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Homework Help: Commutation relationsl, angular momentum

  1. May 5, 2010 #1
    1. The problem statement, all variables and given/known data
    calculate the following commutation relations
    [tex][L_{x}L_{y}]=[/tex]
    [tex][L_{y}L_{z}]=[/tex]
    [tex]][L_{z}L_{x}]=[/tex]




    2. Relevant equations

    [tex] [L_{x},L_{y}]= -\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}][/tex]
    where the expression in the parentheses describes [tex] L_{z}[/tex]
    but i still have [tex] \hbar^2[/tex] can someone explain to me how we get [tex] i \hbar L_{z}[/tex] where did the [tex] \hbar^2[/tex] go? And how did we get i back again?

    here is my calculation
    [tex][L_{x},L_{y}]= L_{x}L_{y}-L_{y}L_{x}=-\hbar^2[(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})-(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})][/tex]


    here is my second commutation
    [tex] [L_{z},L_{x}]= [/tex] in this one i get plus instead of minus......do not know where i make mistake...
    [tex][L_{z},L_{x}]= }=-\hbar^2[(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})-(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})]= -\hbar^2(x\frac{\partial}{\partial z}+z\frac{\partial}{\partial x})[/tex]
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. May 5, 2010 #2
    Ummm, yeah, you're missing the ihbar because you haven't equated the
    [tex]
    [L_{x},L_{y}]= -\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]
    [/tex]
    part to how Lz is supposed to look. You'll hit your head as you notice that to get to Lz you need a factor of ihbar. Same for the others. :)
     
  4. May 5, 2010 #3
    hm i still do not get it.......first it dissapears and then suddenly comes upp again....
    and why in the last commutation the sign is wrong?
     
  5. May 5, 2010 #4

    lanedance

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    Homework Helper

    [tex] [L_{x},L_{y}]
    = -\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]
    = (i^2)\hbar^2[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]
    = (i\hbar) \left(i\hbar[y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}]
    \right)

    [/tex]
     
  6. May 5, 2010 #5

    vela

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    Staff Emeritus
    Science Advisor
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    Education Advisor

    That's your mistake. You're forgetting the factor of [itex]i\hbar[/itex].

    [tex]L_{z} = i\hbar \left(y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}\right) \ne y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}[/tex]
    We can't see where you made the mistake either since you didn't show your work. The set-up looks fine, though.
     
  7. May 6, 2010 #6
    if you dont want things to get messy try to start from the fact that L= rxp
    then, Lx= yp(z) - zp(y) .. Ly= zp(x) - xp(z) .. Lz = xp(y) - yp(x)

    also make use of : [x,y]=0, [x,z]=0, [y,z]=0, [px,y]=0, [px,z]=0, [py,x]=0, [py,z]=0, [pz,x]=0, [pz,y]=0, [px,x]=-ih-par, [py,y]=-ih-par, [pz,z]=-ih-par ..
     
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