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Why no decoherence after reflection from metal film

  1. May 30, 2015 #1
    When photons reflect from a mirror coated with, say, an aluminium film... the reflection is presumably down to their interactions with free electrons in the metal (and each electron is in an unknown state before and after).

    Yet the interference does not go away if a mirror is inserted in one path of the setup. How come?
  2. jcsd
  3. May 30, 2015 #2
    I believe the free electrons form a dense cloud, or plasma, in the metal. In a simple case, they are free to react to the E-field of the incident wave, and accelerate accordingly. In doing so, they re-radiate the wave with opposite phase. This case is for small angles of incidence and when the frequency is not too high.
  4. May 30, 2015 #3
    That's exactly what bothers me. After all, decoherence has to do with an interaction with complex and random stuff that has not been "prepared" in a known state (as I understand it). And the "plasma" or electron gas is just such a thing.
  5. May 30, 2015 #4
    If you are thinking of random motion of the free electrons due to temperature, I am not sure that the free electrons are jumping around, because a perfect conductor does not radiate thermal noise.
  6. May 30, 2015 #5
    Is it the fact the relection time varies that is bothering you?
  7. May 30, 2015 #6
    Not exactly that. It's more about the idea that, after reflection, the photon is entangled with the free electron gas, which is a random part of the universe and not a part of the prepared/measured experiment.

    Reflection off a dielectric surface seems less problematic, since visible photons would not be able to change the states of bound electrons.
    Last edited: May 30, 2015
  8. May 31, 2015 #7


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    Reflection of a classical electromagnetic wave (or a coherent state, if you want to talk about it in a quantum-theoretical language) is microscopically the absorption and induced (re)emission, and induced emission is coherent with the incoming wave.
  9. Jun 1, 2015 #8


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    This question is answered in detail in the book
    in Sec. 10.4. (I highly recommend this book for many other foundational aspects of QM.)

    In short, the point is that the mirror is heavy, much heavier than "mass" (energy/c^2) of the photon. Consequently, the momentum and position of the mirror are not much affected by the impact of the photon. Therefore, the quantum state of the mirror after the photon impact is not much different from the state of mirror before the impact. The difference is so small that the scalar product between these two mirror states is very close to 1. Hence the mirror cannot "know" that a photon reflected from it. In other words, the mirror cannot "measure" the photon, i.e. the mirror cannot induce decoherence.
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  10. Jun 1, 2015 #9
    It seems to me that for low frequencies at least an incoming photon must cause an electron to accelerate, otherwise there would be no re-radiation?
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  11. Jun 1, 2015 #10
    Thanks, I'll try to get a copy. I wonder if there are any online articles etc about this kind of thing.

    I was thinking not so much about the mirror as a whole, but about the electron cloud that does all the reflecting (or re-emitting). Considering it as an ensemble of zillions of electrons, it is amazing that it does not change from one microstate to another after each photon reflection event. Presumably, an event that leaves the free electron cloud in a different microstate would irretrievably "damage" the phase coherence, not to mention (possibly) a tiny change in the frequency of the photon.

    Mind you, I have no problem at all with this whole process as long as I think of it as a classical process. The problem comes when I look at it in terms of "events" involving photons and electrons. That's what comes of spending too much time reading "QED The Strange Theory..." o_O
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  12. Jun 1, 2015 #11


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    For a mirror to work properly, as everybody knows from everyday experience, it is important that the mirror is flat and smooth at distances comparable to the photon wavelength (which is much larger than the size of the atom). This shows that reflection of a photon does not really happen due to interaction with a single electron attached to an atom. Instead, it seems that it is some kind of an collective effect (experts in solid state physics, please correct me if I said something wrong!), in which photon interacts with a large number of conducting electrons at once.
    Last edited: Jun 1, 2015
  13. Jun 1, 2015 #12


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    Feynman in his book states that he is not describing microscopic reality but rather mathematical tool that gives correct prediction for complete event. But one thing about which he deserves blame is that he keeps talking about photons while explicitly stating that they are not photons but rather photon probability amplitudes (abstract mathematical entity). That way he invites confusion.

    But where I find Feynman model contradictory with decoherence is that different photon probability amplitudes can be viewed as coherent due to some sort of "coherent quantum clock" in photon source. And decoherence as a measurement approach sort of says that macroscopic objects loose coherence very quickly so we can't really attribute "coherent quantum clock" to macroscopic experimental equipment (except in some special cases).
  14. Jun 1, 2015 #13
    I think the free electrons are subject to the Brownian Motion, caused by collisions with atoms. Maybe this will cause a reflected wave to be modulated with shot noise. However, at the macro scale this sort of noise seems to average towards zero (a bulk effect as you describe). For instance, the electron space charge from a hot cathode is not itself noisy. Microwave antenna reflectors do not introduce noise. But at the micro scale, with very short wavelengths and if few electrons are involved, maybe the reflected wave will be subject to noise modulation?
  15. Jun 2, 2015 #14


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    Collisions with mirrors can cause decoherence. The formalism is actually much easier to understand if you use two mirrors to form a cavity. If you then go through the formalism you will find that the coherence time T1 is just the classical relaxation time (proportional to to quality factor of the cavity) (T2 would just be 2*T1 in the absence of pure dephasing).
    Hence, two perfect mirrors wold result in an infinite coherence time, i.e. no decoherence. But any imperfection will result in photons being lost or absorbed which (obvously) causes decoherence (this is mathematically equivalent to decoherence of an harmonic oscillator, so it is a nice "toy" system which can be handled analytically).

    This is part of a field called cavity quantum electrodynamics (CQED) and you can find quite a lot of information about this if you google. See also the papers by e.g. Haroche (who eventually won a Noble prize, interestingly many of his papers actually deal with how to make very good "mirrors" for use in his cavities)
  16. Jun 2, 2015 #15
    This is necessairily true, since the reflected wave displays spatial properties of reflection from all points on the reflecting surface.

    I doubt that we can attribute the prescence if decoherence to microscopic surface structure.

    The weight or even mass of the mirror can't actually be relevant either.

    The relevant property must be that free electrons have sufficient degrees of freedom that recoil information is not stored.
    Last edited: Jun 2, 2015
  17. Jun 2, 2015 #16


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    It is well known that the mass of the mirror is highly relevant. Look at all the papers from cavity optomechanics, where mirrors are placed on a cantilever (and thus do not weigh much) and cause optomechanical damping. This will never happen for a fixed mirror which weighs as much as the earth does in the idealized case.
  18. Jun 2, 2015 #17
    Thanks again for all the replies. I shall content myself with a provisional description (based on these discussions), like this: If we compare the electron cloud with a feather pillow, then there is some quirk of solid state physics that says that when light bounces off the pillow, the momentum is exchanged with the whole bed, and the light does not interact with the feathers' internal degrees of freedom.
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