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Comments to the question (v3): The main fields assumption that goes into the proof of the Wick's theorem for fields $\hat{\phi}^i\in{\cal A}$ is that their (super)commutators $$[\hat{\phi}^i,\hat{\phi}^j]~\in~ Z({\cal A}) \tag{1}$$ are central elements of the operator algebra ${\cal A}$, cf. e.g. this Phys.SE post. For free fields $\hat{\phi}^i\in{\cal ... 1 The eigenstates of a time-independent Hamiltonian $$H |\psi_j \rangle = E_j|\psi_j\rangle$$ have the usual rotating-phase time dependence in the Schrödinger picture: $$|\psi_j(t)\rangle = |\psi_j(t_0)\rangle \cdot \exp(E_j(t-t_0)/i\hbar)$$ However, your formula indicates that$H_0$and$V$are time-dependent. So if the state vector is an eigenvector of ... 2 Charged particles can't have Majorana masses of any type because they would violate the charge conservation law. The Majorana mass is really a term that is converting a particle into its antiparticle. It implies that the particle must be considered "physically indistinguishable" from its antiparticle. The Majorana mass term violates the lepton number or its ... 0 The condition on the modulus of$\Omega$ensures that the complex bilinear form$\mu_c$defined as$\mu_c(\cdot,\cdot)=\mu(\cdot,\cdot)+i\Omega(\cdot,\cdot)$is a scalar product. Therefore$(\mathscr{S},\mu_c)$is a complex pre-Hilbert space. Denoting by$\mathscr{H}_{\mu_c}$its completion, it is then possible to define the symmetric Fock space ... 0 Yes or no, it depends what you precisely understand by this short sentence. The word "inelastic" doesn't really mean that we don't "want" to detect something. Physics doesn't say what we should "want". "Deep" means that the quarks (and perhaps the electron) penetrate very close to each other and the scattering studies the behavior of the matter at ... 1 The small number of "conceptually independent types of processes and calculations" is exactly a symptom of the theory's being fundamental! Even in classical physics, all calculations could have been mathematically reduced to the calculation of the final state that evolves from an initial state (or a state that is stationary etc.). In quantum mechanics, this ... 1 The unitary operator$U$is not a real number, so$|U|=1$, whatever$|U|$exactly means, does not imply$U\in \{+1,-1\}$in any way. The unitary operator isn't even a complex number. It's more complicated than that. Moreover, time ordering isn't a "unitary operator". It isn't even an operator in the same sense as unitary operators in quantum mechanics. It ... 4 @GerryHarp's answer contains the gist of the main idea, but there is a point that doesn't quite make sense: There is no such thing as bare ions crystal in solid state physics. To answer your last question: The "bare phonon frequencies" definitely do not refer to phonon frequencies in the absence of electrons. In fact, an ion crystal without electrons is ... 4 Lets start with the bare ions. You set up a linear lattice of protons where the boundaries are held in place by some means. The protons want to get as far apart from each other as possible. So the minimum energy state has the protons spaced on the line with equal distances between them. So yes, the ions form a regular lattice without the electrons. Now ... 4 In the quantum mechanical description of any physical system, including a quantum field or a collection of interacting quantum fields, there is always one state vector – one collection of numbers (probability amplitudes) that generalizes what is referred to as the "wave function" in quantum mechanics of particles. In quantum field theory, a better name is a ... 3 Because we have observed processes where the photon number is not conserved. For example positronium can decay into 2 or 3 (or more) photons. This means that it is not possible to assign a global conserved charge to photons. For neutrinos we can assign lepton number and so far we have not observed a process that would violate total lepton number (lepton ... 1 Almost everything from the wikipedia page you link is just false, or at best very misleading. IMHO, that page was written by someone that doesn't know anything about quantum mechanics beyond what one could find in TV documentaries. "Not even wrong" came into my mind many times as I was reading the article. In quantum physics, a quantum fluctuation (or ... 2 You're right that the Feynman propagator for a spinor field is indeed$(i \gamma^\mu D_\mu - m)^{-1}$. The tricky part is interpreting exactly what the "inverse" means. It doesn't just mean that you invert the gamma matrices (although you do do that). The derivative operator is also being inverted. That is, if we define a function$G(y, x) := (i ...

