Re: "A Snapshot of a Photon"

John Kennaugh wrote:
Tom Roberts wrote:
The "length" of a photon is not well defined. Photons are QUANTUM OBJECTS and do not behave as you naively expect. In particular, a photon is NOT a short wave-train as you seem to think. The best definition I know of what a photon actually is: a specific type of term in the perturbative approximation to an amplitude in QED.

If you attempt to measure the "length" of a photon (e.g via self-interference), you will obtain a value that depends on precisely how it was emitted, and your value will usually be the coherence length of the source. That can vary from microns for an incandescent bulb to many meters for a single-mode laboratory laser.

Surely the coherent length is related to photons 'on mass'.

As I said, the coherence length is related to properties of the SOURCE, not of the photons themselves.


If one assumes a photon has some property analogous to (or actually) 'spin' then the coherent length is the length of a burst containing photons 'spinning' 'in phase' with each other and can in theory be any length.

You must select another word than 'spin' -- photons do indeed have what is known as spin (anbgluar momentum). But I cannot see any relation between the photon's spin and what you say here. This seems like word salad to me.


Any regular electrical waveform can be analysed using Fourier analysis [...] 

Yes, of course.


The question I am exploring is related to your statement:

Note that if you turn a source on and then off quickly enough, you will inherently introduce an uncertainty in the frequency of the pulse (Heisenberg always has the last word). For a conventional RF source you can certainly arrange to emit less than a full wavelength, which results in an enormous frequency spread.

The point I am wrestling with is this. OTOH the frequency spread is simply and a faithful conversion of electrical signal into RF signal. I can't see why, apart from the obvious difficulty of generating it, why it would not be possible to have short bursts of photons all with the same energy/frequency.

Do the Fourier analysis. That is, perform a Fourier integral of (say) a single cycle of a sinewave. The result will be a large spread of frequencies, with a peak at the frequency of the sinewave. To get a delta-function at its frequency you need an INFINITELY-LONG sinewave.


If one considers a very, very dim source of light one can conceive of photons arriving one at a time. 

Yes.

Surely one photon constitutes an extremely short burst

No. See above. Photons are not "short pulse trains", and they are not like any ordinary object you have ever encountered. Thay are QUANTUM OBJECTS.


and we think of a single photon as having a single frequency f = h.nu rather than an enormous frequency spread

Wrong pronoun -- YOU may think of it like that, but physicists don't. Physicists know that the frequency of a photon is uncertain....


Suppose you have a monochromatic light beam which is only just too low in energy to make current flow in a photo electric experiment. Suppose you break up that beam by passing it through a rotating toothed wheel. Does that produce higher energy photons such that current now flows?

Only in a gedanken for which the monochromaticity is perfect and the "just too low" can be made vanishingly small. The energy uncertainty introduced by such a toothed wheel is ENORMOUSLY smaller than the energy scales for electrons. But yes, a pulsed light source cannot be truly monochromatic (see above -- do the Fourier integral).


you cannot say the toothed wheel would necessarily have no other effect than to interrupt the flow of photons - there would be diffraction for a start. 

Yes. And absorbtion. And scattering.


Tom Roberts	tjroberts2lucent.com
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