# Tag Info

0

The theorem says that higher spin particle cannot couple to gravity in a consistent way. You can find some notes on the Weinberg-Witten theorem in David Bailin's website. http://www.phys.susx.ac.uk/~mpfg9/

2

Whilst the question is not a resource request, I would recommend Edward Witten's paper on the topic published in 1988, titled, 2+1 Dimensional Gravity as a Soluble System. In the paper, Witten shows: $2+1$ dimensional gravity with or without $\Lambda$ is soluble classically and at the quantum level $2+1$ dimensional gravity is related to a Yang-Mills ...

2

Classically they are clearly topological. The metric does not appear, and you don't need a metric for integration on manifolds to make sense. Now in dimension 3 you can cast the Einstein-Hilbert action into a Chern-Simons theory as you say. The connection takes it values in the Lie algebra of the Poincare group. In higher dimensions you need to use higher ...

2

The only naturally occurring symmetry breaking radiation of this kind is the CMB. Unless you are talking about charged particles of more than approx. 1e19eV energy (in the CMB rest system), the effects are negligible, as far as I know. For those ultrahigh energy particles, however, this so called Greisen–Zatsepin–Kuzmin limit (GZK limit) forms a cosmic fog ...

2

The gravitational Chern-Simons action is topological, yes. The gauge connection encodes the field of gravity and since it is being integrated over, the result does not depend on a metric. (In the expressions you write maybe the vielbein contribution is missing? Or maybe you mean to have absorbed it in the notation.) Notice that it's just the usual ...

0

This is really straight forward, once you get used to the notation. (Don't you hate it when people say that?) $$[\pi (\vec{x},t), (-\nabla^{2} +m^{2})\phi (\vec{y},t)] ,$$ Here you need to remember that $\nabla^2$ acts on the $\phi(\vec{y},t)$ only, so $\pi$ can pass right through this wave operator. Now when you evaluate the commutator you'll end up with ...

6

I don't think that there would be any more diagrams. Having a total derivative term in the Lagrangian leads to derivative interaction vertex, which after symmetrising gives you something like $$ig \sum_i p_i \ ,$$ where $g$ is some coupling and $p_i$ the momenta of the particles. This vertex, however, vanishes due to momentum ...

