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7

Let me first make a general remark about internal symmetry groups, unrelated to our problem of the correct symmetry group for QCD. The symmetry must act on Hilbert space as a unitary operator for the conservation of probability. Now let us turn to the strong interaction. The most important experimental facts were that Observed hadron spectrum was ...


0

Well, they could develop a vev, depending of the interactions you consider. Take e.g. this term in the lagrangian $L=\frac{1}{\Lambda^2}[\bar{\psi}\psi- v^3]^2$. Of course this would imply that you also break Lorentz, which you may want to avoid.


0

Explanation for the electromagnetism aspect: If the vacuum carried charge under some generator, that would mean that the generator would not annihilate the vacuum. That would mean that even if such a generator corresponds to a symmetry of the theory, the vacuum however is not symmetric under that operation. Then the gauge boson corresponding to this ...


0

I think it is a general fact about grassmannian field, and this has nothing to do with Lorentz invariance or other symmetries (you can invent a lot of QFTs without this kind of symmetry, but the VEV of a fermionic operator will be always zero (in the absence of sources)). In a functional integral formulation, the VEV of a grassmannian field $\psi$ is ...


0

Well, as regards the second part of the first question, "and QCD color neutral", the if the vacuum state of any Standard Model field had a colour charge surely it would be in opposition to the Confinement property of QCD that we observe. That is, we never see colour charge in an experiment, we only ever see colour singlet (white) field states. I'm not sure ...


1

Well, the Logic is reversed compared to the one you are implicitly using. It is the direction of the Higgs'vev that's defining what we call electric charge. In practice SU(2)xU(1) is broken to a certain U(1) that we can always choose to point in a certain direction, and accordingly assign the electric charges afterwards. Since the Higgs can't carry color it ...


3

Why can't fermions have a non-zero vacuum expectation value (VEV)? Lorentz invariance. If anything other than a Lorentz scalar has a non-zero VEV, Lorentz invariance would be spontaneously broken. For example, suppose we have a Lorentz invariant term in a Lagrangian for a vector $$ \mathcal{L} \supset m^2 A_\mu A^\mu. $$ Now suppose the vector obtains a ...


3

You are correct that for a massive spinor, helicity is not Lorentz invariant. For a massless spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant. Helicity defined $$ \hat h = \vec\Sigma \cdot \hat p, $$ commutes with the Hamiltonian, $$ [\hat h, H] = 0, $$ but is clearly not Lorentz invariant, because ...


4

Since $\pi^0$ is a pseudoscalar particle, we have $$\langle 0|J^\mu_{em}|\pi^0 \rangle =0,$$ and the pion cannot decay into two leptons with a simple photon exchange. In the Standard Model, the leading-order contributions for this process are a box diagram and a $Z^0$ exchange, as you can see in fig. 1 of arXiv:0806.4782 (replacing a $c$ quark by a light ...


0

Let me address your questions one by one. Why is it said that lepton number is conserved in Standard Model (SM)? How do I know that lepton number is an Abelian charge? The SM Lagrangian is invariant under the fermion transformations, $$ \psi \to e^{iL\theta}\psi $$ where $L$ is assigned such that $e^-$, $\mu^-$ and $\tau^-$ leptons and ...


-1

Why is it said that in standard model lepton number is conserved? Because the standard model is a mathematical model specifically fit to what the data tells us, and lepton number is conserved according to the data. How do I know that Lepton number is an abelian charge? It is additive in the number of leptons and antileptons reduce the lepton ...


1

I'll only address your first question here. To start off with a sidenote, I think the idea that mass is a fundamental property of a particle has been on shaky ground ever since Einstein showed the equivalence of mass and energy. I can hardly imagine it took very long for people to come to the conclusion that mass cannot be a fundamental property of ...


0

Unfortunately, Bardeen seems to misunderstand the naturalness problem that has nothing to do with quadratic divergences per se. In the strict SM, there is no naturalness problem because the running Higgs mass squared is proportional to itself. But this is not the setup that people care about when talking about the actual naturalness problem that emerges ...


2

In this context, the right-handed neutrino is a singlet under the Standard Model gauge groups. Only the right-handed neutrino is allowed a Majorana mass. The left-handed term is not gauge invariant. If the SM and right-handed neutrino fields were embedded into a Grand Unified gauge group, the right-hand term would break that symmetry. It is expected ...


