# Tag Info

0

I just realized that the answer is just stupid. In the case of two outgoing photons when computing $M^*$ you get the epsilons (not complex conjugated) that in the full $|M|^2$ will allow you to use $\sum\epsilon_{\mu}\epsilon^*_{\nu}\to-\eta_{\mu\nu}$

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The Universe is infinite. If the energy density of the universe is greater than zero, then the total energy must also be infinite.

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According to Grand Unification Theory, protons can decay into electron (even at low energy; just the probability is very low). It doesn't mean you can replace proton with electron.

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One comes to this conclusion due to the fact that the contraction of a symmetric tensor with an antisymmetric one vanishes. Writing down the loop diagrams involves a contraction of both vertices. If you get expressions proportional to $\epsilon_{\mu\nu}\eta^{\mu\nu}$, this will be zero due to the fact that the metric is symmetric and the epsilon tensor is ...

0

The unrenormalized vertex diagram is the dimensionally regularized Feynman integral corresponding to the following single Feynman diagram: It is called ‘unrenormalized’ because it is not accompanied by a diagram in which the momentum loop is replaced by a counterterm vertex generated by the QED Lagrangian. Adding a second Feynman diagram with the ...

0

Define $a=\frac{\sqrt2}{2}(c+d)$, $b=\frac{\sqrt2}{2}(c-d)$. Then the Hamiltonian is diagonalized in this representation. The detail is omitted. Try it yourself.

1

I believe that this may be what you’re looking for. I don’t claim to be an expert in renormalization, so all members of the Physics Stack Exchange Community are welcome to correct my answer. To make the relationship between the two more precise, let’s call the on-shell renormalization scheme the ‘on-shell subtraction scheme’ and the BPHZ renormalization ...

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It depends who you ask. A string theorist would answer that string theory is the idea that the point-like particles of elementary particle physics can also be modeled as one-dimensional objects called strings. According to string theory, strings can oscillate in many ways. On distance scales larger than the string radius, each oscillation mode gives rise to ...

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No. The atoms are protons, electrons and neutrons. The fact that neutrons beta decay into a proton + electron + electron antineutrino does not mean that neutrons are made of a proton and electron and a neutrino.

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No. The decay products of a certain particle are not equivalent to its constituents. This is evident especially in the context of fundamental particles: quarks can decay into other particles, but that does not mean that a quark is not elementary (see my answer to this question). Nuclei are made of neutrons and protons, which in turn consist of quarks and ...

0

User Hindsight has already given a correct answer, but let us here try in different words. I) Given the classical field $$\tag{1} \phi~=~\phi_{-}+\phi_{+},$$ let us assume that the quantum field is of the form $$\tag{2} \hat{\phi}~=~\hat{\phi}_{-}+\hat{\phi}_{+},$$ where the creation/annihilation parts $\hat{\phi}_{\pm}$ do not commute. II) Our ...

-1

For a real particle to be off-shell, it is sufficient to be in an external field of some sort. It is not necessary to be "absorbed". "Absorbed" are virtual particles, for example, "virtual photons" who describe non propagating fields like a Coulomb one. For a Compton scattering, the real electron is "off-shell" during interaction with the real photon, i.e., ...

2

What Qmechanic said in comments is pretty solid, "Lagrangian (2) is not bounded from below because the kinetic term of $A_0$ field has the wrong sign, and hence the theory is not physical in the first place", but I think your Question needs a change of emphasis. Your Lagrangian allows us to construct four equations of motion for four non-interacting fields. ...

3

Confinement, i.e. the fact that bound states in QCD are colour singlets (or in other words: there are no free quarks or gluons) is experimentally well-established. There is no doubt that it is not just a prediction of lattice simulations, but a feature of strongly interacting systems at low energies, as no free quarks (or gluons) have yet been found within ...

0

Field $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$ with a given spin and mass (i.e. field which transforms under irrep of the Poincare group) must satisfy some determined conditions called irreducibility conditions: $$\tag 1 \hat{W}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = -m^{2}\frac{n + m}{2}\left(\frac{n + m}{2} + ... 1 You can think of the continuos formalism as being the limiting case of the discrete-momentum one: if the momentum is taken as a discrete variable (which amounts to constraining the particles to be in some finite volume V) the fourier expansion of the (real, scalar) field is:$$ \tag{1} \varphi(x) = \sum_{\textbf{k}} \frac{1}{\sqrt{2V \omega_{\textbf{k}}}} ...

1

The right way to deal with functions of operators is via the Spectral Theorem. If you have a self-adjoint (or even normal) operator $T$, the spectral theorem states that there is a resolution of the identity $E_T(t)$ such that the operator can be written as $$T=\int_\Bbb{R} t dE_T(t)$$ So, given a measurable function $f$, not nedded to be analytic, we define ...

