# Tag Info

8

Comments to the question: First it should be stressed, as OP does, that the Euler-Lagrange equations (= classical equations of motion = Maxwell's equations) are unaffected by scaling the action $S[A]$ with an overall (non-zero) constant. So classically, one may choose any overall normalization that one would like. As Frederic Brünner mentions a ...

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 ...

5

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 ...

4

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 ...

3

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 in a ...

2

It depends on the non-trivial topological solution of the equations of motion under consideration. The most famous example is the 't Hooft–Polyakov monopole (see http://en.wikipedia.org/wiki/%27t_Hooft%E2%80%93Polyakov_monopole) which has the following solution: $$\phi = h \frac{x^a} {r} t^a$$ A_0=0 ...

2

Your interpretation is not correct. The propagator $D_{\mu\nu}(x-y)$ describes the amplitude for a photonic field perturbation to go from $x$ to $y$, with the implicit picture that you have a "source" $J(x)$, and a "sink" $J(y)$, which are perturbing the vaccuum. However, a field perturbation is not a real particle (for instance, in the photon case, the ...

2

If you want an even more everyday example than Emilio Pisanty's example: "no scattering" would mean that the would be scattering object in question (modelled by the short range potential in Emilio's answer) would beget no change the the forward travelling wave. Otherwise put, an observer sensing the incoming plane wave could not tell whether or not the ...

2

Virtual particles, whether photons or electrons or... are, in the context of QFT, particles that are off-shell, i.e., their associated energy and momentum are not related by the relativistic energy-momentum relation. Please read this to get an idea of how virtual particle exchange can create attractive or repulsive forces. Photons are quanta of the modes ...

2

In 1995 Willis Lamb published a provocative article with the title "Anti-photon", Appl. Phys. B 60, 77-84 (1995). As Lamb was one of the great pioneers of 20th century physics it is not easy to dismiss him as an old crank. He writes in the introductory paragraph: "The photon concepts as used by a high percentage of the laser community have no scientific ...

1

The question here really is: "When are photons not virtual?". As explained in this article: A virtual particle is one that has borrowed energy from the vacuum, briefly shimmering into existence literally from nothing. Virtual particles must pay back the borrowed energy quickly, popping out of existence on a time scale set by Werner Heisenberg's ...

1

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 ...

1

To test whether $|0\rangle$ is physical, you would apply $\partial^\mu A_\mu^+$ to it. So $|0\rangle \in V$. Note that 7.48 does not mean that $\partial_\mu A^\mu = 0$ as an operator identity in this space. It means that all it's matrix elements in this space are zero. You have noticed that the state you created, $\partial_\mu A^\mu |0\rangle$, lives not ...

1

If, by "first principles" you mean without any observation or experimental input, I don't think we have the answer yet. We don't know why we have the particles/fields we do. But if we're given those particular particles/fields, then we know that consistency tells us what the Lagrangian must look like and leaves us little room for fixing coefficients of some ...

1

However, I do not observe any visible light between objects around me, repelling each other when I place them on a desk. They also don't appear to heat up, or emit IR light, from just being placed beside each other either. What type of light is being emitted by this matter constantly? The key point is that those photons are not real photons. They are ...

Only top voted, non community-wiki answers of a minimum length are eligible