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You are seeing particles. However there's more to this than meets the eye so I need to explain exactly what I mean by this. Light is neither a particle nor a wave. Instead it is a quantum field. As a general rule while light is travelling it appears as a wave, but when the light quantum field is exchanging energy with anything it does so in quanta that ...

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The deuterium nucleus is a boson, with spin $\hbar$ (and positive parity). Unlike other stable nuclei, deuterium doesn't have any bound excited states; however if it did they would also have integer spin. The deuterium atom is a fermion, which may have spin $\frac12\hbar$ or $\frac32\hbar$, to be combined with the orbital angular momentum (which is zero in ...

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Yes, it is correct to say that the Higgs boson, just like other elementary particles, get its mass from the interaction with the Higgs boson – which means "with itself" in the case of this particle. More concretely, the mass may be derived from the Higgs potential (energy density) $$V(h) = \frac a4 h^4 - \frac b2 h^2 + c$$ where the additive shift $c$ ...

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It would be physically impossible to be able to "see" light as anything other than a particle (photon). The only time photons, or any other subatomic particle for that matter, can be described as a wave is when we are NOT looking at them.

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A field and a particle are two different concepts and it is well that one should separate them. A field can be classified as a scalar field, a vector field, a spinor field or a tensor field according to whether the value of the field at each point is a scalar, a vector, a spinor or a tensor, respectively. For example, the Newtonian gravitational field is ...

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The talk about "particles" being "excitations" is handwavy, and comes about precisely because of the analogy of the usage of Fourier components that you talk about.1 Perhaps more simply, is the field any different if we have one particle or if we have two particles? "The field" is inaccessible. Some fields aren't even observables (because they may not ...

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You pretty much have the right ideas here. The MC is necessary because the functional dependence of the detector response and the underlying physics is too complex to calculate in any other way. And therein lies the actual analysis problem: it's not enough to simulate the physics based on the standard model (or some other hypothetical interaction). In order ...

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Solar neutrinos have energies under 10 MeV. There isn't enough energy to make heavy leptons. That's really the whole story.

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Roughly sketched, for the quantized Dirac field one has: $$\hat\psi(x)\sim \int d\mathbf{p}\, \sum_r \bigg[ u^{r}(p)\, \hat a^r_\mathbf{p}\,e^{-ipx}+v^{r}(p)\, {\hat b^r_\mathbf{p}}^\dagger e^{ipx}\bigg],$$ where $r=\pm1$ denotes helicity. The ${\hat a^r_\mathbf{p}}^\dagger$ operator creates a helicity-$r$ particle state when it ...

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The Higgs Field is believed to permeate the universe, and the Higgs Boson is just an excitation of one of the four components the Higgs Field! The Higgs field needs high amounts of energy to be excited, so when Higgs Boson is "created", its energy level is usually many orders of magnitude higher than the ground energy level of its surroundings, and hence the ...

