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Mass in QFT

  1. Dec 16, 2015 #1
    I realize that Wiki is not the preferred reference source here, but I'll go ahead with this question anyway... In the latest iteration of the article on the Standard Model is the statement, "We see that the mass-generating interaction is achieved by constant flipping of particle chirality." Is this a generally accepted consensus? On a related note, when talking about mass in QFT, can I assume this refers only to rest energy? That since QFT does not include gravitation, there is no reference to inertial or gravitational mass?
     
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  3. Dec 17, 2015 #2
    For fermions this is the case. After symmetry breaking in the standard model the mass terms look like L*R, where L is the left-handed and R is right-handed. It's just a consequence of Yukawa couplings.
     
  4. Dec 17, 2015 #3
    Yes, I should have indicated I was talking about fermions. Thanks.
     
  5. Dec 18, 2015 #4

    vanhees71

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    Well, that's always true for fermions, but it's still not accurate, because most of the mass of the matter surrounding is does not originate in the Higgs mechanism. At most 2% are (in some pretty vague sense) due to the coupling to the Higgs field. The other 98% are through dynamical symmetry breaking due to the strong interaction (Quantum Chromodynamics). This "mass generation" is far from being fully understood from QCD. We only know from lattice-QCD simulations that it works very well, i.e., the hadronic mass spectrum can be calculated with quite good accuracy from QCD using this ab-initio technique to evaluate QCD on computers.
     
  6. Dec 18, 2015 #5
    If I understand correctly, you mean QCD predicts the mass of hadrons based on the energy of the strong interaction between the constituent quarks, which is by far the greatest contributor. I was wondering more about the mass of elementary fermions (including leptons). Again, if I understand correctly, the mass term comes from the interaction between left- and right- handed fields (if I am using correct terminology)?
     
  7. Dec 19, 2015 #6

    vanhees71

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    That's right, the mass I'm talking about is dynamically generated by the strong interaction. It's a non-perturbative issue and thus hard to fully understand, and in my opinion it is not fully understood yet. That's why we use effective hadronic models instead of QCD to describe many aspects of hadrons, including my own subject of research, relativistic heavy-ion collisions and the description of the hot and dense fireballs created with them. At the highest energies (SPS and LHC at CERN, and RHIC at BNL) it is pretty sure that one creates a socalled quark-gluon plasma, where quarks and gluons (or better said quasi-particles with the quantum numbers of quarks and gluons) become the relevant effective degrees of freedom. Of course these fireballs rapidly expand and cool down. So the medium undergoes a phase transition to a hot hadron (resonance) gas.

    What you are asking, is indeed something different, how the elementary particles of the standard model, i.e., the quarks and leptons get their mass. This is very hard to answer in a B-level thread, because you need a bit of relativistic QFT. Indeed, when a high-energy particle physicists talks about mass, he means the invariant mass. For massive particles that's their rest mass; massless particles cannot be at rest with respect to any inertial reference frame and thus it's better to talk about invariant instead of rest mass or simply about mass, because nobody I know uses the idea of a relativistic mass which depends on the speed of the particle, because that's just energy divided by ##c^2##. So it's much easier to use the relativistic covariant quantities (invariant) mass (scalar) and energy and momentum (making up together a four-vector). In the following I also use natural units, i.e., I set the speed of light ##c=1## and the modified Planck constant ##\hbar=h/(2 \pi)=1##. Then all lengths and times are measured with the same unit (in high-energy physics usually in fermi or femto-metres, i.e., ##1 \text{fm}=10^{-15} \text{m}## and masses, energy, and momentum also in the same unit, namely MeV or GeV (mega or giga electron volts) or inverse fm. To convert from GeV to 1/fm and vice versa you only have to remember that ##\hbar c \simeq 0.197 \text{GeV} \text{fm}##.

    Further, in quantum field theory a particle is well defined only in the sense of asymptotic free particles, i.e., you register particles that are far away from each other so that you can neglect the interactions of these particles with other particles. Then you can define, within the formalism of QFT, single-particle momentum eigenstates. These (asymptotic) free particles also are "on the mass shell", i.e., the momentum eigenstantes are also energy eigenstates, and the energy of the particle is given by ##E=\sqrt{m^2+\vec{p}^2}##.

    In collisions, of course the interaction is what's really interesting, and that's also the diffcult part of QFT, because so far we do not have any full analytic solution of any realistic relativistic quantum field theory yet. So the only way to analytically calculate something from QFT is perturbation theory (sometimes resummed to arbitrary high orders). This means you treat the interaction as a perturbation of the free-particle case. The concept is that you have some asymptotic free particles to begin with (usually made in accelerators to have a pretty well determined momentum), which you let collide at some point in a detector, which then measures the again asymptotic free particles after the collision. These can be the same particles you began with ("elastic scattering") or, more interestingly, other particles created in the collision.

