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Notation: I will write a Poincaré transformation as ${x'}^\mu = {\Lambda^\mu}_\nu x^\nu + a^\mu$, the operator representing this transformation on the Hilbert space is $U(\Lambda, a)$. An infinitesimal transformation with ${\Lambda^\mu}_\nu = \delta^\mu_\nu + {\omega^\mu}_\nu$ and $a^\mu = \epsilon^\mu$ can be expanded as $$U(\delta + \omega, \epsilon) = 1 ... 2 According to these lecture notes (page 18 in the pdf) and a few others I have found, the integral can in fact be evaluated and yields the modified Bessel function of the second kind, of order 1 (see Wikipedia). The integral does not seem converge, but I am not sure if you can say that it diverges, because it oscillates (with increasing amplitude). The way ... 2 This is a particular example of a general theorem in effective field theory: if you have an operator that is proportional to the lowest order equations of motion, you can push that operator to higher order in perturbation theory by a field redefinition. This is especially useful if you are working to a fixed order in perturbation theory, in which case you ... 1 You have found the reason that most 4D QFTs need to have renormalization: The propagator is (UV) divergent! To "cure" the theory, you need to regularize the theory (make the divergences expressible in some simple parameter like a momentum cutoff) - e.g. by dimensional regularization - and then proceed with some renormalization scheme. That the theory is ... 0 I don't think it is exactly the same regulator: In the first method, you integrate \int_{-\infty}^\infty dk^0 \int^\Lambda d^3k, but in the second calculation you integrate \int^\Lambda d^4 k_E. 0 You can't remove brackets |k \rangle\langle k| into the second method which you've reprsesented because there is e^{\frac{k^{2}}{M^{2}}} term; you have to make a summation over all k, which is completely equivalent to the first method (note also that |k\rangle  doesn't coincide with integration variable k) 4 \Pi(x) and \Phi(x) are not operators. They are operator valued distributions (distributions that when acting on the one-particle functions become operators). Therefore, it does not make sense to do their square since it is like doing the square of the \delta function. The normal ordering is a prescription on how to cure this ill definition, and obtain ... 0 I think the statement in Witten's paper, "Quantum Field Theory and the Jones Polynomial" saying the term is topological in the sense it is indeed independent of metric. However, he also mentioned, in order to make sense of this integration, i.e. make it to be a number, you need to choose a trivialization of tangent bundle, i.e. choose a framing. The tricky ... 0 Since you start with a group G which is a symmetry of the theory (ie of Lagrangian) there will be some generators for G. Call them T_1,\ldots,T_N. In the original Lagrangian, (before the Higgs gets a vev) you can do a transformation with any of the T_i's and the Lagrangian will not change. After the field gets a vev, you can try the same and you will ... 2 Yes, \mu can be anything. Usually in renormalization, we measure (or define) the coupling constant g at scale \mu, and then use this information to predict the coupling constant g' at another scale \mu'. We require that g' at \mu' is independent of \mu. What I mean is that you should get the same g' at \mu' even if you use another g ... 0 I think one has to be very careful when talking about "particles popping in and out of existence". This interpretation is only sort of fine in flat-spacetime QFT, where the Minkowski metric is time-invariant, so has a global timeline Killing vector. The definition of a particle depends on the notion of there existing time invariance! Since black hole ... 2 Energy and momentum are conserved at every vertex of a Feynman diagram in quantum field theory. No internal lines in a Feynman diagram associated with a virtual particles violate energy-momentum conservation. It is true, however, that virtual particles are off-shell, that is, they do not satisfy the ordinary equations of motion, such as$$E^2=p^2 + m^2.$$... 0 You are making confusion between operators and their representations on the position basis. Your hypotheses 1) and 2) are wrong (more ill-interpreted): as you can easily notice the left hand sides contain an element in the Hilbert space whereas the right hand sides contain numbers (functions evaluated in a point x, therefore a number). To answer your ... 0 No. If$$a_p(t_0)\left | 0\right>=0$$then$$a_p(t)\left|0\right>=e^{iH(t-t_0)}a_p(t_0)e^{-iH(t-t_0)}\left|0\right>=0$$since we usually think of the vacuum as a energy eigenstate with energy zero (even if it's not zero, the above equation is still true). 2 I would suggest one of the (standard) books (though somehow old) on QFT in curved spacetime, Quantum Fields in Curved Space (Birrel & Davies) Relates to path integral formalism, and covers a lot of topics in QFT on curved spacetime Quantum Field Theory in Curved Spacetime (Parker & Toms) Uses DeWitt notation, a lot based on effective action ... 