# Photons, schmotons

1. Feb 14, 2006

### Ratzinger

1. I found a nice site. http://math.ucr.edu/home/baez/photon/schmoton.htm

2. I also found a nice quote.

"All the fifty years of conscious brooding have brought me no closer to the answer to the question, "what are light quanta?" Of course today every rascal thinks he knows the answer, but he is deluding himself."
Einstein (1951)

3. Then, I also have a question. Is it true that due to time-energy uncertainty only a photon that lived for a very long time has (almost) definite energy and momentum?

4. Matter fact, i have a second question. Do a atoms that had been longer in an excited state produce photons with more precise energy?

2. Feb 14, 2006

### vanesch

Staff Emeritus
3. Feb 16, 2006

### Ratzinger

But what about my two brilliant questions?

4. Feb 16, 2006

### Ratzinger

Alright, then let me try with a less brilliant question. It is actually an embarrassing basic question.

Are electromagnetic waves created by accelerating charges not quantised and have therefore no quantum description? Do we only speak of photons when bound states are involved?

5. Feb 16, 2006

### ZapperZ

Staff Emeritus
Er... isn't this similar to asking if light coming out of an incandenscent light bulb has no photons, but light coming from a He discharge tube, for example, does? All because you see DISCRETE allowed wavelength (or frequency) from a discharge tube?

Synchrotron centers all over the world would be in deep doo doo if this is true.

Zz.

6. Feb 16, 2006

### Physics Monkey

Ratzinger,

Question 1: It is a general property of the fourier transform that as you narrow your spectral range, your temporal width must increase. A perfectly well defined frequency corresponds to a signal that is constant for all time. If you want a laser to provide a very precise frequency of light then you have to stabilize it against fluctuations and keep it on for a long time.

Question 2: There are several mechanisms by which the spectral lines of atoms can be broadened. The relevant mechanism for what your talking about is radiative broadening which originates in spontaneous emission. Vacuum fluctuations give every excited state a finite lifetime, and the spectral line is broadened with a Lorentzian lineshape whose width is the spontaneous emission rate. If the rate of spontaneous emission is very small then the state is long lived, and the spectral line can in principle be fairly sharp. In practice, the spectral line is also considerably broadened by other mechanisms like Doppler and collision broadening.

Hope this helps.

7. Feb 17, 2006

### Ratzinger

Physics Monkey, great post. Very helpful.

Zapper, yes, my question was rather a silly one. But the problem is when I google "photon accelerated charge" I get classical EM physics (where they say EM radiation can be created with osciallating dipoles, another way is when excited bound states give off energy than we need to invoke quantum theory). I just like to know how quantum theory treats it.

Last edited: Feb 17, 2006
8. Feb 17, 2006

### DaTario

Let me put some spicy season on this subject. If you are in the same reference frame of this accelerated charge, you will see no radiation at all.
In the Am. J. of Phys of this month there is a paper on this which says that the radiation of an accelerated charge is outside its light cone.
I didn't follow the details, but it seems to be an interesting result.

Best Regards

DaTario

9. Feb 18, 2006

### Gokul43201

Staff Emeritus
Even the vibrations/rotations of dipoles are quantized. A classical calculation will give you a partial picture, while the quantum calculation (harmonic oscillator/rigid rotor) will provide a more accurate description of the emitted radiation.

10. Feb 18, 2006

### Vixus

Great text! Lots to read though. :D

11. Feb 18, 2006

### Ratzinger

I too heard that accelerating charges give problems with refrence frames and especially when the equivalence principle comes into play. But putting this aside for a moment.

My simple (sounding) question: how many photons does an (linear) accelerating charge give off? Do we need full-fledge quantum field theory to explain this? Or is it best explained in electrodynamics?

And what is meant with a countinous spectrum that's being created? How do photons fit in here?

12. Feb 18, 2006

### Physics Monkey

Hi Ratzinger,

The photon concept is associated with the discreteness of the energy for a particular color of light rather than directly with the discreteness of allowed colors. If you put your electromagnetic field in a box then you indeed find certain discrete allowed frequencies. On the other hand, if you quantize the field in free space then in principle there is a continuum of modes available. Nevertheless, for each mode (color) of light, the energy of that mode comes in lumps, and it is these lumps which are the photons.

