# I Photon energies from Planck-Einstein: confirmed in practice?

Tags:
1. Apr 11, 2016

### DavidReishi

I understand that the photon energies given by the Planck-Einstein relation, though highly precise, are approximations. But have they been confirmed at all experimentally or in practice? If so, across the board or just some of them?

2. Apr 11, 2016

### vanhees71

What do you mean by approximations? The energy-momentum relation for Photons $E=|\vec{p}| c$ is exact by definition (a photon is an asymptotic free one-photon Fock state).

3. Apr 11, 2016

### DavidReishi

I don't mean the energy-momentum relation. I mean the Planck-Einstein relation, i.e. between frequency or wavelength and energy.

4. Apr 11, 2016

### vanhees71

There cannot be anything approximate here, because a photon is by definition a single-particle Fock state of the quantized electromagnetic field, but perhaps I don't understand you question right. Of course, the socalled old quantum theory by de Broglie and Einstein is obsolete. So there's no way to describe photons as particles (they don't even have a position observable in the strict sense) nor is there anything like "wave-particle duality" that makes sense in the context of modern relativistic quantum field theory!

5. Apr 11, 2016

### DavidReishi

Yeah, you misunderstood my question. I don't mean that the Planck-Einstein relation treats photon-energies as approximate. I mean that the values themselves of the photon-energies given by the equation, though very precise, are approximations. That is, the correspondence between frequency and energy given by the equation isn't perfect.

Good to hear, and I find those words very interesting. Can you point me to a good writer or two in whom such views, i.e. more or less rejection of the classical Einstein photon, are expressed most clearly?

6. Apr 11, 2016

### vanhees71

The modern way of describing photons is through the quantization of the free electromagnetic field. The relations $E=\hbar \omega$ and $\vec{p}=\hbar \vec{k}$ follow from the decomposition of the electromagnetic field in momentum eigenmodes you also get the energy-momentum relation from the dispersion relation of electromagnetic waves, i.e., $\omega = c |\vec{k}|$. Multiplying by $\hbar$ leads to $E=c |\vec{p}|$, i.e., in momentum space the photon energy-momentum relation is that of a massless particle.