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Quick answer My question is how does the presence of nonzero $J(x)$ results in a non-trivial spacetime dependent value of $\langle 0|\phi(x)|0\rangle$? The equation $\phi(x)=\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}$ works both for $J=0$ and $J\neq 0$. Therefore, $$\langle \phi(x)\rangle_J= {}_J\langle 0|\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}|0\rangle_J ... 1 This formula for the effective action is valid if (i) you are interested in a slowly varying field (i.e. in the low energy/momenta modes) and (ii) if the effective action not singular in this expansion. If the expansion is well behaved, then you always have the right to do the expansion, which will obviously be valid only for fields varying slowly enough. ... 2 By definition, H|\Omega\rangle = E|\Omega\rangle, so that \langle \Omega |e^{-i T H}|\Omega\rangle=e^{-i T E}. The presence of source terms in the Hamiltonian does not change anything about that. The RHS of the equation 11.43 is just the functional integral rewriting of \langle \Omega| e^{-i T H}|\Omega\rangle. 0 The four vector p is given as (E,{\bf p}) so that$$(p' -p)^2 = (E' - E)^2 - ({\bf p'} - {\bf p})^2 = (E'^2 - {\bf p'}^2) + (E^2 - {\bf p}^2)- 2E'E + 2 {\bf p'} \cdot {\bf p}$$Then put E'E \approx (m + \frac{{\bf p'}^2}{2m})(m + \frac{{\bf p}^2}{2m}) \approx 2m^2 + {\bf p'}^2 + {\bf p}^2 (ignoring the term \frac{{\bf p'}^2{\bf p}^2}{4 m^2}). Using ... 0 That cannot be true, because$$ \left(\frac{1-\gamma_5}2\right)\left(\frac{1+\gamma_5}2\right)=0 $$EDIT: The youtube video from the comments is actually wrong. The correct expression is$$ \bar\psi_L\gamma^\mu\psi_{R,\mu}=\bar\psi P_R\gamma^\mu P_R\psi_{,\mu} $$From this, you should be able to show that this term vanishes. 1 It turns out this diagram (one of eight) was given as an assignment in which we were to uncover the subtleties for ourselves. Considering the free quark propagator in isolation and then looking at the 1PI correction associated with the gluonic correction (the self energy) , we see that this vanishes in dimensional regularisation with the result that the ... 0 If a particle is a wave function describing a probability amplitude distributed through space, what happens when two wave functions meet? Well, I just try to think allowed and one can not portray a 'wave function' picture unless a physical system is at hand to use the picturization or description. Let us think about a hole in the reactor and neutron ... 1 Wavefunctions combine trough tensor products, which is not the addition that one would expect naively. The reason for this is that a wavefunction contains the description of all possible futures of the system at once, so if there are multiple subsystems, then the wavefuntion of the entire system has to describe all possible futures for each part ... 1 If the Fourier transform of f(x) is \tilde{f}(k), then the Fourier transform of df/dx is ik\tilde{f}(k). Proof:$$\frac{df}{dx} = \frac{d}{dx} \int \frac{dk}{2\pi} \tilde{f}(k) e^{ikx} = \int \frac{dk}{2\pi} \left[ik\tilde{f}(k) \right] e^{ikx}.$$This explains the momentum factors, so we've reduced the task to showing$$\Gamma(x, y, z) \sim ...

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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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How does the matter-radiation system, on its own, goes over to the Blackbody distribution? Evolution towards equilibrium (in macroscopic sense) happens when the system matter + radiation is isolated, for example if a piece of matter is inside a cavity that slows down leakage of energy out of the system. For example, a piece of coal in a well reflecting ...

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Phonons follow a wave equation, which is at least in first approximation simply a standard wave equation, the only difference to relativistic particles is that the speed of the waves is not c but the speed of sound $c_s$. But this does not change the mathematics of the equation, so that in general there may be phonons which follow a massless wave equations ...

