# Tag Info

12

Frequency is not quantized, and has a continuous spectrum. As such, a photon can have any energy, as $E=\hbar\omega$. However, quantum mechanically, if a particle is restricted by a potential, i.e. $$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2 + \hat{V}$$ for $V\neq 0$, the energy spectrum is discrete. For example, in the case of the harmonic oscillator, ...

12

There is only one kind of photon. Indeed, when we describe elementary interactions between two electrons for example, we call the photon "virtual" as opposed to a physical photon that might exist outside of this process. Still, these are the same particles, i.e. excitations of the same fundamental field, as the photons that make up light for example. ...

12

Photons do not exhibit the property of virtual particles, but it is not your reasoning that is faulty, you have simply fallen prey to an imprecise use of terminology. Let me start with my view of the wave/particle duality. Most of the images of "particles" and "waves" comes from a time when we really didn't understand the quantum world, and some ...

11

(I henceforth assume $c= \hbar=1$.) It is forbidden by the four-momentum conservation law. Put yourself in the centre of mass reference frame of the couple of massive particles (electron and positron). There $P_{e\overline{e}} = (2E,\vec{0})$ with $E\geq m_e>0$. Just because four momentum is conserved, this four-momentum must be the same as the one of the ...

9

I'm a bit rusty on my qed, but I'll give this a shot. The simplest case would be described by a diagram similar to: But the $e^--e^--\nu_e$ vertex doesn't exist (also note that I can't draw the required arrow on the neutrino) - the vertices of the standard model (with the exception of vertices involving the Higgs and neutrino oscillations) are: With ...

8

The magnetic moment of the electron is a magnetic moment, so the right magnetic field around it is $$\mathbf{B}({\mathbf{r}})=\nabla\times{\mathbf{A}}=\frac{\mu_{0}}{4\pi}\left(\frac{3\mathbf{r}(\mathbf{\mu}\cdot\mathbf{r})}{r^{5}}-\frac{{\mathbf{\mu}}}{r^{3}}\right).$$ The world is quantum mechanical – and so is any viable description of the spin – so we ...

7

Short answer: A virtual particle is not the opposite of a classical particle. While the other answer captures some aspects correctly, there are still a few flaws and inaccuracies which in the following, I will try to set straight. Wave-particle duality Strictly speaking, quantum objects are neither waves or particles. They are entities behaving like ...

7

We assume a square box, because it simplifies the argument. Yes, in the limit of $L_1, L_2, L_3 \to \infty$ this is equivalent to a square box in the limit $L \to \infty$ (we can't measure the difference between infinities). Also, in the limit $L \to \infty$ the quantized momenta will eventually cover all of momentum space, making the distinction ...

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 ...

6

The Lagrangian provided is Maxwell's Lagrangian, supplemented by a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The equations of motion are, $$\partial_\mu F^{\mu\nu} + \partial^\nu (\partial_\mu A^\mu) = \partial_\mu \partial^\mu A^\nu = 0$$ Instead of making a gauge fixing procedure a ...

6

A major difference between real and virtual photons is that virtual particles are not required to have energy and momentum on the "mass shell". That is, virtual photons may have $E^2-p^2 \neq m^2$, while real photons must obey $E^2-p^2=m^2=0$. My memory disagrees with Neuneck (v1): I think that a coherent superposition of real photons is a laser, while ...

6

Yes, you would have to introduce another gauge field. For example in the Standard Model there is gauge invariance under $SU(3)\times SU(2) \times U(1)$, and so there are three gauge fields: the gluons, the $W^\pm, Z$ weak gauge bosons and the photon. In general terms, it is simpler to argue like this: if you have gauge invariance under a Lie group $G$, the ...

6

In the first case, the vertex is a vertex in the common sense (used to construct diagrams). In the second case, the gauge field is not dynamic (in a path integral formulation, you do not integrate over), it is a background field that is fixed. In that case, we are interested on the effect of this non-dynamical field on the electron field. This is useful to ...

5

The QFT for the scalar is considered to be massive for a very good reason: it is infinitely unlikely for the mass to vanish. There is no symmetry principle that would protect the scalar field from acquiring a generic mass. (The gauge symmetry is the principle that protects the masslessness of the photon but the scalar fields can't sacrifice to lose ...

5

Disclaimer: This is answer is given from a mathematical physics point of view, and it is a little bit technical. Any comment or additional answer from other points of view is welcome. The classical limit of quantum theories and quantum field theories is not straightforward. It is now a very active research topic in mathematical physics and analysis. The ...

