# Tag Info

1

Since $\delta \omega_{\rho\sigma}$ is antisymmetric, the part of what's inside the parentheses symmetrized over $\rho$ and $\sigma$ will vanish because if you contract antisymmetric indices with symmetric ones, you get zero. So you can anti-symmetrize over that pair of indices inside the parentheses without changing anything, which is precisely what's ...

3

Maybe I am missing something but $$\sum_{n \neq 0} e^{-i E_n (T-i\epsilon)} |n \rangle \langle n|0\rangle =\sum_{n \neq 0} e^{-i E_n T} e^{-\epsilon E_n } |n \rangle \langle n|0\rangle$$ and not a sum of series. This really allows you then to kill non zero $n$ as $T$ goes to infinity. As for the physical meaning of this trick, ...

1

The Lagrangian density for a Dirac field is $$\mathcal{L} = i\bar\psi\gamma^\mu\partial_\mu\psi -m \bar\psi\psi$$ The Euler-Lagrange equation reads $$\frac{\partial\mathcal{L}}{\partial\psi} - \frac{\partial}{\partial x^\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\right] = 0$$ We treat $\psi$ and $\bar\psi$ as independent dynamical ...

3

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

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 ...

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 ...

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 ...

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 ...

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. ...

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. ...

0

Here is my solution. For the Ramond Ramond Sector, $${m_\rm{I}} = \frac{{\left[ {\hat a_ + ^\mu ,\hat{\tilde a}_ + ^\nu } \right]}}{2}{m_{{\rm{IIB}}}}$$ For the Neveu-Schwarz Neveu-Schwarz Sector, $${m_\rm{I}} = \frac{{\left[ {\hat d_ {-1/2} ^\mu ,\hat{\tilde d}_{-1/2} ^\nu } \right]}}{2}{m_{{\rm{IIB}}}}$$ For the Ramond Neveu-Schwarz Sector or 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 ...

0

Decoherence is more than anything a matter of what you define to be the "environment". The environment is supposed to be external to the system of interest and entangling interaction with it produces decoherence. If the environment in question is a part of the adS space then the subsystem can certainly decohere. If what you are asking is whether the space as ...

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 ...

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 ...

0

Generally, in the MSSM one works with the "minimal flavor violation" paradigm that states that all flavor violation originates in the SM Yukawa sector. This paradigm is ad hoc, but explains why no huge SUSY contributions to FCNC observables are seen. There are models that go beyond minimal flavor violation and some that give an explanation for the ...

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 ...

0

Let us consider the Hamiltonian of an electromagnetic field in a finite region of space. It is possible to find a suitable unitary transformation to change it in a form same as a Hamiltonian of a independent harmonic oscillators. A photon is a quantum of energy of such an ensemble of independent oscillators. All excitations of this ensemble will be ...

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 ...

0

I don't know if I'm right and it's not directly my topic, but naively guessing I think that sea quarks emerge from fluctuations in the vacuum, for example as a pair of a quark and its antiquark. And as such, because of conservation of angular momentum, each of these pairs then should have zero total angular momentum.

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

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 ...

2

The Pion fields are the coordinates of the Stereographic projection: $\phi_i = \frac{2 \pi_i}{1 + \pi^2} , i = 1, ..., n-1$ Where: $\pi^2 = \sum_{i=1}^{N-1} \pi_i\pi_i$ And: $\phi_n = \frac{-1 + \pi^2}{1 + \pi^2}$ As can be seen, this construction solves the constraint equation: $\sum_{a=1}^{N} \phi_a\phi_a= 1$. Substituting in the Lagrangian, we ...

0

If you solve the equation to second order you obtain: $g=\frac{g_0}{1-a g_0 log\lambda}=g_0 +\sum_n g_0^{n+1}a^nlog^n \lambda$ the second part is the contribution of powers of the first loop diagram, for instance try to write g at 1-loop first and then sum all the powers of the first order diagram.

4

Very rougly, the logarithm has the function of transforming a product into a sum, therefore if the partition function is factorizable (i.e. disconnected), it will give a sum on connected parts. For a slighly more rigorous discussion, consider the Hamiltonian of two non interacting systems, the partition function $z(j)=Z(j)/Z(0)$ (without vacuum ...

