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Getting from electroweak W_1,2,3 and B to W+-, Z and photon?

  1. Feb 27, 2016 #1
    I'm trying to understand how the [itex]Z[/itex] and the [itex]\gamma[/itex] are interpreted as combinations of the and [itex]B[/itex] gauge fields. I'm comfortable with where the [itex]W_{1}[/itex], [itex]W_{2}[/itex], [itex]W_{3}[/itex] and [itex]B[/itex] come from, and I get that the [itex]W_{3}[/itex] and [itex]B[/itex] get related to the [itex]Z[/itex] and the [itex]\gamma[/itex] by the 2x2 matrix containing the weak mixing angle [itex]\theta_{W}[/itex], but I'm not sure how this was arrived at.
    From my googling I'm also aware that the explanation is to do with Goldstone bosons being 'eaten'. These things being 'eaten' is something I have read many times without fully understanding. Electroweak symmetry breaking starts with a Higgs doublet with each doublet component having a real and imaginary part, so that's four Higgs degrees of freedom. The textbooks start working in the 'unitary gauge', ie. choosing a vacuum where three of these four Higgs components are set to zero... This must be related to them being 'eaten'...? At this point I'm lost, any takers?

    Thanks.
     
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  3. Feb 27, 2016 #2

    Orodruin

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    When the Higgs takes a vev, the mass matrix of the gauge bosons is not diagonal in the basis with ##B## and ##W_{3}##. In order to find the physical states, ie, the stats with definite mass, you need to diagonalise it. This is what the rotation by the Weinberg angle does and the resulting physical states we call ##A## (photon) and ##Z##.
     
  4. Feb 28, 2016 #3
    OK, so, I found slides with the following:

    higgs_lagrangian.png

    So we have a non-diagonal 4x4 matrix in that last term. This must be what you mean. And when it's diagonalised, the corresponding eigenbasis [itex](W^{+}, W^{-}, Z^{0}, \gamma)[/itex] is related to this [itex](W^{\mu 1}, W^{\mu 2}, W^{\mu 3}, B^{\mu})[/itex] basis by

    [itex] Z^{0} = \frac{g}{\sqrt{g + g'}} W^{\mu 3} - \frac{g'}{\sqrt{g + g'}} B^{\mu} [/itex]
    [itex] \gamma = \frac{g}{\sqrt{g + g'}} W^{\mu 3} + \frac{g'}{\sqrt{g + g'}} B^{\mu} [/itex]

    That seems straightforward enough. But where do 'eaten Goldstone bosons' come in?
     
  5. Feb 28, 2016 #4

    Orodruin

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    If you look at the number of degrees of freedom before and after symmetry breaking, the massless gauge fields essentially have two degrees of freedom each (corresponding to the transversal polarisations). After symmetry breaking, you are left with massive gauge bosons which also have a longitudinal polarisation. Where did this extra degree of freedom come from? Depending on your choice of gauge, you will have to take the goldstone propagators into account in different ways. This is particularly evident in unitary gauge, where the goldstone degrees of freedom vanish and are completely absorbed into the massive gauge boson propagators.
     
  6. Feb 29, 2016 #5

    vanhees71

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    The only disadvantage of the unitary gauge is that manifest renormalizability is gone. That's why you usually take the 't Hooft ##R_{\xi}## gauge, which is a modified Lorenz gauge such that the mixed would-be-Goldstone-gauge-field terms conveniently vanish. Then you have gauge bosons, would-be-Goldstone bosons and Faddeev-Popov ghosts (+all matter fields you like, in QFD the quarks and leptons, but these are irrelevant for the argument). All these conspire such that you get only physical degrees of freedom interacting, i.e., you cancel unphysical degrees of freedom in the gauge fields with the would-be-Goldstones and FP ghosts. It's really beautiful.

    Another question to the experts: Is there somewhere a source, where the background field gauge is applied to a Higgsed gauge theory? In QCD it's a very elegant and convenient gauge choice, having "naive" Ward-Takahashi identities instead of the more complicated Slavnov-Taylor identities from BRST symmetry in the conventional gauges. It's also somewhat more convenient in Abelian gauge theories like QED (but not much) compared to the conventional covariant gauges. I've not yet found a source, where the background field gauge has been applied to a Higgsed gauge theory. Is it not advantageous in some way for this case or why isn't it used in the literature?
     
  7. Feb 29, 2016 #6
    So for [itex]W_{\mu}^{1}, W_{\mu}^{2}, W_{\mu}^{3}, B_{\mu}[/itex] there are eight degrees of freedom in total. I'm not too sure what 'transversal / longitudinal polarization' means in this context though.
     
  8. Feb 29, 2016 #7

    Orodruin

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    For a massless vector boson, only the polarisation vectors which is transversal (ie, orthogonal) to the direction of propagation are physical. For a massive vector boson, also the polarisation vector longitudinal (ie, parallel) to the direction of motion is physical. Where did this degree of freedom come from? (You cannot have more degrees of freedom in the broken state.)
     
