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Wave function collapse by orbital angular momentum operator Lz

  1. Aug 12, 2012 #1
    I have some doubts about the implications of the orbital angular operators and its eigenvectors (maybe the reason is that I have a weak knowledge on QM).
    If we choose the measurement of the z axis and therefore the Lz operator, the are the following spherical harmonics for l=1.

    [itex]Y_{1,0}[/itex]=[itex]\sqrt{3/4\pi}[/itex]cos([itex]\sigma[/itex])

    [itex]Y_{1,\pm1}[/itex]=[itex]\mp[/itex][itex]\sqrt{3/8\pi}[/itex]sin([itex]\sigma[/itex])[itex]e^{\pm i\phi}[/itex]

    It’s evident that in the first case there is no current since there is no phase change and therefore [itex](∇ψ)^{+}[/itex]ψ - [itex](ψ)^{+}[/itex]∇ψ = 0.

    My doubt is: How can we have inertial momentum if there is no current? From my point of view, when l=1 the particle must have orbital momentum over some axis but the Lz operator collapses the wave function and that can destroy its momentum with a probability that might depend on the angle between the real axis of rotation and Z, but as far as I have searched, I haven’t found any reference to confirm or deny this. .

    Taking the previous hypothesis as true, another qüestion arises: Is there any phenomenon in the nature that can create this wave function collapse?

    The spin orbit coupling could be an example of this wave collapse. According to the orientation between spin and orbital momentum we have different energy levels for same number “l”. In fact, what matters is the total momentum “j”, but this is obtained by the sum of the quantum numbers “m” and “sz”, which brings me to the last qüestion:

    Is the spin momentum only well-defined in one direction? That’s plausible since in the Pauli representation only one “Z” coordinate has the eigenvectors (1, 0); (0,1), “X” and “Y” coordinates mix both spin up and spin down.

    I would like to check ask whether my reasoning is OK or instead I’m a mess.

    Thanks in advance.
    Sergio
     
  2. jcsd
  3. Aug 12, 2012 #2

    Bill_K

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    Science Advisor

    USeptim, Many questions at once! -- please let me keep the answer simple and generic.

    One of the features of quantum mechanics is that an observable A can have <A> = 0 and yet <A2> ≠ 0. That's basically what's going on here. Because angular momentum is described by the quantities Lz and L2. When we say L = 1, we're really talking about L2 = Lx2 + Ly2 + Lz2. It's quadratic. And the point is, in the state Lz = 0, even though <current> = 0, there's a <(current)2> ≠ 0, and that's what's related to L2, the angular momentum.
     
  4. Aug 13, 2012 #3
    Thanks for your answer Bill_K.

    The fact that [itex]A^{2}[/itex] [itex]\neq[/itex] 0 when <A>=0 comes because you can be summing or integrating quantities with different sign, it's the classical problem with the average and the variance.

    But is this what is happening with the orbital angular momentum? From the book I have read I get the impression that a single zero spin particle (i.e. a single wave function) can have [itex]L^{2}[/itex] [itex]\neq[/itex] 0 and yet [itex]L_{z}[/itex].

    On the other hand, which formula can we use to get <[itex](current)^{2}[/itex]>?
     
  5. Aug 14, 2012 #4
    I forgot to say that for l=1, m=0 the harmonic not only gives a <current> = 0 but also the current is zero everywhere, so I don't think it could give a <[itex]current^{2}[/itex]>[itex]\neq[/itex]0
     
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