    From such scattering experiments the physicists found out that there are a lot of conservation laws, and in 1918 Emmy Noether (investigating a totally different question, namely general relativity) found one of the most important mathematical theorems for physics ever! Conservation laws are related to symmetries of mathematical structures, in our case the fields, describing particles at relativistic energies. These fields obey equations of motion, which keep their form the same when you change the fields in some specific way. E.g., the Dirac equation, describing charged particles with spin 1/2, does not change its form when you multiply the Dirac field with a phase factor, i.e., you use instead of a Dirac field ##\psi(x)## another Dirac field ##\psi'(x)=\exp(-\mathrm{i} q \alpha) \psi(x)##, where ##\alpha## is a constant. Noether found out that such a symmetry, when described with help of the principle of least action, defines a conserved quantity in terms of a local conservation law. In this case of the symmetry of the Dirac field it defines a four-vector current
    $$j^{\mu}=q \bar{\psi} \gamma^{\mu} \psi,$$
    and if the field solves Dirac's equation, the current fulfills the continuity equation
    $$\partial_{\mu} j^{\mu}=\partial_t j^0+\vec{\nabla} \cdot \vec{j}=0,$$
    which implies that the total charge
    $$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} j^0(t,\vec{x})=\text{const},$$
    i.e., this charge is a conserved quantity.

    Now comes a very clever idea into the game, which goes back to Hermann Weyl, who invented it again in a different context (ironically the idea was physically useless, because it contradicted well-known facts already when Weyl brought it to the attention of his physics colleagues, and he got very sharp criticism from Pauli and (a bit more friendly formulated) Einstein; nevertheless the mathematics of the idea is very solid and important for modern physics). The idea is to ask, how to make the symmetry valid also for the case, when the phase factor is made dependent on time and position. To understand the problem, let's look at Dirac's equation for a free Dirac field (in its quantized version describing, e.g., an electron or a quark):
    $$\mathrm{i} \gamma^{\mu} \partial_{\mu} \psi=m \psi.$$
    Of course, multiplying ##\psi## with any constant doesn't change the form of the equation, and thus it also holds for the above introduced field ##\psi'##, but if we now write
    $$\psi'(x)=\exp[-\mathrm{i} q \alpha(x)] \psi(x),$$
    ##\psi'## no longer fulfills the Dirac equation, because there's an additional term from the derivative of the now space-time dependent phase factor:
    $$\mathrm{i} \gamma^{\mu} \partial_{\mu} \psi'=\mathrm{i} \gamma_{\mu} \exp[-\mathrm{i} q \alpha(x)] [\partial_{\mu} \psi(x) -\psi(x) \mathrm{i} q \partial_{\mu} \alpha(x).$$
    With the Dirac equation for ##\psi## this gives
    $$\mathrm{i} \gamma^{\mu} \partial_{\mu} \psi'=m \psi'+q \gamma^{\mu} \psi'\partial_{\mu} \alpha,$$
    which is not the Dirac equation, but has this nasty additional term ##\propto \partial_{\mu} \alpha##.

    To cure this, one has to introduce a vector field, and write everywhere where a partial derivative acts on the Dirac field
    $$\partial_{\mu} \psi \rightarrow (\partial_{\mu}+\mathrm{i} q A_{\mu}) \psi.$$
    Now we have
    $$(\partial_{\mu} \psi' + \mathrm{i} q A_{\mu}' \psi'=\exp[-\mathrm{i} q \alpha] (\partial_{\mu} \psi + \mathrm{i} q \psi (A_{\mu}'-\partial_{\mu} \alpha) \psi).$$
    If we now define that ##A_{\mu}=A_{\mu}'-\partial_{\mu} \psi## we get
    $$(\partial_{\mu} \psi' + \mathrm{i} q A_{\mu}' \psi'=\exp[-\mathrm{i} q \alpha] (\partial_{\mu} \psi + \mathrm{i} q A_{\mu} \psi).$$
    Thus, the socalled "gauge-covariant" derivative of the Dirac field
    $$\mathrm{D}_{\mu} \psi=\partial_{\mu} \psi-\mathrm{i} A_{\mu} \psi$$
    transforms under multiplication with a space-time dependent phase factor as
    $$\mathrm{D}_{\mu}' \psi'=\exp[-\mathrm{i} q \alpha] \mathrm{D}_{\mu} \psi,$$
    where one defines the transformation to act on both the Dirac field and the vector field ##A_{\mu}## as
    $$\psi'(x)=\exp[-\mathrm{i} q \alpha(x)]\psi(x), \quad A_{\mu}'(x)=A_{\mu}(x)+\partial_{\mu} \alpha(x).$$
    This is called a local gauge transformation, and now the gauge-covariant Dirac equation reads
    $$\mathrm{i} \gamma^{\mu} \mathrm{D}_{\mu} \psi=m \psi,$$
    and under the gauge transformation it's again invariant, because obviously
    $$\mathrm{i} \gamma^{\mu} \mathrm{D}_{\mu}' \psi'=m \psi'$$
    again!

