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From a decoherence point of view, fields are more fundamental as they give rise to particle-like behavior from the wave behavior if interactions with the environment are strong. In the end though, quantum mechanics only describes correlations between macroscopic changes in detectors (or other materials), so whatever kind of ontology you want to take in the ...

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Simply because it is usually taught from historical, heuristic and pragmatic point of view, rarely from axiomatic point of view (e.g. Wightman axioms, as mentioned in a comment by ACuriousMind). This is because it is taught to be useful, as most QFT calculations boil down to scattering and decay amplitudes, and as Sean Carroll said: Heuristic QFT, on ...

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An elementary particle is defined as an irreducible representation of the Poincar\'e group. These were classified by Wigner in 1939. This was done via the little group construction. The important representations are (metric signature $(-,+,+,+)$ $p^2 = 0$, $p^0 < 0$ - The little group is ISO(2). All finite dimensional representations of this group are ...

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I think that the answer is that there is a flavor symmetric octet representation and a flavor antisymmetric octet representatio, while the decuplet is totally symmetric. Therefore, when you consider the spin and flavor wavefunction of a baryon for an octet baryon you have: $\chi(spin)\cdot\phi(flavor)=\frac{1}{\sqrt{2}}(\chi^{1/2}_s\cdot ... 1 I'm trying to give a less technical answer. It's not rigorous but should give you the idea how spin and the regular rotation related. Maxwell's equations say in order to have magnetic field, you need a ring current. This can be achieved by giving angular momentum to charged particles. This can be orbital or simply because the particle is spinning. This was ... 4 It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%. There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds ... 0 1) No, substances almost never completely absorb photons. Otherwise you could not see them. In case a substance would absorb all photons (which is quite hard to achive intentionally) it would be pitch black even if you shine arbitrarily strong light on it (-> black-body). 2) It will be reflected back and forth, but only a finite amount of time. This is ... 1 Almost always, when photons hit matter or interact with it, they are not reflected in the way a billiard ball bounces off a billiard table edge. Rather, they are absorbed, the absorber rises into a metastable state, and then a new photon is emitted on the decay of the metastable state. Sometimes, though, when photons undergo an interaction with a lone ... 2 Usually phase diagrams, e.g. of water, are shown as pressure vs temperature. However, we could just as well write a phase diagram as temperature vs density. This is because the equation of state of matter is a relation between pressure, density and temperature. The above phase diagram should be interpreted as showing: what is the state of matter at a certain ... 2 1) Why don't we consider finite dimensional representations of this group? As you said, we ask (anti)unitarity, so it is impossible to find finite-dimensional representation. 2) Why associate the Lorentz group to fields? The essence of the answer is what Trimok already said in his comment: the "translational part" of the Poincarè group is already ... 1 The SIS accelerators are heavy ion accelerators, and the German for heavy ion accelerator is SchwerIonenSynchrotron (my capitalisation), hence the abbreviation SIS. There is more info in this article. 1 SIS-100/ SIS-300 is an accelerator under construction for FAIR (Facility for Antiproton and Ion Research) in Darmstadt, Germany. see - http://cern.ch/AccelConf/e08/papers/mopc100.pdf I believe, but am not sure, that the -100 and -300 refers to the magnetic rigidity (i.e. Magnetic field * bending radius) of the accelerators, which determines the maximum ... -1 You asked HOW it is that the bonded neutron is stable, but the free neutron is not: 'What happens inside the nucleus that makes neutrons stable?' This is an ontological question and these are the hardest to answer. The best answer you can get in terms of conventional physics is differences in binding energy, as Lagerbaer explained. The Table of Nuclides ... 0 Yes! Neutrons are electrically neutral, but they have a magnetic moment. You can accelerate a bar magnet with a magnetic field, so you can also accelerate a neutron with a magnetic field. For most beams, the change in energy is pretty negligible, but there's a major exception for ultra-cold neutrons (UCN), which just so happen to be my specialty. UCN have ... 2 Because muons don't carry a color charge, and hence don't participate in the strong interaction... 4 For a scalar field$\phi$, the most widely used convention, based on my experience, is to write the Lagrangian with kinetic and potential terms, followed by interactions like so, $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2 - \frac{1}{2}m^2 \phi^2 - \sum_{n \geq 3} \frac{\lambda^n}{n!}\phi^n$$ where$\lambda_n$are coupling constants. (We could not have a ... 0 A perhaps not completely rigorous, but easy to understand derivation: It can be shown that the magnitude of angular momentum of circularly polarized classical radiation of frequency$\omega$and energy$E$is given by$E/\omega$. If we now assume that radiation is quantized in packets of energy$E = \hbar \omega$, we arrive at the conclusion that the ... 