4

Let us consider a real, scalar field theory for simplicity (and metric signature $+ - - -$). In the free theory, one can use the mode expansions of the field $\phi(x)$ and its canonical conjugate momentum $\pi(x)$ to derive the following expressions for the creation and annihilation operators: $$a(p)=i\int d^3x\ ... 2 First, the obvious explanation for the sign is that if J has a minus sign in (1), then there should be a minus sign in (4). For some reason your G turned into \phi_i. Assuming that they are the same thing, then I'm not sure I understand your problem. We didn't use the homogeneous KG equation to get the delta function; we used the inhomogeneous one, ... 1 A wavefunction is (typically understood to be) a complex valued function on configuration space; a wave function assigns a complex number (probability amplitude) to each point in configuration space. For a system of N particles, the system's wavefunction is 3N dimensional. A quantum field, on the other hand, is an operator valued function on physical ... 0 A quick answer is that simple wave functions can not describe the processes of annihilation and creation represented in fields. Fields support multi-particles states in a way that wave functions can not. 0 The error is that \gamma_5 doesn't intrinsically change sign under parity. Also, don't forget that under parity the spatial components of W_\mu change sign. And also \gamma^0 \gamma^\mu \gamma^0 \neq \gamma^\mu. 1 In the standard model the mass for the quarks and other elementary particles comes from from the Higgs mechanism. In the case of the quarks, the masses given in the table of the link are calculated using convoluted theoretical models and data input from scattering experiments.4 At the moment chiral symmetry breaking does not contribute to the quark masses ... 2 In the second approach, you do have the option to treat the short distance physics in a variety of ways. Choosing point interactions will give you some particular values for  \lambda _{ bare} and some particular rules about how to calculate \delta \lambda , and if you choose a different "UV completion" you will get a different value and a different rule. ... 0 We know from translations symmetry that the expectation value of the wavevector operator is constant -- that is,$$ \langle \mathbf{k} \rangle = \langle \mathbf{k}_1 + \mathbf{k}_2 + \ldots \rangle = \text{const.} $$In other words, wavevector is a conserved quantity. If the constant of proportionality between \mathbf{k} and momentum were different for ... 0 Weak interactions with W and Z gauge bosons violate parity simply because the righ-handed and the left-handed fermions coupled differently to W and Z. For example, the W's couple only to the left-handed fields. A parity inviariant dynamics would require that both left- and right- fields couple in the same way to the gauge vector since they get ... 0 When you act \phi(t,\vec{x}) on the vacuum, you get a particle at (\vec{x},t), that is |\vec{x},t>=\phi(t,\vec{x})|0>=\int \frac{d^3 k}{(2\pi)^3}e^{-ikx}a^\dagger _k|0>. It's a Fourier transformation. And it means that if a particle wants to be in (\vec{x},t) then it should have all kinds of momenta, which is actually the uncertainty ... 3 The neutron is not a fundamental particle. It carries no electric charge, yet it can interact with photons as its components - the quarks - carry electric charge and thus couple to photons. Macroscopically/classical, these interactions cancel out since its net charge is zero, but quantumly, there is a very big difference between objects with charged ... 1 This is at the current limit of my understanding of fermionic QFT. It's such a good question that it deserves an answer from an expert. a) Fourier analysing the second quantized Dirac operator field gives, \hat{\psi}^{a}(x), $$\hat{\psi}^{a}(x)=\int \frac{d^{3}p}{(2\pi)^{3/2}}\exp(ix^{r}p^{r})\hat{\psi}^{a}(p)$$ where the label ... 0 I have been able to determine the answer to this question myself. If A is a flat connection, then$$ \partial_z A_{\bar z} - \partial_{\bar z} A_z - i [ A_z , A_{\bar z} ] = 0 \implies A_z = i U^{-1} \partial_z U $$for some scalar U \in {\cal G}. Now, we wish to invert the equation$$ D_z \phi = f(z,{\bar z}) $$To do this, we note$$ D_z \phi = U^{-1} ...

3

Maybe looking at the situation the following way helps to clarify: The field $J$ is indeed another field, but it is not dynamical (there is no $\partial_\mu J \partial^\mu J$ term in the Lagrangian). In particular that means, that $J$ does not propagate, so there should not be lines associated to $J$ in the Feynman diagrams. This should answer your final ...

0

I guess it is because you first of all change sign of $\vec x$ to $- \vec x$ in physical space(this is parity transformation in a nutshell). All this peculiar algebra concerning left and right chirality fields comes from $J = 1/2$ Lorenz group representation, so transformation rules are defined as representatives of parity transformation of physical space.

