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Based on what J Richard Gott writes in " Time Travel and Einstein's Universe" , the vacuum energy density at the point of Unruh radiation absorption is equal and opposite in sign to the Unruh radiation density. In the accelerated frame, vacuum energy density is negative. It has been mentioned above that the energy for Unruh radiation can't come from the ...

6

Operators and derivatives Let $X$ and $P$ be individual operators satisfying $$[X,P]=i. \tag{1}$$ The operator $X$ is just one operator, so there is nothing to differentiate here. Now consider a one-parameter family of operators, with one operator $X(t)$ for each time $t$. This is often called an operator-valued function. That's just a fancy name for ...

1

Presumably OP is worried that the spin-statistics theorem states that spin $1/2$ fields should be described by Grassmann-odd (as opposed to Grassmann-even) fields. Recall that in the Dirac field, the Grassmann-odd nature sits in the creation and annihilation operators. On the other hand, the Dirac spinors $u_s({\bf p})$ and $v_s({\bf p})$, which Ref. 1 ...

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It is a result of integrating over $d^3p_1d^3p_2$ (momenta of the final-state particles). Such integration counts both $$\left(\vec p_1,\vec p_2\right)=\left(\vec a, \vec b\right)$$ and $$\left(\vec p_1,\vec p_2\right)=\left(\vec b, \vec a\right)\;,$$ wile these two situations are physically identical if particles $1$ and $2$ are indistinguishable. Hence, to ...

0

A concept you're going to want to look in to is parallel transport. Long story short - you have some underlying manifold (your space) that you want to move locally defined quantities around in, and that movement process has to be continuous/smooth, and preserve inner products. Everything else follows from those requirements and how the vector space being ...

0

It should be stressed that the fact that the propagator has a pole at the physical mass $k^2=-m_{\rm ph}^2$ is the very definition of the physical mass, i.e. it does not depend on the renormalization scheme, cf. e.g. Ref. 1. The on-shell (OS) renormalization scheme equates the renormalized mass $m\equiv m_r$ and the physical mass $m_{\rm ph}$. The unit ...

2

OP's question seems mainly spurred by the fact that there are different notions of topology. OP is mentioning general topology, while the topology of Feynman diagrams is described by geometric topology.

1

Elementary particles and the composites from them are defined by spin: if spin is a multiple of 1/2 it is a fermion, if it is a whole number it is a boson. This is incorporated in the spin statistics theorem that describes the difference in their wavefunction, leading to the use of different differential equations. The proof is further in the link. The spin–...

1

The problem is a rather fundamental one in fact, and it cannot be overcome so easily, and it is not per se about having different Hilbert spaces, but rather inequivalent representations. All separable, infinite dimensional, Hilbert spaces are isomorphic (they are thus the same mathematical structure essentially). Nonetheless, they accommodate infinitely ...

