Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Photons, schmotons

  1. Feb 14, 2006 #1
    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. jcsd
  3. Feb 14, 2006 #2

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  4. Feb 16, 2006 #3
    But what about my two brilliant questions?
     
  5. Feb 16, 2006 #4
    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?
     
  6. Feb 16, 2006 #5

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    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.
     
  7. Feb 16, 2006 #6

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  8. Feb 17, 2006 #7
    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
  9. Feb 17, 2006 #8
    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
     
  10. Feb 18, 2006 #9

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  11. Feb 18, 2006 #10
    Great text! Lots to read though. :D
     
  12. Feb 18, 2006 #11
    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?
     
  13. Feb 18, 2006 #12

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    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 [tex] n(\omega) = \frac{1}{e^{\beta \hbar \omega} - 1} [/tex]. In the high temperature limit this expression becomes [tex] \frac{kT}{\hbar \omega} [/tex] which makes physical sense, you have a photon for every [tex] \hbar \omega [/tex] in [tex] k T [/tex].
     
    Last edited: Feb 18, 2006
  14. Feb 19, 2006 #13
    Thanks Physics Monkey. I'm becoming a big fan of you.
     
  15. Feb 19, 2006 #14

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member


    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).
     
  16. Feb 20, 2006 #15

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  17. Feb 21, 2006 #16
    Are you saying with this that Classical EM fields does not have a well defined intensity ?

    Best Regards

    DaTario
     
  18. Feb 21, 2006 #17

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  19. Feb 21, 2006 #18

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Say what, vanesch??!!

    :)

    Zz.
     
  20. Feb 21, 2006 #19

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah, not my best phrase of the day, I suppose :redface:
    For my excuse, a collegue is leaving and we had some wine over lunch...
     
  21. Feb 21, 2006 #20

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  22. Feb 21, 2006 #21

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member


    This is actually directly hitting the nail, for me....
    How does one show that it hs a perfectly well defined intensity??
    All the derivations I have seen calculate (in a fixed photon number state) [tex] <E> [/tex] and [tex] (\Delta E)^2 [/tex]. The first is obviously zero and the second is [tex] 1 / 2 (1 + n / 2) [/tex] (in appropriate units). Now, they usually conclude right away that this is a wave of definite amplitude, [tex] \sqrt{ (1 + {n \over 2})} [/tex]!! I don't see how one can say this, just from knowing the dispersion.

    Thanks...

    Pat
     
    Last edited: Feb 21, 2006
  23. Feb 23, 2006 #22

    reilly

    User Avatar
    Science Advisor

    In the modern quantum world, quantization does not care if transitions are discrete or continuous; quanta can have any frequency from 0 to infinity, and can be described by a real number. Quantization stems from the quantum version of canonical commutation rules -- boils down, sort of, to: creating and absorbing a photon involve non-commuting operators, non-commuting processes. When applied to discrete atomic or nuclear transitions, this approach gives precisely the familiar Bohr/Einstein/Planck formula. It's not easy being quantum.........

    As was proved some time ago, for QED, by Bloch and Nordsieck(1937?), a charge undergoing acceleration -- whether speeding up or slowing down -- emits an infinite number of photons, but only a finite amount of energy -- this observation is the key ingrediant of the solution to the infra-red divergence problem, particulalry important in QED. (Many books, Jackson for one, discuss brehmstrahlung or radiation from decellerating charges. It's an old subject, and was of huge importance in particle physics, and radiation physics.)

    As vanesch has pointed out, the concept of the coherent state is about as close as the quantum radiation world can get to the classical world. Glauber showed, a long time ago, that the solution to a quantum E&M field generated by a classical current is a superposition of coherent fields -- which, by the way, can have an infinite number of particles, in the sense that the particle number is indeterminant. It's also the case, that a coherent field is the quantum version of a Poisson process -- the mean equals the standard deviation, as is the case for radioactive decay processes. Poisson processes also govern the transport, quantum or classical, of charges across a surface, as in electric current. In the large, a Poisson process behaves much like a smooth, continuous process, virtually classical in most aspects.

    So, what is a photon? Why photons? Nature gives us phenomena that are passing strange: photoelectric; black body and atomic spectra and radiation, Compton scattering, and more recently lasers. The reality is that our older, classical 19th century notions about nature simply do not work with these quantum-phenomena, as we now call them. I think of photons as a metaphor, a story that rests on top of QED. Because almost all our language is based on these 19th century ideas about the world, it does not do well with quantum theory -- there's no conflict in what we measure, only in how we choose to describe what we measure. (There are, I think, many ironies stemming from the extraordinary flowering of new ideas at the end of the 19th and early 20th century--say 1880-1920. Much of the modernism of those long distant days is still not widely accepted -- Freud, Einstein, Picasso, Braques, Stravinsky and Schonberg, Joyce, .... At the same time, modern society was rapidly coming into being-- electricity, gas light, automobiles, airplanes -- along with revolutionary zeal in Russia, not to mention anarchism; the rise of unions and their conflicts with managment, the end of the US Indian wars (1880s).... )The world in which classical ideas nominally served us well has long vanished. We may live in a modern society, but we tend to cling to outmoded ideas like causality, continuity, a belief that all experience and knowledge should conform to human predjudices, .... Quantum theory, with all its imperfections, waits for us to catch up to it with its messages about reality -- it reflects nature better and more faithfully than any 19th century approach -- I include Bohm's version of QM as basically 19th century. (My remarks about continuity and causality serve only, I hope, to suggest that these ideas and concepts, comfortable and familiar and useful in many circumstances, are not universally applicable to our experience, as nature has shown us.)

    Enough.

    Regards,
    Reilly Atkinson
     
  24. Feb 23, 2006 #23

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Hi nrqed,

    Yes, I am familiar with that handwaving argument. You can in some sense regard the average electric field in a photon number state as a wave with definite amplitude but completely undetermined phase. Of course this undefined phase gives you a wave with no amplitude, so it seems somewhat contradictory. All such handwaving arguments must be taken with a grain of salt, and there is ultimately no real reason for insisting that we give the average value of the electric field some extra special significance within the quantum domain. To show that a photon number state has definite intensity you should simply act on it with the intensity operator which looks something like [tex] \hat{\textbf{E}}^{(-)} \hat{\textbf{E}}^{(+)} [/tex] where [tex] + [/tex] and [tex] - [/tex] refer to the positive and negative frequency components of the field respectively. Note that the intensity is not just the field squared. In the end this intensity operator is just proportional to [tex] a^+ a [/tex] for a single mode.

    Hope this helps.
     
  25. Feb 26, 2006 #24

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi.. Yes, it does help a lot. That makes a whole more sense than the presentations I have seen! Then it is obvious that a state with a definite number of photons has a definite amplitude.
    Of course, the question now becomes: why this choice of definition? What is the motivation?

    Thank you, your comment makes me feel that I am closer to understanding this whole business of coherent states.

    Pat
     
  26. Feb 27, 2006 #25
    The Nobel Laureate Willis E. Lamb wrote a paper on photons where he stated that they don't exist. The paper is called Anti-photons Appl. Phys. B, 60, 77-84 (1985). The abstract to that paper starts out with

    The summary reads

    :)

    No. The so-called time-energy "uncertainty" principle isn't really an uncertainty principle. For it to be one then there would have to be a time-operator and there is no such thing in QM. What this principle means is that the uncertainy of the energy level of an excited state is inversly proportional to the time the system will remain in that state.
    No. It is always possible to determine the energy of a state exactly. always - so long as you're not measuring something whose operator does not commute with the energy operator!

    Pete
     
    Last edited: Feb 27, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook