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Photon Annihilation, where does the energy go?

  1. Jul 28, 2010 #1

    When we look at diffraction, single or double slit, we can obtain the minima and maxima by determining the path lengths and relating that to the wavelength. The idea here is that 1/2 wavelength path difference or 180 deg phase difference result in minima etc.

    does the The minima, or dark spots, corresponds to NO light being there or No light being able to be detected. What happens to the energy from the electric and magnetic fields?

    Consider the Michelson M experiement with beam splitter and perfectly refelcting mirrors.

    A laser is split into two paths and recombined. Lets say its a perfect plane wave even though we can't make one. When the two paths recombine, and if the system is set up for destructive interference, the net magnetic and electric fields combine to give zero at any point in time (for the outgoing pulse).

    How does the energy in equal the energy out?
  2. jcsd
  3. Jul 28, 2010 #2
    Destructive interference for waves do not act with a continuing linear aspect.
    You will find that, on a linear graph, the waves "distance" cycle between constructive and destructive.
    Not sure if that makes any sense to you, or is even helpful.
  4. Jul 28, 2010 #3


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    In a double slit, you have regions of deconstructive and constructive interference. The energy is not destroyed but instead it gets shifted from the areas of minima to the areas of maxima. The darker areas correspond to no light being there, we can do these experiments all the way down to single photon measurements and get the same results. So you can think of the double slit as a way of concentrating the photons along certain paths and discouraging their flow along other paths.
  5. Jul 28, 2010 #4
    Ok... I waas just thinking the same thing before i saw your reply.

    but what about my othe example of a Michelson Morley interferometer.

    Imagine that it is set up so that the light recombines via beam splitter and interferers destrutively. (imagine it for plane waves) - no fringe crap. Now the missing light is going in the same direction. Where is the energy.

    Not realistic experiement but...
  6. Jul 28, 2010 #5
    Uh, I don't think you can set up a Michelson interferometer where the waves interfere destructively without producing the fringe crap. Someone correct me if I'm wrong.

    On a sidenote, photons actually can "annihilate." One example of an interaction that destroys photons is:

    [tex]\gamma\gamma \rightarrow e^+e^-[/tex]

    I don't know if you'd call that an annihilation process. But they say that the photon is its own anti-particle. Maybe this is why.
  7. Jul 28, 2010 #6
    Yes... Annihilation is the wrong word.

    So Im forced to see fringes...

    What if we only send one photon through?

    But then we can't treat it like a particle, we must treat it like a wave that interferes with itself or all possible paths..... and it will statistically fall where those fringes would have been.
  8. Jul 28, 2010 #7


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    When we talk about photons interfering in terms of quantum mechanics, the interference dictates where the photon will end up. The wavefunction describes the probability of observing the photon at a given position. When the wavefunction experiences destructive interference (from say a double slit) this translates to some volumes of spacing have a low value for the wave function and others a high one which indicates that the former areas will have a low probability compared to others of a photon being observed there.

    So we are not destroying the photon, just mearly manipulating the probable paths that it will travel down. So going back to say an instance of picture perfect interference, it really just means that the photon will not be seen in that area but it should end up somewhere else.
  9. Jul 29, 2010 #8
    It is a known fact that from two point like sources you can have almost perfect constructive interference in all space. But almost perfect distructive interference is not possible at all. Conservation of energy forbids it.

    If there is distructive interference you may bet that there are channels of constructive interference.

    Best regards,

  10. Jul 30, 2010 #9
    I think that in this case, the energy has gone back inside the beam source.
  11. Jul 30, 2010 #10
    Pardon my ignorance can someone explain to me what fringe crap are you referring to or maybe point me to a blog where I can find out.
  12. Jul 31, 2010 #11


    Staff: Mentor

    The conservation of energy is a direct consequence of Maxwell's equations:

    In free space destructive interference in one location will always result in constructive interference elsewhere such that the field's energy is conserved. In the presence of matter it is possible to obtain total destructive interference, in which case the missing energy goes into the matter, usually heating it.
  13. Jul 31, 2010 #12
    Now that's a good explanation. Thanks Dale.
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