Hot answers tagged quantum-field-theory
7
One shouldn't imagine the T-duality between the two heterotic strings to be a $Z_2$ group, like in the case of type II string theories' T-duality. In type II string theory, there is only one relevant scalar field, the radius of the circle producing T-duality, and it gets reverted $R\to 1/R$ under T-duality.
In the heterotic case, it's more complicated ...
5
I believe it is the same field for all photons in the universe. Similarly there is one field for all up quarks in the universe, another field for all down quarks in the universe and so on for each particle type. In this way the field theory model explains the uncanny consistency of particles; a neutrino made in a supernova millions of miles away is identical ...
5
Because you can prepare a state with an arbitrarily long wavelength, hence arbitrarily low energy, photon. That's essentially the definition of a massless particle. If you put in an IR regulator, by putting the system in a box for example, a gap appears since there is now a largest possible wavelength. This can be mimicked by giving the photon a small mass. ...
3
Yes it is.
The volume form on any (pseudo-)Riemannian manifold $(M,g)$ of dimension $n$, where $g$ is the metric, is given in local coordinates $(x^1, \dots, x^n)$
$$
\sqrt{|\det (g_{\mu\nu})|}dx^1\wedge \cdots \wedge dx^n
$$
where $\det(g_{\mu\nu})$ is the determinant of the metric in these coordinates. In cartesian coordinates, the determinant of the ...
2
The theory is (even classically) not scale invariant. Just by dimensional analysis, you can note that the scalar field has scaling dimension 1, and the mass (as the name suggests) must also have a scaling dimension of 1. So $m^2$ has a ascaling dimension of 2, which suggests the RG equation which you've written in the question. That essentially says that the ...
2
Hagedorn spectrum just means that the density of states varies exponentially with the energy/mass. $m^2$ (asymptotically) given by the "level" (N) of the state (upto a sqrt). The number of states at level $N$ corresponds to the possible partitions of $N$ into different oscillator modes. That means that the number of states at level $N$ will increase ...
2
The magnetic quadrupole moment tensor is given by
$$m_{ij}=\left\langle \frac{2}{3}\left(\mathbf{r}\times\mathbf{J}\right)_i r_j \right\rangle,$$
in analogy with the magnetic dipole moment vector
$$m_i=\left\langle \frac{1}{2}\left(\mathbf{r}\times\mathbf{J}\right)_i \right\rangle.$$
The magnetic field at a point $\mathbf{R}$ is then, up to quadrupole ...
2
I wouldn't say that "holographic theories are non-local by definition". On the contrary, in AdS/CFT the CFT is completely local and satisfies cluster decomposition.
The cluster decomposition property in AdS can be proved using the CFT bootstrap for all CFTs in $d > 2$ (see http://arxiv.org/abs/arXiv:1212.3616, the proof only requires CFT `axioms', ...
1
There are lots of questions here! I think I can answer at least some...
First of all, you are aware that the fields in $W$ and $K$ are superfields? These contain the entire supermultiplet, so they must be complex in general. This is a short entry but it links to others: http://en.wikipedia.org/wiki/Superfield
As mentioned by Jose in his comment, the ...
1
One way of defining conformal transformations are by a (positive) local scaling of the metric, of the form $e^{2 \phi}$. Such a transformation always preserves the sign of spacetime distances. In particular, the light-cone remains unchanged since null distances map to null distances. What was inside the lightcone stays inside and things outside stay outside. ...
1
Probability of photon emission by an atom depends on the occupation number of already existing photons of this sort. The corresponding occupation number is determined with the "boundary conditions". We cannot take the occupation number of all existing photons in the Universe, so time-space "separated" regions have effectively their own photon fields and ...
1
1PI graphs don't contain singularities at $q^2=0$ because those only arise from propagators that carry the external photon momentum $q$. The external ones are omitted (as factors), as you said, and if the graphs had a single propagator with the momentum $q$, it could be cut to two pieces by cutting this propagator and this is by definition a diagram that is ...
1
Perhaps not a totally satisfactory answer, but a partial clarification of one of the things I was confused about:
In the semiclassical treatment of the Hawking radiation process, there is no need to have an interacting quantum field theory. Therefore the vacuum-vacuum bubble diagrams of interacting perturbation theory are completely irrelevant to the basic ...
1
I would really recommend a study in QFT before going on to study SUSY. QFT has many quirks that make supersymmetry a very interesting expansion of the regular framework. You'd miss out on all that as you just had to believe the facts presented w/o following the thought that lead to the results in detail.
On the Mathematical level you will need Grassmann ...
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