3

In linear algebra an operator is something which acts on a vector and returns another vector. A linear operator can be represented by a matrix. So in this sense you should think of differentiating an operator like differentiating a matrix. Take for example this time dependent rotation matrix: $$\partial_tR=\partial_t\begin{pmatrix} \cos\omega t & \sin\... 2 To define an operator it is sufficient to define its action on an arbitrary state \Psi. The definition of d \hat{x} / d t is as follows:$$ \left( \frac{d \hat{x}}{dt} \right) \left| \Psi \right> = \frac{d}{dt} \left( \hat{x} \left| \Psi \right> \right). $$Equivalently, the matrix elements of the derivative of an operator are derivatives of the ... 2 This is a very general question, and I am not sure whether there is a simple way to answer it. 1) Fluid dynamics: The need to have scale dependent parameters can be seen purely within fluid dynamics, without making explicit reference to an underlying theory. Fluctuation-dissipation relations require that dissipative fluids have thermal fluctuations (just ... 0 There is not one state under a wave function unless the wave function has been determined, which is possible. The universe should have arose in many multiple possible ways even in a ground state. 0 Considering that simple one-loop integrals as evaluated in the paper Scalar one-loop integrals feature functions such as square roots and logarithms for two- and three-point cases, it is clear that three-point amplitudes may develop branch cuts as soon as at one-loop level. 2 The field itself doesn't have any specific value (at a given point in space and at a given moment of time), and doesn't "fluctuate" like a thermodynamical quantity. In a way, it is a "blurred" quantity and is represented as a field operator in QM. If an ideal observer could do some field measurement there, he would find any value from some given set of ... 0 I think the difference between these two explanations is a differing view of a quantum field. I have found that some people define as a quantum field the field itself with the creation and annihilation operators that generate the particles in quantum field theory, while others consider the space filled with a quantum field and the creation and annihilation ... 0 Retrocausality is unnecessary to describe DCQE behavior. Nothing is working backwards in time (at least not anymore than in any other basic QM process, for fans of the Transactional interpretation). In the DCQE, remember that the recovered interference patterns from the “no which-way info” arms of the experiment are complimentary - e.g. the bright bands ... 0 I’m going to expand on Marco’s (correct) answer a bit. In the DCQE, remember that the recovered interference patterns from the “no which-way info” arms of the experiment are complimentary - e.g. the bright bands of one arm would line up with the dark bands of the other arm. This is very important - it means that when you first detect an entangled ... 0 I found a more directly way to solve......$$ \mathrm{R.H.S.}=g^{\mu \rho} W^{\sigma}-g^{\mu \sigma} J^{\rho}=-\frac{1}{2} \epsilon^{\rho \sigma \alpha \beta} \epsilon_{\theta \tau \alpha \beta} g^{\mu \theta} W^{\tau}=-\frac{1}{4} \epsilon^{\rho \sigma \alpha \beta} \epsilon_{\theta \tau \alpha \beta} g^{\mu \theta} \epsilon_{\phantom{\tau}\nu \iota \...

1

No, that is not a safe assumption. The class of experiments you refer to tend to have two measurements taking place at different times, in such a way that it is possible to interpret the later measurement as influencing the outcome of the earlier one. However, there is in each case a reciprocity that allows you equally to consider that the earlier ...

3

Photons are elementary particles, part of the SM, they are traveling at speed c in vacuum, when measured locally. The world line (or worldline) of an object is the path that object traces in 4-dimensional spacetime. It is an important concept in modern physics, and particularly theoretical physics. https://en.wikipedia.org/wiki/World_line Now ...

