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Simple check for operaters communting, think im missing something very obvious!

  1. May 28, 2009 #1
    1. The problem statement, all variables and given/known data

    I am supposed to find if the following commutes: [Lx,Ly]



    2. Relevant equations


    Lx= -i[tex]\hbar[/tex][y([tex]\partial/[/tex][tex]\partial[/tex]z) - z([tex]\partial/[/tex][tex]\partial[/tex]y)]

    Ly= -i[tex]\hbar[/tex][z([tex]\partial/[/tex][tex]\partial[/tex]x) - x([tex]\partial/[/tex][tex]\partial[/tex]z)]

    where [Lx,Ly]=LxLy-LyLx

    If it commutes then [Lx,Ly]=0

    3. The attempt at a solution

    [Lx,Ly]= (i[tex]\hbar[/tex])2{[y([tex]\partial/[/tex][tex]\partial[/tex]z) - z([tex]\partial/[/tex][tex]\partial[/tex]y)[z([tex]\partial/[/tex][tex]\partial[/tex]x) - x([tex]\partial/[/tex][tex]\partial[/tex]z)]}


    After expanding this I got a result of 0. So my solution concluded that they commute.

    The answer however is [Lx,Ly]= i[tex]\hbar[/tex]Lx

    I clearly expanded it wrong. I was hoping if anyone could explain how they expanded the LxLy-LyLx part. In my calculations I cannot seem to figure out how the answer contains a few more parts in the expansion which results in a non-commutation..

    Thanks!
     
  2. jcsd
  3. May 28, 2009 #2

    George Jones

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    Staff Emeritus
    Science Advisor
    Gold Member

    I think you mean [Lx,Ly]= i[tex]\hbar[/tex]Lz.

    Hint: Consider [itex]\left( L_x L_y - L_y L_x \right)f[/itex], where [itex]f[/itex] is an arbitrary function of the spatial coordinates. Take things one step at a time, don't forget about the product rule for differentiation, and remove [itex]f[/itex] at the end.
     
  4. May 28, 2009 #3
    Keep in mind that you're working with operators, so it's not just a matter of simple expansion. To calculate such a commutator, let the whole thing act on a function.

    [tex][L_x,L_y]f(x,y,z)[/tex]

    Since you have all sort of derivatives you will find that the product comes into a play a number of times. For instance, one term gives:
    [tex]y\frac{\partial}{\partial z} \left( z\frac{\partial}{\partial x}\right) f(x,y,z) =
    y \frac{\partial}{\partial x} f(x,y,z) + yz \frac{\partial^2}{\partial x\partialz}f(x,y,z)[/tex]

    One of these terms will be canceled by some other term, while the other one survives.

    EDIT: OK, George beat me to it ;)
     
  5. May 28, 2009 #4

    Cyosis

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    Homework Helper

    Another way to do it is to use the two following commutator identities:

    [tex]
    \begin{align*}
    &[A+B,C]=[A,C]+[B,C]
    \\
    &[A,BC]=[A,B]C+B[A,C]
    \end{align*}
    [/tex]
     
  6. May 28, 2009 #5
    Perfect it all makes sense now! You guys are great thanks!
     
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