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The double slit

  1. Jul 23, 2004 #1
    there's something that is annoying me, because I can't find a explanation
    It's about the classical double slit experiment, where you have two screens, and one of them has 2 narrow slits
    then you launch a photon against the screens, and if there are no detectors in the slits, you observe an interference pattern in the background screen
    But my understanding is that any interaction of the wavefunction with whatever thing collapses it, so I think that the wavefunction should collapse as a result of its interaction with the borders of the slits, even if there's no measuring apparatus in the slits, hence the interference pattern shoudn't be observed in any circunstance
    Comments are appreciated
  2. jcsd
  3. Jul 23, 2004 #2
    View "collapse" in terms of the "making of a record" or - in the case of a "null" measurement - the "absence of a record". The "primary collapse" in which the incoming wavepacket is split into two (upon traversal of the first screen) is a "null" measurement in which the photon has not been absorbed by the first screen. The possibility of a "secondary collapse", about which you are asking, will indeed occur whenever a photon which traverses the first screen "makes a record" at one of the borders of one of the slits. Surely, there will be events of this kind. But the number of such events relative to the number of events in which no such "record is made" will be negligibly small. (The vast majority of photons which traverse the first screen do not even "touch" any border of either slit. On the other hand, any photon which "touches" a border may well become absorbed. In other words, there is a negligibly small chance for a photon to both "traverse the first screen" and also "make a record".)
  4. Jul 23, 2004 #3
    I'm not familiar with the expression "null measurement". Can you explain?
    Note that I'm speaking of launching one single photon, so I do not understand why you say that the vast majority of photons will not touch the slits. The photon, as a spread wave, must touch the slits, I assume that the width of the wave is larger that the separation between the slits
  5. Jul 23, 2004 #4


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    Null measurement, in this case, means no photon observed.

  6. Jul 24, 2004 #5
    The electrons which make it through the first screen are the ones that didn't interact with it. If we place a detector at a slit and the photon interacts with it, it randomizes the phase of the amplitude associated with that slit, and the interference pattern is destroyed.
  7. Jul 25, 2004 #6
    ... Perhaps you have already resolved the difficulty ... nevertheless, I will continue with this post.

    There are three basic "types" of events:

    (o) The photon is blocked at screen1 and does not reach screen2;

    (i) The photon reaches screen2, but with no "photon-screen" interaction having taken place at screen1;

    (ii) The photon reaches screen2, but has done so by scattering off of a slit border.

    In a "type-(i)" event, the wavefunction, so to speak, "interacts" with both slits, but there is no "physical record" of any kind, such as an atomic excitation, to be found in screen1. In other words, to say that the wavefunction "interacts" with screen1 does not imply that anything "physical" has taken place. That is to say, in a "type-(i)" event there is no "making of a record" (not even a transient one) which "records" that the photon has passed through one of the slits and not the other. Hence, the amplitudes for both slits will interfere.

    For a simple example (in the case of a potential barrier) see post#4 in the thread entitled "Potential barriers, wavepackets, probabilities etc.". Here is a quote of the relevant section of that post:


    What I had in mind was a repetition of the experiment many, many times over ... after which we ask a question about the relative proportion of "type-(ii)" events to "type-(i)" events.

    But is the "spread wave" really "the photon"? At the very least, (all of us would agree that) the "spread wave" has an interpretation as the amplitude for finding the photon to be at some place in a "measurement-like" interaction. We might also venture further to say (and I think most of us would agree) that the "spread wave" is "the photon" ... BUT only in a "state" of "possibility" or "tendency" (Heisenberg's words).

    To make this last point clearer (at least, to my understanding (of Heisenberg's words)) ... take the example of an electron in the ground state of the hydrogen atom. There is zero angular momentum, yet the wavefunction is spherically symmetric. To say that the "spread wave" is "the electron" - but only in a "state" of "possibility" or "tendency" - means that "physically" the electric field is really "static" and it really has "zero dipole moment". The problem with testing this hypothesis is (at least to my eye) twofold:

    (a) With the current technology we are unable to measure the electric field of a single hydrogen atom, nor a very small number of such atoms (note: I am not a 100% sure about this statement);

    (b) It is impossible to have a completely isolated hydrogen atom ... at the very least it is always "coupled" to the "vacuum fluctuations" (note: I am unsure as to the precise effect that these "couplings" will have, but I am guessing that it would make the electric field like a sort of "fluctuating dipole", or at least something which spoils the chance of verifying the above "possibility/tendency" interpretation as I have construed it).

    Going back to our double-slit scenario, this idea of the wavefunction as representing a "state" of "possibility" or "tendency" means that, as far as "type-(i)" events are concerned, "the photon" cannot be construed as having passed through one of the slits and not the other ... at least not in the "naïve" sense of Classical Mechanics.


    Have you been able to resolve the difficulty?

    Do you want to hear more about what I am referring to as "type-(ii)" events?
  8. Jul 27, 2004 #7
    I have to disagree with my statements above, for a simple argument shows that they cannot be true.

    Let x = 0 be the position on screen2 of the center of the (would be) interference pattern. Then, if the photon is detected at any point x ≠ 0, there must have been a nonzero transfer of momentum to screen1 given by δpscreen ~ pphotonx/L , where L is the distance from screen1 to screen2, and L >> x.