5

The extra term, in general $$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{2 \xi}(\partial_{\rho}A^{\rho})^2$$ is called gauge fixing term. This term is needed in order to be able to quantize the field $A_\mu$. Without this extra term the photon propagator is ill defined $$D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu ... 4 This is an excellent question. The technical term for this effect is a collinear divergence. When p_i-p_f tends to 0 you get a divergence in the scattering amplitude. So why is this physically reasonable? Well remember that actual physical observables are cross-sections, not scattering amplitudes. Also recall that you cannot prepare a particle with an ... 4 Part b) is a big mathematical physics topic in its own right. The divergent tail of an asymptotic series is not garbage, rather it contains a lot of information that together with some additional information can be used to compute non-perturbative effects. A general introduction to this topic is given here. There are different approaches possible, some ... 4 There are, in general, no closed form solutions (aka formulas) for the spectra of multi-electron atoms. There are reasonably precise formulas for special cases, like approximate values of x-ray transitions from inner shell electrons, though. Unlike in case of hydrogen and Rydberg atoms, which can be treated as a non-relativistic one-body problems (i.e. for ... 4 Virtual particles are not real It's in the name. You may draw Feynman diagrams where there are internal lines, and we call these internal lines virtual particles. They are not real. You will never detect a virtual particle. They are not really exchanged between the real charged particles. Virtual particles are just-so stories designed to explain Feynman ... 4 Feynman diagrams are more than just the Lagrangian. They can be acquired by expanding the path integral of the theory into a perturbative series. There is a priori no reason to assume that all quantities needed in order to produce sensible results are consistent with gauge invariance. One possible issue is the problem of regularization: the way your ... 4 Classical electrodynamics and quantum electrodynamics don't agree with each other in general. They are distinct, inequivalent theories. Your observation that classical electrodynamics is deterministic, unlike QED, is one sufficient proof that they're inequivalent. At most, the expectation values of some observables in quantum electrodynamics obey the same ... 4 For instance, how did he come up with interpreting the propagator as the propagation of particles? The path integral is usually introduced as a matrix element of the time evolution operator$$ \langle x_f\lvert\mathrm e^{-\frac{\mathrm i}{\hbar}\hat{H}(t_f-t_i)}\lvert x_i\rangle, $$which is a measure of the probability of finding a system in final ... 4 If I understand what you're asking, it's false: there are plenty of examples of theories that are asymptotically free and also weakly coupled in the IR. A QCD-like theory with more flavors of quarks would be an example. The phrase to search for is "Banks-Zaks fixed point." For the revised version of the question: there are certainly RG flows that are free ... 3 I think that in some scenarii of the IR limit in Pure Yang-Mills SU(3) (QCD without fermions), the theory is also gaussian (trivial) in the IR (of course, one has to do more than the usual perturbative approach), and thus realize what you are looking for. See for example PhysRevD 84, 045018. 3 The polarization vector is just the vector-valued coefficient in front of the exponential but without the exponential. It's suppose to encode the "internal" degrees of freedom of the particle, not its dependence on space or momentum. The particular vector (0,1,i,0) expresses a circular polarization (either left-handed or right-handed; I guess that the ... 3 I understand that one can measure a single photon being absorbed using a photomultiplier tube or CCD. Can one measure a single photon being emitted by monitoring the current through an LED or the recoil of an emitting ion? The photon is a particle. It will have particle interactions, i.e. scattering off electrons and/or the spill over electric and ... 3 The additional correction to the magnetic moment of the electron, aptly called the 'anomalous magnetic moment,' arises from a one loop Feynman diagram calculation in quantum electrodynamics. To be specific, the Landé g factor is given by,$$g=2[1+F_2(0)]$$where F_2 is a 'form factor.' The electron vertex scattering amplitude is given by, ... 3 Let me anticipate: I am a novice in QFT so, please, do not mock me if I say something stupid. I was taught that the Feynman diagrams arose as a smart way to write down the intricated computations appearing in the perturbative approach to the path integral. The keystone is the well known Wick rule, that allows one to compute standard and Grassmannian ... 3 I) The un-gauge-fixed QED Lagrangian density reads$$\tag{1} {\cal L}_0~:=~-\frac{1}{4}F_{\mu\nu}^2 + \bar{\psi}(iD\!\!\!\!/ \ \ -m)\psi.$$The gauge-fixed QED Lagrangian density in the R_{\xi}-gauge reads$$\tag{2} {\cal L}~=~ {\cal L}_0 +{\cal L}_{FP}-\frac{1}{2\xi}\chi^2 , $$where the Faddeev-Popov term is$$\tag{3} {\cal L}_{FP}~=~ ...

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