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Desructive interference = destroyed energy?

  1. Sep 20, 2008 #1
    I am trying to resolve a paradox in my head, and I call upon the forums to help.

    Consider two identical photons traveling, each with E=hf, so the total energy is 2hf.

    Now, by whatever method, the photons are made to travel right on top of each other, with one being perfectly out of phase with the other. Their amplitude vanishes.

    What happens to the energy? An electromagnetic wave with zero amplitude has zero energy (it is just empty space).

    For that matter, what happens to the photons? Can we expect them to be recovered from such a situation?
     
  2. jcsd
  3. Sep 20, 2008 #2

    atyy

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    Since it would be a disaster if energy were not conserved, what generally happens in destructive interference is that energy is simply redistributed. It usually takes a bit of thinking in a specific situation to see what has happened.

    If you started out an electromagnetic wave of zero amplitude, then energy is automatically conserved.
     
  4. Sep 20, 2008 #3
    Are you suggesting that the energy left the photons and permeated out into space some way in some other form?
     
  5. Sep 20, 2008 #4
    To experimentally quantify how this may be done:

    consider the first photon is bouncing in a loop bounded by partially reflecting surfaces (lets just say its bouncing back and forth between two partially reflecting mirrors).

    The second photon is the fired (by whatever means, say the energy transition of an atom) at PRECISELY the right moment and from precisely the right distance behind one of the mirrors so that when it passes through the first mirror (which would probably take a few tries since its partially reflective, implying a probability of passing though), it joins the first photon, but is perfectly out of phase.

    Now the mirrors are removed and this ghost double-photon with zero amplitude is left to wander the universe.

    The mirrors did not absorb any of the photons' energy.

    The photons apparently have no energy.

    Now - where is it?
     
  6. Sep 20, 2008 #5

    atyy

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    I'm a bit uncertain if the situation you set up contains any energy in the first place. It's a bit difficult to consider photons and classical electromagnetic waves (there aren't a definite number of photons in such a wave). How about just setting up a classical situation where you have two waves from different sources, and have them interfere?
     
  7. Sep 20, 2008 #6

    atyy

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    I think if you have a wave bouncing back and forth between mirrors you set up a standing wave, which is two waves travelling in opposite directions. So if you somehow manage to cancel the wave in one direction, you should boost it in the other direction - or something like that.
     
  8. Sep 20, 2008 #7
    As I feared, introducing the particulars of the thought experiment would result in too much focus on the apparatus.

    Very well:

    Use just one mirror, shoot the first photon at it from far, far away. Then on the opposite side, shoot the other photon such that it will coincide with the reflection of the first photon, and they will be perfectly out of phase.
     
  9. Sep 20, 2008 #8

    atyy

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    If the mirror is perfectly reflecting, the second photon can't pass through it. If it is only partially reflecting, then you again get two waves in opposite directions.
     
  10. Sep 20, 2008 #9
    Discrete photons are not separated into transmitted and reflected portions when they travel through a partially reflecting surface. They simply have a one-time probability of passing through or not passing through. This probability gives the relative proportions of transmission and reflection when a large quantity of photons shine on the surface.
     
  11. Sep 20, 2008 #10

    atyy

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    Well, I'm thinking of classical waves for which we can define an amplitude and phase so that they can interfere, but I think these don't have a well-defined number of photons.

    Edit: Also, I think single photon states don't have a well-defined phase.
     
  12. Sep 20, 2008 #11

    Redbelly98

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    If this is a partially-reflecting mirror, there are two possible outgoing paths:

    Path A: Photon #1 transmitted, Photon #2 reflected
    or
    Path B: Photon #1 reflected, Photon #1 transmitted

    If you arrange the phases so there is total destructive interference in one path, there will be constructive interference in the other, so the energy doesn't just "disappear". This is a pretty general result of any attempt to combine two photons or two laser beams: destructive interference can only occur in some places, but not everywhere that the photons or beams are. Energy remains conserved.

    Edit:
    True. But we could just as easily be talking about combining two laser beams and arranging phases to get interference.
     
  13. Sep 20, 2008 #12

    atyy

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    Is that the same or different from the classical waves I was using above to make the similar argument that interference simply results in a redistribution of energy?

    We got down to single photon states, because of icelevistus's last scenario. But I was somewhat hazy about how interference for a single photon state would be defined.
     
  14. Sep 20, 2008 #13
    Two different photons do not interfere. Interference is always caused by an interference in the probability amplitude. If you combine two laser beams of two identical lasers, then the interference is caused by the fact that each laser contributes to the amplitude of each single photon.

    So, just like in case of the two slit interference experiment where you cannot tell through which slit each individual photon went, in case of two identical lasers, you cannot tell out of which laser each individual photon came. When you do have that information, the interference pattern will vanish, just like when you have the "which path" information in the two slit experiment.
     
  15. Sep 20, 2008 #14

    atyy

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    I'm not sure this is related to the "interference" question, but does the uncertainty principle allow for any momentary violation of energy conservation?
     
  16. Sep 20, 2008 #15

    Redbelly98

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    The same.

    It seems the basic question of the thread is, can destructive interference of light violate conservation of energy. I agree that classical waves provide a better means of thinking about or experimentally demonstrating interference effects than individual photons.
     
  17. Sep 20, 2008 #16
    I don't think so. If you have a state that is a superposition of many energy eigenstates, you can construct a state that will have a probability amplitude at some point which will change fast as a function if time. If you want to let the absolte value square of the amplitude to chsnge very fast in time, you need to use a very broad range of states with different energy eigenvalues. So, the energy of the state will then not be sharply defined. But, you can't use this effect to violate conservation of energy.

    There is a relation with momentum-position iuncertainty relation and interference experiments. Suppose you do an interference experiment by using two mirrors that reflect light from one source in some direction. Then yo could try to extract the which path information by measuring the recoil of the mirrors.

    You could argue that the mirror should always recoil so there seems to be a problem: How can we have an interfernce pattern, if we have access to the which path information? The answer is that the recoil can only be detected if the wavefunction of the mirror in momentum space has a width that is less than the momentum change of the photon. The uncertainty principle then implies that the position of the mirror must be more uncertain than the photon's wavelength. But then you can't have an interference pattern.
     
  18. Sep 21, 2008 #17
    if you fire 2 laser beams at each other then at the point in time at which their electric fields cancel I think their magnetic fields would add. so there would be no point in time at which they completely cancel.
     
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