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Quantum operators algebra

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi,guys. I have a hard time understanding algebra and tricks of operators.
    So i have few examples:
    1)[[itex]\hat{p}[/itex]2x,xn]
    2)[[itex]\hat{l}[/itex]z,x],where [itex]\hat{l}[/itex]z=x[itex]\hat{p}[/itex]y-y[itex]\hat{p}[/itex]x
    2. Relevant equations



    3. The attempt at a solution
    1)[[itex]\hat{p}[/itex]2x,xn]=
    [[itex]\hat{p}[/itex]x [itex]\hat{p}[/itex]x,xn]=
    [itex]\hat{p}[/itex]x[[itex]\hat{p}[/itex]x,xn]+[[itex]\hat{p}[/itex]x,xn][itex]\hat{p}[/itex]x
    So xn is not a operator,i dont know what i should do next?

    2)[[itex]\hat{l}[/itex]z,x],where [itex]\hat{l}[/itex]z=x[itex]\hat{p}[/itex]y-y[itex]\hat{p}[/itex]x
    Some tips or ideas here ?


    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 25, 2014 #2
    Hi.
    What makes you say that x is not an operator? You've probably seen the relation:
    [itex]\hat{x}[/itex]|ψ>= x|ψ> in position-eigenstates basis, so xn doesn't contain
    anything mysterious, it's just the operator x raised to the n-power (applied n times).
    Since you know already the identity: [A,BC] = B[A,C] + [A,B]C, and presumably the values of:
    [xi,pj] and [xi,xj] you should manage to get through the problem...
     
  4. Mar 12, 2014 #3
    Hello,again . Im looking now at 2. example.
    *will not write the operator sign ,its too time consuming .
    So [lz,x]=lzx-xlz=
    now can i simply multiply lz components by x and then i assume that lz components act on x?
    like this:
    =xpy(x)-ypx(x)-xlz ?

    p.s. the original goal of this example is to find comutator. I have seen that in this kind of examples are functions on which operators act ,for example,[lz,x]f(x). Maybe i need to introduce a function here ?
     
  5. Mar 12, 2014 #4
    Ok, if you don't know already the commutation relations between xi and pj then indeed i suggest you derive them by yourself, using a "dummy" function:
    [xi,pj]f(x,y,z) = xi{pjf(x,y,z)}–pj{xif(x,y,z)} = ...

    But once you know them, this problem doesn't require using a function anymore since, for instance:
    [lz,x]=[xpy,x] – [ypx,x] = x[py,x] + [x,x]py – y[px,x] – [y,x]px = ...
     
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