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B Electron in the double-slit experiment

  1. Feb 19, 2017 #1
    Newbie here: Is the (single) electron leaving the "machine" in the famous double-slit experiment the same one hitting the screen? Please give a short explanation on how this is proved, thank you.
     
  2. jcsd
  3. Feb 19, 2017 #2

    PeroK

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    Electrons are indistiguishable. There is no way to mark, label or number an electron to distinguish it from others. Technically, therefore, all you know is that an electron left the machine and an electron hit the screen. In fact, it would make no sense to ask if it was "the same electron".
     
  4. Feb 19, 2017 #3
    Thank you for the fast answer. But isn't this huge, the fact that we don't really know if the electron we fired is the one hitting the screen??
     
  5. Feb 19, 2017 #4

    PeroK

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    It's not that we don't know (whether it's the same electron), it's that it doesn't make sense to ask the question.
     
  6. Feb 19, 2017 #5
    I recognize this type of answer from somewhere. But I'll play along: why does it not make sense to ask the question? :)
     
  7. Feb 19, 2017 #6

    PeroK

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    It's not a question of playing along. All elementary particles, by virtue of what they are, cannot be changed or marked or numbered in any way. The reason you can identify, say, one pool ball from another is that you can change each ball (paint a number on it) and it's still a pool ball.

    Even if you could paint a number on an electron, it would no longer be an electron. If you, say, attach a proton to it, then it becomes a hydrogen atom. And all hydrogen atoms are likewise indistinguishable.

    All you can ever know about an electon is that it is an electron. You cannot identify it further in any way.

    The indistinguishability of electrons is, in fact, at the root of all chemistry, so it is of physically fundamental importance.
     
  8. Feb 19, 2017 #7
    Thank you again. I was thinking about the fact that the fired electron takes all possible trajectories simultaneously before hitting the screen. Instead of saying this, could it maybe be that the fired electron is not taking all the trajectories simultaneously, but rather "bumping" all the other electorns in its straight path, where the last bumped electron hit the screen? Something like the pool ball hitting the other ones and another ball ends up in the pocket. I am starting to feel like there is holes to my theory but I'll still post it so maybe you can add/contract from it. :)
     
  9. Feb 19, 2017 #8

    PeroK

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    What do you mean by the electron takes all possible trajectories simultaneously? Do you think it decides where on the screen it wants to hit first, then gets there by all possible trajectories? Or, if it sets out on all possible trajectories, how does it decide where to hit the screen?
     
  10. Feb 19, 2017 #9
    Agreed.
    Not agreed. If electron emission is slowed down so that each electron is individually detected then it makes perfect sense to think it might be the "same" electron as was emitted if we had detected that electron at emission without significantly disturbing its flight. It doesn't guarantee it, of course, because it might have been annihilated and replaced en route.

    So, in the sense of the OP, where he is contrasting the idea of a single electron to multiple collisions and "bumping" electrons, I think we can say quite clearly that their idea would be inconsistent with a tightly constrained momentum state (and interference pattern) and they should definitely prefer the notion of the "same" electron..

    I think this illustrates how careful we must be with our language when we stray from mathematics.
     
  11. Feb 19, 2017 #10
    I am reading Brian Greene's book "the elegant universe". In there he writes about the Feynmans double-slit experiment and uncertainty principle. Under a figure (which I don't know how to insert, but it is only showing a scematic picture of a standard double-slit experiment) it says:

    "According to Feynman's formulation of quantum mechanics, particles must be viewed as
    travelling from one location to another along every possible path. Here, a few of the infinity of
    trajectories for a single electron travelling from the source to the phosphorescent screen are shown.
    Notice that this one electron actually goes through both slits."

    So it says that the electorn goes through both slits and every possible path simultaneously before hitting the screen. This is impossible for me to wrap my head around, that it takes every possible path at the same time! So I was thinking maybe it doesn't? That the electron hitting the screen is not the one that left the gun? Again, please keep in mind that I am very new to physics and may have misunderstood the whole thing.
     
