# 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 ...

8

Both the wave theory of light and the particle theory of light are approximations to a deeper theory called Quantum Electrodynamics (QED for short). Light is not a wave nor a particle but instead it is an excitation in a quantum field. QED is a complicated theory, so while it is possible to do calculations directly in QED we often find it simpler to use an ...

8

This is a perceptive question. Consider the following from the Wikipedia article "Virtual Particle": As a consequence of quantum mechanical uncertainty, any object or process that exists for a limited time or in a limited volume cannot have a precisely defined energy or momentum. This is the reason that virtual particles — which exist only ...

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

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

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 ...

6

Forward scattering need not be equivalent to "no scattering" - and, indeed, will only rarely be indistinguishable from it. In the usual scattering-theory setup, you have an electron coming in in a plane wave $$\psi(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}=e^{ikz}$$ and impinging on some short-range potential. This will add to the wavefunction a scattered ...

5

Try, for instance, section 9 of Srednicki. The way to do it is to replace the fields in the interaction Lagrangian by functional derivatives with respect to the sources, then write power series for the exponents. Take the first order contribution. Then, use that you need to consider three-point functions where the fields are again replaced by functional ...

5

In this link there exists a mathematical explanation of how an ensemble of photons of frequency $\nu$ and energy $E=h\nu$ end up building coherently the classical electromagnetic wave of frequency $\nu$. It is not simple to follow if one does not have the mathematical background. Conceptually watching the build up of interference fringes from single photons ...

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 ...

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

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

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 $$D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu ... 4 Classical electrodynamics and quantum electrodynamics don't agree with each other in general. They are distinct, inequivalent theories. Your observation that classical electrodynamics is deterministic, unlike QED, is one sufficient proof that they're inequivalent. At most, the expectation values of some observables in quantum electrodynamics obey the same ... 4 For instance, how did he come up with interpreting the propagator as the propagation of particles? The path integral is usually introduced as a matrix element of the time evolution operator$$ \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

Excellent question. There are forces for which a force-carrier particle is not it's own antiparticle (eg: Strong force or the Weak force) so if we accept your explanation, then we might have to abandon Newton's 3rd law for those forces, which is implausible. Edit: Maybe I'm being obtuse. I don't think really matters that the antiparticle is not the same ...

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 ...

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