2

Hint: 1. $\phi(x,t)$ at different times are not independent. 2. $\int{d^4p\delta(p^2-m^2)}=\int{d^4p\frac{\delta(p^0-E_p)}{2p^0}}$. The left side of this equation is Lorentz invariant. This time your question is much clearer. If $\phi(x)$ is an arbitrary function of $x$, there's nothing confusing. If $\phi(x)$ is constrained by the Klein-Gordon ...

0

A function of $p_{\mu}$, $T^{\mu\nu...}p_{\mu}p_{\nu}...$is invariant under Lorentz transformation if and only if $T^{\mu\nu...}$ is a invariant tensor. However, there is only one invariant tensor for Lorentz group. That is $\delta^{\mu}_{\nu}$. Thus, each saclar function of $p_{\mu}$ must be a function of $p^2=\delta^{\mu}_{\nu}p_{\mu}p^{\nu}$. ...

2

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: ...

0

Good question. There are two ways of defining operator functions, which often coincide. 1) As you said, given an analytic function $V(x)=\sum_n V_n \, x^n$ of the scalar $x$ which converges for all $x$ (as $V$ is analytic), you can define the corresponding operator for that potential using the same series expansion, but now with the operator series ...

2

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. $i g_\text{physical} = i (\frac{g}{2N}) \times \text{Sum of all X tree ... 0 Exactly as you mention in the second part of your post. Leptogenesis takes place at temperatures higher than the electroweak scale$T_{EW}\simeq100\,\mbox{GeV}$. Therefore, heavy neutrinos decay into lepton and higgs doublets, whose fields are still all physical. 1 The point is that there is no mass difference between left-handed and right-handed neutrinos. The mass is only defined for the complete field$\nu=\nu_L+\nu_R$. For this reason, I think it's not correct to use the oscillation probability formula to compute oscillations between LH and RH neutrinos. 2 I think I now have an answer. My problem was that I assumed how much neutrinos oscillated depended solely on their level of mixing. With that intuition it seems that neutrinos should oscillate significantly into their right handed counterparts. However, there is more to the story. Oscillations are also dependent on the difference of masses between the mass ... 3 There is no better definition than what Wikipedia offers - in general, a topological excitation is a (field) state, i.e. a localized quantity since fields depend on spacetime, whose integral is a topological invariant. One prime example are Yang-Mills theories in 4D, where the integral$\int \mathrm{Tr}(F\wedge F)$, as essentially the second Chern class of ... 6 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 ... 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 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 ... 0 Your argument is valid for unitary evolution. However turning the time into the complex plane you make it non-unitary. You may say that you introduce small decay for every state $$e^{-iHt}=e^{-iHt-\eta Ht}$$ with the ground energy state the slowest to decay (stable if you set$E_0=0$) 0 The$x$in the measure$D\phi(x)$of the functional integral$\int D\phi(x) F(\phi(x))$is a dummy variable that gets integrated over implicitly, so, no, you should not integrated it again. In fact,$D\phi(x) \propto \prod_x \Delta \phi(x)$, assuming a discretized$x$, where$\Delta\phi(x)$represents the difference between two (close) instances of the ... 0 It is not clear if you are looking for textbooks specifically dedicated to scattering in QFT (the title says so, but then you make reference to Peskin which is not), so going for good expositions inside more generally broaded textbooks: Chapter 17 of Robert D. Klauber, Student Friendly Quantum Field Theory (also see here): Aside for a summary of ... 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, ... 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 ... 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$\epsilon_\lambda$. 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 ... 0 I think that we may add that in the confining phase, the QCD-string descritpion of quarks (say, mesons, which are bound states of quark/anti quarks) is that these particles sit at endpoints of "QCD-strings" (I use "" to distinguish this from the normal superstring which is a well defined object, though it failed for the moment at describing exactly ... 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| ...

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 $u$ & $d$ 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.

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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 ...

0

If $\epsilon>0$, you can start with a small imaginary part in the denominator of the integrand, $\Delta - i\delta$. After finishing the calculation, you take the limitation $\delta \rightarrow 0^+$. Then the result will be $\displaystyle{\int \dfrac{d^{4-\epsilon}l}{(2\pi)^{4-\epsilon}}\dfrac{1}{(l^2 - \Delta + i\delta)^2} = ... 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. ... 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 ... 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 ... 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 ... 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 ] = ... 0 I stumbled upon a pdf, in another question here, where it is stated that a term of the form$( \partial_\mu \Phi )(\partial^\mu \Phi)$is forbidden for spinors, because it leads to a hamiltonian that is unbounded from below. I will update this answer as soon as I have investigated this any further 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. 0 I find things are clearer using the dotted and undotted spinor notation. The L-spinors$\chi_{L}$are dotted vectors$\chi^{\dot{A}}$and the R-spinors$\xi_{R}$are undotted vectors$\xi^{A}$with index$A=1,2$. The parity operator has to be a tensor$P^{\dot{A}}_{B}$and another tensor$P^{A}_{\dot{B}}\$ in order to change the way each type of spinor ...

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