In fact, classical electromagnetic fields do not have a definite number of photons. However, to determine the average number of photons in a given mode, you need only know the amount of energy in that mode. In other words you need some kind of spectral resolution of the energy. The precise way in which you define the energy of a mode will depend on the mode structure (since it's hard to tell photons with very close frequencies apart). For example, think about the thermal electromagnetic field in a box where the average number of photons in each mode is given by the Bose factor $$n(\omega) = \frac{1}{e^{\beta \hbar \omega} - 1}$$. In the high temperature limit this expression becomes $$\frac{kT}{\hbar \omega}$$ which makes physical sense, you have a photon for every $$\hbar \omega$$ in $$k T$$.

Last edited: Feb 18, 2006
13. Feb 19, 2006

### Ratzinger

Thanks Physics Monkey. I'm becoming a big fan of you.

14. Feb 19, 2006

### nrqed

How can one see that? I mean, in principle, one can do a Fourier decomposition of a classical EM wave and therefore write it as a linar combination of waves of well-defined wavelengths. One can calculate the energy carried by a classical wave. So why can't one associate a definite number of photons to a classical wave? I know that people say that a classical state is a coherent state of photons but what is the rationale behind this? (which leads to the equivalent question of why a classical gravitational background must be seen as a coherent state of gravitons).

15. Feb 20, 2006

### vanesch

Staff Emeritus
Yes, but then one would have lost entirely the phase information in the classical wave! What you are doing is to fix the energy density (abs square of fourier transform). If you insist on having a small error on this, the complementary quantity (phase) will be totally arbitrary.

The coherent state comes closest to a small error on the energy density and on the phase. Just as with (in fact no surprise!) a harmonic oscillator. Your argument would be that the "classical" state of a harmonic oscillator must be an energy eigenstate, because from the classical motion, we can precisely calculate the energy (KE + PE), and hence know the energy eigenstate. But then other aspects of the classical state, such as a rather precise position, would be badly represented: an energy eigenstate is totally delocalized. So you do a compromise on the precision of the energy in order to regain some precision on the position. And the "nicest" compromise you can make (smallest combined error on momentum and position) is given by a coherent state, not by an energy eigenstate.
Do this for each mode, and you are in the case of a coherent state of the EM field coming close to the classical field description.

16. Feb 21, 2006

### DaTario

Are you saying with this that Classical EM fields does not have a well defined intensity ?

Best Regards

DaTario

17. Feb 21, 2006

### vanesch

Staff Emeritus
I guess he means that the *best possible quantum approximation of a classical state* does not have a well-defined intensity, but only a rather well-defined intensity.

18. Feb 21, 2006

### ZapperZ

Staff Emeritus
Say what, vanesch??!!

:)

Zz.

19. Feb 21, 2006

### vanesch

Staff Emeritus
Yeah, not my best phrase of the day, I suppose
For my excuse, a collegue is leaving and we had some wine over lunch...

20. Feb 21, 2006

### Physics Monkey

DaTario,

I am certainly not saying that. Classical electromagnetic fields have well defined intensity and phase within classical electromagnetism , just like particles have well defined position and momentum within classical mechanics . However, Ratzinger asked about photons, and the quantum description is appropriate for answering such a question. As such, saying that a classical electromagnetic field doesn't have a definite number of photons must be interpreted as vanesch has done.

To repeat what he said, the states of the quantum electromagnetic field that most resemble what we would call classical electromagnetic fields are the coherent states, and these states do not have definite photon number. It is true, for example, that any classical source interacting with the quantum field produces a coherent state. Furthermore, it isn't hard to show that states with definite photon number always have a vanishing average electric field while at the same time having perfectly well defined intensity. This is a manisfestation of number-phase uncertainty which is a useful conceptual description even if there is not actually a hermitian phase operator.

I hope this helps clarify my meaning for you.