# Tag Info

## New answers tagged quantum-field-theory

0

Below I will use some simple formulas, that's why I must make a distinction between longitudinal and transverse relativistic mass. The transverse relativistic mass of an object has very much to do with the energy of the object: We just multiply the energy by a constant to get the transverse relativistic mass. The energy of an object has very much to do ...

0

You can simply compute the integral using your preferred regularization method (cut-off, dimensional, Pauli-Villars...), and if all goes well (which is not guaranteed), the divergences will not depend on your parameter and they will eventually disappear when you compute physical stuff. If this does not happen, maybe your theory is simply ill-defined. And as ...

5

One of the main reasons the virtual particles are used is that in many contexts we do not have a non-perturbative formulation of quantum field theory. What we can do is compute some amplitudes perturbatively (e.g. for outcomes of particle collisions) using Feynman diagrams. These diagrams have input/output lines in them, usually identified with colliding ...

-1

Virtual particles are used to avoid force at a distance.

0

Is this concept of relativistic mass increase, related to the concept of Doppler effect of matter waves? No. Doppler's effect also happens for classical waves, including "classical matter wave", by which I meant Schroedinger's wave function. The effect is in fact trivial. When you change the reference frame, the momentum of the particle changes. By de ...

0

That depends on what you mean by "made up of". Is static electric field "made up of" photons? Is the stream of water "made up of" waves? To me these are semantic games that have no connection to reality. But if you are willing to answer "yes" to the questions above, then you can safely say that those domain walls are "made up of" Higgs bosons (and also the ...

0

I was re-reading von Neumann's tome in which he recapitulates his views on his own invention, the density matrix. As well as Dirac's inclusion of this in his second edition of his Principles, and the explanation of it in Landau--Lifschitz (second ed)...now, Landau independently had invented it, too. Von Neumann's rationale is that if our information ...

0

Well I can show you that the Weyl equations are relativistic invariant, at least. It relies on an identity I haven't found in any standard QFT books, but it's easy to show. First of all take our gamma matricies in the chiral representation $\gamma^\mu= \left(\begin{array}{cc} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{array}\right)$. Under ...

0

Yes for the $\phi^{3}$ theory the vertex has 3-lines, whereas for the $\phi^{4}$ theory this becomes 4 lines meeting at the vertex. g just refers to the number of vertices in the diagrams, so for $g^{1}$, you're summing all diagrams with one vertex, for $g^{2}$, you're summing all diagrams with two vertices, and so on.

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BCS theory deals with superconductivity in a metal, or basically a finite density of non-interacting fermions. There is a Fermi surface with tons of gapless particle-hole excitations, so you can say it is critical. As long as the Fermi surface has certain symmetry (time-reversal or inversion), the pairing instability is infinitesimal, meaning that the ...

3

Here are two facts - A vacuum expectation value of a quantum field is equal to the minimum of the effective potential (taken from the 1PI effective action). The effective potential takes the general form $$V_{\text{eff}}(\phi) = V_{\text{classical}} (\phi) + \text{quantum corrections}$$ In perturbation theory, where quantum corrections are assumed to be ...

0

So actually I just computed it by hand from the very beginning. Starting from the expression of the field $A_i(\mathbf{x},t)=\sum_s\int\frac{\text{d}^3\mathbf{k}}{\sqrt{(2\pi)^3 2|\mathbf{k}|}}\left[a_s(\mathbf{k})\epsilon_i(\mathbf{k},s)e^{i(|\mathbf{k}|t-\mathbf{k\cdot x})}+\text{H.c.}\right]$ where $\epsilon_i(\mathbf{k},s),\,s=1,2,\, i=1,2$ are ...

2

You write that you do not like the wave-particle duality explanation of the Young experiment, and therefore turn to QFT. Before going further I would like to point out that the double slit experiment is a one-particle effect. That means you only need consider one particle at a time to explain what is happening. Because of this QFT will not buy you much as ...

