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Photon Wave Packet Envelope Question

  1. Jul 11, 2012 #1
    Anyone know how “long” a photon wave packet envelope is? I get the impression from Purcell that it might be only about 1 wavelength long. But then I think I saw on this forum someone mention that it could be “thousands” of wavelengths long.

    How would or could one determine that? I am thinking, for example, of the Kennedy-Thorndike experiment, where the two arms of the interferometer were deliberately and significantly different – maybe by about 10 inches, as that was the “limit of coherence”. Would that be a clue to how long the envelope might be? If they were experimenting with optical wavelengths, then the “thousands” of wavelengths would be correct.

    Any thoughts are appreciated.
     
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  3. Jul 11, 2012 #2

    mfb

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    It depends on the source of light. Femtosecond lasers can produce wave packets with a length similar to the wavelength (or longer, if required), other lasers can produce wave packets with a length of ~1 lightsecond.
    Interference with different lengths of the arms is a good way to measure the coherence length.
     
  4. Jul 11, 2012 #3
    Thanks but we appear to be referring to different things. I would expect all photons of the same wavelength to look the same. Does a single photon wave packet look like the following plot? If so, then how many wavelengths long is the “envelope”? If not, then what would a single photon wave packet look like?
     

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  5. Jul 11, 2012 #4

    mfb

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    You have wave packets, therefore you do not have "a single photon with a specific wavelength". A single packet can (but does not have to) look like the attached image. And as I said, the length depends on the light source. Femtosecond lasers produce wave packets with a length of ~1µm (I think the shortest pulses are somewhere at ~80as, which corresponds to 24nm), other lasers can produce wave packets of ~1light second length, maybe even more.
     
  6. Jul 11, 2012 #5
    I think that the physics of photons is properly understood in terms of Quantum Electrodynamics which I'm not educated in. In the absense of that education I have this to add: photons behave like electrons and in that repsect a photon which is the superposition of many quantum states is represented by a Fourier integral which is an infinite sum of well defined states of well-defined wavelengths. The resultant state does not have a unique wavelength associated with it. However once the momentum of the photon is measured the wavefunction collapses into a well defined momentum state. I imagine that's how compton scattering works, the photon is scattered off an electron and the photon then has a well-defined momentum.

    Is there any truth to my guess?
     
  7. Jul 12, 2012 #6

    vanhees71

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    Indeed, a photon is a quite abstract entity. It is any quantum state of the free quantized electromagnetic field with the sharp photon number 1.

    First of all it is important to note that a single-photon "state", corresponding to a sharp momentum (and thus also sharp frequency) is not such a state in the strict sense since it is not normalizable to 1 but only "to a [itex]\delta[/itex] distribution", i.e.,
    [tex]\langle \vec{k},\lambda|\vec{k}',\lambda' \rangle=\delta_{\lambda \lambda'} \delta^{(3)}(\vec{k}-\vec{k'}).[/tex]
    Here, [itex]\lambda \in \{-1,1\}[/itex] refers to the helicity of the photon (corresponding to left- and right-handed circular waves).

    This is like in non-relativistic quantum theory in the position representation: A momentum eigen"state" of a particle, represented by a plain wave, is not a true state since the plain wave is not a square integrable function. It's a distribution. Note that for photons the notion of a position operator doesn't make sense and also wave-function formulations are misleading (the best). Thus, here I stick to the momentum-helicity representation.

    A true single-photon state with definite helicity is given in terms of a square integrable momentum-space function
    [tex]|A,\lambda \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^{3/2}} A(\vec{k}) |\vec{k},\lambda \rangle.[/tex]

    If [itex]A(\vec{k})[/itex] is sharply peaked around a certain [itex]\vec{k}_0[/itex], this corresponds to a quite long wave train. If it's a broader distribution, it's more like a short wave train.

    A classical electromagnetic wave is, in terms of quantum field theory, represented by a different type of quantum states, socalled coherent states. They are a superposition of photon states with any number of photons and can have any mean number of photons you like. For very large mean numbers this state behaves (almost) like a classical electromagnetic wave. This is a good description of what a laser emits.
     
  8. Jul 26, 2012 #7
    I too am interested in the question of the coherence length of a single photon and I don't think the replies so far have answered exmarine's question.

    As I understand it, the experimental results of relevance are as follows. The coherence of light from a gas discharge tube (as measured by an MZ interferometer with very unequal arms) is of the order of 30 cm. The coherence length of a good laser is many orders of magnitude larger.

    These experiments, however, tell us nothing about the coherence length of a single photon because they both involve huge numbers of photons.

    On the other hand, it is not impossible to conceive of an experiment in which single photons are passed through an interferometer. As you crank up the path difference, sending photons through one at a time, surely you would expect to see the photons oscillate between the two detectors? But how long would this continue? Surely it cannot go on for ever? If the interference pattern disappears at some point (ie the photons begin arriving at the detectors at random) then you have made a measurement of the coherence length of a single photon. Does this make any sense?

    If the coherence length of a photon from a Sodium atom turns out to be 30 cm, say, is it necessarily true that the coherence length of a photon of approximately the same wavelength produced by a Caesium atom has to be the same? Could two photons of the same energy have different coherent lengths?

    There are a lot of fundamental questions to be answered here.
     
  9. Jul 26, 2012 #8

    Cthugha

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    Sure. Such experiments have already been performed. I think the record coherence time for a solid state single photon source is around 22 ns (Matthiesen et al., PRL 108, 093602 (2012)http://prl.aps.org/abstract/PRL/v108/i9/e093602). For atoms coherence times will be longer.