    However, the impulse δpscreen suffered by screen1 will in general be much smaller than the uncertainty, ∆pscreen, in screen1's momentum, as implied by the Uncertainty Principle, if we assume that the uncertainty in screen1's position, ∆xscreen, is small enough to sufficiently "fix" the slit positions so as not to "smear out" the interference pattern.

    Note that, roughly,

    [1] ∆pscreen >~ h/∆xscreen .

    Since we do, in fact, see an interference pattern, it follows that ∆xscreen must be significantly less than the spacing between adjacent interference fringes. That is,

    [2] ∆xscreen << Lλphoton/d = (L/d)(h/pphoton) ,

    where d is the distance between the slits. Putting [2] into [1] gives

    ∆pscreen >> pphotond/L ~ pphotonx/L ~ δpscreen .

    This shows that the uncertainty in screen1's momentum is much greater that the actual momentum which is transferred to screen1. Thus, while "something physical" does take place, there is no "making of a record".
    Last edited: Jul 27, 2004
  9. Jul 28, 2004 #8
    I've also a question about the double slit experiment.

    When both slits are open, and you seperate the space between screen1 and screen2 in two parts (one half for each slit), do you still have inference effects on screen2?

    Is there a minimum distance, regarding L (the distance between screen1 and screen2)?

    The questions handle the place of interaction of a photon (with itself), and time/space that it needs (to find itself again).
  10. Jul 28, 2004 #9
    If you mean to invoke this separation through the agency of an opaque wall, then no there will not be any interference. A necessary condition for interference to occur is that the waves from each slit will overlap.


    This question is too difficult ... why do you ask?
  11. Jul 28, 2004 #10
    While I think about this question some more, here is a diagram for you to consider. Perhaps it will help you to conceptualize the situation from another perspective.

    The reason I asked, "... why do you ask?", is because knowing the reason for the question could be helpful.


    Attached Files:

    Last edited: Jul 28, 2004
  12. Jul 29, 2004 #11
    What makes the double slit experiment so special?

    Thanks, if I may use the analogy of water and water waves, I'll understand that a opaque wall, will prevent the waves from overlap. Nothing special about that.
    I'm interested in the difference between the patterns obtained by transmitting a photon (time on time) through these two slits, and the patterns in the corresponding water expirement. You showed nicely in that diagram the result of both.

    About the distance L. Of course still with L >> x, I don't want it to make it too difficult. ;-) It's the same kind of question: does the distance between screen1 and screen2 have influence about the possibilities to inference?

    I'm asking this questions, because I heard weird stuff about photons 'knowing' the state of others (if they interacted with them in the past). About communication in no time. And I really do not see how they obtain this stuff from this experiment. So, I was asking some questions that would make the situation different from classic wave theory.

    Thank you for your answers, I'm aware that it's very basic stuff...
  13. Jul 29, 2004 #12


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    The things you have heard are about quantum entanglement, and they are true quantum effects which cannot be modeled by classical processes. That is the point of the Bell inqualities, which were verified by the Aspect experiment and later experiments. The words I have italicised can be googled on to get more information.
  14. Jul 30, 2004 #13
    Indeed, the double-slit experiment is explainable on the basis of a single-particle wavefunction, with no obvious requirement for "communication in no time".

    To selfAdjoint's list, I would only consider adding "EPR" and "EPR paradox" (where "EPR" is short for Einstein, Podolsky, Rosen).

    Here are some brief points to help you put things into perspective.

    1) The EPR thought-experiment was intended to show that Quantum Mechanics cannot be a "complete" theory in the sense that there are certain objective "elements of reality" for which the theory makes no account. The EPR argument was based upon an assumption of "locality", whereby Einstein, Podolsky, and Rosen argued [1] that "locality" implies the existence of those objective "elements". Since, at that time physicists considered "locality" to be a fundamental principle, not to be rejected, the EPR claim had considerable force.

    2) Some three decades later, around the mid 60's, Bell was able to show, on the basis of an EPR-type scenario, that Quantum Mechanics itself implies "non-locality". He did so by showing that "locality" implies a certain "condition" ("Bell's inequality") which simply does not hold according to Quantum Mechanics ... and that, therefore, Quantum Mechanics implies "nonlocality". (A minority of physicists have since attempted to refute Bell's proof by arguing that it relies upon more than just an assumption of "locality", and that, instead of "locality", it is one of those other assumptions which is false. The most prominent such alternative assumption is referred to as "CFD", "contrafactual definiteness".)

    3) Since then, there have been a number experiments attempting to verify whether or not a "Bell-type of inequality" is in fact dis-respected (as QM would predict) by nature (the most celebrated of which is the "Aspect Experiment"). To date, none of these experiments has been entirely loophole free.
    [1] It appears to me that what I am saying here is more like an equivalent "paraphrase" of their argument.
    Last edited: Aug 5, 2004
  15. Jul 30, 2004 #14
    The simple answer is: "NO".

    The not simple answer is also: "NO" ... BUT...

  16. Aug 5, 2004 #15
    Thank you both, I've still to learn much I see. :)
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