    The point is that now of course, we have a field equation, which no longer describes a free Dirac field but a Dirac field which interacts in a specific way with a vector field ##A_{\mu}##. So the local gauge invariance of the theory, we have just constructed, also tells us how certain kinds of interactions have to look like in order to make the theory symmetric under local gauge transformations.

    Now all fields describe particles in the corresponding quantized version of the field theory. To make sense, we must also have an equation of motion for the vector field, and also this equation of motion should be invariant under the local gauge transformations. Now the most simple way to form a gauge invariant (and Lorentz covariant) expression out of ##A_{\mu}## one can write down the field
    $$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
    Indeed, because ##A_{\mu}'=A_{\mu}+\partial_{\mu} \alpha## we have
    $$F_{\mu \nu}'=F_{\mu \nu} + \partial_{\mu} \partial_{\nu} \alpha - \partial_{\nu} \partial_{\mu} \alpha=F_{\mu \nu},$$
    because the partial derivatives on the scalar field ##\alpha## commute.

    Using the least-action principle, it turns out that one gets a constistent set of equations of motion, when writing
    $$\partial_{\mu} F^{\mu \nu} = j^{\nu}=q \bar{\psi} \gamma^{\mu} \psi$$
    together with the gauge covariant Dirac equation written down above. When one decomposes this equation in terms of spatial and time components, using
    $$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A},$$
    it turns out that the equation of motion for the vector field are Maxwell's equations for the electromagnetic field created by a current ##j^{\mu}##. Thus the principle of local gauge symmetry has lead to an elegant way to "derive" electromagnetism from a symmetry principle.

    The interesting thing is that more complicated local gauge symmetries (the socalled non-Abelian gauge symmetries) rule the equations of motion describing all so far observable matter, i.e., the Standard Model of elementary particle physics. The real merit of these symmetry principle is that it gives guidance to write down the equations, because it restricts you to such equations of motion that are form invariant under the local gauge symmetry. One restriction is that it is not allowed to make the gauge-vector fields to describe particles with finite mass, i.e., the "natural" prediction is that these gauge fields describe massless vector bosons. Indeed the photon is (at least within the very high accuracy we can measure it) massless, and the photon is nothing else than the quanta of the quantized electromagnetic field, which thus is described by the above constructed (Abelian) local gauge symmetry.

    The same holds true for the gluons, which are the quanta of the gauge field of the strong interaction, which is based on a more complicated non-Abelian gauge group, called SU(3), which describes the strong interaction. It acts on quarks, and these thus carry (in addition to the electric charge) a socalled "color charge", i.e., the charge of the strong interaction, but each quark comes not only with one type of color charge but with 3 (and each anti quark with three anti-colors). Another important difference of the non-abelian gauge theory is that the gauge bosons themselves now also carry color charge, and the math dictates that (in the case of SU(3)) there are eight different charges. So there are 8 gauge fields, and the corresponding quanta, the gluons describe massless vector particles which come with 8 different color charges.

    The most complicated part of the Standard Model is the weak interaction. Also the weak interaction is described by a local gauge symmetry, but with a further constraint, because it is also a socalled chiral symmetry, which not only forbids to give the gauge fields a mass in a simple way but also the Dirac fields (i.e., in the standard model the quarks and leptons). As it turns out, the gauge bosons of the weak interaction should be massive, and it is well known that also the electron (one of the leptons) carries a mass and that's true also for the quarks. Now one envokes in some sense a similar trick as when we introduced the gauge fields, i.e., one adds more fields to the game. In this case we need some scalar fields.

    The weak interaction in the standard model is introduced together with the electromagnetic one. In some loose sense one has a kind of unification of the elecomagnetic and the weak interaction. So one sometimes also calls it the electroweak interaction. The symmetry group of the local gauge symmetry is ##\mathrm{SU}(2) \times U(1)##. The corresponding charges are called the weak isospin and the hypercharge.

