# Tag Info

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To put it simply, a QFT state is a linear superposition of the possibilities of one-particle wavefunction, two-particle wave function, three-particle wavefunction, ad infinitum. Each of those wavefunctions is a map from $\mathbb{R}^3 \rightarrow \mathbb{C}$ just like what you said. The time evolution of that configuration "follows" once you specify a ...

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Quantization is often treated as a mystery, yet it can be seen to arise naturally from classical Hamiltonian mechanics.1 This yields the quantization prescription $$\{\dot{},\dot{}\} \mapsto \frac{1}{{\mathrm{i}\hbar}}[\dot{},\dot{}]$$ for all classical observables on the phase space that are to be turned into quantum operators.2 Therefore, if we want to ...

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Quantization, in the Heisenberg picture is very much related to establishing quantum non-trivial commutators between the main observables of the theory. Originally this is linked to the classical Poisson bracket, which in the Hamiltonian formalism includes the cannonical momentum. Hope I gave some insight, as to why we are led to define the conjugade ...

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So in your calculation, you have set $\epsilon$ to be zero. Since the result should be independent of the gauge you choose, why not calculate under $\epsilon=0$, which is equivalent to take the propagator as $-\frac{g^{\mu\nu}}{k^{2}}$? If you still want to calculate with a brute force, I think the problem may arise from the Feynman parameterization, where ...

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In Hamiltonian formalism, operators such as $a(k)$ and $a^{\dagger}(k)$ can depend on time (in Heisenberg picture), or they can not (in Schroedinger picture). They don't depend on space in sense that they are given for any 3-momentum $k$. But there is a Fourier transform $\tilde{a}(x)$ and $\tilde{a}^{\dagger}(x)$, which is given for any spacial point $x$. ...

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In canonical quantization one constructs the Hamiltonian formalism. Energy conservation is therefore manifest (since Hamiltonian is time-independent and commutes with itself). Quantum-mechanically, the Hamiltonian of the system can be expressed via particle creation-annihilation operators. So, the total energy of the field is also the total energy of all ...

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I find the whole notation here a bit confusing since we are talking about momenta modes while using real space notation (maybe I'm the only one...). To clarify what is going on we can switch to momentum space instead. Consider the quadratic term in the exponential: \begin{align} \int d^4x \phi ^2 & = \int d^4x \int d^4k d^4k' \phi _k \phi ^\ast _{ k ' ...

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I'm not being precise, but morally: Imagine you were integrating out all modes above a frequency $b\Lambda$. Consider $\omega < b\Lambda < 3 \omega$. A mode $\phi$ with frequency $\omega$ when cubed, will have some part of it as mode of frequency $3 \omega$, since: $\sin (3x) = 3 \sin (x) - 4 \sin^3 (x)$. (Easier to see that ${(e^{i \omega t})}^3 = ... 2 It is meaningless to talk about the precise number of virtual particles corresponding to the interaction of a certain particle with other particles. It is simply not well-defined, neither in the mathematical framework of the path integral formalism of quantum field theory, nor if one uses them as a heuristic picture of reality. Virtual particles appear in a ... 0 After reading the section put forth by @gcsantucci I think I kind of understand what was happening, but I'm eager to hear feedback on this. If you break$ SU(2) $using a doublet you indeed break all the generators and get massive gauge bosons. What was confusing is that a linear combination of generators leaves the vacuum invariant (namely,$ \sigma _1 ...

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Firstly, I'd like to recommend Conformal Field Theory by Di-Francesco, it is a comprehensive text which is thorough and contains many applications of conformal field theory. The text is indispensable. In conformal field theory, it is often characteristic of correlation functions to diverge as points of two or more fields coincide. The operator product ...

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I figure that string theory is a new breed of QFT which looks at fields in terms of a network of strings and also incorporates gravity into its module, You should read some reviews of what string theory is . String theory rejects the idea of a point particle as the fundamental constituent of the theory, which is the central concept in quantum field ...

