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Angular momentum commutation relation, extra terms?

  1. Dec 18, 2014 #1
    1. The problem statement, all variables and given/known data
    What is the commutation relation between the x and y componants of angular momentum L = r X P

    2. Relevant equations
    None.

    3. The attempt at a solution
    I do r X p and get the angular momentum componants:


    [tex]L_{x} = (-i \hbar) (y \frac{d}{dz} - z \frac{d}{dy})[/tex]
    [tex]L_{y} = (-i \hbar) (z \frac{d}{dx} - x \frac{d}{dz})[/tex]
    [tex]L_{z} = (-i \hbar) (x \frac{d}{dy} - y \frac{d}{dx})[/tex]

    then when I attempt to put into the commutation relation [Lx,Ly] it comes out very complicated.

    My question:

    I found a derivation BUT it has extra terms in it (circled in red)

    9EVMLQo.jpg

    Why are there 5 terms when there should only be 4 when multiplying out 2 brackets? It's important because those 2 extra terms enable the correct solution. Or have I missed something?

    Thankyou for any help in advance.

    Source of derivation: http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect8.pdf
     
  2. jcsd
  3. Dec 18, 2014 #2

    Orodruin

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    There are four terms when you multiply the brackets. However, one of them can be rewritten with two terms, namely
    $$
    y\partial_z z \partial_x = y \partial_x + yz \partial_z \partial_x.
    $$
    This is just the product rule for differentiation.
     
  4. Dec 18, 2014 #3
    thats great, many thanks for your help!!
     
  5. Dec 19, 2014 #4
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