0 It you are working with second quantization (quantum field theory), then what you wrote under "1." and "2." is incorrect. Your physical degrees of freedom are not coordinates x, but rather field variables \phi. So,$$ \left| 0 \right> \sim \exp \left\{ -\frac{1}{2} \int d^3 x \phi(x)^2 \right\} $$This object should be called wave functional ... 2 If physics isn't an issue, you can add arbitrarily many terms. Once the physics comes in though, you will encounter a few restrictions : As said by Gennaro, it is assumed that the Poincaré symmetry applies. Higher derivative terms (second derivatives and above) are generally bad news. They can cause vacuum instability (energies can be arbitrarily ... 0 1.why is renormalization even necessary? It is because, experiments force us to do that. You suppose to get finite values i.e, for charge or for mass of a particle. 2.if we always get a single finite result, why do we care that there are infinities in the equations at the level of amplitudes? Actually we don't. Your integrals in the loop ... 0 Actually there are no infinities in physics. These "infinities" come from integrating over loop momenta when the momentum goes to infinity. But we don't know what actually happens to particles when they have infinite momentum. Maybe they turn into strings, maybe loop quantum gravity and black holes become relevant, maybe there's infinitely small unicorns ... 4 The upshot is that we need one condition to specify how the operators in the Schrödinger and Heisenberg picture are connected. This is usually done by declaring that the two pictures agree at some fixed instant t_0. To summarize: The Schrödinger operator \phi(\textbf{x},t_0) does not depend on time t, while the Heisenberg operator ... 1 1)They neglect higher powers of \Delta \phi because this effective action desribes dynamics of the fluctuations \Delta \phi above the background fields \phi_0. Namely \Delta \phi is small 2)As you know $$\int d^n r\, e^{-r_i A_{ij}r_j}=\frac {(2\pi)^{n/2}}{(\mbox{det}\, A)^{1/2}},$$ where A is a matrix with positive ... 1 This is perhaps best seen via the dispersion relation after doing a Fourier transformation to the energy-momentum representation. The time derivative \partial_0 corresponds (up to a factor of i) to multiplication with energy E (which is the total energy minus the rest energy that was subtracted in eq. (2).). In the non-relativistic (NR) limit, we are ... 2 Hints: OP's identity follows from standard manipulations in the interaction picture, cf. e.g. Ref. 1. Start with the evolution operator$$\tag{1} U(t_f,t_i)~:=~\exp\left(-\frac{i}{\hbar}H (t_f-t_i) \right), \qquad H~=~H_0+V, $$which satisfies the Schrödinger equation$$\tag{2} i\hbar\frac{\partial}{\partial t_f}U(t_f,t_i)~=~HU(t_f,t_i).$$Define the ... 0 Yes, parity is really violated, even if neutrinos are massive. You seem to be confusing the relationship between parity, helicity, and chirality in the modern standard model with the physical symmetry operation of spatial inversion. Wu's experiment did not measure neutrino helicity. Wu and collaborators prepared a thin layer of a beta-emitting nucleus ... 5 In physics there is no general criterion on how to write down suitable Lagrangians, rather than a posteriori check on the equations of motions: all the Lagrangians generating the same dynamics are equally correct. For example, as an exercise, you may try to write down all the possible Lagrangians giving you back F_j = m \ddot{x}_j. This said, to directly ... 4 It's not quite correct to take \Lambda \rightarrow \infty, even at the end of the calculation. That comes from ancient mistaken notions that the field theory under consideration needs to describe physics upto arbitrarily small distances. The modern way to think about this is that you're making your theory agnostic of the value of \Lambda. A theory is ... 0 This question is subtler than one might think at a first glance. Three points may be worth noting. It will make a difference whether the manifold M is compact or not. The nice answer by Michael Seifert implicitly assumes the non-compact case. Otherwise the infinite Gaussian integrals (over \mathbb{R}^n) will not make sense. In the compact situation ... 0 \newcommand{\ket}[1]{\lvert #1 \rangle}*Unitarity* is a central feature of all quantum theories, fields or no fields. Unitarity is simply the demand that the time evolution operator U(t_1,t_2) from any time t_1 to any time t_2 be unitary, i.e. preserve the inner product (\dot{},\dot{}) of the Hilbert space of states \mathcal{H}, i.e. for any ... 4 You can get a nice expression for the leading-order correction to the flat-manifold result via the use of Riemann normal coordinates. Basically, imagine expanding the metric in a power series at the point x_0:$$ g_{\mu \nu}(x) = g_{\mu \nu} (x_0) + \partial_\rho g_{\mu \nu} (x^\rho - x_0^\rho) + \frac{1}{2} \partial_\rho \partial_\sigma g_{\mu \nu} ...

