Hot answers tagged

6

Yes. Ordinary quantum field theory is as wrong as Newtonian gravity for not including GR effects. That is to say, it is a perfectly fine theory inside its domain of validity, which in this case means pretty much everything below the Planck scale, just as Newtonian mechanics is valid for speed much less than the relativistic scale (the speed of light). ...


3

(1) The operator $U(\Lambda,a)$ is a unitary "rotation" in the Hilbert space corresponding to an inhomogeneous Lorentz transformation of the spacetime coordinates. When $U(\Lambda,a)=\exp(iH\tau)$, it is an operator that adjusts the clock forward by $\tau$. Conceptually this is not a physical time evolution of the system. (2) A unitary rotation $U$ in the ...


3

Just one comment to the higgsss answer. Formally from the Wigner theorem we have that if there exist time shift symmetry, for which the scalar product of quantum mechanical rays is conserved, $$ \tag 1 |\langle \psi (t)|\kappa (t)\rangle| = |\langle \psi{'}(t+\tau)| \kappa{'}(t+\tau) \rangle|, $$ then the symmetry transformation acts on $|\psi\rangle$ as ...


2

(An answer from a student also studying Weinberg's book) Basically I'll be repeating what @Robin Ekman has said in his comment, but with some clarifications. First of all, here Weinberg is talking about the states of a single massive particle. That is, $\Psi_{k,~\sigma}$ and $\Psi_{k',~\sigma'}$ are two possible momentum eigenstates of this particle, and ...


2

1) As you know, $$ \tag 1 \theta\epsilon^{\mu \nu \alpha \beta}F_{\mu \nu}F_{\alpha \beta} = \theta\partial_{\mu}K^{\mu}, $$ where $K_{\mu}$ is the so-called Chern-Simons class. The Feynman diagrams method tells us that the term $(1)$ defines the diagram which corresponds to the two-photon (or two-three-four non-abelian bosons) vertex $V_{A}$ (where the ...


2

First let us address "emty space". Empty space is a theoretical concept, a space where there is no matter and no energy. In our universe, no matter how far away one goes in space, it is not empty. It contains the cosmic microwave background radiation, cool photons, which is at a temperature of 2.7 K . Within quantum mechanics and elementary paricles, the ...


1

Rather tautologically, a non-perturbative effect is one that is invisible to perturbation theory. An effect is invisible to perturbation theory exactly if it is in a non-analytic part of the partition function with respect to the coupling constant $g$. Observe that perturbation theory is essentially the Taylor expansion of the partition function $Z$ (or ...


1

Disclaimer: Renormalization is a huge subject with many facets, such as, e.g. overlapping divergences of subgraphs, regularization, renormalization group, etc. Here we will only elaborate on OP's quote from Ref. 1. Ref. 1 is considering a Feynman diagram ${\cal F}(q_1, \ldots, q_E)$ in momentum Fourier space, with external 4-momenta $(q_1, \ldots, q_E)$, ...


1

Even if a theory is naively (classically) scale invariant (eg: the scalar theory with $\lambda \phi^4$ interaction therm), quantum mechanically, the 4-point scattering amplitude depends on the energy of the scattering particles (as can be shown by a one-loop computation. Tree level computations are the classical approximation). Suppose the scattering ...


1

That $g^{-1}\mathrm{d} g$ is Liealgebra-valued for a Lie group-valued function $g$ has nothing to do with the particular model or with physics, it is true for all matrix groups. Write $g(x) = \exp(k(x))$, where $k(x)$ is now Lie algebra-valued and $\exp$ is the usual power series in the case of a matrix group. Then $\partial_\mu g = \partial_\mu k\exp(k)$ by ...


1

Can I simplify this to pick out of the summation only the mode that matches the transition energy $\hbar \omega_{21}$? This would be because photons are emitted/absorbed entirely anyway, so the energies have to match, and the photons which do not have the right energy simply won't interact with the atom That's exactly right. Let's see it in detail. ...


1

If $\mathbf{A}$ does not depend on time it follows from Maxwell's equations that, excluding a linear growth of $\mathbf{E}$ in time, that it is also time-independent. We can try to quantise $\mathbf{A}$, \begin{equation*} \mathbf{A(x)=}\sum_{j}\int d\mathbf{k}\frac{1}{\sqrt{2k}}\{a_{\mathbf{k}% j}^{\ast ...


1

this paper might help.$^1$ It's written pedagogically and hence is easier to read. It goes on to discuss a lot more than just color decomposition too. $^1$ Scattering Amplitudes, Henriette Elvang, Yu-tin Huang.



Only top voted, non community-wiki answers of a minimum length are eligible