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For relativistic fermions in 2+1 and 6+1 (mod 8) dimensions a Dirac mass changes sign under time reversal. This is because we do usually do time reversal using the matrix ${\mathcal T}$ that obeys $${\mathcal T}\gamma^\mu {\mathcal T}^{-1} = (\gamma^\mu)^T$$ and setting $${\mathfrak I^{-1}} \psi(x,t) {\mathfrak I}= \eta_T {\mathcal T}\psi(x,-t).$$ Here ...

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Don't know if anyone is still expecting a reply to this, but here is how I managed to find the result. Peskin&Schroeder found result (3.50) by doing the calculation in a specialized frame (boost along z-axis) which resulted in (3.49). If you apply the high energy limit in that same frame by setting E = p3 and rewrite it in a covariant form (using a dot ...

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If one measures an observable of some system, the wave function collapses to an eigenstate of the Hamiltonian of this system. The Hamiltonian does not change under the measurement. The resulting wave function is an eigenfunction of the Hamiltonian. Thus, for your question, $\phi$ solves $H$ after the measurement since $\phi$ is an eigenfunction of $H$. In ...

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These[1][2] review papers contain good introductions to fractons. Generally, there is no uncertainty principle for such systems. This is because these fractons are usually emergent particles. For example, we can think of a domain wall excitation in a 1D Ising model as a particle, but this does not have any uncertainty principle associated to it. The ...

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The notion of the virtual particle originates from the quantum mechanical old-fashioned perturbation theory. In the latter, time is a dedicated quantity. Within this theory, the matrix element of the transition $|i\rangle\to |f\rangle$ behaves as $$\mathcal{M}_{i\to f} = \overbrace{V_{if}}^{\text{1st order}} +\overbrace{\sum_{n \neq i}\frac{V_{in}V_{nf}}{E_{... 3 A short answer is: virtual particles come as a convenient and pictorial description of a perturbative expansion in QFT. As such, they do not depend whether it is a relativistic or non-relativistic QFT. by definition, it is not possible to observe a virtual particle. However it is always possible to interpret some experimental findings in term of them. But ... 1 I think it is easy to say that the charge is conserved, but might be hard to measure. The cart is before the horse in this sentence. It is because the measurements completely agree with charge conservation that charge conservation is imposed in the mathematical models describing particle data. Or lepton and baryon conservation. At elementary particle ... 0 A recent book by François Gelis is in my opinion a fantastic book on QFT. The first half of the book is all the basics up to and including non-abelian gauge theory. While the second half introduces unrelated modern concepts in each chapter. For example it has a very nice review on the spinor helicity formalism, etc Lots of exercises and solutions at the end. ... 3 Actually... There is some evidence of CFTs in d>6. In [1] they construct a solution in AdS_8 implies the existence of a CFT in d=7. This is not a definitive answer because there are still some issues about the solution. One has to prove full nonperturbative stability and also there is a region in spacetime where the coupling becomes big and one has ... 4 I would say that the claim that there are no non-trivial CFTs in d>6 is just a speculation for which there isn't much evidence. The belief is that above six spacetime dimensions, the only unitary CFTs are simply free theories and that all non-trivial fixed points can be described by mean field theory. In addition to what you said, that there are fewer ... 1 It is not clear what do you mean in statement "QED is very predictive...": QED does not contain any particles except electrons & photons. QED describes only electron-photon proccesses and when you include additional particles (for instance, muon) you obtain a different theory (strictly speaking). In Standard Model (SM) there is the following ... 2 Good question! Their is lot to unpack to so I will do my best here. I think I can give you a global answer (or try to anyway), but a true answer to your question would involve knowning what process you are looking at. I am going to assume that you are thinking about W_\pm/Z-boson mediated interaction and not a more complicated process with W boson loops ... 0 For what it is worth, the seagull term in scalar QED$$ \exp\left(i\Gamma_{\text{1-loop}}[\phi_{\rm cl}\!=\!0; A_{\rm bg}]\right)~=~\int \!{\cal D}\phi~\exp\left[i\int\!d^dx\left\{-\left|\left(\frac{1}{i}\partial_{\mu} - qA_{{\rm bg},\mu} \right) \phi\right|^2 - m^2 |\phi|^2\right\}\right] $$has a formal analog in the Hamiltonian worldline (WL) formalism$$\...