4 The sun gets its energy from the pp-chain. The first step is the two protons forming the diproton (Helium-2): $$\,^1_1H+\,^1_1H\to\,^2_2He+\gamma$$ where the$\gamma$is the photon (of energy about half an MeV). This quickly$\beta^+$-decays into a deuterium by converting a proton into a neutron: $$\,^2_2He\to\,^2_1D+e^++\nu_e$$ where$e^+$is the ... -2 Perhaps this helps: https://de.wikipedia.org/wiki/Benutzer:HolgerFiedler/Strukturen/Summary It explains how the photons electric field changes to the magnetic and back and there are no points of zero energy. But what you have to keep in mind is that radio waves are cliffs rent waves, they are omposed from a lot of photons and the radio wave is zero at all ... 8 the only difference between matter and anti-matter is simply charge Nope. All "quantum numbers" reverse for them. In particular, for the example you consider, there's Baryon number, which is different for baryons and anti-baryons, and in particular, for a neutron and its anti-particle, an antineutron. It is in fact possible to differentiate between the ... 1 It's all about spin. The conservation you mentioned is the key to it - you could conserve energy and/or momentum by tweaking the resulting KE, but you're still left with unconserved spin. Historically, observations showed an upper limit to the electrons which was not high enough to include the missing energy that is found if you don't include neutrinos. 4 The momentum is harder to deal with than the energy. If a stationary neutron decays into an electron with momentum$(a,b)$and a proton with momentum$(a,-b)$, then there is no way to conserve 3-momentum without the creation of a third particle with momentum$(-2a, 0)$And while real numbers wouldn't work out this nicely, it would be obvious that the ... 3 An electron is a charged particle, charge conservation would not work as the neutron has zero charge. In addition it would have been detected with its interaction as its energy would be similar to the energy of the other electron seen. The neutrino was posited as a weakly interacting particle exactly because it was not caught by the detectors, and because ... 0 Light particle is a vague term actually. There are two ways I can think of a particle being "light". If you mean has least mass than photons have zero mass so it is indeed lightest. The exact mass of lightest neutrino is not yet known. It is actually wrong to talk about "A photon" because it is not a single particle. Photon is a "quanta" of ... 2 How is it possible to accelerate a neutron? Neutrons have a dipole moment, so they may be 'accelerated' insofar as they will turn in a magnetic field – that is their primary interaction with the electromagnetic field. It is possible to accelerate a charged particle in an electric field, how is it possible to accelerate a neutron? Neutrons also ... 1 The neutron can be attached to a proton via the Strong Force by colliding a high-energy proton with the neutron, and then the proton-neutron atom can be accelerated with a regular electric field. Gravity can also accelerate a neutron. 14 Although a neutron is electrically neutral, it has a non-zero magnetic dipole moment. It interacts with a magnetic field to give a potential $$U = \vec{\mu} \cdot \vec{B}$$ A gradient of magnetic field strength will give a force $$\vec{F} = \nabla|\vec{\mu} \cdot \vec{B} |$$ It's not possible to produce large, sustained field gradients, nor is it ... 27 Basically, the answer is no, it's not possible. When we produce neutrons for research purposes, we have to produce them using nuclear reactions. They come out of the nuclear reactions with energies that are determined by the reaction, are not otherwise under our control, and that are on the MeV energy scale of nuclear physics. Examples of a neutron source ... 2 Within the limits of accuracy of the experiments, no negative energy particles, with negative invariant mass, have been leaving a signature. Particle accelerators deliver beams at fixed energies. All the event ordering algorithms depend crucially on energy and momentum conservation, that is what missing energy and missing transverse momentum are all ... 3 Interesting question! This may not directly address your question, but I think it might be relevant, and it's too long to fit in a comment. Some questions that arise here are: What do we mean by a negative-energy object? Do we have any reason to think that they're worth searching for? If so, is there any method that we know should, at least in principle, ... 2 How would you define a negative energy particle? Is that one that, when it hits your detector, takes a fixed amount of energy out of it? That's trivially forbidden by the third law of thermodynamics, otherwise you could construct a fridge that can make negative temperatures. The only way to escape that would be by requiring, that the particle can measure the ... 1 At the beginning of page 6 he talks about the specific cross section of a resonance interesting to study. Those tends to scale with$E^{-2}$, look at figure 49.5 in the PDG and compare the$J/\Psi$, the$\Upsilon$and the$Z$. This is the reason why he would like to increase the luminosity with$E^2$. The total inelastic cross section,$\sigma$, is assumed ... 4 I addressed this a little at your other question but this one is more like physics. Yes, they're decay product lists. Beware that if all the modes are only "seen" or "not seen," you are sort of looking at the hairy edge of what's experimentally accessible. The indented lines are subtypes of the same reaction. Using your example, a neutral particle can ... 5 Briefly, but perhaps less briefly than the footnotes to the tables, and reflecting my own personal biases: The strings like$I^G(J^P) = 1^-(0^-)$represent the quantum numbers of the particle.$I$is isospin: the pion, with isospin 1, is a member of an isospin triplet ($\pi^+,\pi^0,\pi^-$).$J$is spin: the pion is spinless.