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Consider the toy integral $$Z(\lambda)= \int_{-\infty}^{\infty} e^{-x^2-\lambda x^4}dx$$ Lets expand in powers of $\lambda$. We find $$Z(\lambda)= \int_{-\infty}^{\infty}dx e^{-x^2} \sum_{n=1}^\infty (-1)^n\frac 1{n!} \lambda^n x^{4n}\\ = \sum_{n=1}^\infty (-1)^n\frac 1{n!} \lambda^n\int_{-\infty}^{\infty}dx e^{-x^2} x^{4n}\\ = \sum_{n=1}^\infty (-1)^... 2 Quantum field theory on a cylinder? I think you are talking about conformal field theory in two dimensions. You can learn almost everything about this huge field from the book Conformal Field Theory by Philippe Francesco. Conformal field theory is a quantum field theory that has conformal invariance. If you are interested in quantum field theories that has ... 0 You should check out basically any introductions to string theory. QFT on S^1 \times R is exactly what you study when looking at closed strings. Ibanez and Uranga have published a Text book that I liked - you could also start by checking out these lecture notes: https://www.thphys.uni-heidelberg.de/~weigand/Skript-strings11-12/Strings.pdf 1 Questions 1. and 2. are intimately related. To try to make sense out of the \det = \exp \rm Tr \ln formula, the meaning of the trace has to be elaborated, I quote QFT and the Standard Model by Schwartz, line 30.58:$$\det U = \exp \rm Tr \log U = \exp \int d^4x \langle{x}|\rm tr \ln U |x \rangle{}$$Meaning that the \rm Tr written with uppercase ... 1 FWIW, OP's expression (5) does become the RHS of eq. (2) if we first take the limit \alpha\to 0 and then take the limit L\to\infty. In contrast, if we take the limits of OP's expression (5) in the opposite order we get zero. 3 There isn't a sense in which the extra terms obtained by using the renormalization group correspond to any specific subset of Feynman diagrams. As you have already pointed out, for \phi^4 theory, it is not true that you just sum the "bubble" diagrams; you need to calculate all of the corrections, and then these corrections will contain the correct \log^2(... 2 The important quantities in dimensional regularization are precisely the poles you will obtain in the limit \omega \rightarrow 2 and their associated residues. In other words, your bare correlation functions will involve integrals with some divergences,$$ I = \sum_{n = 1}^m\frac{a_n}{(\omega - 2)^n} + \mathrm{finite}, $$and the renormalization of your ... 0 The decomposition of the a and b that you refer to come from writing the most general solution to the (classical) equations of motion. This allows the identification of the physical degrees of freedom (a and b being essentially the Fourier coefficients of the solutions) which should be used for quantisation. 0 Maybe I understand a little better now. For a real physical process, one requires that q^2 is large and time-like (by evaluating q^2 in the CM frame of reference and thus in any frame of reference). However, in order to evaluate the perturbative QCD contributions relatively easily, it requires that q^2 is large. In principle, if one only considers ... -1 The first view is mostly wrong, a handwaving popularization that leads to wrong conclusions. Virtual particles exist only in the Feynman diagrams of any reaction one can think of, and they are always internal lines in the diagram. External lines represent a particle with a given identity, and a four vector on the mass shell of the particle. The identity ... 7 I suspect you are relying on the modern language, which is yet controverted by the effective theory community these days, if I am not too cut off from recent developments... I believe it is all hiding behind the receding obsession with renormalizability, and thus minimal coupling, obviated by the Wilson revolution. The point is the minimal-coupling ... 1 Equation 40 should read (a correct version can be found can be found in Shankars book)$$ e^Ae^B = :e^{A+B}: e^{\left\langle AB + \frac{1}{2}\left(A^2+B^2\right)\right\rangle} $$The procedure is essentiall laid out in the paragraphs under eq.(40); given A and B commute with [A,B], we have eq.(41) (this can be derived as a special case of the Baker-... 1 First of all there is a typo on the left-hand side of eq. (40) in Ref. 1. It should read$$T(e^Ae^B)~=~\ldots.\tag{40'}$$Here the time-order T is often not explicitly written in the notation. E.g. there is also an implicitly written time-order$$G_{\beta}~\equiv~\langle T\left(\ldots \right)\rangle\tag{39'}$$inside the correlator function (39). Under the ... 4 The "Mexican hat" Higgs potential is$$V(H)=\lambda(|H|^2-v^2)^2$$where v is the Higgs vacuum expectation value, 246 GeV, and the dimensionless Higgs coupling \lambda is about 0.0323. To get this value for \lambda, look at the quadratic "mass term" when expanding the Higgs field around its VEV, H=v+h. It is 4\lambda v^2 h^2, so the mass of the ... 0 The Hamiltonian, by construction singles out the time component,$$H = p \dot{q} - L, \quad \text{with}\quad p = \frac{\partial L}{\partial \dot{q}}meaning it breaks Lorentz symmetry explicitly or in other words the Hamiltonian is not a Lorentz covariant quantity. As it is written it makes little sense since you cannot operate with \Lambda_\mu^\nu on a ... 2 As it stands, the right-hand side of (3) is indeed zero, because it corresponds to summing over all of the degenerate vacua. In other words, the |0\rangle on the left-hand side of (3) is actually |0\rangle=\int dc\ |0(c)\rangle, where |0(c)\rangle is the vacuum state in which the expectation value of \phi_n is c_n. These different vacuum states ... 0 Negele & Orland is a good book for the basics. Covers the standard formalism in great detail, with all the derivations step by step. It has very few applications and actual calculations though. Altland & Simons has a wide range of problems,calculations and covers some more advanced topics, although often glosses over the details. Prerequisites ... 1 Here's the symmetric energy-momentum tensor of the free Dirac field: \tag{1} {T}_{\mu \nu}^{\textsf{D}} = i \, \frac{\hbar c}{4} \, \big( \, \bar{\Psi} \: \gamma_{\mu} \, (\, \partial_{\nu} \, \Psi \,) + \bar{\Psi} \: \gamma_{\nu} \, ( \, \partial_{\mu} \, \Psi \,) - (\, \partial_{\mu} \, \bar{\Psi} \,) \, \gamma_{\nu} \, \Psi - (\, \partial_{... 0 First, neutrinos are now known to have mass, meaning "sterile" right-handed neutrinos with zero electroweak interaction likely exist. Second, chirality is only conserved when it coincides with helicity. What's really conserved is angular momentum. A left-handed neutrino + a left-handed anti-neutrino cannot annihilate under the standard model. \nu_L + \... 1 First, what is he calling the number of excitations for the 2-atom system? There are two atoms coupled to a radiation mode, i.e. to a harmonic oscillator. The states |n\rangle refer to the harmonic oscillator, the states |S, m_s\rangle refer to the two atoms. (Each atom is assumed to have only two energy levels and thus behaves like a spin-1/2. Two spin-... 0 There is an unresolved problem with the equivalence principle. Acceleration thought experiments do not involve any curvature of spacetime. As far as I know all explanations of gravitational time dilation do. This does not necessarily invalidate the equivalence principle. It merely says that something somewhere is not right. The direct observation of ... 0 A generic four-vector field contains four degrees of freedom that upon quantization describe a spin-1 and a spin-0 particle. In the action for the Proca field, the spin-0 component is projected out by using an antisymmetric kinetic term. So the answer to your question really is: there are only three degrees of freedom in the Proca field since we want it to ... 1 The fact that D(x) = D(-x) follows immediately from the fact that the both the integrand and the integration measure are invariant under Lorentz transformations, and recalling that x \to -x (parity transformation) is indeed a Lorentz transformation. Alternatively, it's not difficult to prove it explicitly. \begin{align} D(-x) &= \int_{R^4}\frac{d^4k}... 0 An opérator may have different définitions. Here Derezinski defines P as the orthogonal projection on the normed coherent state \Phi_0 centered on 0. So P = |\Phi_0 \rangle \langle \Phi_0| He proves in the link i gave above that P is equal to the integralP = (2 \pi) ^{-d} \int e^{-\eta^2 /4} e^{-q^2 /4} e^{i q \eta /2 } e^{i \eta x } e^{i q D } d\...