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In learning about the duality of quantum particles, I wonder if a quantum wave stretches out into the distance, essentially forever? Not necessarily although in many cases: yes. Consider the educational case of the particle in a 1D box (or a 2D or 3D box, for that matter) Here the particle is strictly confined to the domain $[0,L]$ with $\text{zero}$ ...

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At first sight the consequences for physics would be earth-shattering: they have showed that there exists a Bell inequality for which the commuting Tsirelson bound is strictly larger than the tensor-product Tsirelson bound, so the commuting bound cannot even be approximated by finite-dimensional systems. If we could reach the commuting Tsirelson bound ...

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Gauge bosons are part of the quantum mechanical models built in order to fit an enormous amount of data gathered with particle physics experiments. In that sense they have been measured in experiments by that data since the establishment as mainstream of the standard model of physics. All the particles in the table are considered measured, because with out ...

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Mass renormalization (in QED) does not involve virtual electron-positron pairs. It involves the emission and reabsorption of a virtual photon by an electron in motion. This is a quantum-mechanical interaction between the electron and the electron’s own electromagnetic field. The intuitive idea of mass renormalization actually can be explained without ...

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It is very good and deep question. Let me try to answer and clarify some subtleties. In QED we have: $$e^2_{phys}(\Lambda) = e_0^2\left(1-\frac{e_0^2}{6\pi^2}\ln \frac{\Lambda}{m_0}+\dots\right) = \frac{e^2_0}{1+\frac{e_0^2}{6\pi^2}\ln \frac{\Lambda}{m_0}}$$ $$m_{run}(\Lambda) = m_0\left(1+ \frac{3e_0^2}{8\pi^2}\ln \frac{\Lambda}{m_0}+\dots\right) = m_0 \... 0 Indeed non-stardard interactions can modify neutrino mixing. Here's one paper that discusses a few types of NSI and their effect on the solar survival probability. https://arxiv.org/pdf/1305.5835.pdf I'm not personally aware of any paper that discusses the affect of Majorons in particular on neutrino mixing though. 0 In short yes. I believe this idea arose when people were thinking of the symmetry breaking from a bigger symmetry group (for example some hypothetical GUT scale group), down to the standard model as the universe cooled down. Morally speaking, the same story of the Higgs mechanism applies. That is we have a theory with symmetry group G in which the vacuum ... 0 Note that the transformation \xi\rightarrow-\xi corresponds to a change of q_{in}^2\leftrightarrow q_{out}^2 (Eq. (4)), which means that p_1\leftrightarrow -p_3 (Eq. (1), the minus sign is needed to have momentum conservation), which changes (p_1+p_2)^2\leftrightarrow (p_2-p_3)^2 in other words s\leftrightarrow u. Also remember that s+t+u=0 ... 0 P&S are considering all (possibly disconnected) diagrams that contribute to the 2-point correlation function G(x,y) with the condition that the 2 external points x and y are assumed to always be connected, i.e. belong to the same connected component. Here (V_i)_{i\in I} are all possible connected vacuum bubbles. The equation mention by OP then ... 0 All theories are incomplete, but it does not mean they all need renormalizations. See, for example, this explanation here. 1 One way to understand the concept of a UV completion is in terms of fixed points in the space theories generated by the renormalization group. The RG flow, interpreted as a course-graining procedure, generates a vector field that is determined by the criterion that as we move along the vector field flow, we obtain theories that better and better describe the ... 0 Let us start from QED. In this theory, vacuum polarization in the lowest order is given by one-fermion loop correction to bare photon propagator (note that photon has zero mass). It is quite straightforward (or I can provide a derivation) that this correction modifies charge but not mass. For me, physical ground becomes clear after calculation. If you ... 