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Momentum conservation, the double slit and Heisenberg

  1. Jan 21, 2015 #1
    One thing that's been troubling me lately is the idea that a quantum experiment can start with the same initial conditions, but the outcome is probabilistic, not deterministic, and how this fits in with the conservation of momentum.

    I was thinking about the classic double-slit experiment, in which a single electron is fired from an electron gun at two slits and ends up hitting a screen in one particular place, exhibiting particle-like behaviour both at the beginning and end of the experiment. But what about in the middle, when the electron is in transit and is not being measured? When this experiment is repeated many times, the electrons build up an interference pattern on the screen, as if each electron spread out into a superposition of states described by a wavefunction that interfered with itself before the wavefunction collapsed and the electron finally ended up at one particular point upon contact with the screen. The exact point on the screen where each electron will land cannot be determined, but we can use the wavefunction to give the probability that each electron will end up in any given place. Eventually, with enough electrons, we build up a pattern on the screen. For obvious reasons, the pattern of dots will be densest where the most electrons hit and therefore where it was most likely any given electron would hit. Since the probability of an electron being in a particular place is described by a wavefunction that passes through both slits at once and interferes with itself, the pattern of dots that builds up on the wall will be an interference pattern. This explains the so-called "wave-particle duality" - the "wave" is simply a probability wave that describes the probability of finding the electron in a particular place once the wavefunction collapses. Before that collapse, the electron exists in a superposition of many states at once, and the evolution of that superposition through time as the electron travels is described by the changing wavefunction in the Schrödinger equation. Someone correct me if I'm wrong!

    The key point of all this is that the trajectory of any one electron is not deterministic - starting out with the exact same initial conditions, we cannot predict exactly where the electron will end up. We can give probabilities, of course, but not certainties. Carrying out the same experiment with identical initial conditions can result in different outcomes - outcomes we can't predict deterministically, but can assign a probability to (and so can still be tested empirically if we run the experiment enough times to see if the results match our statistical prediction).

    I thought to myself that this would be a little like hitting a baseball, then having the baseball spread itself out into a superposition of many positions described by a wavefunction, and finally having the baseball hit something in one particular place. This is non-deterministic, so even if the batter swings at the baseball with exactly the same velocity, force, etc., where the baseball will end up hitting cannot be determined - only the probability of it hitting any particular place.

    Of course this doesn't happen with real baseballs, because a) Planck's constant is tiny, so the de Broglie wavelength for baseballs is so incredibly small that they exhibit purely particle-like behaviour within experimental errors, and b) at that kind of scale, there would be so many interactions/"measurements" (the air molecules hitting the baseball, photons of light hitting the baseball, etc.) that the superposition would quickly collapse - superpositions simply can't be maintained for huge objects. However, my understanding is that this is a fair analogy for quantum particles, like electrons or photons, and is pretty much what happens during the double-slit experiment. (Again, someone correct me if I'm wrong!)

    However, I'm having a conceptual problem understanding this because of the conservation of momentum. In Newtonian physics, if I hit that baseball with a particular momentum, it will shoot off in a particular direction at a particular speed, which will always be predictable because momentum is conserved. Moreover, because of Noether's theorem, the conservation of momentum is a consequence of the symmetry of space, which doesn't change in quantum mechanics, so momentum conservation should still hold true in that theory even if the rest of classical physics goes out the window. So surely if I "hit" a stationary electron with a certain momentum, that momentum will be transferred to that electron and it will fly off with a particular momentum, i.e. in a particular direction at a particular speed? And surely that direction and speed are non-negotiable, because of momentum conservation? So how can the electron's motion be probabilistic and not deterministic? In other words, how can the electron's motion be probabilistic while momentum conservation is still true?

    At first I thought a possible answer might be the Heisenberg Uncertainty Principle. If I hit a baseball with a baseball bat, because they are both very large objects, the Heisenberg uncertainty in their momenta will be much tinier than any experimental measurement, and that means I can say (for all practical purposes) that the baseball was hit with a particular, unique momentum and will therefore fly off at a particular momentum that I can calculate classically using momentum conservation. But if electron A was hit by electron B, electron B's mass is so small that the Heisenberg uncertainty will be pretty large. Since momentum of electron B cannot be determined exactly (thanks to Heisenberg), all momentum conservation will give me is a wide range of possible momenta that were transferred from electron B to A during the collision rather than any one particular, unique momentum - allowing the electron A's movement after the collision to still be probabilistic rather than deterministic and its exact trajectory to remain undetermined. Is this the correct answer to my question?