0

try the lecture notes by van Hees from Frankfurt. Check out p. 114 http://faculty.ksu.edu.sa/djdou/Lectures%20Notes%20PHY556/Introduction%20to%20Q.F.T.pdf

3

The NS-NS sector is the same in type IIA and IIB, but the R-NS and NS-R sectors differ. The type IIA theory is non-chiral, so the R-NS and NS-R fields transform in $\mathbf{8}_s \otimes \mathbf{8}_v$ and $\mathbf{8}_v \otimes \mathbf{8}_s'$, where $\mathbf{8}_s$ and $\mathbf{8}_s'$ are the two eight-dimensional spinor representations of $SO(8)$. Type IIB, on ...

4

The only argument I can find for this is a couple pages earlier, where they say ...which in turn implies that the photon self-energy diagrams have the structure $= i(q^2 g^{\mu\nu} - q^\mu q^\nu)\Pi(q^2)$ The only divergence possible is a logarithmically divergent contribution to $\Pi(q^2)$. In non-Abelian gauge theories, (16.57) still holds, ...

2

The most physical and understandable definition of Nekrasov's partition function to me uses five-dimensional gauge theories. Namely, any 4d N=2 susy gauge theory has a 5d version with the same matter content, so that compactifying it on a small $S^1$ brings it back to the original 4d theory. Then we put the theory on the so-called Omega background: it is ...

4

For simplicity, let's restrict the discussion to that of a single particle moving in one dimension. Path integrals can be performed in much broader contexts like quantum field theory, but I think that would conceptually obscure the issue at this point. Let $H$ denote the (time-independent) quantum hamiltonian. Then the time evolution on the system is ...

0

What your friend actually meant is that you can obtain all desirable correlation functions. Assuming you're talking about a non-relativistic electron, consider a source term added to your action $$S'[x] = S[x] + \int dt \, J(t) x(t)\,.$$ Now you can write any correlation function as a derivative of $\ln Z$ calculated at $J = 0$, i.e. $$\langle B| ... 2 In QM we have the coordinate representation. In this case the basis consists of eigenfunctions of the position operator r_i, where i runs over the discrete set of degrees of freedom. Now, when we go to QFT, i generalizes to the continious space-time position x, and r generalizes to the field \phi:$$ r\rightarrow\phi\\ i\rightarrow x\\ ...

2

I am not sure that one will appreciate this answer, but I will try. I think that the punchline may not always be the case. It seems that you want some symmetry which forbids some counterterm. This means that this counterterm would have some unappreciated behaviour under the symmetry. Then you want the same symmetry to forbid the divergence. Why? Naively, ...

0

There's no related exception for tachyons. Tachyons' statistics must be determined a priori. Most typically, tachyons have to be bosons – and under certain additional assumptions, they have to be scalar (spin-zero) bosons. They differ from massive bosons just by the fact that the mass term $m^2\phi^2/2$ has the opposite sign – opposite sign of their $m^2$. ...

