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When beginning by calculate transition amplitudes in position space, and taking the Fourier transform of these amplitudes, to get the transition amplitude in momentum space, you get terms (for instance in a $2 \to 2$ interaction) in $\int d^4v e ^{-i(p_1+p_2-p_3-p_4)v}$, and this is equals to $(2\pi)^4 \delta^4(p_1+p_2-p_3-p_4)$ An example of such an ...

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I suppose you talk about the standard $2\pi$ that appears in the rules for Fourier transform. The factor of $2\pi$ or $1/2\pi$ or two factors of $1/\sqrt{2\pi}$ have to appear "somewhere" in the Fourier transform rules because this is what the mathematics implies. At any rate, if this is your question, it is a mathematical question and you may learn it in ...

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As Trimok said, the probability of scattering of some nicely focused packets will still go like the cross section and like $|{\mathcal M}|^2$. For bosons, there are no energy-dependent extra factors, so $|{\mathcal M}|^2$ itself has to be smaller than a number of order one for the probability to stay smaller than one. This is related to $T^\dagger ... 1 I think there is a long list of things we know about but don't understand yet. For example, we know neutrinos have mass but we don't yet know what those masses are or even exactly how they acquire mass (but it's probably the Majorana process). There is a long list of things that we're still trying to figure out but that's not the central part of my point. ... 0 [Caveat emptor: this is slightly speculative suggestion from a position of relative ignorance.] There's also another scale in the game in "ordinary" AdS/CFT: while$\lambda$sets string length,$N$sets Planck length. Large$N$suppresses quantum effects, while large$\lambda$suppresses stringy effects. Stringy (higher derivative) effects have no obvious ... 7 Note carefully Nick's comment. Suppose I send two plane EM waves on some collision course so they interfere. The waves will pass through the region where they meet, generating some interference pattern in that region, then they will exit that region and continue on their separate ways unchanged. In other words neither the energy nor the momentum of the waves ... 3 There is no need for high order mechanism. It is simply because a single photon can interfere with itself. If you remember the double slit experiment, they are indeed looking for a single photon passing through a slit and interfere with itself. Now if, instead we have billions of billions photons, the same single photon interference still happen ... 0 The universality of the coupling of the photon to charged particles exhibited by this formula is only valid in the limit of ultrasoft photons. This is also known as the eikonal approximation, in which the photon couples only to the charge x velocity of the charged particle. 2 Both$\mathbb{R}^3$and$S^3$are rank 1 symmetric spaces explicitly, as a homogeneous spaces they are given by: $$\mathbb{R}^3 = ISO(3)/SO(3)$$ and $$S^3 = SO(4)/SO(3)$$ The significance of their being rank-1 symmetric spaces is that there is only one "two-point" invariant on them, i.e., any function of two points$r_1$and$r_2$invariant under the ... 0 The general two-particle state will look like$\displaystyle \int dp_1 dp_2 \psi(p_1,p_2) a^\dagger_{p_1} a^\dagger_{p_2}| 0\rangle $Here$\psi(p_1,p_2)$is the momentum-space wavefunction. Since the creation operators commute, only the symmetric part matters, so we may as well take$\psi(p_1,p_2)=\psi(p_2,p_1)$(there would be a minus sign if they were ... 3 If basic symmetry and homogeneity assumptions about the Universe hold, then yes, all massless real particles (see Anna V's answer for virtual particles must travel at a universal constant$c$, the speed of a massless particle, in all frames of reference. Given these basic symmetry and homogeneity assumptions, one can derive the possible co-ordinate ... 5 Another way to say this: Speed of photon, graviton, gluon all equal to c? or Whether all massless particles necessarily have the same speed? You must not have been introduced to the concept of a virtual particle: In physics, a virtual particle is a transient fluctuation that exhibits many of the characteristics of an ordinary particle, but that ... 