# I Why isn't the photon energy h nu over two?

1. Jul 29, 2017

### DaveLush

Can anyone counter my argument that the photon energy is properly half of the generally accepted value? It is a short argument, as follows:

Accepting the Planck hypothesis that the energy of the standing-wave electromagnetic modes of a cavity oscillator at thermodynamic equilibrium is E = n h nu (with n a non-negative integer, h the Planck constant, and nu the frequency of the mode), then recognizing that standing waves can't have momentum, but a photon always has momentum, and also that a standing wave is representable as a superposition of two oppositely-traveling waves of equal amplitude, then the minimum number of photons needed to change the energy of a standing wave by h nu is two. Therefore the energy of a single photon can be at most h nu / 2.

I realize there is a lot of observation that seems to back up that the photon energy is h nu, so if that is correct, there must be something wrong with my argument. What is it that's wrong, then?

2. Jul 30, 2017

### bhobba

Well first of all all that historical stuff you wrote about Plank's analysis etc is well and truly outdated (ie basically wrong ie their really is no wave particle duality ie standing wave nodes is not the correct way of looking at it and deducing it has quantised energy) has been replaced by QM and QFT.

The QFT explanation is quite simple. Photons are excitation's of an underlying quantum quantum EM Field that permeates all space. This means photons literally are exactly the same - you interchange two photons (ie excitation's) and its exactly the same - there is no difference, That is what separates them from normal classical particles. It applies of course to any elementary particle but photons, as predicted by QFT can occupy the same state as other photons - this is different to Fermions. Proving this - while not really germane to this thread has a long and convoluted history involving great luminaries like Feynman and Pauli:
http://www.worldscientific.com/worldscibooks/10.1142/3457

But anyway the key point is they are indistinguishable - as you would except being excitation's of a field. That's all you need to derive the Black-Body radiation. Its really a problem in statistical modelling and doing the counting right.

What the early pioneers wrote is a mishmash of classical and quantum ideas and you can't really trust any conclusions drawn from it.

Thanks
Bill

3. Jul 31, 2017

### Staff: Mentor

Sure; you're using the wrong definition of "photon"; you're implicitly assuming that it's a "particle" that always moves at the speed of light. The correct definition is that a "photon" in your standing wave scenario is a standing wave. More generally, the term "photon" is just a shorthand for "a mode of the electromagnetic field" (or, mathematically, a solution of Maxwell's Equations with appropriate boundary conditions).

4. Aug 1, 2017

### vanhees71

You get the energy eigenvalues by solving the eigenvalue equation of the Hamiltonian. For the em. field the Hamiltonian reads
$$\hat{H}=\sum_{\lambda=\pm 1} \int \mathrm{d}^3 \vec{k} \hbar |\vec{k}| \hat{N}(\vec{k},\lambda).$$
For a photon of wave number $\vec{k}$ thus the energy is $\hbar \omega$ with $\omega=|\vec{k}|$.