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Uncertainty principle puzzle

  1. Apr 24, 2008 #1
    I have a puzzle I seem to be unable to answer satisfactorily. It concerns the 'Uncertainty Principle': If you can locate a particle (any kind of quanta) at a given location within an uncertainty of deltaX then the uncertainty in its momentum deltaP must be at least h-bar/deltaX. (Please don't quibble about factors of 2, or pi, because that does not matter.)

    Here is the gedanken-experiment setup. We prepare some near pure material and we implant a single atom of a known impurity in the crystal. The crystal (being near pure) does not have many natural impurities in it, and we will also assume that these impurities are of different type from the atom we purposely implanted. Now using various techniques like X-ray crystallography or neutron scattering, the location of the atom we purposely implanted can be determined to an accuracy comparable to the size of a typical atom in the crystal. So, let's take deltaX = 5.0E-8 cm.

    Now we are ready to do the experiment: The experiment consists of shining light on this material, of which wavelength is carefully chosen. The chosen wavelength must NOT excite the atoms of the crystal, or the other impurities in it, but it will excite the implanted atom, when absorbed. Most likely the energy of the photon will be a few eV. It could be an optical or at most an ultraviolet photon. Anyway, after shining the light for sufficiently long, the impurity will absorb one of the photons, jump to an excited state, and then make a cascade of transitions down to its ground state, which we can observe. In other words, when the implanted atom jumps to the excited state we know for sure that it absorbed one of the photons we have been hitting the
    crystal with. Since we know the impurity location to an accuracy of 5E-8 cm, we also know that the inaccuracy of the position of the absorbed photon is also 5E-8. (No other atom in the crystal can absorb it.) Therefore the uncertainty in the momentum of the photon when absorbed within the atomic dimensions must be deltaP >= h-bar/5.0E-8 cm, or deltaP >= 400 eV/c. But this is sheer nonsense (because the photon has no mass and the uncertainty in its energy must be 400 eV.)

    The problem is perfectly OK for electrons. You can locate an electron within a few angstroms with the same deltaP of 400 eV/c. Because of the electron mass, this translates into 0.32 eV of kinetic energy, not 400 eV. But we cannot locate a photon as accurately as 5 angstroms, without incurring an uncertainty of 400 eV in its energy. And yet, this gendanken-experiment will work in reality, right??

    So where did I go wrong????:confused:
  2. jcsd
  3. Apr 24, 2008 #2
    Maybe I did not pay enough attention, but
    [tex]\hbar = 200[/tex] MeV fm = 0.2e-6 eV m
    [tex]\frac{\hbar}{0.5\text{nm}}=4[/tex]meV ?
  4. Apr 24, 2008 #3
    Yes I agree hbar c = 0.2E-6 ev m. However, dividing that by 0.5nm = 0.5E-9 m, I get 0.4E+3 eV, or 400 eV just like I said. But in a way, it is good that I did not screw up the simple arithmetic. If 400 eV/c was really 4 MeV/c instead like you said, the puzzle would only get worse, not better. We want the uncertainty in the photon energy to be no worse than a few eV, preferably a million times less than that. The photon energy to excite the right atomic levels must a few eV with an uncertainty of millions of times less that due to decay width. It's just standard atomic physics type numbers.

    PS: How do you get to type the formulas so nicely? I am afraid I know no better than old-fashioned text.
  5. Apr 24, 2008 #4

    Ken G

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    You went wrong associating a vector momentum uncertainty with a scalar energy uncertainty. The uncertainty in the photon momentum is very large because the photon that interacts with the atom can be scattered into any direction. Indeed, this localization in its position sets a minimum to the amount of spreading in the scattering angle. But you can know the magnitude of the momentum precisely, and hence know the energy precisely, and still not know the vector momentum precisely because you don't know the direction. A similar thing happens when you diffract a photon through a slit that is as wide as the wavelength-- the diffraction angle becomes very unknown, so the transverse momentum is unknown, but you still know the energy (and magnitude of momentum) very precisely.

