There is formula relating power radiated and acceleration of charged particle. But i want to know the amount of energy is carried by the photon at a particular time by given acceleration by using relativistic quantum electrodynamics.
A classical radiation expressions, expressed in terms of a frequency (ω) distribution, can -- as a semi-classical interpretation -- be related to a mean distribution of emitted photons with various energies ħω.
Examples of such frequency distributions can f.i. be found in sections 14.5-6 in J.D.Jackson, Classical Electrodynamics (2nd edition).
Quantum electrodynamics does not operate with the notion of acceleration . And no moment can be ascribed to an emission. The classical Bremsstrahlung is well described.
Quantum electrodynamics, with the "dynamics of electrons" replaced by an appropriate classical current mimicking quantum scattering of charged particles, can usefully model the soft radiation (bremsstrahlung) from such processes. This leads to results in agreement with the usual (infinite order) QED perturbation treatment of bremsstrahlung, but such a treatment in addition provides a natural and reasonable photon energy cutoff. The quantum description only involves scattering parameters of the charged particles (incoming and outgoing 4-momenta), not the intermediate dynamics. But classically the scattering parameters are sufficient to uniquely define the full dynamics --- at least for simple systems.
This paper: https://www.rand.org/content/dam/rand/pubs/research_memoranda/2006/RM2820.pdf might be useful-though classical. Though if it isn't understood, there's no point discussing QED.
The energy isn't carried off by a single photon, however, except in some approximation.
@Stam Nicolis: I have to read it. And it look very nice paper..Thank you!
@Igor: And now i do understand why radiation is not dependent on acceleration in quantum mechanics. Theoretically we know, a electron in a LHO do not radiate if a electron is in ground state of LHO, because if it did, it wont be a ground state of electron, though electron do have angular velocity(w) and energy hw/2 in its ground state and in LHO a electron have to accelerate. Using Einstein relation for A & B coefficient we find, the value of A (spontaneous rate). And then finding dipole moment term (using quantum mechanics). Then using both A & p, we get the value of A in terms of w. And we know P(power)=Ahw so,
P=.(q^2w^2/6pie(o)mc^3)*(E-hw/2). When, E=hw/2 (ground state) then there is no radiation. However, classically we use larmors formula, then using acceleration in LHO, in lamors formula we get quite same result except the hw/2 term. Classically we get q^2w^2E/6pie(o)mc^3. So even classically (in limit h tends to zero) we are getting approximately same result.
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