    Even two photons from the same kind of atom can have different coherence lengths/times. A typical way to create single photons is resonance fluorescence. Here the coherence time is limited by the antibunching timescale which in turn can depend on how strongly you drive the system. If you drive the system very weakly, you may even end up in the Heitler regime which is basically coherent single photon scattering of the incoming light beam. Here the coherence time of the single photons is limited only by the coherence time of the laser used for excitation.
     
  10. Jul 26, 2012 #9
    Wow! that's really interesting, Cthugha. 22ns translates into nearly 7 metres.

    I didn't understand that bit about antibunching and I certainly don't want to end up in a 'Heitler regime' but could you just explain one more thing. I had understood that a photon was a kind of ideal form of energy with just four physical properties - energy and three momentum components. Period.

    If I encountered a photon in deep space, having no idea where it came from. The HUP allows me to measure all of its energy and three momentum components with infinite precision given a) enough time and b) enough space. Are you saying that there is something else I could measure too?
     
  11. Jul 26, 2012 #10
    Eggzackly. Thanks for keeping this going.
     
  12. Jul 26, 2012 #11

    DrChinese

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    Don't forget position and spin (polarization) properties. :smile: Of course, it doesn't have all properties at all times, as per the Heisenberg relations.
     
  13. Jul 27, 2012 #12
    Along with position, etc., does it have a "length" (my original question)? Width? Height? Volume??
     
  14. Jul 29, 2012 #13
    I hope exmarine has not stumped our esteemed science advisers on this one. Mind you, I can understand their reluctance to commit themselves as anything they say has potential to be used in evidence against them.

    For example, Cthugha says
    Now if coherence length is truly a property of a photon (and not just a property of the way it is measured) then the Uncertainty Principle is in deep trouble. The HUP states that it is possible, in principle, to measure the energy (and hence frequency and wavelength) of a photon with arbitrary accuracy. But if the coherence length of a photon is finite, it is impossible to measure its wavelength with arbitrary accuracy.

    Also, if coherence length is a real property of a photon, the wave/particle paradox is solved. Photons are definitely waves. Coherence length has no counterpart in the language of particles.

    In fact, all of modern quantum theory is in peril. QED works by summing the potential effects of an infinite series of possible interactions between electrons and photons. If (virtual) photons have a finite length, the series is finite.

    The stakes could not be higher.
     
  15. Jul 29, 2012 #14

    mfb

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    I think you can measure the energy of the detected photon with arbitrary precision. But you cannot be sure to measure the same value again if you repeat the experiment - this comes from the finite width of the photon in its frequency distribution.

    You have a similar situation in electron orbitals: You can measure the position of an electron with fm precision (at least in theory). But if you repeat the experiment, you will see that the measured position has a distribution of the order of 10^5fm.

    Coherence length is a property of a superposition of different "QED photons" (which have a single frequency).
     
  16. Oct 24, 2012 #15
    Do you know if anyone attempted to measure the dependence of maximum coherence length of single photons from wavelength of the photon? This would be is just the experiment described by JollyOlly, but performed over a range of wavelengths. Or maybe somebody knows a reference where this question is considered theoretically?
     
  17. Oct 25, 2012 #16
    Thanks Kkonst for keeping this thread going. The issue is too important to drop and I keep hoping that someone who really knows what they are talking about will spot this discussion and give us an authoritative answer.

    My own feeling is that it is meaningless to talk of the coherence length of a single photon because of the objections I raised in my post of July 29. If so, you have got to explain in QT terms the results of an interference experiment such as the one I suggested in July 26.

    The best way of interpreting the HUP is in terms of wave packets. The more waves you measure, the more accurately you can measure the wavelength. But this does not mean that a photon is a wave packet. All it means is that if you make the measurement in a finite time [itex]\Delta[/itex]t you will get a result which has a finite uncertainty [itex]\Delta[/itex]E.

    If you send a single photon into an interferometer whose path difference is d the photon will emerge at either one detector or the other depending on how many wavelengths there are in d. Now, according to the standard Copenhagen interpretation of QT, while the photon is inside the apparatus, we are not even allowed to ask the question 'which way is the photon going?'. All we are allowed to do is open up the apparatus after a finite time a ask 'which detector did it strike?' Since a finite time is involved, there must be a finite uncertainty in the wavelength of the photon and the photon will appear to have a finite coherence length (ie if d is big enough, the photons will appear at the two detectors at random.)

    The implication of this is that, if you do this experiment many times repeatedly, if you perform the experiment quickly (ie you look at the detectors after a short time) you will measure a relatively short coherence length; but if you leave the interferometer on its own for a long time, thereby giving the wave equation time to develop, the measured coherence length will be longer. In the limit, if you never look at the detectors at all, you will be able, with absolute certainty, to predict which detector the photon will hit, regardless of the path difference d.

    This bizarre result has just the sort of wicked self-consistency as to render it entirely plausible.
     
  18. Oct 25, 2012 #17

    mfb

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    The detectors themself "look" - it does not matter whether you want to see the results or not.

    There is no physical limit on the coherence length, just experimental issues.
     
  19. Oct 25, 2012 #18

    kith

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    Why? I can just use a big detector and measure the energy of the photon. This of course tells me only what energy the photon has after the measurement process but this is the best I can get. Before measurement, the energy -like all observable quantities in QM- is not necessarily well-defined.

    That's nonsense. Although QM may be replaced by a better theory some day, nothing will change about the fact that it has predicted a great variety of experimental results correctly. Also, it doesn't help to claim science advisors are wrong without citing references, because it is much more probable that you didn't understand what they are telling you. Doing so will in many cases only lead to them leaving the thread.
     
    Last edited: Oct 25, 2012
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