-1

Okay, I think I have an idea why the terminology is used, but I think this argument makes little sense: The Lagrangian term describing weak interactions is of the form $$\bar \Psi \gamma_\mu P_L \Psi W^\mu$$ Under parity transformations $\Psi \rightarrow \gamma_0 \Psi$ and $\bar \Psi \rightarrow \bar \Psi \gamma_0$, therefore $$\bar \Psi ... 2 For anyone with similar problems: The following observation has helped me immensly: We have in fact four particles directly related to an electron: A left-chiral electron \chi_L, with isospin -\frac{1}{2} and electric charge -e, A right-chiral anti-left-chiral-electron (\chi_L)^c=\chi_R with isospin \frac{1}{2}, electric charge +e A ... 3 The Higgs mass does not stem from eating Goldstone bosons, since the Higgs is not a gauge field. Since we are breaking an \mathrm{SU}(2) \subset \mathrm{SU}(2)_L \times \mathrm{U}(1)_Y completely, we have three Goldstone bosons, which are eaten by three of the four electroweak gauge bosons to form the massive W^\pm,Z with the photon remaining massless. ... 1 This is a kind of optical amplifier -- see 1, 2. Probably the main effect of the molecules on the beam will be scattering / refraction, just like if you put anything into of a beam of light. But a secondary effect, especially if the beam is much much larger than the clump of molecules, is that the profile of the beam will change a bit. The part of the beam ... 1 Lorentz spinors appear as irreducible representations of the group SL(2,C). Elements of the group are 2x2 matrices with complex entries and unity determinant. A Lorentz spinor is a two component vector \psi^{A},\chi^{A}\in V_{2} with A=1,2. The Levi-Civita tensor \epsilon_{AB} is an invariant tensor under SL(2,C). This means that if we have an irrep ... 1 This is simply a Fourier transform. The wave equation, Hamiltonian, and Green's functions are all simpler in momentum space than in position space. This is because, for the free, theory, the different momenta decouple, and you can create a particle with momentum k with a^\dagger_k \left|0\right\rangle, and that state will evolve in a nice way. This is ... 0 What you did seems fine to me, and indeed you can show that \frac{\partial}{\partial{p}^k}(a_\vec{p}^\dagger{a}_\vec{p})=0, just take the definitions$$a_\vec{p}=\int{d^3x}\,e^{-i\vec{p}\cdot\vec{x}}\left[\frac{E_\vec{p}}{2}\phi(\vec{x})+\frac{i}{E_\vec{p}}\dot\phi(\vec{x})\right]\\ ...

1

In particle mechanics you integrate along a path, which is bounded by points, but in field theory you integrate over a spacetime volume, so your boundary is a hypersurface, not just points. For a typical quantum field theory process (at least the way it's formulated for calculations), there is some initial state consisting of noninteracting wavepackets, ...

1

In field theory the values of the field at every point in space are independent degrees of freedom, just like the positions of different particles in a multi-particle system. So, AFAIK to specify the initial and final configurations for an action integral you have to give the values of the field at every point in space at the initial and final times. The ...

3

Roughly sketched, for the quantized Dirac field one has: $$\hat\psi(x)\sim \int d\mathbf{p}\, \sum_r \bigg[ u^{r}(p)\, \hat a^r_\mathbf{p}\,e^{-ipx}+v^{r}(p)\, {\hat b^r_\mathbf{p}}^\dagger e^{ipx}\bigg],$$ where $r=\pm1$ denotes helicity. The ${\hat a^r_\mathbf{p}}^\dagger$ operator creates a helicity-$r$ particle state when it ...

0

All the experimentally observed particles that are believed to be fundamental particles, but for neutrinos, in the Standard Model get a mass from the interaction with the Higgs boson. Please note that this aspect of the dynamics of the Standard Model is still under experimental investigation, despite there is data that clearly points towards this direction. ...

2

The Dirac operator, as we know it, is $D_\mu\gamma^\mu$ with $D$ as the gauge covariant derivative. Using the Fujikawa method of deriving the Adler-Bell-Jackiw or chiral anomaly, one finds that the anomaly of the chiral current is given by $$\partial_\mu \langle (j^5)^\mu\rangle = 2 \mathrm{i}A(x)$$ where $$A(x) = \int \sum_n \psi_n^\dagger \gamma^5 ... 1 Neutrino definitely have anti-particles. Another question whether an anti-particle differs from its "particle" or not. The charge conjugation operator may generally change a given "neutrino state" because a neutrino solution is not completely determined with just Dirac's equation. 2 We then talk about a left-chiral electron we do it in an informal way, you are correct that a massive particle cannot be inherently chiral. To see this, let us remember that he handedness of an elementary particle depends on the correlation between its spin and its momentum (helicity). If the spin and momentum are parallel, the particle can be said to be ... 1 One usually starts from the CCR for the creation/annihilation operators and derives from there the commutation rules for the fields. However, one can start from either (see for example here about this). Suppose we want then to start from the equal-time anticommutation rules for a Dirac field \psi_\alpha(x):$$ \tag{1} \{ \psi_\alpha(\textbf{x}), ...

