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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.


4

Virtual particles refer to actual, nonzero features in the quantum fields of real objects, but they are features that are not particles in many ways so you should not expect anything from their being named "particle". Basically, the idea of virtual particles was invented as a device for when you want to hold on to the particle picture while doing quantum ...


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 ...


3

It's Lie theory. When we have got a continuous symmetry, the means the symmetry group is a Lie group $G$. Saying that the Lagrangian $\mathcal{L}$ is invariant under $G$ means that $\rho(g)(\mathcal{L}) = \mathcal{L}\;\forall\; g \in G$, where $\rho$ is the representation the Lagrangian transforms in. The "expansion to first order" physicists so often do ...


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 ...


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

Conservation of helicity requires that $$ \sum_{i=1}^n h_i = 0 $$ in any amplitude. If the above equation is not satisfied, then conservation of helicity requires that the corresponding amplitude vanish. Maximally violating helicity conservation implies that the above sum take its maximum possible value with its corresponding amplitude being non-zero. Now, ...


2

It basically boils down to the term $e^{-\frac i\hbar E t}$, where the minus can either be included in the energy, making it negative, or into the time. But a negative charge moving backwards in time is exactly the same as a positive charge moving forwards in time, and that is much more sensible than negative energy.


2

The $\psi_i$ in your sum do not need to be delta functions. You can think for example as them being energy eigenfunctions $$ \mathcal{H} \psi_i(r) = E_i \psi_i(r) $$ thus creating a particle at $r$ means that you obtain a superposition of all the possible ways a particle can be at $r$ (in this particular choise of basis): $$ \underbrace{\psi^\dagger ( r ...


2

Virtual particles are not observable by definition. They represent "internal lines" in Feynman diagrams. For example, this diagram: Here two electrons move toward each other, interact, then move away from each other. The external lines represent "real" electrons which we can measure/observe. The internal line here is an excitation of the electromagnetic ...


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| ...


1

The following was meant to be a comment rather than an answer. However, since it was a bit long for a comment so I am writing it in the answer box. In the case of a field theory, states can be thought of as functions on the space of boundary conditions on a spatial slice. This is so because the space of boundary conditions on a spatial slice is the ...


1

I would recommend the first chapter of Scully and Zubairy, "Quantum Optics". Maxwell's equations are to the photon what the Dirac equation is to the electron. Indeed, one can write down Maxwell's equations in a form that is identical to a zero mass Dirac equation. In quantum optics the electric and magnetic field vectors become vector valued quantum ...


1

Either you incorrectly think that $1\cdot U=1$ or more probably have forgotten $U$ in the first term: $$P_L' \Psi' = \frac{1}{2}U\Psi^{{\mathrm{Weyl}}}-\frac{1}{2}\gamma_5'U\Psi^{{\mathrm{Weyl}}} = \frac{U-U i \gamma_0 \gamma_1 \gamma_2 \gamma_3 }{2}\Psi^{{\mathrm{Weyl}}}=U\Psi^{{\mathrm{Weyl}}}_{{\mathrm{left}}}=\Psi'_L$$


1

First, caveat -- I am still in the learning phases of QFT. Math skills used and needed: Linear Algebra, vectors in Hilbert Space, Hamiltonians, Lagrangians (just like regular QM). Tensor notation, 4-vectors, special relativity, metric tensors at times. Feynman Path Integrals. Calculus of Variations. Fourier Analysis. And, certainly this list is not ...


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

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

Think of it as a change of basis. $a_i^\dagger$ creates a particle in the state $|i\rangle$. Now, this state $|i\rangle$ can be written in terms of the position states $|r\rangle$ as $$ |i\rangle=\int dr\, \psi_i(r)|r\rangle,$$ thus creating a particle in this state is equivalent to create a particle in a superposition of position state with the ...



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