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I guess instead of me trying to explain, it is way better if you read section "Non Abelian Examples" of Peskin and Schroeder, in chapter 20. It explains exactly what you are asking for. It's actually the predecessor of the Standard model. Georgi and Glashow proposed this model before the SM, because they didn't know about the Z boson. So it is exactly what ...

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Question 1. "What does it mean for a field to be excited?" It means to be in something other than the quantum ground state. Modern physics conceives of the universe as being made of a handful of quantum fields. Space and time aren't just filled with quantum fields, they themselves are made of quantum fields. "Empty space" is simply a shorthand for what ...

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For a quick (and somewhat dirty) way of deriving the bound, do the following: Recall, that in Poincaré coordinates, the metric of $AdS_{d+1}$ is $$ds=\frac{1}{z^2}\left(dz^2+\eta^{\mu\nu}dx^\mu dx^\nu\right).$$ The equation of motion for a scalar is $$(\Box-m^2)\,\phi(z,x)=0,$$ where $\Box=\frac{1}{\sqrt{-g}}\partial_a \sqrt{-g}g^{ab}\partial_b$. ...

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Comment to the question (v4): OP seems to effectively conflate spacetime symmetries and internal gauge symmetries. They act in different representations, or more precisely as a tensor product of representations. For instance the fermion $\psi$ carries two types of indices, say $\psi^{\alpha i}$, $\alpha=1,2,3,4,$ and $i=1,2$. The fermion acts as a ...

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One may only talk about a discrete sum over $k^\mu$ vectors if all the spacetime directions are compact. In that case, $k^\mu$ is quantized. If the spacetime is a periodic box with periodicities $L_x,L_y,L_z,L_t$, then $V=L_x L_y L_z$ and $T=L_t$. The component $k^\mu$ in such a spacetime is a multiple of $2\pi \hbar / L_\mu$ (I added $\hbar$ to allow any ...

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If photons transmit the electromagnetic force, which is observable: the photon or the electron? Do we ever directly measure a photon, or do we only measure it's effect on electrons. For example suppose I shine a laser at a wall Let us clear up that photons ( and also electrons) are quantum mechanical elementary particles, and classical electromagnetic ...

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Fields were first invented as a tool for "book-keeping" or keeping track of how a configuration of charges would effect a charge at a distant point, but it was soon realized that the field itself is much more "real" than simply a convenient mathematical tool. The first reason we treat the field as a separate real thing is the theoretical fact that you don't ...

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Both Photon and electron are real. Electron has rest mass while photon is a form of energy which gains some mass when it travels at speed of light. "For example suppose I move some electrons around in my laser, these electrons move the electrons around on the wall, and then these electrons move around some electrons in my eyes, so I see a red spot on the ...

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In order to understand the significance of center symmetry, consider Yang-Mills theory with dynamical gluons and static test quarks. Gluons are neutral with respect to the charge corresponding to center symmetry transformations, while test quarks are charged. One can now characterize the phase structure of the theory as follows: In the deconfined phase, ...

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The whole point of virtual particles is that they do not obey the on-shell physical laws, they are just computational crutches in Feynman diagrams. You should not take the idea of them being exchanged or "filling the vacuum" too seriously, there is no reality to them (hence the name). The energy-time uncertainty relation is one of the most misused results ...

