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I) OP is asking about the case where the infinitesimal symmetry transformations $$\delta\phi^a~=~t^a{}_b \phi^a \tag{A}$$ are linear in the fields in the path integral$^1$ $$Z[J] ~=~\exp\left[\frac{i}{\hbar}W_c[J]\right]~=~\int \! {\cal D}\phi~\exp\left[\frac{i}{\hbar}\left( S[\phi]+J_a\phi^a \right)\right]. \tag{B}$$ Recall the Legendre transformation ...

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Normally, when combining Lagrangians, we often leave the constant multiplying factor to be determined by experiment. For example, if $\mathcal{L}_{k}$ is the kinetic term (for a system of charges and the electromagnetic field), and we choose to describe the electromagnetic coupling by $\mathcal{L}_{int} = A_\mu J^\mu$, then we combine them as ...

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Here are a couple quick and dirty ways to count these operators: Compute the conformal block expansion of the four-point function $\langle \phi\phi\phi\phi\rangle$. This will only contain blocks with $\Delta-\ell=d-2$. This is done in http://arxiv.org/abs/1009.5985, equation 64. Compute the character of the conformal group acting on operators in the ...

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Gauge theories become constrained Hamiltonian systems when passing from the Lagrangian $L(q,\dot{q},t)$ to the Hamiltonian $H(q,p,t)$ where $p = \frac{\partial L}{\partial \dot{q}}$. Generically, you get a constrained Hamiltonian system whenever the matrix/operator with components $$\frac{\partial^2 L}{\partial \dot{q}^i\partial\dot{q}^j}$$ is singular, ...

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OP wrote (v3): Is there anything in particular I should be careful of? Yes. Watch out for secondary constraints, cf. e.g. this Phys.SE post. Below follows a brief partial derivation. Let Greek letters $\mu,\nu,\ldots$ denote spacetime indices, while Roman letters $i,j,\ldots$ denote only spatial indices. The Lagrangian density {\cal L}~=~ ... 4 This is how we taylor expand the determinant to first order \begin{align} \mathcal{J}&= \text{det}\left(\frac{\partial x'^j}{\partial x^i} \right) \\ &= \text{det}\left(\delta _i ^j + \partial_i\delta x^j\right) \\ &= \text{exp}\left( \text{tr log }\left(\delta _i ^j + \partial_i\delta x^j\right)\right) \\ &= \text{exp}\left( \text{tr ... 1 Ali moh's is a wonderfully full and descriptive answer, but unfortunately you are not looking for a purely mathematical answer in which case I will try to give the kind of answer you are looking for. 1) The reasons associated with the variations being described in terms of these sums have to do with perturbation theory (which goes beyond the scope of this ... 2 First of all \delta and \frac{\partial}{\partial x} don't commute because \delta \left(\frac{\partial}{\partial x}\phi\right) = \delta\left( \frac{\partial}{\partial x}\right)\phi + \frac{\partial}{\partial x}\left(\delta\phi\right) $$Second we divide \delta\phi_I = \bar{\delta}\phi_I + X^k_n \delta \omega_n \partial_k \phi_I because the total ... 1 i searched for the exact same problem recently after a debate with one of my colleagues. In my opinion, you already gave the answer to your question yourself. A source dipole is the flow field resulting from a sink and a source brought together. In a sink, all streamlines point radially inward to the singularity at the origin, in a source, all point ... 1 For z=2 there is a study here http://arxiv.org/abs/quant-ph/0404026 in the context of ferromagnetic spin chain. The result is basically \log L. Swingle and Senthil argued in http://arxiv.org/abs/1112.1069 that "generally" the violation of area law for EE is at most L^{d-1}\log L where d is the space dimension. However, http://arxiv.org/abs/1408.1657 ... 1 Before going to field theory, it seems instructive to first ask the same questions in point mechanics: Can the Lagrangian L(q,v,t) depend on time explicitly? Yes. The Lagrangian L(q,v,t) can depend explicitly on time. E.g. there could be external sources. On the other hand, if the Lagrangian does not depend on time explicitly, then the ... 1 OP wrote (v1): Why the Lagrangian density becomes a function of the spatial derivatives? Well, one intuitive answer is, that if the theory is supposed to be relativistic, and if the Lagrangian density has temporal field derivatives, then it must also contain spatial field derivatives. Another answer is that if the theory is supposed to be local, this ... 1 If we write A_\mu(x)=\varepsilon_\mu(p)e^{ipx}, the polarization vector should satisfy \varepsilon_\mu p^\mu=0, which is a Lorentiz-invariant relation, and is necessary to make sure that we have an irreducible representation of the Lorentz group (actually, the little group that leaves the momentum invariant). This knocks down the number of D.O.F to 3. ... 3 The longitudinal mode decouples from all physical processes as a consequence of gauge invariance, which in turn forces the Ward identity$$ k^\mu \mathcal{M}_\mu = 0 where the S-matrix element decomposition $\mathcal{M}^\mu$ is obtained from the polarization vector $\epsilon^\mu(k)$ by $\mathcal{M} = \epsilon^\mu(k) \mathcal{M}_\mu$. This decoupling (and, ...

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To justify giving mass to a would-be massless particle, scientists were forced to do something out of the ordinary. They assumed that vacuums (empty space) actually had energy, and that way, if a particle that we think of as massless were to enter it, the energy from the vacuum would be transferred into that particle, giving it mass. A mathematician named ...

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