New answers tagged

3

I'll address your issues with definition (1): $E$ is a function of $\vec p$ because $\lvert \lambda_{\vec p}\rangle\sim\lvert \lambda_0\rangle$ where by $\sim$ I mean that they are related by a Lorentz boost. That is, to "construct" these states, you actually first sort out all the states $\lvert \lambda_0\rangle,\lambda_0\in \Lambda$ ($\Lambda$ now denotes ...


1

Worldsheet supersymmetry is the fermionic symmetry of the worldsheet RNS action under the worldsheet supsresymmetry transformations that look like $$ \sqrt{\frac{2}{\alpha'}}X \mapsto \sqrt{\frac{2}{\alpha'}}X + \mathrm{i}\bar{\epsilon}\psi^\mu$$ and which I'm too lazy to type out for all fields (and which also depend on whether or not we're looking at the ...


1

Mathematically, the R-matrix is an (invertible) element of a quasi-triangular Hopf algebra. The R-matrix there is what "controls" the failure of the cocommutativity of your Hopf-algebra, and the Yang-Baxter equation is a consequence of all that. You can interprete it as a "braid-like" equation. See the wiki-pages on these subjects for more precisions. In ...


0

There is a fabulous explanation in Schwartz QFT and the standard model pg 153. The absense of MASSLESS particles with spin > 2 is a consequence of little group invariance and charge conservation. For massless particles you can take the soft limit in the scattering matrix elements Lorentz invariance implies the matrix numbers should be the same in ...


1

It looks like the sum appears because of the completeness relation $$\mathbf{1} = \sum_{\alpha} \left| \alpha \right> \left< \alpha \right|$$ taken over all the $\{ \alpha_j \}$ to give $$\mathbf{1} = \sum_{\{ \alpha_j \}} \left| \alpha_j \right> \left< \alpha_j \right|$$ Starting with the first expression, which I believe you corrected in ...


3

Right now, I have no idea about what you actually mean by gauge fields, or what anyone ever means by gauge fields. According to Danu, the gauge field is the connection itself, while if I remember well, Bleecker defines gauge fields as sections of those vector bundles that are associated to principal bundles. However, I can, I think, give a definitive answer ...


3

Well, the main point is that Yang-Mills theory is just one out of many gauge theories, cf. e.g. this Phys.SE post. E.g. SUGRA is a gauge theory. In fact, theoretically there are relativistic gauge theories with gauge fields transforming in virtually any possible representation of the Lorentz group. Whether they are realized in Nature is another story.


2

If the graviton exists it's not a 4 vector, but a tensor.


0

A somewhat meaningful and useful concept for massive particles is the Compton wavelength: https://en.wikipedia.org/wiki/Compton_wavelength#Limitation_on_measurement The Compton wavelength of a particle is determined by its rest mass. The position of a particle cannot be measured with a precision less than half its reduced Compton wavelength. In this sense, ...


1

A point particle is an idealization of a particle. It simplifies calculations by using a 0 dimensional object instead of a normal particle in calculations where size, shape, and structure are irrelevant. For example, in the theory of, say, electromagnetism, scientists will talk about a point charge - a particle represented by a point that has a non-zero ...


2

The vacuum is defined as the state where the Higgs field (and the rest of quantum fields) have no excitations. The mass of the Higgs boson is the minimum energy that you have to supply to the Higgs field in order to create an excitation. The huge mass for the Higgs field means that it is more difficult to create excitations, so there is no problem to the ...


2

Comments to the post (v2): Ref. 1 is considering the $d$-dimensional real Euclidean space $(\mathbb{R}^d,|\cdot|^2)$ with the standard norm $$|x|^2~:=~\sum_{\mu=1}^d (x^{\mu})^2~=~\sum_{\mu,\nu=1}^d x^{\mu}\eta_{\mu\nu}x^{\nu}, \qquad \eta_{\mu\nu} ~=~{\rm diag}(1,\ldots, 1),\tag{A}$$ and inner product $$\langle x ,y\rangle~:=~\sum_{\mu,\nu=1}^d x^{\mu}\...