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Zee just wants to distinguish the integration $\int \mathrm{d}^4x$ from the integration $\int \mathcal{D}\phi$ when calling this "integration by parts under $\int \mathrm{d}^4x$". As for how the integration by parts itself works, observe that, for any function $f$ vanishing at $\pm\infty$: $$\int f'(x)f'(x)\mathrm{d}x = f(x)f'(x)\rvert^\infty_{-\infty} - ... 3 You have stumbled upon the spinor helicity formalism. The idea is as follows. Massless spinors have many useful relationships among them that we can use. Thus instead of using the polarization vector, \epsilon_\mu or massive spinors lets just map these objects into massless spinors and then use the convenient relationships between them to simplify ... 0 There is no sacred difference between the way a gauge boson and a Goldstone boson couple to a matter field. Goldstone bosons are just ordinary bosons that are massless due to the breaking of a global symmetry. To avoid Lorentz violating these have to be scalar fields so that does turn out to restrict the possible coupling that they could have, but the fact ... 0 Sorry but the analysis of what commutators mean (in the given link) although very good, does not provide intuition and does not generalise to anti-commutators. Commutators used for Bose particles make the Klein-Gordon equation have bounded energy (a necessary physical condition, which anti-commutators do not do). On the other hand anti-commutators make ... 0 I am really just motivating a few links. I also hope a few comments may be added to the mix. UPDATE: See Qmechanic's links for extended details relevant to qft. Well I was not sure if I should dare write anything here, but I am sure there are some flaws in my thinking, so someone can correct me. OK. Let's begin with some ... 2 Let us start with a simple analogy that helps grasping the concept. Say, we need to calculate a electric field of a charged cylinder. What is the first thing you do? You choose your z-axis along the axis of the cylinder. Why won't you choose your axes differently, like pointing whatever they want? Well, you could, but that will make the problem ... 7 The advantage of unitary gauge is that it completely removes unphysical fields, while adding additional degrees of freedom to the gauge bosons, which consequently become massive. This gauge works well for tree-level calculations, but complications arise when considering loops: The propagators of gauge fields and ghosts (which are needed to impose the ... 0 Besides the need of having a Lorentz invariant term, there is another "against" that comes only when we consider classic fields. If we restore units, a Planck constant appears in the kinetic term of the fermion field, telling that it will disappear in the limit \hbar \to 0. But of course this "non-go" argument is bypassed by any macroscopic electrical ... 2 I) It is difficult to comment without seeing the textbook, but one interpretation is that it is essentially just a matter of assigning appropriate representations as follows. Let G be a Lie group with the corresponding Lie algebra L. Let \exp: L\to G be the exponential map. Let t_a\in L be a Lie algebra generator. Let A be an algebra with a set ... 3 It depends on your definition of force. Force means a change in momentum, ~dp/dt , so any change in momentum in a Feynman diagram is a force. For example this diagram for compton scattering says yes. If one is talking of gauge theories and exchanged bosons , because those are the ones that build up the three, electromagnetic, weak, strong ( maybe ... 0 It depends what you would accept as a valid answer but two bosons could feel a small force resulting from a virtual fermion loop for example. Does that count? 3 A small comment first: usually people call this theory as a Chern-Simons theory in (2+1)d, while the BF theory usually refers to a similar theory in (3+1)d. But anyways, this naming is not important. The U(1) Chern-Simons theory in (2+1)d is always formulated in the following general form$$S=\int\frac{\mathrm{i}}{4\pi}K^{IJ} a^I\wedge \mathrm{d}a^J.$$The ... 6 This issue is a bit confused in textbooks, however the statement of the professor is physically wrong (mathematically all the procedure can be rigorously justified using the theory of distributions). The point is that the claimed position operator is not the position operator because it is not even self-adjoint (nor Hermitian) in the relevant Hilbert space ... 2 Here is the answer (I will not consider the constants on the denominator of your Fourier transform for simplicity, however they are there ;-) ). When you write the operator \hat{\phi} you have to be careful. I will drop the hats, because it will be clearer I think (maybe here the hat stands for an operator and not for the fourier transform). Your operator ... 0 If we want to study the decay of an excited atomic state, the atom must be coupled to the radiation field. This can be done (and usually is) non-relativistically. The overall Hamiltonian then consists of the sum of a free atomic part, a free field part and their interaction. Now the atomic excited states are no longer bound states but turn into "resonances". ... 1 Perhaps it is now only a curiosity, but there is a classical answer to this question that can give order of magnitude estimates that are close to the answers provided by quantum mechanics in many cases. If an electron is in a circular orbit (i.e. thinking of the now obsolete Bohr theory), then if we look at it from a direction that is in the orbital plane, ... 2 Both the ground state and the excited state are eigenfunctions of the time independant Schrodinger equation. That means they are also time independant so the excited state will never decay to the ground state i.e. the transition time would be infinite. However the excited state doesn't decay to just the ground state, it decays to the ground state plus a ... 0 As far as I know, the very idea of normal-ordering implies making modifications to the quantity. For example, consider the hamiltonian of the free real field.$$ H = \int \frac {d^3 p} {(2 \pi)^3} \frac{\omega_p}{2} \left( a_p a^{\dagger}_p + a^{\dagger}_p a_p \right) = \int \frac {d^3 p} {(2 \pi)^3} \omega_p \left(a^{\dagger}_p a_p + \frac{1}{2} ...

