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Destroying interference pattern of classical waves

  1. Jun 12, 2008 #1
    In a two slit interference experiment, if you measure through which slit the particles move, you'll destroy the interference pattern. You can also do this in the classical regime, but usually this has a trivial classical interpretation. E.g. you could give the polarization of photons depend on which slit they move through. Then the polarization state of the photon carries the "which path information" with it. You then don't get an interference pattern. In the classical interpretation, you can simply say that the electromagnetic field components of the light coming from the two slits are orthogonal w.r.t. each other.

    Instead of giving the photons a polarization one could, in principle, collect the "which path information" without affecting the state of the photons. If you could do this fast enough, you could have classical electromagnetic waves which according to classical physics should interfere, but they will fail to do so.
  2. jcsd
  3. Jun 12, 2008 #2


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    The only way you can tell that a wave (C or QM) has passed through one slit and not the other (without polarization) is to collimate it enough to miss one slit. Then there will be no two slit interference (C or QM).
  4. Jun 13, 2008 #3
    It's not clear to me how is possible to collect the "which path information" without affecting the state of the photons.
    Can you elucidate it?
    Thank you.
  5. Jun 14, 2008 #4
    Ok, let's consider the following experiment. We do a two slit experiment, in which we replace the two slits by mirrors that reflect the light in some direction to a screen. We'll then get an interference pattern as usual. But now consider letting the two mirrors float using magnetic fields.

    The mirrors are in some effective potential. Initially they are in the ground state. When a mirror absorbs the momentum of a photon, it will go into some excited state. We denote the ground state by |0,0>, the state in which the left mirror is excited as |1,0> and the state in which the right mirror is in the excited state as |0,1>.

    If we denote the state of the photon arriving at the screen (just before it hits the screen) via mirror 1 as |1> and via mirror 2 as |2>, then the state of the photon/mirror system will be:

    |psi> = 1/sqrt(2) [|1,0>|1> + |0,1>|2>]

    The probability that the photon will arive at some position x on the screen is given by:


    If the ground state and the excited state of the mirror in the potential are orthogonal, then the probability becomes:

    1/2 [|<x|1>|^2 + |<x|2>|^2]

    So, we don't get any interference.

    If we fix the two mirrors then one could still say that there is a very small recoil and that the final state does depend on which mirror the photon reflected off. But the two states of the "rest of the universe" |1,0> and |0,1> are then almost identical and thus far from orthogonal. The probability then becomes:

    1/2 [|<x|1>|^2 + |<x|2>|^2] + Re[<x|1><x|2>*<0,1|1,0> ] =

    1/2 [|<x|1>|^2 + |<x|2>|^2] + Re[<x|1><x|2>*]

    and we then do have the usual interference term.

    The absense of interference in case of the two floating mirrors cannot be explained clasically. In case of the experiment with the polarization filter, we could explain the absense of the interference by saying that the electric field component of the wave that moves through slit 1 is orthogonal to the electric field component of the wave that moves through slit 2.

    However, quantum mechanicaly, what is relevant is that the states |1,0> and |0,1> which for the polarization case would be the two polariaztion states of the photon are orthogonal. In case of the two floating mirrors, these two states describe the mirrors.
  6. Jun 15, 2008 #5
    Maybe you could say, classically, that when the mirrors are floating, the phases of the reflected em waves are not specified anylonger (one respect to the other) = no coherence, so this destroyes interference.
  7. Jun 16, 2008 #6
    Yes, you are right. If we want the mirror to be able to be ble to recoil when it reflects the photon, then we must allow for a certain spread in the mirror's wavefunctions. If we imagine the mirror as confined in some volume, freely floating inside this volume, then the uncertainty relation (or just the standard result about the number of different momentum eigenstates in a volume) will tell you that the mirror's wavefunction wil have a spread of the order of the photon's wavelength.

    So, it cannot be done in this way. But is it impossible? I don't think so, because what we want to do is implement a unitary operation.

    Denote by state |n,m> the state corresponding to the photon moving through slit n and a detector (we leave the question how to implement it aside) in state m. The detector is supposed to tell us through which slit the photon moved.

    We initialize the detector to state m = 0. We want the detector-photon system to undergo to following transitions:

    |1,0> ------> |1,1>

    |2,0> -------> |2,2>

    After a photon hits the screen, the detector will be reset to 0.

    The question is, can we write down a unitary transformation that realizes the above transition? Of course, we can:

    |n,m> -------> |n, n + m Mod 3>

    This is a unitary transform. So, in principle one can destroy interference even if the photons arrive without their state being affected when they arrive at the screen. The detector must be reset every time, but if you can do this fast enough you can let the photons stream out of the laser faster and faster, so can then define a classical electromagnetic field which nevertheless failes to produce an interference pattern.
  8. Jul 12, 2008 #7
    Actually, there is no need to reset the detector every time. Anyway, I found a more practical ways to achieve the desired result which have been experimentally verified (copied from my posting on another thread).

    This article mentions some ways to achieve the desired result in a practical way. Take e.g. the entangled state described by Eq. 27 on page 13:

    1/sqrt(2) [|a>|a'> + |b>|b'>]

    Here |a> and |b> are spatial modes for a photon and the primed states are different spatial modes for another photon. Then you don't have single photon interference, but you do have interference when you measure two photon correlations. All this is quite obvious.

    But if you try to interpret the result classically, you get exactly what I wanted to show: You don't get interference even though the light from the spatial modes a' and b' should interfere classically. In case of the state:

    1/sqrt(2) [|a'> + |b'>]

    Then there would be inteference. In the classical picture, you cannot see the difference between the two cases. So, it is possible to create classical waves that fail to behave as predicted by classical theory.

    This is true, in principle, for all classical wave phenomena. You could theoretically create two sources of sound waves such that they should interfere, yet you can make them fail to interfere if the phonons are in certain entangled states (whith other phonons or with some other degrees of freedom)

    So, the conclusion must be that you don't necessarily get classical behavior in the classical limit. Or, perhaps one should say that classical wave phenomena like electromagnetic waves are in fact macroscopic quantum phenomena...
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