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Can anyone help? in quantum mechanics commutator prove [L^x,L^y] = ihL^z

  1. Feb 25, 2007 #1
    can anyone help?? in quantum mechanics commutator prove [L^x,L^y] = ihL^z

    given

    :L^x =(y^(pz)^-z^(py)^)
    :L^y =(z^(px)^-x^(pz)^)
    :L^z =(x^(py)^-y^(px)^) where ^ is just showing its operator


    prove comutator [L^x,L^y] = ihL^z

    I am swamped at every hurdle and cant seem to get my head around this question to find the answe of ihL^z . any help would be very much appreciated:surprised
     
  2. jcsd
  3. Feb 25, 2007 #2
    You're aware of the form of the commutation relation, correct?

    [tex][\hat{L}_x,\hat{L}_y]\psi = \hat{L}_x (\hat{L}_y\psi) - \hat{L}_y (\hat{L}_x\psi) [/tex]

    Solve this for the angular momentum operators [tex]\hat{L}_x[/tex] and [tex]\hat{L}_y[/tex] - which you know. The result should cancel down to the form of the angular momentum operator [tex]\hat{L}_z[/tex].
     
  4. Feb 25, 2007 #3

    dextercioby

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    I like the nice, geometric proof of the commutation relations for angular momentum that Sakurai, for example, gives in his delightful book. Unfortunately not mathematically rigorous, but very well looking.
     
  5. Feb 25, 2007 #4
    yeah im aware commutator rules but i just cant prove the relation ,i am confused about how to rearange and put in that form. i.e my maths might not be up to scratch but if u can help me with this little proof i might be able to crack the rest... here it goes ...
    z(py)x(pz) - x(pz)z(py) ,how do i break this down mathematically without breaking any rules to get ih{x(py)} if [z,pz] = ih..
     
  6. Feb 25, 2007 #5

    nrqed

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    Write

    [tex] [ L_x, L_y] = [Y P_z - Z P_y, Z P_x - X P_z ] [/tex]

    and then use the fact that the commutator of sums is equal to the sum of commutators

    [tex] = [Y P_z, Z P_x] - [Z P_y, Z P_x] - [Y P_z , X P_z] + [ Z P_y , X P_z] [/tex]

    Now do each of those commutators.

    (My favorite trick is to use directly [AB,CD] = A[B,C]D + B[A,D]C + AC[B,D] +[A,C] DB which can be proven by simply expanding or starting from the simpler and obvious [AB,C] = A[B,C] + [A,C] B )

    Patrick
     
  7. Feb 25, 2007 #6
    Lx=yPz-zPy
    Ly=zPx-XPz
    Pk=ih d/dk use these relations, careful derivatives and then eliminate.
    and result is [Lx,Ly]=ihLz .
     
  8. Feb 25, 2007 #7
    thanks patrick , do i use [AB,CD] relation for each 4 commutators?,like [YPz,ZPx] and also [ZPy,ZPx] etc
     
  9. Feb 25, 2007 #8

    nrqed

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    Yes. And you use [itex] [X_i, X_j]= [P_i,P_j] =0 [/itex] and [itex] [X_i,P_j] = i \hbar \delta_{ij} [/itex]. you'll see, you will get [itex] i \hbar L_z [/itex].

    You are welcome

    Patrick
     
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