0 The amplitude for the decay is:$$M = \cfrac{g^{αβ}g^2 v}{2}ε^{*}_{α}(p)ε^{*}_{β}(q) $$You can obtain the vertex factor from the assosiated Lagrangian. ε here are the polarisations of the outgoing spin 1 bosons and p and q denote the momenta. You want to calculate:$$Γ = \frac{p_{f}}{32π^2 m_{H}} \int |M^2|dΩ$$For the M^2 you use the fact that W... 2 It depends on what inner product you are using to define the \dagger operation. If you are including integrals over x as part of the inner product, then (\partial^\mu)^\dagger = - \partial^\mu,just as it is with the momentum operator in QM. If \phi is an operator on a Hilbert space and you are only taking the adjoint on that space, then the \... 1 I think you must be right, unless I'm also misunderstanding something, this doesn't even hold for the other off-diagonal elements in the Gram matrix at level 2 and 3, e.g. \langle{h}|L_1^2 L_{-2}|{h}\rangle = O(h) at level 2, \langle{h}|L_2 L_1 L_{-1}^3|{h}\rangle = O(h^2) and \langle{h}|L_3 L_{-1} L_{-2}|{h}\rangle = O(h) at level 3, (your nice ... 3 The answer is yes: in general, it is possible to have operators sharing the same scaling dimension and spin. However, this is a very peculiar situation, and in practice it only happens when there is some kind of symmetry in the system. For instance: The linear sigma model in d = 2 dimensions (see the other answer by Prof. Legolasov); in this case the ... 1 The expression for \lambda in terms of \lambda_R is just perturbation theory. Given an expression for \lambda_R in terms of \lambda$$ \lambda_R = \lambda + \frac{\lambda^2}{32\pi^2}\ln\frac{s_0}{\Lambda^2}+... $$we solve perturbatively, assuming that \lambda can be written as a power expansion in \lambda_R. That is, we guess a solution of the ... 0 I haven't seen the lecture, but if you take the time axis to be the same in each diagram, the processes are not equivalent, and you cannot "rotate" one diagram into the other. The key is that although space and time both are a part of Minkowski space and "mix" under Lorentz transformations, they are not entirely equivalent: there is no ... 3 First off, I'd want to point out that most to all the complicated mathematical difficulties that go into "in practice" quantum field theory are really more to do with trying to deal with fields that are interacting with each other, and thus don't really have so much to do with how and what I believe you're asking about, which is to get to know the ... 2 A decay doesn't have a cross section associated with it, only a decay width. Most likely they are referring to the production cross section for the process \rm X+X\to H\to W^+W^- at some specific collider. Note that this depends on the identity of the initial particles and their momenta. It can also depend on other properties such as the polarization of ... 6 Simplest example that comes to mind: a linear sigma model (the kind you encounter in bosonic string theory). Its central charge c is a positive integer and there are c different primaries \partial X^{\mu} (\mu ranging from 0 to c-1). These primaries are different fields, but they all have conformal weights of (1, 0) hence the same scaling ... 2 Since that is a disconnected diagram, the issue is not really with the two-point function, but with the one-point function. The diagram you have drawn obviously just factorizes into two of the problematic one-point functions. The issue is that \phi^{3} theory is unstable, and that means that zero-momentum particles can disappear. (Note that the one-point ... 2 According to the answer linked above, I think that quantum fields, i.e. operator-valued distributions on the spacetime manifold, are supposed to be the objects representing an observable, in comparison to quantum theory where observable are represented by self-adjoint operators. This is a misconception. quantum fields in particle physics , they are used ... 8 A little caveat: The quantum fields are a way of organizing the generators of the algebra of observables. They might not be observables themselves. The actual observables are things like \phi(f), rather than \phi. Regarding your main question: You are correct. States are unit vectors in the Hilbert space upon which the observables act. They are not ... 0 @mike stone 's answer is impeccable, but requires emphasizing that the answer therefore depends on the representation of the Higgs multiplet under that group. For most of the groups used in physics, this was done in a legendary paper Group theory of the spontaneously broken gauge symmetries, Ling-Fong Li (1974), Phys Rev D9 1723, routinely used as a "... 1 That part follows from the condition that the residue at the pole should be -i (or one, depends if you include the i in the numerator of the propagator). In fact it's easy to see that defining$$\Pi(q^2) = \Pi_2(q^2)-\Pi_2(0)$$one directly gets that, at the pole$$q^2=0 \to \Pi(0)=0$$and so that the residue at the pole is$$\underset{q^2=0}{\text{Res}}\...