4

The proof is unnecessarily confusing. It is hard to follow, and so it is hard to tell whether it is correct or not. But anyway, here is the simplest approach. In fact, let us do something slightly more general. Let $D=D(i\partial)$ be an arbitrary differential operator, and consider the expression $$\mathscr D:=\langle x|D|y\rangle$$ We will ultimately ...

1

I think I've managed to clear up my confusion and would like to summarize the above comments in an answer. Pions are described in QFT using three different fields, all satisfying the K-G equation. Due to their (almost) equal mass the three fields correspond to the same isospin number $T=1$ and form a basis $\{\pi^+,\pi^0,\pi^-\}$ for the $SU(2)$ ...

-1

Yes, I think you can use the equivalence principle within the realm of special relativity if, at least, the observer(s) is (are) inertial. In short, a person located at the center of a rotating disc, for a small compartment located a distance $r$ away from the center which orbits the observer, measures an anti-gravity (gravity outward the center). His ...

0

Yours and his expressions are equal. Remember that the sum over $i$ in your series run from $i=1$ to $i=N$, only. The other part of the combinatorics is just $$e^{x+y}= e^xe^y$$ or $$\sum_s \frac 1{s!} (x+y)^s= \sum_{n=1}^\infty\sum_{m=1}^\infty \frac 1{n!} \frac 1{m!}x^n y^m$$

1

Yes, in quantum field theory a equal-time (super)commutator relation, say, $$[\hat{\phi}({\bf x},t),\hat{\pi}({\bf y},t)]=i\hbar \hat{\bf 1}~\delta^3({\bf x}\!-\!{\bf y}),$$ can be differentiated on both sides wrt. spatial derivatives.

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