0
$\begingroup$

(Correct where applicable)

Light is a wave on the electromagnetic field (or the electromagnetic field can be interpreted as a representation of the magnitude and direction of the force photons will have on another charge).

When we concentrate light, we are essentially changing the paths of the electromagnetic waves or photons so they are closer to each other. Since electromagnetic waves can be said to follow field lines, and in the photon interpretation, the path a photon takes is the field line (and yes, in quantum physics a photon takes all possible paths, so the field lines of a single photon are all over the place), and the electromagnetic field intensity increases when field lines are closer together, does concentrating light also increase electromagnetic field intensity?

$\endgroup$
0
$\begingroup$

Yes there is a bigger EM field when photons are concentrated. But the field lines of a photon are perpendicular to the line of travel. Note that the field is quantized in energy, so even though it looks like a bigger wave only part of it corresponding to the single photon energy can be absorbed at a time. Also although the photon sees many paths as its EM field radiates out, most of the energy in concentrated in a small space, but yes a photon going through a hula hoop still gets diffracted slightly.

$\endgroup$
  • $\begingroup$ "the field lines of a photon are perpendicular to the line of travel". See my comments under anna v's answer. $\endgroup$ – Abdul Moiz Qureshi Nov 18 '18 at 17:59
0
$\begingroup$

Do not confuse the path integral over the photon field with the classical field lines.

"so the field lines of a single photon are all over the place"

No they are not , a photon does not have a field ,electric or magnetic, except in its non measurable wavefunction,it is a quantum mechanical particle in the table of elementary particless.

photwavf

Measureable are only its energy=h*nu and its spin .

The zillions of wavefunctions added up in superposition, give a new total wavefunction for the beam that they build up, and the complex conjugate square of this will give the probability of interaction of the beam as a whole. How this happens in QED can be seen here.

The addition of the fields will depend on the coherence of the beam, and there is also quantum mechanical "squeezing ". Quantum mechanics is not simple additions.

$\endgroup$
  • $\begingroup$ Is it wrong to say that in the photon interpretation of light, the photons are fundamental and create the field, while in the wave interpretation, the field is fundamental and creates the waves? If so, the field lines in the photon interpretation would be the path of photons, and due to photons taking all possible paths, be all over the place. $\endgroup$ – Abdul Moiz Qureshi Nov 18 '18 at 16:37
  • $\begingroup$ Within the standard model of particle physics, which is very well validated, it is wrong. In the field interpretation it is the creation and annihilatio operators that generate and destroy photons. $\endgroup$ – anna v Nov 18 '18 at 16:42
  • $\begingroup$ If I am correct, then "a photon does not have a field ,electric or magnetic" is not incompatible with me saying that the electromagnetic field is made out of photons. So basically the inverse of your statement. $\endgroup$ – Abdul Moiz Qureshi Nov 18 '18 at 16:42
  • $\begingroup$ see this link:van.physics.illinois.edu/qa/listing.php?id=414 ------- since a pure magnetic field does not exist and is always accompanied by an electric field, then the electromagnetic field is also made of photons. So why can't we say the electromagnetic field lines are the paths the photons take, and the magnitude and direction of the electromagnetic field intensity is directly proportional to the magnitude and direction of the photon at that point? $\endgroup$ – Abdul Moiz Qureshi Nov 18 '18 at 16:57
  • $\begingroup$ because that is not the way field theory works in QED. The field is like a coordinate system, and the creation operators propagate the photons , and a free photon is modeled as a wave packet on the field. $\endgroup$ – anna v Nov 18 '18 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.