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Unitarity stems for "conservation of probability". It should be quite clear that quantum-mechanical evolution, as prescribed by Schroedinger equation, is due to the strongly unitary group generated by the Hamiltonian (when time-independent) or, more generally, by the corrisponding unitary propagator (when the Hamiltonian is time-dependent). In each case, ...

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There are some possibilities to produce a statement in QFT similar to the one valid for QM. In this case $X$ and $P$ must be replaced by the analogous objects in QFT, the field operator and its conjugate momentum. Consider a quantum scalar field $\phi$ and the equal time CCR: $$[\phi(t, \vec{x}), \pi(t, \vec{y})] = i\hbar \delta(\vec{x}-\vec{y}) I \:.$$ The ...

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The eigenvalues (or measurable values) of each spin operator are ±1, corresponding to the only two mutually exclusive outcomes that could occur after a measurement of an electron's spin has been made. The mutual exclusivity is an experimental fact, but follows from the definition of the eigenstates above.In classical mechanics an electron has a magnetic ...

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Quantum Field Theory is essentially modelled on top of the theory of Quantum Mechanics for finitely many degrees of freedom. With the creation and annihilation operators one can define the analogue of the position and momentum operators $q$ and $p$ as the closures of $$q_0(x) = \frac1{\sqrt2}[a(x) - a(x)^*],\qquad p_0(x) = \frac i{\sqrt2}[a(x)+a(x)^*]$$ ...

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The UV cutoff procedure that is used to regulate Feynman diagrams must be chosen so that the region of integration involving large values of $p^\mu$ (all components individually) is cut-off. A Lorentz invariant cut-off such as $p^2 < \Lambda^2$ does not achieve this. However, what we can do is to Wick rotate $p^0 \to - i p^0$ and then impose a cut-off ...

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Theoretically it can. Adler wrote a book Quaternionic Quantum Mechanics and Quantum Fields, where the details are worked out. See also Arbab's recent paper Quaternionic Quantum Mechanics. However, it is unclear what advantages quaternionic theory offers over the complex one, and analytic issues do not work out very well. Already Hamilton encountered ...

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One reason for the box is the Fourier expansion of field in stable macroscopic condition (thermal radiation, cavity oscillations) works well only for finite volume. For infinite volume, the Fourier integral of such stationary field is problematic, because the field function is not L2 integrable.

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In principle it is not necessary to restrict the classical space to the set of solutions of the classical dynamics. The usual "basic" requirement for the classical space $X$ is the following: $(X,b)$ is a couple with $X$ a real vector space and $b:X\times X\to \mathbb{R}$ a non-degenerate skew-symmetric bilinear form (symplectic form). This is ...

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The exponential factor is necessary (from a particle-centric perspective) for the fields to transform covariantly under spacetime translations so they can be used to define local Lorentz invariant interactions, or (from a field-centric perspective) for the $a_p$ and $a_p^\dagger$ operators to be interpreted as particle creation and annihilation operators ...