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There are pro-forma classical solutions available mixing on one side of potential step particles with, on the other, antiparticles. Such situation implies the need for understanding of what is the physics problem one is solving. This situation was addressed and studied in depth in early-mid seventies and detailed studies were published as a part of a Physics ...

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You can start from Schwinger parametrization, $$\frac{1}{k^2+r}=\frac{1}{2}\int_0^{\infty}d\alpha\,\exp\left(-\frac{\alpha(k^2+r)}{2}\right),$$ which gives you $$G(x)=\int_0^{\infty}d\alpha\int_k\exp\left[-\frac{\alpha(k^2+r)}{2}+ik\cdot x\right],\quad\int_k=\int\frac{d^dk}{(2\pi)^d}$$ and then just complete square, perform simple Gaussian integration and ...

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You're supposed to think of $\mu^2$ as a parameter, and there's no need to consider if it $>0$ or $<0$. You proceed by minimising the potential and then seeing that the nature of vacuum/vacua is different for $\mu^2>0$ and $\mu^2<0$. As @CosmasZachos mentioned in the comments, it is certainly a function of $T$. The exact function can be ...

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As @knzhou noted, we first Wick-rotate so $\color{blue}{k}\in\Bbb R^4$ is Euclidean. Then $\int_{\Bbb R^4}f(\color{blue}{k}^2)d^4\color{blue}{k}=2\pi^2\int_0^\infty f(\color{red}{k}^2)\color{red}{k}^3d\color{red}{k}$, where $\color{red}{k}\in[0,\,\infty)$ is the radius of $\color{blue}{k}$, and the proportionality constant is the solid angle in $4$ ...

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In general, such cutoffs only have meaning when you Wick rotate to Euclidean signature. In that context, $k$ is a vector in $\mathbb{R}^4$, and the condition is that its magnitude is less than $\Lambda$.

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Fields in quantum mechanics are not measurable quantities, and also wavefunctions are not measurable. What are measurable are interactions between elementary particles or composites of elementary particles, like atoms and nuclei. Fields and wavefunctions are mathematical representations that allow a predictive modeling of these interactions . The predictions ...

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Now, my hypothesis is that the above infinities should go away when you consider a more realistic measurement scenario, like measuring the average value of the field in some small region. When it come to treating quantum field as "the average value of the field in some small region", two guys named H Epstein and V Glaser beat you to the punch. In ...

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Here's my shot at an explanation—would love anyone's feedback. In the above, I argued that since we normalized the fields to create a single particle state in the free theory (and satisfy the canonical commutation relations), we should expect them to do so in the interacting theory as well. As far as I can tell, this is just wrong. It assumes that the ...

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Using the representation of the Kronecker delta $$\delta_{l,l'} = \frac{1}{N} \sum_{n=1}^N e^{2\pi i (l'-l) n/N}$$ we have $$\sum_n \sum_{l'} \alpha_{l'} e^{-i(k_l-k_{l'})na} = \sum_{l'} \alpha_{l'} N \delta_{l,l'} = N \alpha_l$$ and $$\sum_n \sum_{l'} \alpha^\dagger_{l'} e^{-i(k_l+k_{l'})na} = \sum_{l'} \alpha^\dagger_{l'} N \delta_{-l,l'} = N \alpha^\... 2 It seems relevant to mention the Nambu-Goto action for p-branes$$S_{NG}[\phi]~=~\int_{\Sigma}{\rm dvol} (\phi^{\ast}(g)),\tag{A} which is the volume of the pullback $\phi^{\ast}(g)$ of the target space metric to the $p$-brane world volume $\Sigma$ of dimension $p+1$. In local coordinates, the Lagrangian density is the square root of the determinant of ...

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Each column of Φ is an $SU_L(2)$ doublet, since $U_L$ only scrambles rows. Each row of Φ is an $SU_R(2)$ doublet, since $U_R$ only scrambles columns. N.B. In response to your comment, $\Phi \sigma_x =\begin{bmatrix} c & a \\ d & b \end{bmatrix}$. So the first column has been interchanged with the second one, rigidly; in that sense it has been ...

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