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A calculation involving differential operators

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data
    I have the following expressions for angular momentum components: [itex]L_1 = x_2\frac{\partial}{\partial x_3} - x_3\frac{\partial}{\partial x_2}[/itex], [itex]L_2 = x_3\frac{\partial}{\partial x_1} - x_1\frac{\partial}{\partial x_3}[/itex], [itex]L_3 = x_1\frac{\partial}{\partial x_2} - x_2\frac{\partial}{\partial x_1}[/itex], and I simply need to work out [itex]L^2 = L_1^2 + L_2^2 + L_3^2[/itex].

    2. Relevant equations
    N/A

    3. The attempt at a solution
    Well, the way I expand it gives [tex]L_1^2 = (x_2\frac{\partial}{\partial x_3} - x_3\frac{\partial}{\partial x_2})(x_2\frac{\partial}{\partial x_3} - x_3\frac{\partial}{\partial x_2}) = x_2\frac{\partial}{\partial x_3}x_2\frac{\partial}{\partial x_3} - x_2\frac{\partial}{\partial x_3}x_3\frac{\partial}{\partial x_2} - x_3\frac{\partial}{\partial x_2}x_2\frac{\partial}{\partial x_3} + x_3\frac{\partial}{\partial x_2}x_3\frac{\partial}{\partial x_2} = -x_2\frac{\partial}{\partial x_2} - x_3\frac{\partial}{\partial x_3},[/tex] and similarly [tex]L_2^2 = -x_1\frac{\partial}{\partial x_1} - x_3\frac{\partial}{\partial x_3}[/tex] and [tex]L_3^2 = -x_1\frac{\partial}{\partial x_1} - x_2\frac{\partial}{\partial x_2},[/tex] so that [tex]L^2 = -2x_1\frac{\partial}{\partial x_1} - 2x_2\frac{\partial}{\partial x_2} - 2x_3\frac{\partial}{\partial x_3}[/tex] But this is not the expression for [itex]L^2[/itex] that I'm supposed to get! So I must be doing something wrong. Can anyone help?
     
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  3. Sep 7, 2011 #2

    Dick

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    How can the derivative expression for L_1^2 not have any second derivatives in it?
     
  4. Sep 7, 2011 #3
    That's a good question. I guess it should. When I simplify a term like [itex]x_2\frac{\partial}{\partial x_3}x_2\frac{\partial}{\partial x_3}[/itex], do I have to use the product rule? That would get the second derivatives.
     
  5. Sep 7, 2011 #4

    Dick

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    Yes, you do! I see what you were doing. E.g. d/dx(x*d/dy)f=d/dx(x*df/dy). Sure, it's the derivative of a product. Not the same as (dx/dx)df/dy.
     
    Last edited: Sep 7, 2011
  6. Sep 7, 2011 #5

    lanedance

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    also, hopefully not to distract from teh question at hand, but do you know about index notation with einstein summation convention? Definitely worth learning and wiould simplify this problem

    it that notation you could write the problem as:
    [tex] L_i = \epsilon_{ijk}x_j \frac{\partial}{\partial x_k}[/tex]

    [tex] L^2 = L_i-L_j \delta_{ij}= L_iL_i [/tex]
     
  7. Sep 8, 2011 #6

    HallsofIvy

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    And that is why we always use the "comutator", LM- ML, which will NOT have second differential operators.
     
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