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The uncertainty in any particular measurement is $\sigma_E$. Resolution for these devices is almost always stated in relative terms as here, but take it like this because it depends on the energy measured. So just multiply by the energy. That is, express your signal in $\mathrm{GeV}$ and then find \begin{align} \sigma_E = \left(\frac{0.1}{\sqrt{E}} ... 3 Skobeltzyn recounts his research on particle physics in his text "The early stages of cosmic ray particle research", which has been mentioned in another answer. I found it in the book The Birth of Particle Physics edited by Laurie M. Brown and Lillian Hoddeson in 1985. It looks like he was the first one to try using Wilson chamber to detect tracks of ... 3 It is instructive to look at chart of isotopes , number of neutrons on the x axis and protons on the y. The stable (black) isotopes diverge from the diagonal, more neutrons are needed to neutralize the coulomb repulsion of the protons, for stability. The main forces are the coulomb force (repulsive) and the strong force (attractive) , but the specific ... 2 We then talk about a left-chiral electron we do it in an informal way, you are correct that a massive particle cannot be inherently chiral. To see this, let us remember that he handedness of an elementary particle depends on the correlation between its spin and its momentum (helicity). If the spin and momentum are parallel, the particle can be said to be ... 2 According to Bazilevskaya's Skobeltsyn and the early years of cosmic particle physics in the Soviet Union paper (emphasis mine), Skobeltsyn demonstrated a series of photographs with the cosmic ray tracks at the Cambridge conference presided by Ernest Rutherford on 23–27 July 1928, where they made a strong impression on the audience. The comprehensive ... 2 Electron positron annihilations can give mu and tau neutrinos as well as electron neutrinos. For a calculation of the probabilities see for example Mu and Tau Neutrino Thermalization and Production in Supernovae: Processes and Timescales. You might also be interested to read DavidZ's answer to Why does electron-positron annihilation prefer to emit photons?. ... 2 I think there can be confusion around what 'binding energy' and 'mass excess' mean. The Wikipedia entry on Deuterium has links to explanation them, which may clear it up if the following doesn't. IF you could start with isolated protons and neutrons and assemble your own nucleus, the mass (energy) balance of the result would be the sum of the isolated ... 2 Even if the the strong nuclear force is the most powerful force over subatomic distances. The electrostatic force is almost always significant, and, in the case of beta decay (which also involves a transformation of a proton into a neutron), the weak nuclear force is also involved. But the basic answer is yes, you cannot two protons together (Helium 2 ... 2 The interaction between two charged partices occurs through a change of momentum. All very well, but the next question is how we calculate the momentum exchange, and this is where quantum field theory comes in. The interaction between two electrons is described as a disturbance in the quantum fields involved. Quantum field theory gives us an expression for ... 2 For anyone with similar problems: The following observation has helped me immensly: We have in fact four particles directly related to an electron: A left-chiral electron \chi_L, with isospin -\frac{1}{2} and electric charge -e, A right-chiral anti-left-chiral-electron (\chi_L)^c=\chi_R with isospin \frac{1}{2}, electric charge +e A ... 2 I think that you have to neglect the s-quark mass not its momenta in your expression. Anyway, if you perform the standard procedure for calculating this integral you will find an expression which is a function of all the possible external momenta and the metric tensor as well. Then you can carry the limit p_s \to 0 you will find the formula which are you ... 1 In answer to the main question, matter does, in fact, "pass" through other matter. Starting from the macro scale (stars , galaxies), down to the micro scale (atoms), it happens all the time. The "free" movement of matter starts to get impeded, as the atoms start making latices (solids, crystals). But even at this scale, as Rutherford demonstrated, matter ... 1 All you need to do is conserve energy and momentum in the lab frame. Firstly you conserve energy in lab frame: $$E_{\gamma 1} + E_{\gamma2} = E_{\pi} = 1.3GeV$$ Then you work out what the pion's momentum was (still in the lab frame) using the mass-energy-momentum relation where the E_\pi is the total kinetic and mass energy: ... 1 The definition of s is the following: s=(p_1+p_2)^2, where $p_1$ and $p_2$ are the 4-momenta of each colliding proton. For head on collision of particles with same energy and momentum (like in LHC) these explicitly read $p_1=(E_p,\vec{p})$ and $p_2=(E_p,-\vec{p})$. Plug this in the definition: \begin{eqnarray} s &=& p_1^2 + p_2^2 +2p_1\cdot ...

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1) There exists the classical electromagnetic wave as described by Maxwell's equations. 2) The photoelectric effect showed that these electromagnetic waves are composed of photons, with energy E=h*nu , where nu is the frequency of the classical wave. Single photon experiments have been performed by limiting the intensity of the beam to one photon at the ...

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Two ways to think about this: instead of thinking of one photon, think of the Quantized Electromagnetic Field. This quantum field is spread throughout all space and time and is everything electromagnetic. An elementary way to think about it is as a collection of quantum harmonic oscillators, one for each of the classical modes of Maxwell's equations. If you ...

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You might want to look at D. SKOBELTZYN's paper The Angular Distribution of Compton Recoil Electrons Nature 123, 411-412 (16 March 1929) | doi:10.1038/123411a0; but it is behind a 'paywall' and only a short abstract is available for free. (FWIW, Nature, Lond., 1929, v. 123, No. 3098, p.411) EDIT (11/27/2014): See also: Dimitry V. SKOBELTZYN, THE EARLY ...

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It is mainly measurement and detector errors that make up the width in the plots you show. The Monte Carlo simulates the detector resolution and folds in the theoretical values when it says that the width agrees. The real width is expected to be much smaller. In this we see that the real width is only given as a bound by the experiments the CMS ...

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This is misconception that light is some kind of 'mix' of waves and particles. Instead, It actually IS both waves and particles at the same time, you can't separate them from each other. So probably, the answer could be: you see particles as well as waves.

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If you assume that whatever generates the mixing patterns of quarks and leptons (beyond the SM) has no underlying symmetry and that nature chose $V^{CKM}$ and $V^{PMNS}$ randomly within the set of $3\times3$ unitary matrices, then it is natural to expect mixing between families because the probability of randomly selecting $V^{CKM}=V^{PMNS}={\mathbb 1}$ is ...

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