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The commutator [L,p]

  1. Feb 22, 2007 #1
    How do i compute the commutator [L,p]?
     
  2. jcsd
  3. Feb 22, 2007 #2

    Mentz114

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    Read the forum guidelines.
     
  4. Feb 22, 2007 #3
    You should know from your class that the commutator [x, y] = xy - yx

    you can express the L operator in terms of the coordinates x,y,z and the momentum operator p. Apply the commutator to a wavefunction psi and simplify!

    Hope that gave you a clue.
     
  5. Feb 22, 2007 #4

    Meir Achuz

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    use L=rXp in the commkuator.
     
  6. Feb 22, 2007 #5
    I find 2ihp, is that correct? do you know the correct answer?
     
  7. Feb 23, 2007 #6

    dextercioby

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    First of all, both L and p are vectors, so the commutator should be computed componentwise. Next, you need to find a common dense everywhere domain for the commutator, it's not difficult to see that on the Schwartz space over R^3 both the momentum and the angular momentum operators are essentially self-adjoint and the invariance conditions are met. Therefore,

    [tex] [L_{i},p_{j}]_{-}\psi (\vec{r})=... [/tex]

    and , without doing any specific calculations (derivatives i mean), using the fundamental comm. relations (also valid on the Schwartz space) and some simple Levi-Civita pseudotensor manipulations, you can find the answer.
     
  8. Feb 23, 2007 #7

    Meir Achuz

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    No its more complicated than that. Use Cartesian coordinates with
    [x,px]=i and (rXp)_i=epsilon_ijk x_ip_j.
     
  9. Feb 23, 2007 #8
    r = (x, y, z) and p = (px, py, pz).

    I assume you know how to take a cross product. The only other thing is that p = -i\hbarh\del which acts on the wavefunction \Psi, and you can't exchange r and p (ie. rxp is not the same as pxr)

    I hope that helps
     
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