1 I thought the same thing for a long time. I wondered why gluons don't fly out of the nucleus at the speed of$c$. The difference is that photons don't interact with other photons and gravitons don't interact with other gravitons. They can move around and pass through each other. On the other hand, gluons do interact with each other. In fact, gluons form ... 3 I think the answer is it depends on distance (relative to the size of your system). Another well known example of a boson which is comprised of fermionic components is the helium-4 atom, which has integer spin (both the nucleus and the neutral atom itself). Fermionic or bosonic behavior of a composite particle (or system) is only seen at large (compared ... 1 I can answer (1) and (2). The answer is: NO. Passing form classical mechanics to quantum one requires, in general, to add more information. There is no rigorous machinery allowing one to write the quantum corresponding of a classical object. Physically speaking, this is because quantum structures are more fundamental in Nature than classical ones. ... 0 You have to put the first term (which in not in normal order, hence the dots) in normal order. The vacuum mapping is the identity matrix so you simple CAN put it. 0 Consider the non-abelian phase factor around a closed path$C$, $$\psi(C) = \mathrm{P} e^{\oint A_\mu dx^\mu} = \mathrm{P} e^{\int_0^{2\pi} dt \, A_\nu(x(t)) \dot{x}^\nu(t) }$$ Let us take the functional derivative with respect to$x^\mu(s)\begin{align} \frac{\delta}{\delta x^\mu(s)} \psi(C) %& = \int_0^{2\pi} dt \, ... 2 Your interpretation is not correct. The propagatorD_{\mu\nu}(x-y)$describes the amplitude for a photonic field perturbation to go from$x$to$y$, with the implicit picture that you have a "source"$J(x)$, and a "sink"$J(y)$, which are perturbing the vaccuum. However, a field perturbation is not a real particle (for instance, in the photon case, the ... 1 Try, for instance, section 9 of Srednicki. The way to do it is to replace the fields in the interaction Lagrangian by functional derivatives with respect to the sources, then write power series for the exponents. Take the first order contribution. Then, use that you need to consider three-point functions where the fields are again replaced by functional ... 1 I'd like to add a really elementary level illustration of some of David Bar Moshe's answer when he says: ...They give two major examples of non separable Hilbert spaces: 1) The infinite tensor product of harmonic oscillators, ... The example below is clearly well below the level the OP is looking for, but hopefully it shows to a wider audience what a ... 1 I find this much better motivated in Weinberg's text "Quantum Theory of Fields", which starts from the idea of particles, and what we measure, rather than from a Lagrangian formulation. Essentially, what we are after is a Poincar\'e invariant and unitary S-matrix. One crucial thing that we require is a Hamiltonian density, transforming as a scalar, and ... 1 In fact, you have$\{\psi^a(x), \bar \psi_b(y)\} = 0$, as an operator, for a space-like interval$(x-y)^2 <0$(stricly speaking, this is a distribution, for instance, at$x_0=y_0$, this is the distribution$\delta^a_b \delta^3(\vec x-\vec y))$, together with relations$\{\psi^a(x), \psi_b(y)\} =\{\bar \psi^a(x), \bar \psi_b(y)\} = 0$Now, if you look ... 0 I wonder if there is not an error in the right hand side of$6.174$, where it would be more logic (to understand the following lines...) to have a$(n+1)h ~z^n$term instead of$(n+1)h$. If this hypothesis is correct (???), then it is just the application of the Lebnitz rule for derivatives, that is :$z^{n+1}\partial_z (z^{h_p-2h} f(z)) \\ = z^{n+1} ...

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From the viewpoint of the Wightman axioms, the separability assumption on the Hilbert space can be actually derived from a few of the other axioms if you adopt a formulation using $n$-point functions. The reasoning goes as follows. A (say, scalar) quantum field theory on $\mathbb{R}^d$ can be thought of as being specified by a sequence of $n$-point ...

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First, let me state that the notion of a Hilbert space is not fundamental in quantum theory. One can realize the same quantum physical system using different Hilbert spaces. This is because quantum states (which are really the objects which physically matter) are only weakly connected to vectors on a Hilbert space. It is true that pure states correspond to ...

