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Your question is not specific to inflation, and really applies to any case where a bosonic quantum field behaves semiclassically due to macroscopically large occupation numbers. One very simple example of this is the Stark effect in quantum mechanics, where a Hyrodgen atom is placed in a uniform electric field. The atom is treated as a quantum mechanical ...

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Theorem: let $L$ be a homogeneous function of degree $k$; then the on-shell lagrangian is a total derivative. Proof: according to the Euler's homogeneous function theorem, $$k\ L(q,\dot q)=q\frac{\partial L}{\partial q}+\dot q\frac{\partial L}{\partial \dot q}\tag{1}$$ On the other hand, because of the Euler-Lagrange equations, $$(1)=q\frac{\partial ... 6 A particle is not a wavepacket. And there are no particle states for interacting theories. We define particle states in QFT by expanding the free field into its Fourier modes and using these modes as creation/annihilation operators for particle states - the mode of momentum p creates the particle state \lvert p\rangle with momentum p. The Hilbert ... 4 Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, ... 2 Yes. You are correct. A non-relativistic theory would be invariant under the Galilean group. Lorentz invariance (specifically, invariance under Lorentz boosts) is what defines a relativistic theory. 2 Consider a map$$S \ni\phi \mapsto F[\phi] \in \mathbb R$$defined on a class S of smooth functions \phi defined on the compact set \Omega \subset \mathbb R^n obtained by taking the closure of an open set with regular boundary \partial \Omega. Thus the map F associates a real number F[\phi] to each function \phi\in S. We say that the ... 2 Yes, there is systematic way called the Noether procedure. Simply you write down all possible 2-derivative fermionic terms with arbitrary coefficients and vary the action using the SUSY transformation rules. Then, you fix the coefficients to obtain the invariance up to a total derivative. When you have 4-derivatives, there are two cases: a. Off-Shell ... 2 Well, you cannot take any ol' matter theory in flat Minkowski space and stick in a curved metric tensor g_{\mu\nu} in the matter action as you like, if that's what you're implying. The caveat is that the resulting matter action S_{\rm m}[\Phi, g] should be a general relativistic diffeomorphism-invariant functional. Then the Hilbert stress-energy-momentum ... 2 There are two kinds of derivatives we should differentiate:$$ \frac{\mathrm d\mathcal L}{\mathrm dx}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x+h),\phi'(x+h),x+h)-\mathcal L(\phi(x),\phi'(x),x)\big]\tag{1} $$and$$ \frac{\partial\mathcal L}{\partial x}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x),\phi'(x),x+h)-\mathcal ...

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This is a simple application of the chain rule $$\frac{\partial}{\partial x^\mu} \phi \big( \Lambda_\nu{}^\mu x^\nu \big) = \frac{\partial}{\partial x^\mu} \big( \Lambda_\rho{}^\nu x^\rho \big) \left[ \frac{ \partial }{ \partial y^\nu } \phi (y) \right]_{y^\mu = \Lambda_\nu{}^\mu x^\nu} = \Lambda_\mu{}^\nu \left[ \frac{ \partial }{ \partial y^\nu } ... 2 My 2 cents on it is that in QM (be it "standard" QM or QFT) one describes only the state of a particle. Having said that, the most general state for a single particle is indeed a wave packet. Now, if you localise certainly a particle at some point in time, then later on it will be associated with a spreading wave packet because of Heisenberg indeterminacy ... 2 The first answer to such a question must always be: A gauge symmetry has no "physical" meaning, it is an artifact of our choice for the coordinates/fields with which we describe the system (cf. Gauge symmetry is not a symmetry?, What is the importance of vector potential not being unique?, "Quantization of gauge systems" by Henneaux and Teitelboim). Any ... 2 This is a very broad question, so there are many ways to answer it. Here is one interpretation. A principal distinction between gauge symmetries and global symmetries is that gauge symmetries lead to long-range interactions between charged particles; the gauge symmetry demands the existence of a massless field which can propagate over arbitrarily long ... 1 No, not every force field you can imagine is conservative. A force field is conservative if its line integral is path-independent, that is to say,$$\int_C \vec F \cdot d \vec x = U(a)-U(b)$$For every curve C with end points a,b. U is some scalar function. If the force field is the gradient of a scalar field, this is automatically verified: ... 1 There are two different values represented by v in this problem: lowercase v is the velocity of the electron uppercase V is the voltage that accelerates the electron In both the force equation F=qvB and the radius equation r = mv/qB, v refers to the velocity of the electron. I believe this is where your mistake was, since you said that the ... 1 It is, a priori, completely correct to add both primary and secondary constraints to the Hamiltonian density by Lagrange multipliers. What is not correct is how you determined the equations of motion: There is no "F^{i0}" in the Hamiltonian theory! It is called \pi^i there and it is not dependent on \partial_i A_0, it is an independent canonical ... 1 A general advice: Before trying to understand Hamiltonian field theory, make sure you understand Lagrangian field theory. Before trying to understand Lagrangian field theory, make sure you understand Lagrangian point mechanics. In Lagrangian point mechanics, the functional derivative of the action is$$\tag{1} \frac{\delta S}{\delta q(t)} ...

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And regarding why it's called a "free" theory, it's not specific to a momentum-space formulation. It's "free" because the Lagrangian is quadratic in the fields, and therefore the equations of motion (what you get from plugging the Lagrangian into the Euler-Lagrange equation) are linear in the fields. Therefore you can superpose different classical ...

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Take for example $q$ a vector field: $$q^a=A^\mu$$ where $a=\mu$ is a vector index. The conjugate momentum is $$\frac{\partial\mathcal L}{\partial A^\mu}$$ and, as it is an upper index in the denominator, it makes sense to write it as $\pi_\mu$. Also, you can use the definition of vectors and covectors to prove that $\pi_\mu$ transforms as a covector, ...

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