1

First, your explanation is...sort-of-right. What's travelling is a quantum object, not a particle, not a wave. The probability of detecting a particle-like localized blip with some sort of detector is given by a probability density $\rho$, which is the "sqaured amplitude" of a "wavefunction" $\psi$. For free particles, the Schrödinger equation that $\psi$ ...

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In the Young double slit experiment it is possible to detect the arrival of individual photons as well as an interference pattern. Yes, I rather like this picture on this Hitachi webpage myself. It's electrons rather than photons, but no matter: It doesn't makes much sense to me that something could be either a particle either a wave, neither the ...

1

To begin with, The Amplituhedron formalism only works for a specific theory, N=4 SYM in the planar limit (only planar Feynmann diagrams are considered). Because of supersymmetry, you can classify scattering processes with two parameters: $n$ and $k$. n is the number of particles involved, and k is, roughly speaking, the number of spin flips in the process. ...

0

Hamurabi, I assume you are asking for the order of the wave equation. Try thinking about intramolecular vs. intermolecular forces. A wave equation is necessary to describe the energy of any quantum occurrence, so far as I recall. It may take an approximation for any non-ideal system, but exact or not the wave equation describes the moment in time when the ...

2

Quantum fields cannot be turned on or off. The field itself exists for all time and space. It is possible to excite various modes of a quantum field at various spacetime points. These field excitations are interpreted as particles. When no excitations are present (i.e. no particles are present) the quantum field is in the vacuum state. Particles do not act ...

0

Nicest question since a long time. Your argumentation is excellent. You are the only one who give a comment to my last question - about the measured maximum distance of the electron influence. I suppose here, that electric fields are finite. That follows immediately if one agree that the electric field is quantized. A quanta has to have a finite energy and ...

1

The answer to your question requires some knowledge on group theory and tensor analysis, but I will try to make it as simple as possible with out going into too much of technicalities. Your question consists of basically two completely disjoint parts, they are: why the gauge bosons(leptoquarks) of Pati-Salam Group do not mediate proton decay. Which is a ...

2

It gets easier if you use the result from part 1. Then you also don't have to deal with the $\mathcal O(\omega^2)$ (see my answer to your other question). In your calculation, you transformed $\bar\psi$ and $\psi$, but not $\gamma^\lambda$. This is correct, as I will show in the end, but I will take another point of view which is really helpful here: ...