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Phonons are indeed massless, as you can see from their dispersion relation or from the fact that they are Goldstone bosons. The phonon dispersion relation that you wrote down tells us that we can excite a phonon mode, with some finite momentum, using an arbitrarily small amount of energy, hence they have no rest mass (in condensed matter language, they are ...

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Both states $\Psi_{k,\sigma}$ and $\Psi_{k',\sigma'}$ are meant to be states of the same particle species i.e. they have the same values of the squared mass $k^2$. The inner product of one-particle states from different species $s$ is zero which one might indicate by additional $s,s'$ labels and a Kronecker symbol $\delta_{s,s'}$. Weinberg claims about the ...

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Newton's third law is a statement that momentum is conserved, so it is equivalent to the law of conservation of momentum. Conservation of momentum follows from a fundamental symmetry (of the action) called space shift symmetry and as far as we know this applies to all our physical theories. So Newton's law is still valid but has to be treated with some care ...

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The convolution $G_{ret}*f$ of the retarded propagator $G_{ret}$ with a source term $f$ which vanishes sufficiently far in the past is the unique solution of the inhomogeneous Klein-Gordon equation with source term $f$ which vanishes in the far past. It is necessarily a nontrivial superposition of positive and negative energy solutions at all times when it ...

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What exactly happens to these massless degrees of freedom in the Higgs potential after symmetry breaking, i.e. after we expand the Higgs fields about the vev? You will make a redefinition of the vector fields $V_{\mu}' = V_{\mu} - \partial_{\mu} \eta$, where $\eta$ was your goldstone, Doing that you can see that a term like $$\nu^{2} (V_{\mu} + ... 0 For some counterarguments against Lubos Motl's argumentation against de Broglie-Bohm theory see http://ilja-schmelzer.de/forum/forumdisplay.php?fid=6 and http://ilja-schmelzer.de/realism/Motl.php The first proposal for a Bohmian variant of a relativistic quantum field theory has been made in Bohm's original 1952 paper, for the EM field. For a possibility ... 1 Conventionally in QFT, particles and antiparticles are defined with positive energy k^0\geq 0 only. (Recall that would-be negative energy states are reinterpreted as matter/antimatter of the opposite kind in order to make the vacuum stable.) 0 Hello handsome poster (why thank you kind stranger). The answer to your question it turns out is in Rovelli's "Quantum Gravity", at least insofar as the free scalar field is concerned. This is done in the following way. As you may recall (from Feynman and Hibbs), through various arguments about doing the path integral as a perturbation on the classical path, ... 3 For the specific case of a fixed number of interacting spinless point particles, there is a Bohmian recipe that works fine: you start with solutions to the Schrodinger equation, construct trajectories from the gradient of the probability current, and assign a probability measure to those trajectories according to the Born rule. That gives you a "classical" ... -3 Just an idea I'm playing with, There is no dark energy/dark matter. The universe is not expanding into “infinite nothing”, but instead reducing into the amount of space it occupied at inception. Matter is reducing in scale because there is nothing in the universe requiring it to maintain a static size at the subatomic level. The gravitational effect of ... 3 As you say, loop corrections change the value of the quartic coupling \lambda is modified by loop corrections. If the renormalization group changed its value so \lambda \gg 1, the perturbative interpretation of the theory would break apart. Note that the condition \lambda < 1 (or \lambda/(4\pi) < 1) is only a hint of the perturbativity, the ... -4 deBroglie developed double solution theory, which is the most relevant description of photon and the Bohm / orthodox quantum mechanics are just high/low energy limits of that model. It should be pointed out that de Broglie disagreed with Bohmian mechanics. 0 SO(3) symmetry means that the amplitudes |\langle \psi|\phi\rangle|^2 is invariant under rotations in the rays. Remember that a ray is specified by a family of vectors e^{i\phi}|\psi\rangle. This means that the linear or anti linear operators that describes how the vectors are changed by symmetric transformations obeys an projective representation. ... 0 A wave function is the projection of a state in the coordinate basis, where the coordinate basis is defined as the basis that diagonalizes the position operator {\hat x}. In QFT, the analog of this is the fundamental field operator {\hat \Psi} itself. Eigenvalues of this are defined as$$ {\hat \Psi} (t,\vec{x}) | \phi \rangle = \phi(\vec{x}) | \phi ...