$P$is spatial parity: the pion ... 1 If you are just looking to simulate the data produced at the LHC then the right program to use is Madgraph. It is by far the most popular Monte Carlo generator to use for simulations at the LHC and can produce events for any process you want. Madgraph will then also invoke Pythia and PGS to simulate hadronization and detector simulations if you wish. There ... 1 Geant4 is probably the closest thing you could conceivably get. It's a detector simulation software package with a steep learning curve that's heavily used in the HEP community, and, as far as I know, is available for anyone to download and build. Full disclosure - I did HEP experiment once upon a time but that was 20+ years ago. So if someone currently ... 1 The Higgs is part of a complex scalar doublet in the standard model. It. carries both hypercharge and weak charge. So we have discovered charged scalars. Now perhaps you are only interested in ELECTRIC charge. So does the Higgs doublet carry this? Well once the Higgs picks up a vev, then some parts do and some parts don't. The parts that do carry electric ... 3 The Lagrangian $$\mathcal L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \mathcal L_\textrm{free} + eA_\mu J^\mu \tag{1}$$ where$A_\mu$is the 4-potential,$F_{\mu\nu} = \partial_{[\nu}A_{\mu]}$is the field tensor,$\mathcal L_\textrm{free}$describes fields other than$A_\mu$, and$J^\mu$is the 4-current density expressed in these other fields, describes a ... 1 You have a few basic formulas for solving this kind of stuff: $$\vec{x}_f=\vec{x}_i+\vec{v}_it+\frac{1}{2}\vec{a}t^2,$$ $$\vec{v}_f=\vec{v}_i+at.$$ These are vector formulas, but all you're doing with the vectors is adding/subtracting them. When you add vectors you add the individual components, i.e. $$\vec{a}=(a_1,a_2,a_3)$$ $$\vec{b}=(b_1,b_2,b_3)$$ ... 9 The standard model is very successful in its group structure in ordering all observed particles. To introduce a particle with charge and zero spin, you will need a different model that would also accommodate the symmetries observed experimentally and fitted by the standard model. So the answer to "why" is "because" we have not seen any and can model well ... 0 Martin wrote: ", hence they can only interact via the weak force, which is, as the name says, weak." Makes me wonder, is "weak" a good name? Or is this interaction better described as short, and still full of surprises? I'm thinking of {Dirac, Weyl, Majorana} spinors, rates of chiral oscillation, inertial-mass... ... which makes me wonder about the ... 0 I was confused about this but it turns out that the context of this equation is important. The energy change from an electron in one energy state to another can be determined by using the formula$\Delta E = h * \nu$which will give the energy difference of the two states or the energy of the emitted or absorbed photon. However if you are just determining ... 1 For simplicity let's imagine a hydrogen atom, which has one proton and one electron. The energy difference$\Delta E$you speak of is the difference in energy of the two initial and final states of the electron-proton system. Initially the system is in some state$\psi_i$with energy$E_i$. Afterward, it's in a different state$\psi_f$with energy$E_f$. ... 5 It's a matter of history. When George Stoney developed Stoney units in 1881, or when Robert Millikan performed the oil drop experiment in 1909, it wasn't yet known that it was possible for anything to have a charge smaller in magnitude than the charge of an electron. By the time the quark model was proposed, in 1964, the use of the "elementary charge" being ... 2 The important point is the fact that such a mass term breaks the gauge symmetry (Edit: I am assming that you want to build the Majorana mass term using SM available fields -- no extension considered -- of which there is only$\nu_L\$). Namely, the desired term is (one generation suffices): $$\frac{1}{2}\,M\, \nu_L^T \,\mathcal{C}^\dagger\,\nu_L\, +\, ... 4 The Standard Model Yukawa interactions must be SU(3)\times SU(2) \times U(1)_Y gauge invariant. The down-type Yukawa interaction is$$ \mathcal{L} \supset -y_d \bar Q \phi d_R + \text{h.c.}. $$This is indeed gauge invariant. The \bar Q d_R form a colour singlet (3^* \times 3), the \bar Q \phi form an SU(2) singlet (2^*\times2), and the whole ... 5 The reason is that the SU(2) invariant in \mathbf{2}\otimes\mathbf{2} (or in their complex conjugate \mathbf{2}^*\otimes \mathbf{2}^*) is given by contracting the two \mathbf{2} with the anti-symmetric 2\times 2 matrix \epsilon_{ab}, as i\tau_2 is. In the case at hand the two \mathbf{2}^* are \bar{Q} and the \Phi^*. You could form another ... -2 The photon is its own antiparticle, even though it has two spin states. Wikipedia write about neutrinos: "Whether or not the neutrino and its corresponding antineutrino are identical particles has not yet been resolved, even though the antineutrino has an opposite chirality to the neutrino." So it is not unreasonable that the neutron decay could be ... 2 If you are considering massless neutrinos there is no such a diagram since all interactions would preserve flavor. If you take instead massive neutrinos, you are probing lepton flavor violation within the SM since the new interaction with \varphi respects flavor. It is thus very very small, being controlled by the neutrino masses. In turn, it is therefore ... 3 Lets analyse the Majorana condition and the Majorana mass term. A massive Majorana neutrino \chi_j (a Majorana spin 1/2 fermion) having mass m_j>0 can be described in a local quantum field theory (eg. the standard model) by a four component spin 1/2 field \chi_j(x) which satisfies the Dirac equation and the Majorana condition which reads:$$ ...

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