2

It is actually possible, and not too difficult, to prove this without expanding the exponentials to first order only. What you are trying to prove is $S^\dagger \gamma^0 = \gamma^0 S^{-1}$, this is equivalent to $$\gamma^0 S^\dagger \gamma^0 = S^{-1}$$ because $( \gamma^0 )^2 = 1$. Expand $S^\dagger = \sum_n \frac{1}{n!} \left( \frac i 4 \omega_{\mu\nu} ... 3 Given the four point function$\langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle$, the conformal block expansion depends on what operators you replace by the OPE. So if you insert the OPE for$\phi(x_1)\phi(x_2)$and the OPE for$\phi(x_3)\phi(x_4)$then this corresponds to the s channel---one can also call this the (12)(34) channel---.. The t channel is ... 0 Let's introduce a bit more notation because I think you're confusing yourself with the$\to$notation: Let$\theta : \mathbb{R}^4\to\mathbb{R}be any function. Then the gauge transformed fields are \begin{align} \phi^\theta & := \mathrm{e}^{-\mathrm{i}\theta}\phi \\ A^\theta& := A - \frac{1}{q}\mathrm{d}\theta \end{align} and a gauge ... 1 Inserting the expansion $$\psi=\int\frac{d^3p}{(2\pi)^32\omega_p}(a_pe^{-ipx}+b_p^\dagger e^{ipx})$$ into the expression for the Hamiltonian $$H=\int d^3x(\dot{\psi}^\dagger\dot{\psi}+\nabla\psi^\dagger\cdot\nabla\psi+m^2\psi^\dagger\psi)$$ we get $$H=\int d^3x\int\int\frac{d^3p}{(2\pi)^32\omega_p}\frac{d^3p^{\prime}}{(2\pi)^32\omega_p^{\prime}}(A+B+C) ... 0 For the complex momentum,$$ T^{\mu\nu}= \partial^{\mu}\phi^{\dagger}(x)\partial^{\nu}\phi(x) + \partial^{\nu}\phi^{\dagger}(x)\partial^{\mu}\phi(x) - g^{\mu}_{\nu}\mathcal{L} $$Now you can consider two separate cases: T^{0i} which gives the 3-momentum, P^{i} i.e. g^{0}_{i} = (0,0,0) T^{00} which gives the hamiltonian, H = P^{0} i.e. ... 1 Instead of integrating over the region in moduli space where \Im(\tau) \to 0 (corresponding to the UV limit), we can perform a modular transformation and integrate over the region where \Im(\tau) \to \infty (corresponding to the IR limit). I think that's all he wants to say, there is (as far as I understand it) nothing deeper behind it. But it is not ... 2 First of all, there are a few problems with your question: J_{ab}^0 = \pi^a \epsilon^{ab} \Phi^b is not a valid expression, since there is a summation on the right hand side of the equation, but a and b are free indices on the left hand side. Your definition of \epsilon is a bit weird, too. What you mean is$$ J_{ab}^0 = \pi^i \epsilon_{ab}^{ij} ... 9 The Born rule (and hence any discussion of collapse in the sense of the Copenhagen interpretation) is relevant only when an observer has made a distinction between a (tiny, observed) system and its (huge, observing) environment (= everything else, containing in particular the measurement equipment). This distinction (not present in relativistic QFT itself) ... 5 I take a minimal interpretation of QFT in a Copenhagen style to seek to make a connection between a classical description of/model for an experimental apparatus and classical records of its measurement results and a QFT model for the same apparatus. Classically, a modern measurement device is most often a thermodynamically metastable system that we engineer ... 1 I would like to make a comment, which may clarify and simplify the things a little bit. In complex analysis [see e.g. Introduction to Complex Analysis" by B.V. Shabat] by definition derivatives over the complex variablesz$and$\bar z\$ are given by: $$\mbox{def:} \quad \partial_z \equiv \frac{1}{2} \left(\partial_{\rm a} - i \partial_{b}\right) ... 2 Let us consider the corresponding Hamiltonian theory, so that we have a notion of a commutator that we can use to form a Lie algebra bracket. Moreover, let us consider the classical theory for simplicity. Then the Poisson bracket$$\tag{1} \{\Phi^a({\bf x}),\Pi_b({\bf y})\}_{PB}~=~\delta^a_b~\delta^3({\bf x}-{\bf y}), \qquad \text{etc},$$plays the role ... 0 There is no 1/2 because such factors arise whenever there are products of the same field. For example, you probably have seen the interaction term in \phi^4 theory written as \frac{\lambda}{4!}\phi^4. This is because when taking functional derivatives to get the Feynman rules, these combinatoric factors will arise from the derivative hitting any of the ... -1 It is just an integration by parts considering that boundary terms vanish. 0 Generally people define these operators so they follow these rules. It's a requirement for them being number operators and hence of any use. 2 The reason your logic fails is because \psi is not simply a Grassmann variable; it is a four-component vector of complex Grassmann numbers (in four dimensions): $$\psi=\left(\begin{array}{c} \theta_1 \\ \theta_2 \\ \theta_3 \\ \theta_4 \end{array}\right)$$ With this knowledge, try computing \bar{\psi}\psi\bar{\psi}\psi and ... 