Single photons: Is there a 90° offset of the electric to the magnetic component in the direction of propagation?

Question

score 1 · Answer 1 · edited May 29 '14 at 09:02

1

We have to make clear that a photon is an elementary particle, the quantum of the electromagnetic field, it has a zero mass and a spin of 1, and its energy is given by $E=h\nu$.

An electromagnetic wave is composed of photons, a photon is not a chopped up electromagnetic wave. The electromagnetic field emerges from a huge ensemble of photons in a smooth way that goes from the microworld of quantum mechanics to the macroworld of classical physics. The how is described in this link. In a nutshell, it is the electromagnetic potential ${\mathbf A}$ that enters in the quantum mechanical wave function of the photon, which in tandem with zillions of photons constructs the macroscopic with frequency $\nu$ wave and builds the electric and magnetic fields of the classical wave.

edited May 29 '14 at 09:02

John Rennie

355,118

answered May 29 '14 at 08:31

anna v

233,453

I see a big difference between the electromagnetic emission from single elementary particles such electrons or protons (from gamma to terahertz) and the radio waves which we get accelerating electrons in a metall bar. 2. Following the graphics above it's easy to find a concept about photon spin. There are exact two possibilities to draw this graphic. You have to redirect the E- or the B-vector and you get it. No more possibilities.

HolgerFiedler

May 29 '14 at 09:26

@HolgerFiedler The diagram is the macroscopic effect, the cause is the individual photons, and you cannot assign an E and B field to a single photon because the E and B fields emerge from the ensemble of photons, there is no quantum of electric field or magnetic field . The diagram shows the macroscopic wave, not a photon, which is a point elementary particle. It is putting the cart in front of the horse. – anna v May 29 '14 at 09:53

@HolgerFiedler Have a look at this quantized form of circular polarization http://en.wikipedia.org/wiki/Spin_angular_momentum_of_light#Mathematical_expression – anna v May 29 '14 at 09:59

The circular polarization you get then the source gives the angular momentum, means the source is rotating. Or the photon goes through a medium like some minerals with cristall characteristics near the wave length of the photons. – HolgerFiedler May 29 '14 at 10:13

The photon does not need a medium – anna v May 29 '14 at 10:28

But if the photon goes through a medium there could be be a influence from the medium. And in some case the result will be a rotation of the photon. But this is not the spin. Übecause the spin if a photon has only two values. The rotation you mean can be with a lot of values. – HolgerFiedler May 29 '14 at 12:48

nevertheless the collective action emerges as +/- 1 . You have to acquire a quantum mechanical intuition/view. Classical does not work with elementary particles, and the photon is an elementary particle. if you do not worry about photons, the classical description is fine. you cannot fit the photon into a classical description except as a collective effect, as shown in the link in the answer. – anna v May 29 '14 at 13:02

We started with a question about a single photon. The value of the spin for a single photon is exact -1 and +1 – HolgerFiedler May 29 '14 at 16:19

yes, but its relationship to the electric and magnetic fields of the classical wave is complex. The simplest relationship classical/ quantum I have seen is in this circular polarization formula – anna v May 29 '14 at 18:53

Anna V is of course right here but the main question, which might be still answerable, (even when we know that as she correctly stated that the phenomenon can be successfully described in QM) on the individual photon level, is whether when we shoot a single photon (it is possible now as we know), will that single photon at a certain position of the propagation axis be the excitation of both E and M fields, (and so the photon itself would have two fields coupled) or is it just the excitation for one of them and then 'changes' to being the excitation of the other one as it propagates? – Árpád Szendrei Nov 25 '16 at 19:05

@ÁrpádSzendrei The photon is a point particle described quantum mechanically by a complex wavefunction which has the A potential in its sinusoidal functions, in some formulations, or E and B in others. Thus it is the probability of finding it that is "waving" as a function of E and B . https://www.nist.gov/sites/default/files/documents/pml/div684/fcdc/preprint.pdf – anna v Nov 25 '16 at 19:37

score 0 · Accepted Answer · answered May 29 '14 at 07:53

If you ask whether there is a phase difference of 90° between the electric field and the magnetic field, the answer is yes. The electric field and the magnetic field oscillate in quadrature. You can see that from the conservation of the total electromagnetic energy during one oscillation cycle, or as you mentioned it, to conserve the energy of a "single photon" during a cycle.

Plug quickly $$E = E_0 cos(\omega t)$$ and $$B = B_0 sin(\omega t)$$ in the instantaneous electromagnetic energy density), remembering that $$cos(\omega t)^2 + sin(\omega t)^2 = 1 $$ and $$ B_0 = E_0/c $$ in the present case of transverse electromagnetic waves propagating in free-space.

While this is true, I don't think it addresses how the classical picture of $E$ and $B$ fields at $\pi/2$ emerges from superpositions of photons. — John Rennie, May 29 '14 at 08:59

Single photons: Is there a 90° offset of the electric to the magnetic component in the direction of propagation?

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