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Quantizing in a finite volume is not specific to the electromagnetic field, and it is not a necessity, neither for the electromagnetic field nor for any other. It is generally more well-behaved to quantize in a finite volume because no infrared-like divergences appear from allowing arbitrarily low momenta (since no arbitrarily long wavelengths fit into the ...

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Wick's theorem tells us that $$\mathcal{T}(\phi_1\dots\phi_N) =\ :\phi_1\dots\phi_N: + :\text{pairwise contractions}:$$ where $:\ :$ is normal ordering. Immediately from the definition of normal ordering (all annihilators to the right, all creators to the left), the expectation value of anything that is normal-ordered and not a constant vanishes because the ...

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In classical field theory, the ground state is also called minimizer (of the energy functional); and just to prove its existence is already a quite difficult task, from a mathematical standpoint (as you can imagine, much more difficult is to write eventually its explicit form). Often you can only have minimizing sequences, i.e. sequences of classical states ...

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Okay, I think I have an answer but I would interested to hear some feedback on this. I initially thought that by writing the left and right handed fields in 2 component notation ($q_L$ and $q_R^c$), they were 'equivalent' objects and can freely rotated between one another. However, I realized now this is not the case. The left and right handed fields are ...

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From your comments, it sounds like you're looking for Deconfinement. At higher energies the quarks and antiquarks are no longer bound, they are asymptotically free, and thus chiral symmetry is restored.

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To study Higgs mechanism, we can use a Lagrangian of the form: \begin{align} \mathcal{L}=(D_{\mu}\phi)^2-\frac{1}{4} F_{\mu\nu}F^{\mu\nu}-V(|\phi|) \end{align} Where: \begin{align} V(|\phi|)&=-2v^2|\phi|^2+|\phi|^4 \\ &=(|\phi|^2-v^2)^2-v^4 \\ D_\mu \phi&=\partial_{\mu}\phi+\mathrm{i}e A_{\mu} \phi \\ F_{\mu\nu}&=\partial_{\mu} ...

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The branching ratio for a certain channel $i$ is given by the ratio of its partial decay with $\Gamma_i$ and the sum of all partial decay widths: $$BR(H \rightarrow i) = \dfrac{\Gamma_i}{\sum_j \Gamma_j}$$ where the $\Gamma_i$ depend on the Higgs mass. If a new channel opens up or becomes important (such as the decay to a pair of W bosons at around ...

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Polarization and gauge symmetry In QFT, the dynamical varible is the four-potential $A_\mu$. The electromagnetic field is defined by $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$, an antysymmetric tensor wich six independent components: 3 for the electric field and 3 for the magnetic field $E^i = - F^{0i}$, $B^i = ... 0 I don't think it's proper to say that the spontaneous breaking of anything is attributed to the Higgs mechanism. It's a postulate of the theory that the potential takes a certain shape, and that shape leads to a symmetry, and that the symmetry is spontaneously broken. The Higgs mechanism is a consequence of that. The symmetry that is broken is the local ... 2 There are a couple of points to be precise: in four spacetime dimensions there is no scalar relativistic interacting field theory that can be rigorously defined (i.e. in which the unitary dynamics can be constructed), at least for the moment. This does not mean it is not possible, but we have not the mathematical tools to do it. The physical calculations ... 1 From right to left: outgoing electron$e^-$spinor:$\bar{u}(p_2)$QED vertex:$ie\gamma^\nu$outgoing photon:$e^*(k_2)$electron propagator:$\frac{i}{\not{q}-m+i0}$incoming photon:$e(k_1)$QED vertex:$ie\gamma^\mu$incoming electron$e^-$spinor:$u(p_1)\$ There is no photon propagator in this process, and also, only one electron propagator. You ...

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There is a great introduction called "This is How Quantum Field Theory Works" which, I think, is exactly what you are looking for. All essential concepts are introduced and the basic idea how one gets from the fundamental equations to cross sections, i.e. quantities that can be measured in experiments is sketched.

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