  12. Feb 19, 2017 #11
    I hope I am not changing the subject too much...but in that experiment is the electrons momentum presumed to be well known, ie, in the preparation state the electrons momentum is placed in a well defined state? (If it is too much off topic I understand if this post is deleted)
     
  13. Feb 19, 2017 #12
    The OP's context was an interference experiment which implies a closely defined momentum.
     
  14. Feb 19, 2017 #13

    PeroK

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    No, you haven't misunderstood. But, Brian Greene is being very imprecise. The only way you can know where an electron is at any time is to measure it. If you don't measure an electron until it hits the screen, then you cannot say how it got there (in terms of a classical trajectory). The Feynman formulation provides a way to calculate the probability of where on the screen the electron hits by considering all possiblities. But, in fact, it's evolution of the electron's wave-function which is considered.

    In particular, you can never say an electron went through both slits unless you looked for it at both slits and then you would find it only went through one.
     
    Last edited: Feb 19, 2017
  15. Feb 19, 2017 #14
    Okay? So why does he write it like this then? Is it a "popular writing" thing?

    Another thing about the uncertainty principle. Could it be that it is uncertain where exactly the electron will hit the screen, because an electron spins around an atom and if it is "to the left" of an atom at the point it hits the screen, then it will hit to the left of an imagimanry centre point on the screen? The center point meaning an imaginary straight line between the gun and screen which ends in a point on the screen where the electro should hit when fired.
     
  16. Feb 19, 2017 #15

    PeroK

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    Yes, partly because even mention of the wave-function requires more time and effort, so it's easier just to talk about the particle. And, partly, perhaps because it sounds interesting and exciting to think of an electron taking an infinite number of paths simultaneously.

    No, nothing like this. You could try watching the Feynman lecture here:

    http://www.cornell.edu/video/richar...ty-uncertainty-quantum-mechanical-view-nature
     
  17. Feb 19, 2017 #16
    All right I'll give it a go. Thanks for the help :)
     
  18. Feb 19, 2017 #17
    If we don't look for it at either slit can we say it went through both slits? If I did say that, would I be right or would I be wrong?

    I think I know the answer to this, and that is the wave function has not collapsed so therefore, the electron did go through both slits. Why doesn't going through the slit(without a detector) collapse the wave function? I think I know the answer to this also...it just doesn't. When we do the experiment, that is what we observe. We have a theory that matches observations and agrees with experiment but does not really tell us a whole lot of what is fundamentally taking place, or if it is telling us what is fundamentally taking place, we don't understand it yet.
     
    Last edited: Feb 19, 2017
  19. Feb 19, 2017 #18
    Wrong. We just don't know which slit it went through. Uncertainty in QM is expressed as "superpositions". So, without having detectors at the slits, we have what is effectively a superposition of "slit states" imposed on the superposition of impact locations at the screen and this results in additive/cancelling terms in the projections onto impact locations. This looks like "interference" because the addition/cancellation is done before computing the intensity at the screen rather than after.

    IMO you would do better to forget about wave-functions entirely. Think of superpositions instead. You may not understand everything, but you'll be nearer to how physicists think through the math.
     
  20. Feb 19, 2017 #19

    Nugatory

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    The quick response is that of course we know - if there's no detector to interact with, then there's no interaction to decohere the wave function. A good layman-friendly reference would be David Lindley's book "Where does the weirdness go?".

    The longer answer is that the question is poorly formed because wavefunction collapse is not a required concept in quantum mechanics. It is something that appears in some interpretations (that is, suggestions/opinions about what is really going on) of quantum mechanics but doesn't appear in the mathematical formalism, and the more you study QM the less helpful you will find it.
     
    Last edited: Feb 20, 2017
  21. Feb 19, 2017 #20
    I don't think that is quite right. I think it's another example of the problem of translating math into physics. It seems clear to me that the electron interacts with the slit -- it produces (single slit) electron diffraction for instance and this one reason why the observer/collapse scenario is problematic. The difference a detector makes is that the information is recorded. One of the abiding principles of physics that seems to hold up in a QM context is that nature doesn't seek to deceive us. Once information is available and recorded so that it can be subsequently read then any future observations must be consistent with that unless the information is lost (e.g. by a quantum eraser).
     
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