1

Quantum field theory is based on quantum mechanics. The ground state which describes the fields is the free particle solution of the corresponding Dirac/KleinGordo/Maxwell equation. QFT is a theory developed to be able to calculate the many body interactions, seen even in the simplest feynman diagram. It posits fields of each type of particle described by ...


-1

As far as I know, QFT does modify QM. Single particle description is problematic when you want to incooperate with Special Relativity, e.g. negative probability. Hence a quantum field theory is needed. I recommend the book Quantum Field Theory for the Gifted Amateur, which is very useful for learning QFT.


4

It is not a postulate that the Hilbert space for QFT is a Fock space. In fact, for interacting theories is often almost surely not a Fock space. The requirements for a Hilbert space to be the space of a QFT is that the Wightman axioms are satisfied. For free theories, a suitable Fock representation of the canonical commutation relations satisfies the ...


3

It is easy to make solitons in water. See this video and the video coming after it with the bubbles. They are the nonlinear solutions of the differential equations governing flow.Also I recomend the one with bubbles . Ball lightning is proposed to be a soliton solution.


5

He inserts the identity $I = \sum_{S'}|S'\rangle\langle S'|$. This gives $\langle S|Q_\alpha Q_\alpha^\dagger|S\rangle = \sum_{S'} \langle S|Q_\alpha |S'\rangle \langle S'|Q_\alpha^\dagger|S\rangle =\sum_{S'} |\langle S|Q_\alpha |S'\rangle|^2 $.


0

So, I actually found this post because I've been trying to answer the same question. Here's what I understand from various sources I've found. These vectors represent a polarization basis. For example, two orthogonal linear polarizations (which I've seen written before as $e_{\textbf{k}\lambda}$ for $\lambda=1,2$), which I think is the more traditional ...


0

Grassmann-odd variables provide a classical description of Grassmann-odd quantum operators in the same way that Grassmann-even variables provide a classical description of Grassmann-even quantum operators. The classical super-Poisson bracket $$\{\psi^i,\psi^j\}_{PB} ~=~ -i (T^{-1})^{ij} \tag{A} $$ is related to the super-commutator$^1$ $$\hat{\psi}^i\hat{\...


1

I would like to stress the difference between 1) Perturbative Renormalization 2) Non-perturbative Renormalization By Perturbative Renormalization I mean removing infinities from the computation of a correlator/amplitude, order by order. This is done by introducing counterterms, i.e. re-writing the bare parameters of the lagrangian as $\lambda_{Bare} = \...


0

You have the following identity $$\gamma^{\nu}\gamma^{\mu}\gamma^{\rho}\gamma^{\sigma}\gamma_{\nu}=-2\gamma^{\sigma}\gamma^{\rho}\gamma^{\mu}$$ This gives you that $$\gamma^{\nu}\not{k^{'}}\gamma^{\rho}\not{k}\gamma_{\nu}=-2\not{k}\gamma^{\rho}\not{k^{'}}$$ Which is exactly what you need to get the above expression.


0

From a physicists point of view, I would start with the following notes, which are Chapter 9 in John Preskill's Quantum Computing lecture notes: http://www.theory.caltech.edu/~preskill/ph219/topological.pdf, as well as the references within. I would also mention Kitaev's paper https://arxiv.org/abs/cond-mat/0506438 as a specially influential reference. ...


0

Whatever you're supposed to do on this, you first need to understand quantum theory in curved spacetimes, and the ADM method in general relativity (which is basically the Hamiltonian approach to general relativity), and there are always issues as to what is the time coordinate you choose, and since it is not invariant what does it mean. In FRW the time is ...


0

I found this paper in Numdam's (a mathematical journal compilation) archive, which encompasses all that you talked about and I found it clear, with some references to Witten as well. This paper goes much further in detail, but I did not read all of it. And this might help if you aren't bothered by learning by forum posts?


1

One has to keep clearly in mind the structure of present day physics. Quantum mechanics is the theory that started as non relativistic with the Schrodinger equation for potentials, and became relativistic with the Dirac and Klein Gordon and quantized Maxwwell equations. Quantum mechanics has postulates which used with the solutions of the differential ...


