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14

Quarks as we know them are fundamental particles, which means that they do not have smaller constituents. This however does not imply that they cannot decay. A particle in quantum field theory does not need to have constituents to decay into, it can in principle decay into any particle its corresponding field couples to (interacts with), as long as it obeys ...


6

The current understanding of quarks is, that they are a fundamental particle. This means for the energy scales currently available in particle accelerators all quarks have behaved like point-like particles. Due to the strange nature of the color-field (the energy stored in it increases with distance instead of decreasing) if you break a proton apart (which ...


6

The $u$ & $d$ quarks decay into $d$ & $u$ quarks and bosons (e.g., W bosons)--this is effectively what happens to the hadrons in weak interactions. This (incomplete) chart shows, for instance, $$ u\to d+W^+\\ d\to u+W^- $$ There isn't anything sub-quark, as far as the standard model goes.


5

Neutrinos interact in the Standard Model only through their left-handed component, via electroweak interactions. However, the propagating neutrinos, which are mass eigenstates, are described by a field that is a Dirac spinor, i.e. with both chiralities $$ \nu=\nu_L+\nu_R. $$ Therefore, when neutrinos are created or measured, the Dirac spinor is projected ...


4

As Prahar said in a comment, dim. reg. allows us to see that two functions are equal in a half-plane, and if one of them is analytic in the whole plane, the other function may be analytically continued as well. So the volume of the spacetime is indeed zero in dimensional regularization. More generally, any power law divergence is set to zero in dimensional ...


3

In realistic QFT, fields or their interaction law mostly correspond to irreducible representations of some symmetry groups. If we assume free theory, there is only one important symmetry: it's Poincare symmetry (it is the most important symmetry - in flat spacetime each field theory must satisfy it). Poincare symmetry leads to the statement that free ...


3

The term virtual is used in other places in physics. For example in virtual images in a a mirror : we see an object in great verisimilitude, even ourselves. Why is the image called virtual and not real? Because it has the optical properties of the imaged object but not a large number of other attributes, mass being the simplest. In addition, its existence ...


2

Yes, you are right. The four momentum of a virtual photon needn't to lie on the mass shell. Thus the zeroth component of the four momentum of a virtual photon is independent of its spatial components. The reason for this is that the zeroth component of the four momentum of a virtual photon arises from the Fourier transform of the step function. See S. ...


2

When you introduce an auxiliary variable, such as a regularization parameter, at the end of the calculation you have to take the limit that sets the expression back to the original one. If you introduce multiple auxiliary variables, you have to do this for all of them. Otherwise you're just doing a different integral. In this case specifically, ...


2

Particles do not constantly appear out of nothing and disappear shortly after that. This is simply a picture that emerged from taking Feynman diagrams literally. Calculating the energy of the ground state of the field, i.e. the vacuum, involves calculating its so-called vacuum expectation value. In perturbation theory, you achieve this by adding up Feynman ...


2

There is no better definition than what Wikipedia offers - in general, a topological excitation is a (field) state, i.e. a localized quantity since fields depend on spacetime, whose integral is a topological invariant. One prime example are Yang-Mills theories in 4D, where the integral $\int \mathrm{Tr}(F\wedge F)$, as essentially the second Chern class of ...


2

I think I now have an answer. My problem was that I assumed how much neutrinos oscillated depended solely on their level of mixing. With that intuition it seems that neutrinos should oscillate significantly into their right handed counterparts. However, there is more to the story. Oscillations are also dependent on the difference of masses between the mass ...


2

I think, from the way you formulated the question, you lost the context of this trick, and then it indeed doesn't make a lot of sense. The point is that in QFT, you want to compute quantities corresponding to the full interacting Hamiltonian, $H$. In practice, however, we only know the eigenstates of the free Hamiltonian $H_{0}$: the plane waves ...


1

Maybe the answer is connected with the fact that propagator is the inverse of Lagrangian operator. An action of free theory may be written as $$ S[\psi ] = \int d^{4}x (\psi^{a})^{*}\Delta_{ab}\psi^{b}. $$ Here $()^{*}$ means conjugation which leaves $(\psi^{a})^{*}\Delta_{ab}\psi^{b}$-form lorentz-invariant. For example, $$ S_{KG}[\varphi ] = ...


1

The coefficient in front of the interaction terms is a bit arbitrary, in the sense that in principle, you could have written anything, for example $\frac{g}{N^2}$. The way to figure out the 'natural' numerical factor is by considering the renormalized coupling constant at tree level. $i g_\text{physical} = i (\frac{g}{2N}) \times \text{Sum of all X tree ...


1

What you want is that your function does not transform under Lorentz transformations that take $$ p^\mu \to {\Lambda^\mu}_\nu p^\nu.$$ To build invariants from one vector there is only the possibility to construct the invariant product with itself $$ p^2 \equiv p^\mu p_\mu \to p^\mu p_\mu.$$ There is one more thing though. The Lorentz group has two branches: ...


1

$E$ is not the energy per se of the particle, it is a Fourier parameter. What gives you the possible energy accessible to the particle is the spectral function $$\rho(E)=-{\rm Im}(G(E))\propto\delta(E-\epsilon_\lambda),$$ which is peaked at the accessible energy of the free particle $\epsilon_\lambda$.


1

In QFT it is not to get down to the ground state, but to choose a correct propagator (amongst variety of Green's function). In other words, it is applying or taking into account the boundary conditions. However, for "incomplete" systems, decaying may really mean getting down to the ground state due to interaction of some sort like irreversibly absorbing ...


1

Not really, at least not if you want to stay with properties you would normally associate to particles. That is because particles are not the fundamental objects of quantum field theories, but fields.1 There's more to the theory than charges and masses. For every symmetry group of the theory, a field must transform in a representation of it. Now, you can ...


1

Propergator do not have derivative. The interaction term is the vertex in Feynman diagram. In the following I use the notation convention of Sredniski. Peskin's convention would cause some addtional minus sign. For example, $\phi ^3 \partial^2 \phi$, the vertex is $\langle 0| \phi ^3 \partial^2 \phi | k_1 k_2 k_3k_4 \rangle = \partial^2_4 \langle0| ...



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