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Commutator with r, p_r and angular momentum

  1. Sep 25, 2007 #1
    Hi, guys..This is my first time to post. and I gotta aplogize for my bad English..I`m a novice..;;

    anyway..here`s my curiosity..

    From Paul Dirac`s Principles of Quantum Mechanics..p.153
    (section of Motion in a central field of force)

    It says that

    The angular momentum L of the ptl about the orgin ..and its magnitude L^2 commute with r and p_r since they are scalars...

    It`s not hard to verify that [L, r] = [L, p_r] = 0

    (L_x=x*p_y-y*p_x, r=(x^2+y^2+z^2)^(1/2) etc.)

    But I just wanna understand the underlying physics..

    commutator(Quantum Poisson`s Bracket) with L zero?

    Are they concerned with conservation of angular momentum? or else?

  2. jcsd
  3. Sep 25, 2007 #2


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    L^2 is a Casimir operator for the Lie algebra of angular momentum. It just means that r and p_r are \phi independent, and \phi is the canonically conjugate variable of angular momentum.
  4. Sep 26, 2007 #3
    I don`t have any idea what Casimir operator is..

    But canonically conjugate variable of angular momentum phi is surely indep. of p_r, r

    thx a lot!
  5. Sep 26, 2007 #4

    Operators of angular momentum [itex] (L_x, L_y, L_z) [/itex] play the role of generators of rotations (actually, this property can be used as a definition of [itex] \mathbf{L} [/itex]). This means that if [itex] r [/itex] is distance measured in the reference frame O, then

    [tex] r' = \exp(\frac{i}{\hbar} L_x \phi )r \exp(-\frac{i}{\hbar} L_x \phi ) [/tex].....(1)

    is the distance in the reference frame O' that is rotated with respect to O by the angle [itex] \phi [/itex] around the x-axis. Since [itex] r [/itex] is a scalar, it is invariant with respect to rotations ([itex] r'=r [/itex]). Then, by Taylor expanding exponents in the right hand side of (1) one can show that this invariance is equivalent to the vanishing commutator

    [tex] [L_x, r] = 0 [/itex]

    Commutators are closely related to inertial transformations in other examples as well. For example, the Hamiltonian is the generator of time translations, i.e., for any operator [itex] F [/itex] its time dependence is given by

    [tex] F(t) = \exp(-\frac{i}{\hbar} Ht )F\exp(\frac{i}{\hbar} Ht ) [/tex]

    and [itex] F [/itex] does not depend on time if and only if [itex] F [/itex] commutes with the Hamiltonian.

    Last edited: Sep 26, 2007
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