- #1
Q-reeus
- 1,115
- 3
The rms EM field within a cavity resonator, excited in a single resonant mode by a current probe of fixed rms amplitude, is known to rise linearly with time; eg. http://www.jpier.org/PIER/pier78/15.07090605.Wen.pdf", sections 5.2, 5.3, fig's. 6, 10. (Normally it is a fixed driving voltage, increasingly opposed by a rising internal cavity field, that applies, resulting in an exponentially decaying rms current. Over a relatively short time interval that can always be compensated by ramping up driving voltage, and we assume that applies here - fixed rms current simplifies the picture). As the probe is pumping in a fixed rms photon rate, it follows that photon density d and field rms amplitude A rise in direct proportion. Field rms energy density W is proportional to A2, so simple maths requires average energy per photon W/d is not the fixed value E = hf for an 'isolated' photon but rises linearly with time to typically a far higher value. The specific time independent proportionality factor will be a function of such things as frequency and cavity mode.
This evidently necessary finding conflicts sharply with standard QED, where the energy in an EM wave is simply the sum of the individual quanta (photons), eg. http://math.ucr.edu/home/baez/photon/odd-ques.htm" , where John Baez proposes:
"First let's think of light as being classical. Then the energy density is (E2 + B2)/2,...So the energy density is something like A2...On the other hand, let's think of light as being made of photons. The energy of a photon is E = hbar w where hbar is Planck's constant and w is the angular frequency,.. So the energy density should be hbar w d, where d is the average number of photons per cubic meter. So we should have A2 = hbar w d
Does that make sense? The lower the frequency, the more photons we need to get a particular energy, so that part makes sense. It's sort of odd how the number of photons is proportional to the square of the amplitude; you mighta thunk it would just be proportional."
Having a layman's knowledge of QM and QED, the circuitous route then taken is one I could not follow at all well.
That this standard QED relation applies generally - to both free-space radiation and fields confined within a cavity resonator was gleaned from "Cavity Quantum Electrodynamics; The Strange World Of Light In A Box" - S.Dutra. A 'plain english' statement on p48, following eq'n 2.148, relating there to a plane wave, reads:"...just as in the cavity case, we will get a Hamiltonian that equals the sum of the energies of each photon plus an infinite term..." (The latter is the zero-point field, irrelevant in context). In the context of say a dipole antenna radiating into free-space, this standard picture makes perfect sense - photon numbers and power are in proportion, and both are conserved overall. Hence energy per photon is invariant IN THIS SETTING.
But as was found earlier, the wheels fall off in the context of an excited cavity - energy per photon is anything but fixed. It should be emphasized at this stage - conservation of energy is never in question in any setting discussed here. PARTITIONING of energy is at issue. Whether or not 'coherent state' has any explanation here, there is a further consequence that seems truly bizarre. Suppose the cavity is a length of hollow waveguide several wavelengths long. Let the probe current be switched off after a time when the computed average energy per photon is very much greater than E = hf. Let one end of the waveguide be suddenly opened, turning it into an antenna. In a little more than twice the guide wave traversal time, the internal cavity field will have emptied into free-space, creating a burst of radiation propagating away in an increasingly spherical wavefront. Total wave energy and photon count are in this free-space setting both invariant, so that energy per photon is fixed but much greater than the 'usual' value. Something is very wrong - at very large distance from the source where only individual photons of energy E = hf might normally be encountered, what will actually occur? Will 'super energetic' photons be found, or will they actually fission en route to eventually leave 'normal' photons only. Or does the concept of photon lose validity in such a case. Certainly there is never a dilemma in the classical EM setting. It only arises when trying to fit a consistent field-as-photons picture. Or am I missing something basic in the original cavity setting? There's too much evidence in favor of the photon as real to ditch it. Need help - what's the answer?
This evidently necessary finding conflicts sharply with standard QED, where the energy in an EM wave is simply the sum of the individual quanta (photons), eg. http://math.ucr.edu/home/baez/photon/odd-ques.htm" , where John Baez proposes:
"First let's think of light as being classical. Then the energy density is (E2 + B2)/2,...So the energy density is something like A2...On the other hand, let's think of light as being made of photons. The energy of a photon is E = hbar w where hbar is Planck's constant and w is the angular frequency,.. So the energy density should be hbar w d, where d is the average number of photons per cubic meter. So we should have A2 = hbar w d
Does that make sense? The lower the frequency, the more photons we need to get a particular energy, so that part makes sense. It's sort of odd how the number of photons is proportional to the square of the amplitude; you mighta thunk it would just be proportional."
Having a layman's knowledge of QM and QED, the circuitous route then taken is one I could not follow at all well.
That this standard QED relation applies generally - to both free-space radiation and fields confined within a cavity resonator was gleaned from "Cavity Quantum Electrodynamics; The Strange World Of Light In A Box" - S.Dutra. A 'plain english' statement on p48, following eq'n 2.148, relating there to a plane wave, reads:"...just as in the cavity case, we will get a Hamiltonian that equals the sum of the energies of each photon plus an infinite term..." (The latter is the zero-point field, irrelevant in context). In the context of say a dipole antenna radiating into free-space, this standard picture makes perfect sense - photon numbers and power are in proportion, and both are conserved overall. Hence energy per photon is invariant IN THIS SETTING.
But as was found earlier, the wheels fall off in the context of an excited cavity - energy per photon is anything but fixed. It should be emphasized at this stage - conservation of energy is never in question in any setting discussed here. PARTITIONING of energy is at issue. Whether or not 'coherent state' has any explanation here, there is a further consequence that seems truly bizarre. Suppose the cavity is a length of hollow waveguide several wavelengths long. Let the probe current be switched off after a time when the computed average energy per photon is very much greater than E = hf. Let one end of the waveguide be suddenly opened, turning it into an antenna. In a little more than twice the guide wave traversal time, the internal cavity field will have emptied into free-space, creating a burst of radiation propagating away in an increasingly spherical wavefront. Total wave energy and photon count are in this free-space setting both invariant, so that energy per photon is fixed but much greater than the 'usual' value. Something is very wrong - at very large distance from the source where only individual photons of energy E = hf might normally be encountered, what will actually occur? Will 'super energetic' photons be found, or will they actually fission en route to eventually leave 'normal' photons only. Or does the concept of photon lose validity in such a case. Certainly there is never a dilemma in the classical EM setting. It only arises when trying to fit a consistent field-as-photons picture. Or am I missing something basic in the original cavity setting? There's too much evidence in favor of the photon as real to ditch it. Need help - what's the answer?
Last edited by a moderator: