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Homework Help: Orbital angular momentum operator

  1. Jun 26, 2008 #1
    Hello, sorry I am new to this forum, I hope I found the right category. I have a question about the momentum operator as in Sakurai's "modern quantum mechanics" on p. 196

    If I let

    [tex] 1-\frac{i}{\hbar} d\phi L_{z} = 1-\frac{i}{\hbar} d\phi (xp_{y}-yp_{x})[/tex]

    act on an eigenket [itex]| x,y,z \rangle [/itex]

    why do I get [itex]| x-yd\phi,y+xd\phi,z \rangle [/itex]

    and not [itex]| x+yd\phi,y-xd\phi,z \rangle [/itex] ,

    with the momentum operators

    [tex] p_{x}=\frac{\hbar}{i}\frac{\partial}{\partial x} , p_{y}=\frac{\hbar}{i}\frac{\partial}{\partial y}[/tex]

    Thanks for your help!
  2. jcsd
  3. Jun 27, 2008 #2


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    can you show us why you think that would yeild:

    [itex]| x+yd\phi,y-xd\phi,z \rangle [/itex]

  4. Jun 27, 2008 #3
    I just use the operator on each component:

    [tex] [1-\frac{i}{\hbar} d\phi (xp_{y}-yp_{x})] | x,y,z \rangle =[/tex]

    [tex] [1-d\phi (x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x})] | x,y,z \rangle =[/tex]

    [tex] |x-d\phi (x \frac{\partial x}{\partial y}-y \frac{\partial x}{\partial x}),y-d\phi (x \frac{\partial y}{\partial y}-y \frac{\partial y}{\partial x}),z-d\phi (x \frac{\partial z}{\partial y}-y \frac{\partial z}{\partial x}) \rangle =[/tex]

    [tex] |x-d\phi (0-y),y-d\phi (x-0),z-d\phi (0-0) \rangle =[/tex]

    [tex] | x+yd\phi,y-xd\phi,z \rangle [/tex]
  5. Jun 28, 2008 #4


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    Isn't that the correct answer?
  6. Jun 28, 2008 #5
    Well, I think the calculation is correct, but then I did a backwards rotation, which I didn't intend to do.
    The rotation matrix for an infinitesimal rotation about the z-axis is (if I rotate the vector, not the system)

    R_{z}(d\phi) = \left(\begin{array}{ccc}
    1 & -d\phi & 0\\
    d\phi & 1 & 0 \\
    0 & 0 & 1 \end{array}\right), R_{z}(d\phi)^{-1} = \left(\begin{array}{ccc}
    1 & d\phi & 0\\
    -d\phi & 1 & 0 \\
    0 & 0 & 1 \end{array}\right)[/tex]

    So [tex] | x+yd\phi,y-xd\phi,z \rangle [/tex] = [tex] R_{z}(d\phi)^{-1}| x,y,z \rangle [/tex]

    Yet if you try to determine the quantum mechanical operator for an infinitesimal rotation around the z-axis, starting with

    [tex]\hat{R}| x,y,z \rangle = | x-yd\phi,y+xd\phi,z \rangle [/tex]

    (as done e.g. here: http://en.wikipedia.org/wiki/Rotation_operator" [Broken], you find

    [tex]\hat{R} = 1-\frac{i}{\hbar} d\phi L_{z}[/tex]

    And then inserting this result for [itex]\hat{R}[/itex] leads me back to my problem...
    Last edited by a moderator: May 3, 2017
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