3 The argument is false in four dimensional space. The error is the assumption that you get one Grassman number per spinor. In fact, you get one Grassman number per spinor component! In 4d, spinors have multiple components. (Both Weyl spinors have 2 components, and Dirac spinors have 4.) In 1d space, this is a correct argument. In 2d, it is correct for ... 0 There are really two main ways of thinking about [electrons]. Quantum Mechanics describes an electron by a wave function who's squared magnitude gives the probability of finding the electron in a certain position or with a certain momentum. QFT ... describes the electron as an excitation of the electron field. Both of these models describe the ... 3 The connection with current expectations and current-current correlations is loose only. There is no comprehensive survey article; key papers are by Goldin and coworkers. http://scitation.aip.org/content/aip/journal/jmp/12/3/10.1063/1.1665610 http://scitation.aip.org/content/aip/journal/jmp/22/8/10.1063/1.525110 You can use scholar.google.com to find ... 0 No. The Uncertainty Principle has to do with the act of measuring. Basically, you cannot simultaneously measure both position and momentum to an arbitrary degree of accuracy. The more accurately you meausre one, the less accurate your measurement of the other becomes. The uncertainty in momentum , as far as I know, won't result from your not knowing when ... 0 For a real scalar field I think what you have written is correct..But if you want to describe a complex scalar field then we need to distinguish between \phi and \phi^{\dagger}... 0 A real scalar field has one degree of freedom...Here we have two degrees of freedom (two real scalar field) and we treat \phi and \phi^{*} as independent field. When we put the lagrangian in Euler-Lagrange's equation we generate a factor of 2 and 1/2 cancel it,e.g. m^{2}\partial_{\phi}(\phi^{2})=2m^{2}\phi..and similar for the derivative term... But ... -3 Is the many-worlds interpretation (MWI) of QM inconsistent with quantum field theory? I think it is. Quantum field theory starts with QED, where the photon is described as as an excitation of the photon field, and the electron is described as an excitation of the electron field. This is compatible with weak measurement, classical electromagnetism and ... 8 You can trivially apply MWI to QFT, so just do it and ignore nay sayers. Or if you want to talk to them, then the burden is on them to argue why you can't do something that you are literally doing. After all, it is just too hard to predict what imaginary concerns they have. Now, in your case a specific concern was listed: Because in QFT time is ... 5 The terminology of a mode of a free quantum field \phi(x) comes from writing it as a Fourier transform, often also called mode expansion:$$ \phi(\vec x) = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left(a(\vec p)\mathrm{e}^{\mathrm{i}\vec x\cdot\vec p} + b(\vec p)^\dagger\mathrm{e}^{-\mathrm{i}\vec x\cdot\vec p}\right)$$where for a ... 2 The term mode is used to define a particular state of a system and may refer for instance to its spin, wavevector, polarisation, charge etc. If we wanted to create a boson at position x with an up-spin and with wavevector k, we may use the field operator \hat{a}^\dagger(x, k, \uparrow) on the vacuum state \vert0\rangle. The most clear distinction ... 4 If an electron is an excitation of the electron field, what causes the excitation to be stable? I think the best way to say it is to take a tip from topological quantum field theory: "Although TQFTs were invented by physicists, they are also of mathematical interest, being related to, among other things, knot theory and the theory of four-manifolds in ... 15 The electron is stable because there is no allowed process in the quantum field theory it can undergo that would lead to its decay. Its mass is the smallest among the electron/muon/tauon, so it doesn't have enough energy on its own to turn into one of those, and all other processes you could imagine are forbidden by conservation laws - either those of energy ... 0 Well, I think that first of all, you should understand, that the reason why you can decompose free field into plain waves is that equations of motions are linear. In case of interaction, equations are motions are nonlinear, so any linear combination of its solutions is no longer a solution. In the second place, about your second question: it depends on ... 1 TL;DR In general, no. A longer but possibly irrelevant discussion follows. Consulting the classic review RevModPhys.58.323 by Rammer and Smith, the quantities you are considering are defined as (Eq. 2.5):$$G^{<}(\boldsymbol x_1,t_1,\boldsymbol x_{1'},t_{1'})=\mp i\langle \psi^\dagger_{\mathcal H}(\boldsymbol x_1,t_1) \psi_{\mathcal H}(\boldsymbol ...

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