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A short answer is that, to calculate process amplitudes, you have to take in account the interacting Lagrangian part, and use it to establish the rules to calculate Feynman diagrams. So, let's have just a taste of it. For QED, The interacting Lagrangian density term $L_{int}$ corresponds to expressions: $j^\mu(x) A_\mu(x) \sim \bar \psi(x) \gamma^\mu ... 2 [By statistical mechanics I mean classical statistical mechanics throughout this answer. If you are curious to think about the complications added with making the statistical side of the story quantum mechanical, that sounds like a very good exercise.] The analogy between Euclidean quantum field theories and equilibrium statistical mechanics is exact, once ... 2 If you are sure that$f$is continuous and does not vanish in the integration domain, it is by no means necessary making use of regularization theory of distributions. Consider the initial integral: $$F:= \int_0^1 \mathrm{d}x\int_0^{1-x}\mathrm{d}y \frac{1}{f(x,y)+\mathrm{i}\epsilon}.$$ It can be re-written as: $$F:= \int_{T} ... 4 You are right, in this case, scalar means Lorentz invariant field. But it is not invariant under the transformations of SU(2)xU(1) of the electroweak model. And it is a scalar under the SU(3) of QCD. So the four real components of the Higgs are indeed invariant under space-time transformations. Physicists are usually not very clear in these distinctions, ... 5 Spin 2 just means that the gravitational field is given by a metric field and general covariance, which is the nonlinear expression of a massless spin 2 representation of the Poincare group. The latter appears when linearizing around the Minkowski metric and dropping all interactions. See the classical paper by S. Weinberg, Phys.Rev. 138 (1965), B988-B1002 ... 2 "Schroedinger equation" unfortunately is a bit ambiguous word. It could refer to$$i\hbar\frac{d \psi(t)}{dt} = H_t\psi(t) \quad (1)$$but also to a more precise form like this:$$ i\hbar\frac{d \psi(t)}{dt} = \left(-\frac{1}{2m}{\bf P}^2 + V_t\right)\psi(t) \:.$$The former version does not depend on the quantum physical system you are dealing with. So, in ... 2 If you want an even more everyday example than Emilio Pisanty's example: "no scattering" would mean that the would be scattering object in question (modelled by the short range potential in Emilio's answer) would beget no change the the forward travelling wave. Otherwise put, an observer sensing the incoming plane wave could not tell whether or not the ... 1 The only way we know that quarks exist is by a series of deep inelastic scatterings with leptons and with protons. This is a reconstructed event at LEP quark antiquark at 12:00 o'clock and 4:00 o'clock gluon the third one. The lepton colliders have the advantage that most of the energy taking part in the collision can be detected in four pi ... 4 The Schrödinger equation you use in non-relativistic quantum mechanics describes the evolution of the wave-function for a single particle, or at least, a fixed number. So you can think of it as being the "wave equation" for a one particle wave-function. Nice, neat interpretation. Also incomplete. But the Schrödinger can also be viewed as the defining ... 4 Forward scattering need not be equivalent to "no scattering" - and, indeed, will only rarely be indistinguishable from it. In the usual scattering-theory setup, you have an electron coming in in a plane wave$$\psi(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}=e^{ikz}$$and impinging on some short-range potential. This will add to the wavefunction a scattered ... 1 There are two ways to deal with a linear term in \phi: Complete the square, as was suggested in the comments. This is very often possible, but sometimes you do not want to do that. Interpret it as an interaction term with a \phi particle popping out of the vacuum or vanishing. This will lead to non-zero tadpoles in your Feynman diagrams, so additional ... 0 Usually, Cronin effect is given in terms of the central-to-peripheral nuclear modification factor for dAu collisions at midrapidity$$ R^h_{CP}(p_t)= > \frac{(1/N^C_{coll})dN^h/p_tdp_t(C)}{(1/N^P_{coll})dN^h/p_tdp_t(P)} $$where C central, P reipheral, N_{coll} the average number of inelastic NN collisions.If hadronization is ... 