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Commutator relations

  1. Jan 16, 2014 #1
    I'm doing something horribly wrong in something that should be very easy. I want to show that:

    [itex][L^2, x^2] = 0[/itex]

    So:

    [itex][L^2, x x] = [L^2, x] x + x [L^2, x][/itex]

    [itex]L^2 = L_x^2 + L_y^2 + L_z^2[/itex]

    Therefore: [itex][L^2, x] = [L_x^2 + L_y^2 + L_z^2, x] = [L_x^2, x] + [L_y^2, x] + [L_z^2, x][/itex]

    = [itex]L_y [L_y, x] + [L_y, x] L_y + L_z [L_z, x] + [L_z, x] L_z[/itex]

    [itex]= -i h L_y z - ih z L_y + i h L_z y + i h y L_z[/itex]

    So

    [itex][L^2, x x] = -i h L_y z x - ih z L_y x + i h L_z y x + i h y L_z x
    + -i h x L_y z - ih x z L_y + i h x L_z y + i h x y L_z[/itex]

    And now?

    This ought to be a lot easier.
     
  2. jcsd
  3. Jan 16, 2014 #2

    strangerep

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    Science Advisor

    Where did this exercise come from? I suspect you misunderstand what ##x^2## means here, but I can't be sure without seeing the original unedited question.

    (BTW, you're supposed to use the homework template when posting in this forum -- else you risk attracting the Wrath of the Mentors.)
     
  4. Jan 16, 2014 #3
    Ughy. You are right. I looked it up again and it's not actually x^2 as in "position operator squared" but X^2 = x^2 + y^2 + z^2.

    I didn't sleep for 29 hours now, working for some exams. I'm slightly confused. My bad. Next time I'll use the exact template. :redface:
     
  5. Jan 16, 2014 #4

    strangerep

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    Science Advisor

    This is a really bad technique. You are probably more confused than you realize, maybe even approaching a delirium state. Non-sleeping is counter-productive. Things that take ages to understand or perform while tired tend to be much quicker when you're fresh.

    Go and sleep (even if just for a few hours -- set the alarm clock accordingly).
     
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