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I tried a search and could not find it in a simple format. Like if the wavelength halves than the magnetic field falls off like $~\frac{1}{\lambda^2}$.

I mean the maximum value of the sinusoid for a single photon. I have read about this some where, but no idea where.

A photon has a well defined energy and wavelength and they can come as single particle. So i just wondered how one photon would induce a magnetic field. They use photons in laser traps and then make a beautfl picture of the electric field as all photons are added. So i was thinking how would the field of a single one be. I mean we can divide by the big number of photons, that are present.

I know that in QED, there is just a chance of the photon coupling to some other particle at some time. And there is little interest in what is happening between.

But in the laser trap we have both pictures.

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    $\begingroup$ What magnetic field are you talking about? Also, the relationship between a single photon and an electromagnetic wave is a bit difficult, see also What is the relation between electromagnetic wave and photon? $\endgroup$ – ACuriousMind Sep 5 '16 at 10:46
  • $\begingroup$ Hi, Thanks, Curious Mind. Well, a photon has a well defined energy and wavelength and they can come as single particle. So i just wondered how one photon would induce a magnetic field. They use photons in laser traps and then make a beautfl picture of the electric field as all photons are added. So i was thinking how would the field of a single one be. I mean we can divide by the big number of photons, that are present. $\endgroup$ – ArthurX Sep 5 '16 at 16:21
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There is no relationship of the form you have in mind.

The main point is that photons with the same wavelength do not have to be identical. For example, if an atom or nucleus emits a photon, we can further classify the photon according to its angular momentum and polarization.

In a semiclassical description, suppose that a classical sinusoidal plane wave packet has a certain given energy. We can't use that information to find the amplitude of the wave, because it depends on the dimensions of the wave train. We could have a low energy density over a large volume, or a high energy density over a small volume.

It is possible to cook up a semiclassical estimate for the greatest field that any photon of a given wavelength can have. Suppose that the photon is compressed to a volume $~\lambda^3$, which is on the order of the smallest volume a wave can have, for that wavelength. Then multiplying the energy density of the electric field by that volume and setting it equal to $hc/\lambda$ gives the estimate $|\textbf{E}|\sim \lambda^{-2}(khc)^{1/2}$, where $k$ is the Coulomb constant. (And in SI units, $|\textbf{B}|=|\textbf{E}|/c$.) For visible light, this is something like $10^5\ \text{V}/\text{m}$. But that should not be interpreted as "the" intensity of the field. E.g., for a photon of visible light from a laser, a realistic estimate of the volume is many orders of magnitude greater than $\lambda^3$, due to the long coherence length, so the field strength would be many orders of magnitude lower than this upper bound.

I tried a search and could not find it in a simple format. Like if the wavelength halves than the magnetic field falls off like $~\frac{1}{\lambda^2}$.

Yes, the estimate does fall off with that exponent -- but only the estimate for the maximum intensity that any photon can have. E.g., for a wave packet with fixed dimensions, the exponent would $-1/2$ rather then $-2$.

They use photons in laser traps and then make a beautfl picture of the electric field as all photons are added. So i was thinking how would the field of a single one be. I mean we can divide by the big number of photons, that are present.

For example, suppose we have a standing wave in a resonant microwave cavity, with volume $\sim\lambda^3$. Then the estimate above is of the right order of magnitude, and replacing the single-photon energy $hc/\lambda$ with $nhc/\lambda$ gives $|\textbf{E}|\sim \lambda^{-2}(nkhc)^{1/2}$. That is, the scaling is like $\sqrt{n}$, not $n$, because the photons are coherent.

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The exact value of the amplitudes of the magnetic as well as the electric field components of the photon is unknown. Perhaps it would be a good approximation to reference to the slits width in polarizer, slit experiments or the distance between atoms / molecules in crystals.

It is obvious that through a polarizer a range of wavelengths passes through without be polarised, some range will be polarised and some range will be blocked at all. Call it the cross section or the amplitude of a photon, in any way you can find the distance in which the photon interacts with the edges of an obstacle.

The above mentioned holds for photons as indivisible units. But there is an other definition of what a photon is. The photon is an excitation of an overall existing electromagnetic field and this excitation is infinit spread out. For this definition there is some trouble because on an obstacle some extend of the excitation stays behind the obstacle (because nothing travels faster than light) and some (infinite) extend has to adding to the photon after it passes the obstacle. In this case the square of the amplitude is defined only as the probability to find the photon in some volume at a given time. Any amplitude in reference of an amplitude of an water wave or a swinging rope is not defined.

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  • $\begingroup$ Hi Holger. Well, a photon has a well defined energy and wavelength and they can come as single particle. So i just wondered how one photon would induce a magnetic field. They use photons in laser traps and then make a beautfl picture of the electric field as all photons are added. So i was thinking how would the field of a single one be. I mean we can divide by the big number of photons, that are present. $\endgroup$ – ArthurX Sep 5 '16 at 16:24
  • $\begingroup$ This answer reads to me like total nonsense. $\endgroup$ – user4552 Mar 25 '18 at 15:44
  • $\begingroup$ @Ben From which point of my answer you feel it is nonsense? $\endgroup$ – HolgerFiedler Mar 25 '18 at 16:06
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The size of the magnetic field for a photon wouldn't depend its wavelength. The amplitude of the fields do not depend on the frequency or wavelength, it depends on the number density of photons (and then there is the distinction between coherent and incoherent sources to consider).

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  • $\begingroup$ This doesn't really make sense. If you claim that the amplitude only depends on the number of photons and not on the wavelength, then the amplitude should have a fixed value for one photon -- but it doesn't. $\endgroup$ – user4552 Mar 25 '18 at 15:08

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