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Scattering experiments can be used to determine the size of a particle. The results for an extended object are different than that of a point particle. But all of these scattering experiments depend on getting the probe particle "close" to the scattering object. In the case of electrons, that means launching the probe with enough energy to overcome the ...

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In the Schrödinger treatment of the Hydrogen atom, the excited states are true eigenstates, that is, their lifetime is infinite. This means that, if the Hamiltonian is $$H=\frac{P^2}{2m}+\frac{\alpha}{r}$$ then $\tau=\infty$ for all the energy levels. What this description misses is the interaction of electrons with the electromagnetic field, that is, ...

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In most of your derivations, you have used the symbol $\psi$ for the genuine field operator (operator distribution). But your equation $$\gamma^5 \psi = -\psi$$ clearly doesn't work for any Dirac field. This equation an operator equation equivalent to $$(1+\gamma^5) \psi = 0$$ which says that one-half of the components of $\psi$ are zero as operators. But ...

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$t=0$ is completely consistent. It is simply too boring since it implies that in any unitary theory, $Q_\alpha |\psi\rangle = 0$ for any state $|\psi\rangle$. This is because in this case, we have $$0 = \langle\psi|\{ Q_\alpha , Q^\dagger_\alpha \} | \psi \rangle = \| Q^\dagger_\alpha |\psi\rangle \|^2 + \| Q_\alpha |\psi\rangle \|^2$$ Positivity of ...

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So the entire electromagnetic force can be described as having these objects which interact by exchanging virtual photons. Photons -- light -- in some sense are the electromagnetic force. It is therefore unsurprising that light can be absorbed by a particle -- an electron, say -- as a sudden "push" which launches the electron in some new direction with ...

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The stress tensor for $h_{ab}$ is not Lorentz covariant, despite the fact that it looks like it is. This is because $h_{ab}$ itself is not a Lorentz tensor. Rather under Lorentz transformations $$h_{ab} \to \Lambda_a{}^c \Lambda_b{}^d h_{cd} + \partial_a \zeta_b + \partial_b \zeta_a ~.$$ The extra term is present to make up for the fact that $h_{ab}$ is ...

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Pauli-Fierz theory does not violate the Weinberg-Witten theorem because the stress-energy tensor you constructed is not gauge invariant under the infinitesimal transformation $h_{\mu\nu}\mapsto h_{\mu\nu}+\partial_\mu\chi_\nu + \partial_\nu\chi_\mu$ (this is the Lie derivative $\mathcal{L}_\xi h$ of the metric along the vector field $\chi$, i.e. the ...

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We do not start from the gauge fixed path integral in the BRST construction. What you describe (once one adds the missing Faddeev-Popov determinant) is the original Faddeev-Popov trick to get the ghosts, not the systematic BRST construction. The (Hamiltonian) BRST construction crucially first introduces the ghosts as parts of the extended phase space, and ...

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The BRST symmetry cannot be seen without introducing auxiliary variables. The fastest way to realize the BRST symmetry is to "exponentiate" the delta function $$\delta(G)~=~\int \!{\cal D} B ~\exp\left[iB_{\alpha}G^{\alpha}\right]$$ and the Faddeev-Popov (FP) determinant \det\Delta ~=~\int \!{\cal D} c ~{\cal D} \bar{c} ~\exp\left[\bar{c}_{\alpha} ...

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A standard reference packed to the brim with other references to everything you ever wanted to know about anomalies is "Anomalies in Quantum Field Theory" by Bertlmann. This particular topic is what comprises part of chapter 11 there. I'll highlight the main points, but this is a technical topic for which you'll have to go to the references and follow all of ...

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Welcome here. From your profile I see that you are at the beginning stages of learning physics. This is an arduous process that needs a lot of elbow grease in solving problems and/or doing experiments in order to get a basic intuition for the subject. Here is a simplified answer to your questions: Why do most theories about what Dark energy and Dark ...

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