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Trick for commutators problem

  1. Jan 30, 2008 #1

    Fys

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    [tex] \left[L_{x},L_{y}\right]=\left[yp_{z}-zp_{y},zp_{x}-xp_{z}\right]
    =\left[yp_{z},zp_{x}\right]-\left[zp_{y},zp_{x}\right]-\left[yp_{z},xp_{z}\right]+\left[zp_{y},xp_{z}\right]
    [/tex]

    How next?

    My book is not of much help
    I Tried
    [tex] \left[A,BC\right]=\left[A,B\right]C+B\left[A,C\right] [/tex]

    But that is too much work
    Does anybody knows a useful trick or something
    not only for this case, but for al commutators

    Thanks guys
     
  2. jcsd
  3. Jan 30, 2008 #2

    malawi_glenn

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    Sometimes you must do the hard work. It depends from case to case regarding what trick you can use, no gereneral rule.

    But the general formula for angular momenta commutators is:
    [tex] [J_i , J_j ] = i\hbar \epsilon _{ijk} J_k [/tex]
     
  4. Jan 30, 2008 #3

    Fys

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    Yes thanks
    but how can i work this out?
     
  5. Jan 30, 2008 #4

    malawi_glenn

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    by working out all partial commutators and using the commutator between position and momentumoperators. Or starting from group theory of rotations.. but I assume this is introductory QM course, so start with the first option ;)

    [tex] \left[yp_{z},zp_{x}\right]-\left[zp_{y},zp_{x}\right]-\left[yp_{z},xp_{z}\right]+\left[zp_{y},xp_{z}\right] [/tex]

    Start with [tex] \left[A,BC\right]=\left[A,B\right]C+B\left[A,C\right] [/tex]
    a couple of times til you get things like [x,p_y] ; [y,z] etc

    I got the same task in my intro QM course, took me a couple of hours ;)
     
  6. Jan 30, 2008 #5

    Fys

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    oke thanks :P
     
  7. Jan 30, 2008 #6

    Avodyne

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    Easier: first work out [L_i,x_j] and [L_i,p_j] using the explicit form of L_i. Then use the [A,BC] formula to get [L_i, x_j p_k].
     
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