    Another way you can think about that is, because you don't know exactly when the photon interacted with the atom, you don't know where the photon "was" in the longitudinal direction when it interacted. This is because the photon wave packet will have a bandwidth, whose inverse corresponds to a time uncertainty as to when it interacts (or if you prefer, a longitudinal stretching of the photon, which can interact with the atom at any time the atom is in that length, if you have to pick a specific time of interaction). So you really only have the accuracy you imagine when you are talking about where the photon was transversely to its direction, so the uncertainty is in tranverse momentum-- not magnitude of momentum. That is, the uncertainty is in the direction of the scattered photon-- there is a fundamental limit to how "forward beamed" the scattering could be.

    You actually see this in "Mie scattering theory", where dust particles tend to forward scatter light, but less and less forward scattering occurs as the particles get smaller and you move into the Rayleigh regime. The uncertainty principle limits how forwardly scattered the beam can be, but not its energy resolution. Good question!
    Last edited: Apr 24, 2008
  6. Apr 25, 2008 #5
    Hey, I realized once in bed my mistake but was too lazy to get out :smile:
    You were right from the begining. I was finding 4e-3 eV out of my own stupidity.

    Apart from that, I was thinking in more or less the same lines as Ken G. Your incident photon really is spherical wave with spheres of constant phase large enough to be described by planes. So before interaction, it was "everywhere" along the beam.
    Last edited: Apr 25, 2008
  7. Apr 25, 2008 #6

    Andy Resnick

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    I think you went wrong in understanding how crystallographic methods work- they do not give locations of anything. Having an impurity in an otheriwse perfect crystal will (ever so slightly) broaden the scattering peaks. So the locaton is not determined, only the effect on crystal symmetry.

    Also, a single photon will illuminate the entire crystal- so you do not need to know the exact location of an atom in order to excite it. This falls back to the difference between momentum and energy.
  8. Apr 25, 2008 #7


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    The photon can indeed interact with the lattice; as in inelastic scattering, which produces a phonon or two; "elastic" scattering is possible; refraction, if you will. For the latter, a secondary photon could interact with the impurity. That is, there are lots of phenomena that make your argument not so plausible; as a good experimentalist, you'd have lots of corrections to make.

    More interesting, I think, is to work all this out for a Gaussian wave packet, heavily localized. Your namesake would have worked all this out on the back of an envelope; the rest of us would need more time and more space.
    Reilly Atkinson
  9. Apr 25, 2008 #8
    Thank you for the reply, but I don't quite understand the above paragraph: even if the direction of the photon momentum is completely uncertain, the uncertainty of the momentum in any one direction cannot get up to 400 eV/c. Just like you said, the photon energy is known with high accuracy, therefore the magnitude of the photon momentum is known with high accuracy. Even when the direction of the momentum is completely uncertain, the certainty in any direction cannot be larger than the uncertainty in the magnitude of the momentum. This is still in the order of a few eV/c, a far cry from 400 eV/c.

    In addition, in this gedanken experiment the photon we are interested in is not scattered by the implanted atom. The crystal and its impurities scatter photons, but we ignore those events, they are not relevant here. Most of the photons we shine on the crystal will be scattered in some fashion. We are only interested in the one photon that gets absorbed by the implanted atom, and puts it in an excited state. The energy of the photons we shine on the crystal were chosen precisely so that they can induce an atomic excitation with resonant absorption.
  10. Apr 25, 2008 #9

    Ken G

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    Why not? If the direction is uncertain, the momentum in some direction could be 400 eV/c, or it could be -400 eV/c, for a difference of 800 eV/c. That's what I mean about momentum being a vector quantity.

    Correct, but the momentum uncertainty does not refer to its magnitude, it refers to its components-- see above.

    That answer is still given by the "longitudinal uncertainty" of the location of the photon when it was absorbed. In other words, just because the atom was at a known spot does not mean you know where the photon was "when it was absorbed". The photon absorption took time, given by the inverse of the frequency bandwidth of the photon wavepacket (so a pure plane wave, with no momentum uncertainty, requires effectively an infinite time to absorb). You cannot specify the longitudinal position of a photon to better than that time times c, even by placing an atom in its path.

    You can either picture that as the photon having an actual longitudinal extension, and rams gradually into the atom like a train hitting a wall (so by the time it has no longitudinal extension, it no longer exists), or you can imagine that the photon itself is just a point particle but it can be absorbed by that atom when it is anywhere within that longitudinal distance from the atom (because its location is governed by a wave function that has a physical extension). Either way, you never determine the longitudinal position of the photon to better than c/bandwidth by absorbing it with an atom.
    Last edited: Apr 25, 2008
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