1

First, a reference article, by Witten, http://arxiv.org/pdf/hep-th/9802150v2.pdf I'll try to expose the basic idea, with a flat space-time. Suppose you have a relativistic scalar field theory, on a flat space-time domain, with boundary. The equation of the field is : $$\square \Phi(x) = 0$$ (fields on-shell) Now, define the partition function $$Z = ... 0 I asked my professor and in a discussion we came up with the following. The process of establishing the effective action for a fluctuation Lagrangian to consist of the functional determinant of the initial differential operator involved, relies on the equality:$$\det(A)=e^{Tr(\log(A))}$$for a matrix A, which is only true for diagonalizable ... 1 The pole corresponds to an on-shell particle going from one point to another. Then, the residue effectively tells you how many of those particles are being transmitted. Since in your physical/renormalized theory, the propagator should correspond to 1 quantum of the renormalized field being transmitted, you set the residue at the pole to 1. 0 TO have an instanton solution, you need to map the (euclideanized) "spacetime at infinity" to the group manifold. In the case of SU(2), both the spacetime at infinity and the group manifold are S^3 and instantons are characterized by the integers. I hope you understand that much, at least for SU(2). If you're interested in 4d instantons, they are ... 2 Yes, suppose [g] = \delta. By dimensional analysis only we can write that a loop diagram contributes$$ \sim g^{n} \int \frac{d^4 k}{k^{4-n\delta}} $$If \delta=0, this diverges logarithmically, but can be re-normalized. If \delta is less than zero, it diverges by simple power counting. This is VERY informal. Technically, you should study the ... 1 The OS condition that$$ \frac{\partial\Sigma}{\partial p^2}|_{p^2=-m^2} = 0 $$implies that the residue in the propagator remains equal to one. Suppose that we used a different renormalization scheme in which our counter-terms contain no finite parts (e.g. MS scheme). In the OS scheme, we removed finite parts which were logarithmic in our artificial ... 3 For Poincare algebra there are (as far as I know) two different approaches to find its representations. In first approach one begins from a finite dimensional representation of (complexified) Lorentz algebra, and using it one constructs a representation on space of some fields on Minkoski space. Representation so obtained is usually not irreducible and an ... 0 Gauge symmetries are redundancies of the description, not a part of physics; gauge fields have surplus structures (e.g., non-physical polarizations) one brings in to describe the system more conveniently (for instance in a local and manifestly Lorentz covariant form). You can describe a gauge system in a language that does not have gauge symmetry at all. A ... 1 First of all, we don't usually talk about the direction of propagation of a plane wave in QFT. Plane waves are said to exist at all spacetime coordinates with a certain internal momentum, k. And, in reference to some of the comments, in QFT, we don't normally operate with wavefunctions. We promote wavefunctions to operators and act on states. But in this ... 4 With respect to the discussion of momentum-eigenstates and the following derivation in Weinberg's book, \sigma is just a label that denotes any degree of freedom that is not momentum. Even though it can be identified with spin, its nature is not relevant for the discussion at hand. 1 The statement can be understood in terms of the GKPW-formula (named after Gubser, Klebanov, Polyakov and Witten), which does exactly that: it relates correlation functions on the CFT side (boundary) to string amplitudes on the AdS side (bulk). Assume that \phi(\vec{x},z) is some field in the bulk, where z is the so-called "holographic coordinate", which ... 2 The the easiest way to see that time reversal transforms electrons into positrons relies on the fact that PCT (parity, charge conjugation and time reversal) combined are a symmetry of every Lorentz-invant QFT. Using P^{-1} = P, C^{-1} = C, T^{-1} = T, i.e. a parity transformation is undone by a second parity transformation etc. you can see that$$1 = ...

0

This link has a very easy to follow derivation of the scalar field Feynman propagator. Once you understand that, you should be able to do it for vector particles like photons too. The correct answer is given at the end of the chart in this link.

7

I'll give you enough hints to complete the proof yourself. If you're desperate, I'm following the notes by Zuber, which are available online, IIRC. Let's start with some notation: pick some basis $\{t_a\}$ of your Lie algebra, then $$[t_a,t_b] = C_{ab}{}^c t_c$$ defines the structure constants. If you define $$g_{ab} = C_{ad}{}^e C_{be}{}^d,$$ then this ...

7

I) The Casimir invariants of a Lie algebra $L$ over a field $\mathbb{F}$ are the central elements of the universal enveloping algebra $U(L)$. Example: The angular momentum square $\vec{J}^2$ is a quadratic Casimir invariant of the Lie algebra $L=sl(2,\mathbb{C})$. II) Given a bilinear associative/invariant form $B:L\times L\to \mathbb{F}$, one can create ...

1

You can write an infinitesimal transformation, with generator $J$, as $$R(\delta\theta) = 1 + iJ\delta\theta$$ A finite transformation is a succession of $N\to\infty$ infinitesimal transformations, $$R(\theta) = (1 + iJ\theta/N)^N = e^{iJ\theta}$$ The rotations $O(3)$ are isomorphic to $SU(2)$, with generators $J = \sigma/2$. The Lorentz ...

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