0

A simple (and quite accurate) answer is that quantum fluctuations are the fluctuations that exist at zero temperature. What it means is that even at zero temperature, there might be fluctuations in the measurements of observables, which does not happen for classical systems at zero temperature, due to the non-commutativity of the dynamical and potential ...

3

The first and last terms strictly and completely disappear. They are equal to zero because $k_i a_k a_{-k}$ (just like its hermitian conjugate) is an odd function of $\vec k$ – a function obeying $f(-\vec k) = -f(\vec k)$ – and the integral of an odd function vanishes because the contribution from the region around $\vec k$ is exactly cancelled by the ...

1

As far as I know, the number of points to not have any influence on divergent behaviour. The infinite vacuum energy comes from the fact that we allow arbitrary frequencies for our quantum fields. There is no difference if we sum or integrate a constant from zero to infinity, the result is still infinite. $$\sum_{k=0}^\infty {1\over 2} \sqrt{k^2+m^2} ... 1 The Lamb shift According to the hydrogen Shrodinger equation solution, the energy levels of the hydrogen electron should depend only on the principal quantum number n. In 1951, Willis Lamb discovered that this was not so - that the 2p(1/2) state is slightly lower than the 2s(1/2) state resulting in a slight shift of the corresponding spectral line (the ... 2 You're close, but wrong. Indeed, one should always consider both possible directions of propagation - from x to y and from y to x. But we do not multiply amplitudes for different processes, we add them. The total amplitude for a propagation process is given by$$ D(x - y) + D^\dagger(x -y) = D(x - y) + D(y - x)$$for D(x - y) = \langle ... 1 from The Reference Frame: Another wrong expectation that a beginner could have - and usually has - is that if you allow the summation over all trajectories of a particle, the typical particles will move faster than light most of the time and this will automatically result in a violation of the special theory of relativity. So an overzealous ... 1 Consider the partition function$$Z = \int D\phi ~ e^{-S_0 - S_I},$$where S_0 is the Gaussian/free part and S_I is the interaction part of the action. Within a perturbative framework we may aim to systematically include the contributions of fast modes to the (effective) action for slow modes. For this we expand in the interaction strength as$$Z = \int ...

0

We are interested in computing $$\Pi^{\mu\nu,ab}(x) = \langle V^{\mu,a}(x) V^{\nu,b}(0) \rangle = \langle : {\bar \psi}(x) \gamma^\mu M^a \psi(x) : : {\bar \psi}(0) \gamma^\nu M^b \psi(0) : \rangle$$ Several structures of this quantity can be directly computed. Firstly, let us study the index structure. Firstly, due to translational invariance we have ...

1

In my understanding $:\hat{f}_2\hat{f}_1^\dagger\hat{f}_2^\dagger:$ is defined to be equal to $\hat{f}_1^\dagger\hat{f}_2^\dagger\hat{f}_2$. Otherwise there would be no need to write the two "$:$" and give this operation a special name. What your are doing is applying the ordinary commutator relations. Therefore  ...

0

I just realized that the answer is just stupid. In the case of two outgoing photons when computing $M^*$ you get the epsilons (not complex conjugated) that in the full $|M|^2$ will allow you to use $\sum\epsilon_{\mu}\epsilon^*_{\nu}\to-\eta_{\mu\nu}$

1

The Universe is infinite. If the energy density of the universe is greater than zero, then the total energy must also be infinite.

1

According to Grand Unification Theory, protons can decay into electron (even at low energy; just the probability is very low). It doesn't mean you can replace proton with electron.

1

One comes to this conclusion due to the fact that the contraction of a symmetric tensor with an antisymmetric one vanishes. Writing down the loop diagrams involves a contraction of both vertices. If you get expressions proportional to $\epsilon_{\mu\nu}\eta^{\mu\nu}$, this will be zero due to the fact that the metric is symmetric and the epsilon tensor is ...

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