaining only first order terms in the fields, you can convince yourself that
    [tex]\sqrt{2} \phi^{'}(x) = U(x) \begin{pmatrix} \rho_{2}(x) + i \rho_{1}(x) \\
    v + h(x) + i \rho_{3}(x) \end{pmatrix} = \begin{pmatrix} 0 \\ v + h(x) \end{pmatrix} [/tex]
    This tells you that the (Higgs) field [itex]h(x)[/itex] is a quantum fluctuation on the physical vacuum.
    Now, if you look carefully at (7), you realize that we still have a residual [itex]U(1)[/itex] gauge invariance generated by [itex](I_{2} + \tau^{3})[/itex]. Indeed, gauge transformation with
    [tex]\alpha_{1}(x) = \alpha_{2}(x) = 0 , \ \ \alpha_{3}(x) = \frac{2}{gv}\beta(x) , \ \ \theta(x) = \frac{2}{y \bar{g} v} \beta(x) , [/tex]
    leaves [itex]\phi^{'}(x)[/itex] invariant:
    [tex]\sqrt{2}\phi^{'}(x) = \exp \left[- i \frac{\beta (x)}{v}(I_{2} + \tau^{3}) \right] \begin{pmatrix} 0 \\ v + h(x) \end{pmatrix} = \begin{pmatrix} 0 \\ v + h(x) \end{pmatrix}[/tex]
    From (3a) and (3b), the corresponding transformation laws for the neutral fields [itex]W_{3}^{\mu}[/itex] and [itex]B^{\mu}[/itex] are
    [tex]W_{3}^{\mu} \to W_{3}^{\mu} + \frac{2}{gv} \partial^{\mu}\beta (x) ,[/tex] [tex]B^{\mu} \to B^{\mu} + \frac{2}{\bar{g}v} \partial^{\mu} \beta (x) .[/tex]
    Now, if we redefine the neutral fields as
    [tex]Z^{\mu} = W_{3}^{\mu}\cos \theta_{w} - B^{\mu} \sin \theta_{w},[/tex] [tex]A^{\mu} = W_{3}^{\mu} \sin \theta_{w} + B^{\mu} \cos \theta_{w} ,[/tex]
    and choose [itex]\bar{g}\cos \theta_{w} = g \sin \theta_{w} = e[/itex], the above residual transformations become
    [tex]Z^{\mu} \to Z^{\mu} , \ \ \ A^{\mu} \to A^{\mu} + \frac{2}{v e} \partial^{\mu} \beta .[/tex]
    And these are nothing but the electromagnetic [itex]U_{em}(1)[/itex] gauge transformations.
    I hope this will leave you with no confusion.
     
    Last edited: Apr 5, 2016
  13. Apr 7, 2016 #12

    CAF123

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    I think I understand why it is just the neutral component that is spontaneously symmetry broken but why is the ground state necessarily invariant under electromagnetic U(1) and not some other ##U(1) \simeq SO(2)##?

    (from one of your other threads)

    Just to check: From what you write this means a complex vector field always corresponds to physically a charged field while a complex scalar field ##\phi^0 = v+ h(x) + i \rho_3 (x) \in \mathbb{C}## is not?

    So this choice of the ##\alpha_i## corresponds to a particular choice of the gauge parameters (the unitary gauge), thereby removing gauge symmetry with the resulting effect that we now have no Goldstone bosons? Why does it tell us its a quantum fluctuation over a classical fluctuation?

    Thanks!
     
  14. Apr 7, 2016 #13

    samalkhaiat

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    If you do not start with the hyper-charge assignment [itex]y(\phi) = 1[/itex], then the pattern of symmetry breaking is just [itex]SU(2) \times U_{1}(1) \to U_{2}(1)[/itex] (this is what I have done in the other thread). However, once we choose Weinberg’s angle to satisfy [itex]e = g \sin \theta_{w} = \bar{g} \cos \theta_{w}[/itex], we hit 2 birds by one stone: [itex]U_{1}(1) = U_{Y}(1)[/itex] and [itex]U_{2}(1) = U_{em}(1)[/itex].
    Are you confusing redefining-the-neutral-fields with gauge transformation? Weinberg’s angle is not a gauge parameter.
    I don’t think you understood what I said in the other thread or what I did in this thread.
    1) Any complex field (scalar, vector, spinor, …) can carry a [itex]U(1)[/itex] charge, because its Lagrangian is naturally invariant under global phase transformation. Single real field of any kind cannot have a [itex]U(1)[/itex] charge because its Lagrangian is not invariant under global phase transformation.
    2) The parameterization
    [tex]\sqrt{2} \phi (x) = \begin{pmatrix} \rho_{2}(x) + i\rho_{1}(x) \\ v + h(x) + i\rho_{3}(x) \end{pmatrix} ,[/tex]
    is the most general parameterization of a doublet consisting of two complex scalar fields. In making that parameterization, I did not assume the particular hyper-charge assignment [itex]y(\phi) = 1[/itex], i.e., I did not assume that the lower component of [itex]\phi[/itex] was neutral.
    3) Even with the [itex]y(\phi) = 1[/itex], the complex combination [itex]v + h(x) + i\rho_{3}[/itex] is neutral because [itex]h(x)[/itex] and [itex]\rho_{3}(x)[/itex] have different masses.
    When, the Lagrangian is gauge invariant, you are free to gauge-“rotate” all the fields in the Lagrangian. This does not change the fact that your Lagrangian is still gauge invariant. Gauge symmetry of the Lagrangian cannot be removed.
    What does “classical fluctuation” mean? The classical vacuum does not fluctuate.
     
    Last edited: Apr 8, 2016
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