  9. Feb 29, 2016 #8
    What sort of 'polarization' are we referring to here though? Spin polarization? Wave polarization?
     
  10. Feb 29, 2016 #9
    For massive particles, it is the spin quantum number, but for massless particles, it is technically the particles helicity - i.e., the projection of spin on momentum. These two quantities are linked with the classical wave polarization. Loosely speaking, the helicity of a photon (massless boson) can be either +1 or -1, which is tied to the classical E&M wave being a transverse wave and having only 2 possible polarizations. In contrast, a massive spin-1 boson can have 3 possible spin projections: +1,0,-1 (ms). Classically, this means the wave can have a longitudinal mode as well as two transverse modes. This is why the presence of an ms = 0 state is called a "longitudinal degree of freedom".
     
  11. Feb 29, 2016 #10
    OK, a couple of things:

    - Does this mean a fermion can be left-handed, right-handed or neither-handed? I never thought about the third option before but if Lorentz boosting can change a reference frame enough to reverse a particle's handedness, then boosting into this particle's own frame I suppose makes its handedness disappear... so that gets you three possible spin projections for a massive particle? But a normal observer could never reach or exceed c, so that leaves only reference frames that can have two possible spin projections for the photon?

    - This is strange to me. I'm thinking along the lines of the EM wave having the E part oscillating in one plane, and the M part in another... so that's the two transverse modes? But, I must be mistaken in what you're referring to here because, as far as I'm aware, Z boson, say, doesn't have two components oscillating in planes offset by pi/2 like an EM wave... So how can it be said to have more than one transverse mode?
     
  12. Feb 29, 2016 #11
    Fermions with spin 1/2, unlike bosons, can't have spin projection of 0, on any axis. Including the axis of their momentum. So, no, fermions can't be "neither-handed".
     
  13. Feb 29, 2016 #12
    For your first part, that sounds correct: helicity for massive particles (be it fermion or boson) is not an intrinsic property and can be reversed by appropriate boosts. Just to make sure, the "handedness" of a particle in terms of helicity is something utterly different than the "handedness" in terms of chirality.

    For the second part, the magnetic vector potential has the requirements that puAu = 0. To form the most general plane wave solution, you then have to make a superposition with any two polarization vectors in the plane perpendicular/transverse to the direction of propagation.

    I apologize for not being very familiar with LaTex; otherwise, I would try to be more explicit with relevant equations.

    But, here is an article i came across based on Weinberg's book that explains in detail about this. If you have access to Weinberg's book, then I suggest that (relevant sections are listed in article) . http://phys.columbia.edu/~nicolis/GR_from_LI.pdf
     
  14. Mar 1, 2016 #13

    vanhees71

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    You have to distinguish between "handedness=chirality" and "helicity". Handedness is a pretty abstract notion about certain representations of the proper orthochronous Lorentz group. Each representation of the Lorentz group is characterized by two "pseudo-spin" quantum numbers, which are defined through the Lie algebra of the Lorentz group, which consists of 6 basis elements that are the infinitesimal generators of boosts and rotations. For a fermion with spin 1/2 (the spin tells you the representation of the rotation group of a massive particle in its rest frame) there are two representations of the proper orthochronous Lorentz group, called (1/2,0) and (0,1/2). Each of these representations is on two-dimensional spinor space and these spinors are called Weyl fermions. If you like to have a theory that is also invariant under spatial reflections, you have to use both representations and link them together in a direct sum of the spinor spaces. This leads you to Dirac fermions, which are four-component objects consisting of the two types of Weyl fermions. This representation of the Lorentzgroup, containing space reflections in addition to boosts and rotations of the proper orthochronous group, is called ##(1/2,0) \oplus (0,1/2)##. In this way to describe these Dirac spinors a space reflection just exchanges the two Weyl spinors which compose the Dirac spinor, and that's why you call one of these Weyl spinors the left- and the other the right-handed part.

    Helicity is the projection of the total (!!!) angular momentum of a particle to the direction of its momentum. For a massive particle it's obvious that this is not a Lorentz invariant quantum number, because a massive particle always goes with a velocity less than the speed of light, so that you can overtake it, which flips its helicity. That means that a particle which has positive helicity for one observer can have negative helicity for another observer boosted against the first.

    This changes for massless particles. They always go with the speed of light in any frame of reference, and you cannot overtake them. Thus all observer agree upon the sign of the helicity (supposed the particle is prepared with definite helicity, i.e., in a specific polarization state). The mathematical formalism shows that for massless Dirac fermions helicity and chirality are the same, but our argument above shows that for massive particles that's not the case: While chirality is a Lorentz invariant quantum number chirality can change by boosts.

    Of course particles do not necessarily have definite chirality or helicity, but you have to prepare the particles in such a state if you want to have them in such a state. That's not such an easy task. E.g., the Relativistic Heavy Ion Collider (RHIC) at the Brookhave National Lab does experiments with polarized protons to learn more details about the inner structure of them (in this case about the socalled generalized parton distribution functions and how protons get their spin 1/2 in terms of its "constituents", quarks and gluons).