4

"A bunch of cool complex analysis stuff popped up and solved my problem" is about as honest as it gets. But physicists do this from more or less their first differential equation: using $e^{i \omega t}$ to track both solutions via the cool-ness of complex analysis. There's no a priori or manifestly physics-based reason to do it that way. In the "original" ...


0

As you said in your question, quantum field theory is very important; it takes the ideas of quantum mechanics and applies them to fields, such as the electromagnetic force (in fact, quantum electrodynamics was the beginning of quantum field theory). Quantum field theory has plenty of evidence to support it, and it is still an ongoing work. String theory, ...


0

According to the abstract of a paper at https://arxiv.org/abs/hep-th/0604072, Breakdown of local physics in string theory at distances longer than the string scale is investigated. Such nonlocality would be expected to be visible in ultrahigh-energy scattering. The results of various approaches to such scattering are collected and examined. No evidence ...


0

Hint: OP's formula follows from a Wick-type theorem $$ \tag{1} T(f(\hat{A})) ~=~ \exp\left(\frac{1}{2}\hat{C}\frac{\partial}{\partial\hat{A}}\frac{\partial}{\partial\hat{A}} \right):f(\hat{A}): $$ between time-ordering $T$ and normal ordering $::$. Here $$\tag{2} \hat{C}~=~T(\hat{A}\hat{A})~-~:\hat{A}\hat{A}:$$ is a contraction. See e.g. this Phys.SE post ...


0

I found a paper that helped me a bit in understanding how things work. So here is what I understood. Given a random variable we define a formal power series and a formal derivation such that: \begin{equation} \frac{\partial}{\partial U}\left(\sum_{n=0}^\infty a_nU^n\right)=\sum_{n=0}^\infty (n+1)a_{n+1}U^n \end{equation} The "usual" Wick product $:\ :$ is ...


0

Purely mathematically, a function $f(x)$ (that can be differentiated enough times and so on) can be Taylor expanded around a point $a$ as \begin{equation} f(x) = f(a) + (x-a) f'(a)+\frac{(x-a)^2}{2}f''(a)+.. \end{equation} Now if we're describing a physical system with $f$ and the point $a$ is an equilibrium of the system, $f'(a)=0$. Then we have \begin{...


2

Answer posted by Lubos Motl in the comments; I reproduce most of it here. This answer was posted in order to remove this question from the "unanswered" list. Some (sketches of) answers to your questions, one by one: Physical states have to be invariant under gauge symmetries, so all of them are singlets and there are no nontrivial representations, (and 3....


1

This might help you out a bit: In what sense is a quantum field an infinite set of harmonic oscillators? From my understanding, most people think it provides a useful way to conceptualize uncoupled quantum fields physically. It doesn't, however, work for coupled quantum fields. The main problem with this seems to be that infinite harmonic oscillators give ...


3

If the eigenfunctions $\phi_n(x)$ are an orthonormal basis of the Hilbert space, then the sum $$\sum_n\phi_n(x)\phi_n(y)^*$$ is the integral kernel for the identity operator. That is, when we multiply this by any function $f(y)$ and integrate over $y$ we get $$\int \sum_n \phi_n(x)\phi_n(y)^*f(y)dy = \sum_n\phi_n(x)\int \phi_n(y)^*f(y)dy = \sum_n c_n\phi_n(x)...


2

I expect this to be a result of the Wick theorem. If you are considering an equilibrium situation with a quadratic Hamiltonian all odd moments will vanish. Thus you remain with even powers of your operator. If you now count the number of possible contractions you should get the correct result. P.S.: I just tried and it works indeed. Note the helpful ...


1

0) Note that in general, there is no need for the lattice spacings in different direction to be the same, as long as they all go to zero in the continuum limit. We can, for example, take the lattice spacing in the $x$ direction to be $a$, and the spacing in the $y$ direction to be $2a$. This is frequently done to achieve better resolution without the ...


2

Couplings that change (run) with scale occur in most quantum field theories! If you want a simple example, take a scalar theory with a quadratic and a quartic interaction (https://en.wikipedia.org/wiki/Quartic_interaction). In this theory, as you go to lower energies, the mass grows and the quartic coupling goes to zero. This theory is important both in ...