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If you have another model, then use it to generate a set of predictions, and let's compare those to the predictions we get in the world we see. If you think there's some ether filling the universe, then write down some properties it has, and make some predictions with those properties. Talking in vague generalities will only lead you down a rabbit hole of ...

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As @RobinEkman mentioned, the kinetic term changes as well. This can be easily computed \begin{split} {\cal L}_D = i {\bar \psi} \gamma^\mu \partial_\mu \psi &\to i {\bar \psi} e^{- i \alpha} \gamma^\mu \partial_\mu \left( e^{i \alpha} \psi \right) = i {\bar \psi} \gamma^\mu \partial_\mu \psi - \partial_\mu \alpha ( {\bar \psi} ...

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The kinetic term $$\mathcal{L}_\text{fermion} = i\overline{\psi}\gamma^\mu \partial_\mu \psi$$ changes too, by precisely the right quantity to cancel the change in the interaction term. Thus, the total Lagrangian is invariant, and this is what matters.

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I think this would be tricky, since any force mediator (at least from conventional thinking) must have a three-valent vertex, two of which are the charged object and one of them is the force carrier. If the force carrier is a fermion, I don't think this combination can be Lorentz invariant (spin zero combination).

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The Weyl transformation (unlike diffeomorphisms) does not affect physical fields, only the geometry of space-time. Probably the best way to show that your equation is Weyl-invariant is to build an action which (when varied) yields the equation. In your example it would be $$S[\Psi] = \int d^4 x \: \sqrt{-g} \: \bar{\Psi} \gamma^a e^{\mu}_a D_{\mu} \Psi.$$ ...

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'Square root of geometry' A Dirac spinor field $\psi^\alpha(x)$ under Lorentz transformations behaves as, $$\psi^\alpha(x)\to S[\Lambda]^\alpha_\beta \psi^\beta(\Lambda^{-1}x)$$ where $\Lambda = \exp(\frac{1}{2} \Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$ and $S[\Lambda] = \exp(\frac{1}{2}\Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$, where ...

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Notice that in $[\partial_xO(x),O(x')]$, the partial derivative only acts on $x$, not on $x'$. So we can pull the partial derivative operator out of the bracket and get $\partial_x[O(x),O(x')]$.

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As others pointed out, the statement probably means that if you want to have a nonzero contribution from creation and annihilation operators acting on vacuum, you need to apply the creation operator first, since $\hat{a}|0\rangle = 0$, in other words, you cannot annihilate if you have nothing. In the question you link to, the situation is different, as that ...

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That's a proverb in the beginning of Chapter I.8 (in both editions). Zee undoubtedly included it in the book because it reminded him of the fact that an annihilation operator annihilates the vacuum ket: $$\hat{a}|\Omega \rangle =0,$$ while this is not so for the creation operator $$\hat{a}^{\dagger}|\Omega \rangle \neq 0.$$ The proverb is not meant as a ...

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No. Feynman diagrams are made by summing over the perturbative contributions of quantum amplitudes. They cannot hold non-perturbative information.

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