    Now comes also chirality in. The Dirac fields can be split in left- and righthanded parts, using the ##\gamma^5## matrix,
    $$\psi_{L/R}=\frac{1 \pm \gamma^5}{2} \psi.$$
    Now the wiso-SU(2) acts only on the left-handed parts of the spinors (leptons and quarks), while the right-handed parts are singlets. This means that the left-handed parts alone must be symmetric under (local) wiso rotations, and this can only be true if there are no direct mass terms for the fermions. Also the wiso gauge bosons must be massless to begin with. Now one also introduces the scalar fields, and in the most simple version one introduces a wiso doublet, which consists of two complex, i.e., four real scalar field-degrees of freedom. For these fields a mass term is allowed, but one chooses "the wrong sign", i.e., one makes ##m^2<0##. Together with the potential term, describing the self-interaction of the scalar fields, this implies that the scalar fields have a non-vanishing vacuum expectation value. Now you can couple these scalar fields also in a gauge invariant way to the quarks and leptons, and they are coupled to the gauge bosons through the gauge-covariant derivatives anyway. The non-vanishing vacuum-expectation value leads to mass terms for the quarks and leptons as well as the gauge bosons. Thereby three of the four real scalar field degrees of freedom provide the additional field components for each of the wiso gauge fields, necessary to make the massless vector bosons (which have only two polarization degrees of freedom as the electromagnetic field) to massive vector bosons (which have three polarization degrees of freedom). The gauge group ##\mathrm{SU}(2) \times \mathrm{U}(1)## has four gauge fields, but only three scalar fields are absorbed. So one gauge field stays massless, and that should be so, because that's now the photon, which is to a high accuracy known to be massless (as already explained above). Last but not least, one of the scalar fields is now left over, and the QFT formalism tells us that besides providing the vacuum expecation value (the "Higgs field"), its excitations around the stable minimum in the potential must appear as a scalar massive particle (the "Higgs boson").

    The Higgs boson was for quite a while the last building block of the standard model that was not observed (the last particle before that predicted to exist by the model was the ##\tau##-neutrino which was discovered in 2000). The missing Higgs and also to determine its mass, which was also not well known before it could be measured, motivated to build the largest accelerator so far, the Large Hadron Collider at CERN, and indeed on Independence Day 2012, ATLAS and CMS could announce the discovery of the Higgs boson, leading immediately to a Nobel prize in 2013 for Higgs and Englert, two of the many fathers of the theory behind it.

    Of course, I left out a lot in my try to explain the ideas behind the Standard Model, just focussing on the question concerning the masses of the elementary building blocks in the Standard Model, e.g., the very important mixing of quarks, the charge pattern of the model which finally is responsible for the whole edifice to be consistent, because the three color-degrees of freedom of the quarks, carrying charges ##-1/3## and ##2/3## together with the leptons (one carrying carge -1 and the corresponding neutrino a 0 charge) conspire in such a way that dangerous terms in the quantum version of the theory exactly cancel. Quantization of a field theory with chiral symmetry can lead to the violation of this symmetry, which is established in the classical theory one starts with. This is called an anomaly, and if a local gauge symmetry is broken by such an anomaly, the whole model gets useless, because then it cannot be properly interpreted in a physical way anymore.

    So far the Standard Model is very successful. Sometimes one feels it's even too successful, because it is not completely satisfactorial. E.g., renormalization theory predicts a large contribution to the vacuum energy (also known as "dark energy") of the universe, which completely contradicts the observation by about 100 orders of magnitude! One has to fine-tune the parameters of the Standard Model by hand to make it fit, and that's not very elegant. One would like to have some explanation for why the dark energy content of the universe is so much smaller rather than have to finetune it. So far, nobody could solve this puzzle in a satisfactorial way, and one hopes to find hints for extensions of the standard model, like a whole bunch of new particles predicted by socalled supersymmetry extensions of the standard model. These particles could also be candidates to make up the socalled dark matter, which should be there because of the gravitational interaction one observes, e.g., in the motion of stars in galaxys, which is due to matter that is otherwise completely invisible, i.e., particularly not electromagnetically interacting and thus really "dark", i.e., it cannot emit light (or any other frequency range of the electromagnetic spectrum).

    The only clear case of physics beyond the standard model is the discovery of neutrino oscillations and the implied fact that neutrinos must also have a mass, while within the Standard model they stay massless. For this discovery this year's Nobel prize was given to Takaaki Kajita (SuperKamiokande) and Arthur B. McDonald (SNO).
     
  8. Dec 19, 2015 #7
    Wow! Did you just write that? Thank you so much for such a clear and detailed explanation to a noob question. Just one thing, I am not familiar with the term "wiso." Weak Interaction Symmetry Operation?
     
  9. Dec 19, 2015 #8

    A. Neumaier

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    Do you have a recent reference?
     
  10. Dec 19, 2015 #9

    ShayanJ

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    Weak IsoSpin
     
  11. Dec 19, 2015 #10
    Oops ... thanks.
     
  12. Dec 19, 2015 #11

    vanhees71

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