    But thinking about this further, I thought - what if something small like an electron was given momentum by a large object? Say we hit a stationary electron with a baseball bat. Before the collision the electron is stationary and has zero momentum. The baseball bat is very, very massive compared to the electron, so the Heisenberg uncertainty in its momentum is tiny. However, doesn't this restrict the range of momenta the electron can be given by the collision? If the uncertainty in the baseball bat's momentum is tiny, and momentum is conserved in the collision with the electron, then surely the electron's momentum has the exact same tiny uncertainty after that collision, and it should therefore behave classically? So if the uncertainty principle isn't the answer to my question "how can the electron's motion be probabilistic while momentum conservation is still true?", what is?
     
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  3. Jan 21, 2015 #2

    TeethWhitener

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    There's your problem right there. Because of the uncertainty principle, the electron cannot be stationary and have zero momentum: [tex]\Delta x \Delta p \geq \frac{\hbar}{2}[/tex]

    [Edit]: Sorry, let me clarify. The exact position and exact momentum of the electron cannot be known simultaneously. So if you know that the electron has exactly zero momentum, you know nothing about its position; i.e., the electron wavefunction is spread out over all space (plane wave).

    [Further edit]:
    Again, the smaller the momentum uncertainty of the bat, the larger its position uncertainty. The bat has to play by the same rules as everything else. Even if the position and momentum uncertainty of the bat have been minimized (let's pretend the bat is just an extremely massive Gaussian wavepacket), there's still a spread of x and p values in light of the uncertainty principle.
     
    Last edited: Jan 21, 2015
  4. Jan 21, 2015 #3
    D'oh, of course, thank you! I already knew the uncertainty principle prohibited things from being completely still, that's why helium is liquid even as close to 0K as possible, right? So the concept of a stationary electron doesn't work, thanks. I'm still confused about the momentum uncertainty of the electron, though. If the bat's momentum gets (partially) transferred to the electron, the uncertainty on the bat's momentum is small, and momentum is a conserved vector quantity, then how can we still have a wide spread of possible, unpredictable directions for the electron to move in?
     
  5. Jan 21, 2015 #4

    bhobba

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    You are still thinking classically. In QM it doesn't have a momentum or any other property until its observed to have that property.

    Your bat analogy is a problem in classical physics. The subatomic particles of the bat and ball are bound in those systems and as a whole behave entirely classically - that such is the case is an assumption of the usual Copenhagen interpretation of QM. Since QM is a theory about observations that appear in an assumed classical world its a bit of an issue exactly how such a theory explains that classical world. A lot of progress has been made in that - but a few issues remain without detailing exactly what they are (eg the so called factorisation problem).

    At your level I think the following may help:
    http://motls.blogspot.com.au/2011/05/copenhagen-interpretation-of-quantum.html

    Thanks
    Bill
     
  6. Jan 21, 2015 #5

    Nugatory

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    The uncertainty in the bat's momentum is small compared with the bat's momentum. However, in absolute terms it is still enormous compared with the the momentum of the post-collision electron, so the electron can head off in pretty much any direction while the system remains within the bounds of the uncertainty principle.

    As an aside: in principle it is possible to measure both the pre-collision and post-collision momenta of both the ball and the bat to arbitrary precision, to as many decimal places as we wish (we have to accept greater uncertainty in the position, of course, so although we can measure the actual momentum as precisely as we wish, we cannot predict it ahead of time). If we do this, we we will always find that conservation of momentum is strictly obeyed - the pre-collision sum of momentum will equal the post-collision sum of momentum.
     
  7. Jan 22, 2015 #6

    vanhees71

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    I think a lot of the trouble people have understanding quantum theory comes from sentences like this. It's very easy to make it clearer: "In QM it doesn't have a momentum or any other property until it is prepared to have that property."

    Just to measure it doesn't tell you what's happening with the observed object afterwards. E.g., you use an electromagnetic calorimeter at the LHC to measure photons you get their energy but the photon is absorbed by the material. Then you don't have a photon with that energy left.

    The quantum theoretical states give the probabilities to find a certain value for an observable. That's it. There's not more to what's called a state in quantum theory.
     
  8. Jan 22, 2015 #7
    Ahhh, that sounds so obvious now you've said it... thank you! I feel extremely silly for having missed something so glaringly obvious... the absolute uncertainty is tiny compared to the momentum of the bat, but huge compared to the momentum of the electron... of course...
     
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