1. Jan 30, 2008

### Euclid

I am looking in KK Thermal Physics ch4 at what I assume to be the standard derivation of the SB law of radiation and I notice something peculiar.
On the one hand, they model the photon as a 1D SHO with energy given by
$$\epsilon = n \hbar \omega$$
On the ohter hand, the distribution of the modes (omega) is given by the condition for a standing EM wave in a 3D box ($$\omega =\pi c \sqrt{ n^2 + m^2 + l^2} /L$$). My question is, why does one not assume a 3D SHO model for the photon with
$$\epsilon = (n+m+l) \hbar \omega$$?
It seems odd to model the photon as a SHO, but only partially. What's the full story?

2. Jan 30, 2008

### vanesch

Staff Emeritus
You are confusing two different "quantisations" here. EACH individual FIELD MODE is an independent 1-D SHO. So the whole system is not a 3-D SHO, but a multi billion-fold dimensional SHO (infinite, in fact). In free space, all plane waves are field modes. But in a box, with boundary conditions, the field modes are quantized (in classical EM). Each of these modes can be described classically by a "harmonic oscillator" with a certain frequency (fixed by the mode) and a certain amplitude/phase (which is free in classical physics). It is THIS harmonic oscillator which will be quantized. So for EACH field mode, we have an oscillator, which, after quantization will take on the famous
E = n(mode) omega(mode) x hbar.

We say that n(mode) is the NUMBER OF PHOTONS in this mode.

So a photon (of a certain type = associated with a certain classical mode of oscillation of the EM field) is nothing else but a quantization step of the associated SHO.

So one quantization is classical, and gives you the modes (and hence the omega(mode)) ,and the other quantization is quantum-mechanical, and gives you the ladder of the oscillator associated with the mode. You have a quantum-mechanical oscillator PER MODE.

Edit:
such an infinite set of harmonic oscillators, associated to classical field modes, is called a QUANTUM FIELD.

3. Jan 30, 2008

### Euclid

Very cool. Thanks for the reply.

Since there are an infinite number of modes, won't the ground state of that system be infinite? It's interesting that KK ignores the zero level energy...

4. Jan 30, 2008

### vanesch

Staff Emeritus
YES.

So what people do is: they subtract this ground level. It's a first taste of renormalization... in quantum field theory, we don't stop subtracting infinities from infinities...

5. Jan 30, 2008

### Euclid

This is very interesting. It seems totally ad hoc. But the renormalization process works?

6. Jan 31, 2008

### vanesch

Staff Emeritus
Yes... it is not *totally* ad hoc, but it is not very clean either. Quantum field theory is mathematically not sound, but as you say, it works. That is, the fundamental mathematical constructions can be shown not to exist (!), but the derived calculational procedures work quite amazingly well. That's why people then said that the actual theory was the "set of calculational procedures" and that the (non-existing) objects one was trying to calculate was just an inspiration. And then it turns out that even these calculational procedures are mathematically ill-defined, except for the first approximations.

However, these first approximations give amazingly accurate numerical results.