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Angular momentum and its uncertainty

  1. Apr 13, 2012 #1

    fluidistic

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    [itex][L_x,L_y]=i \hbar L_z \neq 0[/itex]. In fact it seems we can know only the modulus squared of the angular momentum and one component, at a same time.
    However if I take an electron say in the fundamental state in the hydrogen atom, L=0. Since the modulus squared is equal to 0, it means that all components are worth 0 or I'm missing something? Wouldn't that mean that we can know "the 3 components of the angular momentum without any uncertainty when it's worth 0"? Where does the Heisenberg's uncertainty principle applies here?
     
  2. jcsd
  3. Apr 14, 2012 #2
    The uncertainty relation in general states that [itex]\Delta A\Delta B\geq\frac{1}{2}\left|\left\langle \left[\hat{A},\hat{B}\right]\right\rangle\right|[/itex]. For angular momentum, this takes the form [itex]\Delta L_{x}\Delta L_{y}\geq\frac{\hbar}{2}\left|\left\langle L_{z}\right\rangle\right|[/itex]. Thus, if the expectation value of angular momentum in the z-direction is zero, the uncertainties in angular momentum in the x and y directions can be zero. By similar reasoning reasoning, you can see that if expectation values of angular momentum are zero in all three directions, then uncertainty in angular momentum can also be zero in all three directions, and thus the system can be in a state of definite angular momentum in all three directions as long as the angular momentum is zero.
     
  4. Apr 14, 2012 #3

    fluidistic

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    Ok thank you very much for the information. :)
     
  5. Apr 14, 2012 #4
    The more counterintuitive thing about the L=0 states is that even though the electron has no angular momenum, it doesn't fall into the nucleus. Do you know why?
     
  6. Apr 14, 2012 #5

    fluidistic

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    No, I don't really know why. But I've learned that in QM a central force does not imply a motion in a plane unlike in classical mechanics. It implies a motion in 3 dimensions, so it does not seem strange to me that a vanishing angular momentum doesn't imply the electron to "fall into the nucleous".
    Adding the fact that the electron hasn't a well defined position, stopping to think things classically would help.
    To answer your question mathematically, I think that the Schrödinger's equation has a stationary solution with L=0. But physically I don't know the answer. I would love to read it though :)
     
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