    Massless fields with spin ##\geq 1## are not so easy to understand, because they are represented as socalled gauge fields. This you know from classical electrodynamics. From the point of view of the four-vector potential the very same physical situation is represented by a large set of fields, i.e., if ##A_{\mu}## is a four-vector potential describing a physical situation, with any scalar field ##\chi## the vector potential ##A_{\mu}'=A_{\mu}+\partial_{\mu} \chi## discribes exactly the same situation as ##A_{\mu}##.

    In classical electromagnetism it's very simple to come to a unique description of the physical situation. You simply don't look at the four-vector potential which is just a mathematical simplification to solve the Maxwell equations, but only at the physically observable electromagnetic field. Relativistically it's described by the Faraday tensor ##F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}##, and thus changing from ##A_{\mu}## to ##A_{\mu}'## doesn't change ##F_{\mu \nu}##. That's gauge invariance: For the physics the concrete gauge of the four-potential is irrelevant and thus the four-potential is only defined up to a gauge transformation, while the observable electromagnetic field ##F_{\mu \nu}## uniquely describes your physical situation.

    Now solving the Maxwell equations for the free electromagnetic field, i.e., in absence of charges and currents, you find that the plane waves are a complete basis in the sense that you can describe all electromagnetic fields in terms of Fourier integrals. It turns out that of the original four independent field-degrees of freedom ##A_{\mu}## or the six independent field-degrees of freedom ##F_{\mu \nu}## there's only freedom for two independent field degrees of freedom.

    That's easier to see in the non-covariant (3+1)-dimensional formalism, working with ##\vec{E}## and ##\vec{B}## as the electric and magnetic components of the electromagnetic field. You start with the electric field components making the plane-wave ansatz
    $$\vec{E}(t,\vec{x})=\text{Re} \left [\vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}) \right],$$
    where ##\vec{E}_0## is an arbitrary complex ##\mathbb{C}^3## constant vector. Of course the field itself is real, i.e., we describe it as the real part only. It's, however simpler to work with the complex field, and we are allowed to do so, because the Maxwell equations are linear real equations. So that we can calculate with the complex field and take the real part at the very end of the calculation.

    Now let's see, which constraints the parameters ##\vec{E}_0##, ##\omega##, and ##\vec{k}## follow from Maxwell's equations. The most simple is Gauß's Law for the electric field. Since we assume that there are no charges present we get
    $$\vec{\nabla} \cdot \vec{E}=0, \quad \vec{k} \cdot \vec{E}_0=0.$$
    That means that for any ##\vec{k} \in \mathbb{R}^3## the amplitude ##\vec{E}_0## is perpendicular to ##\vec{k}##; ##\vec{k}## describes the direction of the wave propation, and that means that electromagnetic waves are transverse. So out of the three field-degrees of freedom ##\vec{E}## only two are independent.

    Now we need to fulfill also the other free Maxwell equations (Heaviside-Lorentz units with ##c=1##),
    $$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0, \quad \vec{\nabla} \times \vec{B} - \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$
    These are the Faraday, the Ampere-Maxwell, and Gauß's Law for the magnetic field, respectively.

    To disentangle ##\vec{E}## and ##\vec{B}## we take th curl of the Faraday Law:
    $$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})+\vec{\nabla} \times \dot{\vec{B}}=\vec{\nabla} (\vec{\nabla} \cdot \vec{E})-\Delta \vec{E} +\vec{\nabla} \times \dot{\vec{B}}=-\Delta \vec{E} +\vec{\nabla} \times \dot{\vec{B}}=0,$$
    where in the last step we have used Gauß's Law for the electric field and ##\rho=0## again.

    The time derivative of the Ampere-Maxwell Law reads
    $$\vec{\nabla} \times \dot{\vec{B}}=\partial_t^2 \vec{E}. \qquad (*)$$
    This we plug in the previous equation, leading to the wave equation
    $$\partial_t^2 \vec{E}-\Delta \vec{E}=0,$$
    and for our plane-wave ansatz this gives the dispersion relation
    $$\omega^2=\vec{k}^2,$$
    as it should be.

    The magnetic field is now determined from Faraday's Law. With the plane-wave ansatz
    $$\vec{B}=\text{Re} \left [\vec{B}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}) \right]$$
    we find for the complex field
    $$\partial_t \vec{B}=-\mathrm{i} \omega \vec{B}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}) = -\vec{\nabla} \times \vec{E}=-\mathrm{i} \vec{k} \times \vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}).$$
    This means
    $$\vec{B}_0=\frac{\vec{k}}{\omega} \times \vec{E}_0,$$
    i.e., the amplitude of the magnetic field is completely determined by that of the electric field, and it's also transverse. The vectors ##\vec{k}##, ##\vec{E}## and ##\vec{B}## build a orthogonal righthanded set of vectors.

    All together we have only two independent field-degrees of freedom for the massless vector field.

    You can do the same exercise for a massive vector field, leading to the Proca equation. The key point is that a massive vector field is not necessarily a gauge field. The Proca equations lead to three independent field degrees of freedom rather than 2.
     
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