2 Axial charge'' refers to the (isovector) axial coupling constant g_A of the nucleon$$ \langle p|A_\mu^a|p\rangle = g_A \bar{u}(p)\gamma_\mu\gamma_5\tau^a u(p) $$where A_\mu^a=\bar{\psi}\gamma_\mu\gamma_5\tau^a\psi is the QCD axial current, |p\rangle is a nucleon state with momentum p, u(p) is a free nucleon spinor, and \tau^a is an isospin ... 0 A partial answer, is that supposing the gamma matrices, block-diagonal , as \begin{pmatrix}A&\\&\epsilon A\end{pmatrix}, \begin{pmatrix}&A\\\epsilon A&\end{pmatrix}, where A is hermitian or anti-hermitian, and \epsilon =\pm1, give constraints on A and \epsilon due to (\gamma^0)^2= \mathbb Id_4, (\gamma^i)^2= - \mathbb Id_4. For ... 2 How about just testing the two different cases? I.e. if \mu\not=0 then the LHS becomes $$(\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i$$ while the RHS becomes $$(\gamma^\mu)^\dagger=\gamma^0\gamma^i\gamma^0 = -\gamma^0\gamma^0\gamma^i=-\gamma^i~~~~~~~~ (\text{OK}).$$ For \mu=0, the case ... 0 If you have free fields which obey the same equation, the propagators are the same. So these are these propagators that you are going to use in an interacting theory to establish Feynman diagrams. For instance, if you have an interacting term \lambda \phi_1^2\phi_2^2 in the Lagrangian, you will obtain, at order \lambda, a 4- point Green function as : ... 2 OK, let us start from your example. I think that it is too pathological to be considered as a safe starting point for this discussion, which is worth and interesting however. Nevertheless I would like to spend some words about this case since it permits to introduce some general issue useful in the second part of my answer. AdS_n is not globally hyperbolic. ... 0 Read the papers (and recent references citing them) by S. J. Summers and R. Werner in 1987: "Bell's Inequalities and Quantum Field Theory. I. General Setting," Journal of Mathematical Physics 28: 2440-2447, 1987; "Bell's Inequalities and Quantum Field Theory. II. Bell's Inequalities are Maximally Violated," Journal of Mathematical Physics 28: 2448-2456, ... 2 It seems that OP is pondering about the notion of supernumbers, and the generalization of Fubini and Tonnelli's theorems for integration over superdomains and supermanifolds. See e.g. this Phys.SE post and the references listed therein for details. Example: Consider the integral$$\tag{1} ... 3 Yes, you can show this using only the fact that the Clifford Algebra has a unique representation up to similarity transformation in any dimension. This is shown in the first few pages of http://arxiv.org/pdf/hep-th/9811101.pdf Then you observe that if$\gamma^\mu$obeys the clifford algebra, then so does$-(\gamma^\mu)^T$.$\mathcal{C}$is then defined as ... 7 Defining a Lie algebra by the commutation relations$[T^a,T^b]=if^{abc} T^c$, the adjoint representation is defined by$(T_{adj}^a)^{bc}= if^{abc}$. Now it turns, that in the special case of$so(3)=su(2)$, you have$f^{abc} = \epsilon^{abc}$, where$\epsilon^{abc}$is the totally antisymmetric tensor. So, your representation is the adjoint representation. ... 0 It looks like the following article is relevant: http://arxiv.org/abs/hep-ph/0402256 (published in Nucl. Phys. A). (The phrase you quote is probably from lectures http://www.physik.uni-bielefeld.de/~borghini/Teaching/HIC-Seminar/SoSe2013/Francois_SPhT2006-1.pdf by one of the authors of the article): "The Cronin effect was discovered in proton-nucleus ... 2 It is not an answer, but maybe some information which could be useful : In an other post, it has been also noticed that the commutators of the$R$-symmetry generators with supercharge generators are:$[R^a_b,Q^c_{\alpha}]=\delta^c_bQ^a_{\alpha}-\frac{1}{4}\delta^a_bQ^c_{\alpha}$So, taking the trace (on$a,b$), with$\mathcal N=4$, gives a null ... 5 Only the$su(4)$generators appear on the right hand side of the$u(2,2|4)$commutation relations, so superconformal invariance does not prevent an anomaly in the$u(1)$reducing the symmetry to$su(2,2|4)$. In$\mathcal{N}=4$SYM the central charge is furthermore zero, so the actual symmetry is$psu(2,2|4)$. The breaking of the generator with non-zero ... 2 I think 't Hooft and Kugo are solving different problems. 't Hooft addresses the issue that the anomaly involves a topological term. As a result, in perturbation theory there is no theta dependence and the anomaly equation by itself does not solve the$U(1)\$ problem. He shows that topological objects (semi-classically, instantons) generate theta ...

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