3

The correct keyword for this is "renormalization". And, as you said yourself, high energy physics shares this field with condensed matter. In condensed matter, you always encounter renormalization group of some kind if you are interested in critical phase transitions (which are scale invariant). Good references for this are Wikipedia and this book. ...


0

To understand why renormalization may work you might first consider simpler situations in Classical/Quantum Mechanics. In this case there are explicitly solvable toy models where one can see exactly what happens and why. See my paper "A Toy Model of Renormalization and Reformulation" on arXiv. About how to cope with growing terms, see my short note here.


4

Topological terms of all types are always required not to depend on the metric, so their integrals will correspond to topological invariants, which serve as topological charges in quantum field theory. However, it is important to distinguish between two the types of topological terms mentioned in the question, because they lead to different physical ...


0

If you want to consider this in terms of QED, then it is easy. Virtual pair production-annihilation is described by a vacuum bubble diagram where two propagators a contracted into a loop and there are no external legs. These kinds of diagrams are usually thrown away when we consider the partition function of the theory. And there is a good reason for that, ...


0

Your confusion is elementary. It is specified that the wavepackets have a well-defined momentum, and that at $t=0$ they are located at the origin. Now since the wave packets have a well-defined momentum, they move at a constant velocity. An object which is at the origin at $t=0$ and moves at a constant velocity will be very far from the origin as $t \to \...


8

Renormalization is always needed when the Hamiltonian is singular. Singular means that the formal expression for the Hamiltonian resulting from the interaction specified is not a self-adjoint operator in a dense domain. Then the dynamics is formally ill-defined and must be renormalized by taking care to represent everything properly as a limit that makes ...


0

Q^2 is the Mandelstam variable t, i.e. the four momentum transfer squared where the s channel is the x axis in the feynman diagram. where p1 p2 are incoming. In the same link it is seen that at the relativistic limit : The dot product for t (Q^2) is p transverse to the incoming beam direction of p1 , p2 are incoming. etc. Is this Q the same if ...


0

There are flavor changing weak interactions mediated by the charged $W^\pm$. In the cases you cite I would agree that flavor symmetry would not permit these interactions $uu\rightarrow cc$ etc.


2

Well while it has similarities with the OPE, it is more than that. In fact, it satisfies the OPE limit when $z\to w_j$ for any $j$, since the OPE you are talking about tells you only the singular terms, while there are also infinitely many non-singular terms, i.e., schematically $$ T(z)\phi(w,\bar w)=\frac{h_\phi}{(z-w)^2}\phi(w,\bar w)+\frac{1}{z-w}\...


2

The trick is in the introduction of a renormalization scale. Once the perturbation theory has been regularized, one obtains a momentum (and cut-off) dependent interaction of the (schematic) form in 4D $$\lambda(p)=\lambda_0+\alpha\lambda_0^2 \ln(\Lambda^2/p^2), $$ where $\lambda_0$ is the bare interaction, and $\alpha$ some numerical factors. What one ...


2

Before I try to answer your question, one thing: Does Ryder really calculate the $\mathcal{O}(\lambda^2)$ to the propagator as the first contribution in perturbation theory, because there is actually a $\mathcal{O}(\lambda^1)$ to the propagator and the $\mathcal{O}(\lambda^2)$-loop is as far as I am concerned a two-loop diagram, i.e. having two loop momenta,...


4

User joriki has already provided a correct answer. Here we just add details and make some clarifying remarks. I) Let $n\geq 3$ be an integer. Consider the set ${\cal T}(n)$ of connected trees with cubic vertices only and with $E_n=n$ external labelled lines with labels $1,\ldots, n$, and no labelling of internal lines and vertices. (We stress that loops ...


2

A straightforward way is to use the Bogolyubov transformation in unitary form. Since I don't know your exact transformation, consider a general one of the form $$ {\hat c}_j = (\cosh\theta_j) {\hat a}_j + (\sinh\theta_j) {\hat b}^\dagger_j\\ {\hat d}^\dagger_j = (\sinh\theta_j) {\hat a}_j + (\cosh\theta_j) {\hat b}^\dagger_j $$ Then $$ {\hat c}_j = {\hat U}^...



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