0

As i see it a gravitons would interfere if gravity waves interfere. And I certainly think that a gravity wave would interfere. I guess that they are transverse waves that can interfere like any other waves.


4

I go through the calculation below. However, I won't calculate the integral myself since its very impractical and not what you want to do in practice. You need a quick formula to simplify your integrals. Thanksfully, such a formula is provided in any standard textbook in QFT. You should derive this formula once and then move on. I will do the calculation ...


1

This is a special case of a more general phenomena. Conserved currents never acquire anomalous dimensions, they are protected by the symmetry. If you have a conserved current, you have a symmetry algebra for the theory \begin{equation} [Q^a,Q^b]=if^{abc}Q_c \end{equation} For this to hold, the charges need to be dimensionless. But the charges are given by ...


1

These type of integrals are very common in 1-loop (and higher order) calculations, so they were categorized according to the number of legs attached to the loop. The convention is to label them starting with $A$, so one-point function is $A(m)$, two point function is $B(p,m_1,m_2)$, etc. Depending on the numerator of the integrand they might also have tensor ...


0

Since electrons don't interact through the strong interaction, an electron-quark "atom" is on the face of it the same as an electron-proton atom. (Except maybe weak interaction decays, I'm not entirely sure.) However: a free quark has never been observed in experiments, and it is widely believed - but not proved - that the theory of strong interactions does ...


3

spin 1/2 fermions (electron, proton, neutron, muon, tau, quarks) have +1 parity (by convention as pointed out in Anna's comment). The corresponding anti-fermions have -1 parity. Bosons and their anti-particles have the same parity. See this and this lecture for more information on parity.


0

For the Benfords' law to apply, one needs a set of unrelated numbers that span many many orders of magnitude. The "natural" comparison scale for these very different numbers turns out to be the logarithmic scale. If the numbers are really unrelated, then you can use Pascal's principle of indifference to say that the probability to have a given number in the ...


1

You must also take in account the diagram where a muon is the intermediate state, although it is not dominant. This particle will connect the two Higgs vertices.


1

This isn't a complete answer to your question, just two comments on two specific points that came up that were too long for a comment: WRT the Higgs having a spatially varying background because of inhomogeneities in the universe: It's important to keep in mind that the Higgs field is VERY MASSIVE when it has a nonzero VEV (as it does in the universe ...


5

An obvious difference between the two ways of thinking about it you mention is that in the case of the Higgs mechanism, there is an observable particle excitation of the field associated with it, which was found recently. Furthermore it should be noted that the Higgs mechanism only concerns the mass generation of some elementary particles. The mass of ...


1

In general, derivative couplings lead to momentum-dependencies in scattering amplitudes. This can be seen from the fact that the Fourier transform of a derivative operator corresponds to a multiplication by the relevant momentum. A mass dependence is implicit through by having a momentum, since the momentum of a fermion depends on its mass. In this case, ...


0

You first ask whether there could be interference effects between virtual gravitons. Assuming something like the standard theory for particle interactions (quantum field theory) would apply to a quantum theory of gravitons, there would indeed be interference between virtual gravitons. The total probability of a scattering event involving gravitons, say ...


1

Question 1. Can virtual particles, in particular gravitons, interfere? Gravitons are the hypothetical carriers of gravity, corresponding to the photon for electromagnetism, with a very weak coupling to matter in comparison to all other forces: With the strong force coupling at 1, the electromagnetic is at 1/137 , the weak 10^-6, the gravitational ...


2

A very good introduction is "Introduction to Elementary Particles" by David Griffiths. Then, if you really want to get into the nitty gritty, jump to a text on quantum field theory, such as: "Quantum Field Theory" by Franz Mandl and Graham Shaw "Quantum Field Theory in a Nutshell" by A. Zee, specially if you have any background in the path integral ...


3

The algebraic formulation of geometry as it appears in Connes's spectral formulation of geometry is in fact well-known elsewhere in physics, even if for some reason it is rarely highlighted as being the same (but see the references below): the way a spectral triple encodes a (non-commutative or classical) spacetime geometry is just the same mechanism by ...



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