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In QM operators, why [YPz,YPx]=0, [ZPy,ZPy]=0 [X,Py]=0 [Y,Pz]=0, etc

  1. Nov 14, 2007 #1
    in the quantum mechanical operators :
    why :
    [YPz,YPx]=0
    [ZPy,ZPy]=0
    [X,Py]=0
    [Y,Pz]=0
    [Z,Py]=0
     
  2. jcsd
  3. Nov 14, 2007 #2

    CompuChip

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    For the second:
    What is the commutator of any operator with itself?

    For the last three:
    Can you prove that [X Px] = i h ?
    Now do the same argument for [X Py].

    For the first one:
    Reduce it to Y^2 [Pz Px] using that [Y Pz] = [Y Px] = 0 (which you have just proven) and calculate [Pz Px] in the same way.
     
    Last edited: Nov 14, 2007
  4. Nov 14, 2007 #3
    Isn't [X Px] = i h an axiom, CompuChip?
     
  5. Nov 14, 2007 #4

    OOO

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    Usually one thinks of this as something that nature implements (through de Broglie's relation). If one accepts it as a mathematical axiom a priori then the question "why?" wouldn't make much sense indeed.
     
  6. Nov 14, 2007 #5
    thank
     
  7. Nov 15, 2007 #6

    CompuChip

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    Hmm, yes it is usually taken as an axiom isn't it.
    But actually the way I first learned it was: the p-operator in x-space can be written as [tex]- i \hbar \partial/\partial x[/tex]
    and from this you can show that (indeed)
    [tex][x p_x] f = - i \hbar x \partial f/\partial x + i \hbar \partial/\partial x (x f) = \cdots = i \hbar f[/tex]
    and it was this reasoning that I wanted muydos to continue to [X Py] (which would then of course give 0).
     
  8. Nov 16, 2007 #7
    [Y,Px] = YPx-PxY
    PxY=
    [tex]P_x Y f(x) = i \hbar \partial /\partial x Y f(x)[/tex]
    [tex]P_x Y f(x) = i \hbar Y \partial f/\partial x + i \hbar \partial Y/\partial x f(x)[/tex]
    YPx=
    [tex]Y P_x f(x) = i \hbar Y \partial f/\partial x [/tex]

    [Y,Px] = YPx-PxY
    [tex]= i \hbar Y \partial f/\partial x - i \hbar Y \partial f/\partial x - i \hbar \partial Y/\partial x f(x) [/tex]

    [Y,Px] = [tex] - i \hbar \partial Y/\partial x f(x) [/tex]
    but
    [Y,Px] = 0

    [tex]- i \hbar \partial Y/\partial x f(x) = ?[/tex] is thes =0 . why?
     
  9. Nov 16, 2007 #8
    why [Y,Px] = 0

    [Y,Px] = YPx-PxY
    PxY=
    [tex]P_x Y f(x) = i \hbar \partial /\partial x Y f(x)[/tex]
    [tex]P_x Y f(x) = i \hbar Y \partial f/\partial x + i \hbar \partial Y/\partial x f(x)[/tex]
    YPx=
    [tex]Y P_x f(x) = i \hbar Y \partial f/\partial x [/tex]

    [Y,Px] = YPx-PxY
    [tex]= i \hbar Y \partial f/\partial x - i \hbar Y \partial f/\partial x - i \hbar \partial Y/\partial x f(x) [/tex]

    [Y,Px] = [tex] - i \hbar \partial Y/\partial x f(x) [/tex]
    but
    [Y,Px] = 0

    [tex]- i \hbar \partial Y/\partial x f(x) = ?[/tex] is thes =0 . why?
     
  10. Nov 16, 2007 #9

    ZapperZ

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    dY/dx is zero because this "Y" is a position operator for the y coordinate. In other words, one can think of "Y" as a function of only y, and no other variable. Thus, its derivative with respect to other variable, such as x, is zero.

    Zz.
     
  11. Nov 16, 2007 #10
    thak you
     
  12. Nov 16, 2007 #11
    physically..if certain obsevables commute, there exist states for which they all have particular values..(there exist eigenstate of p_x, y, |p_x`>|y`> having their eigenvalues p_x` and y` respectively..) So a mesurement of p_x, y would give only particular value p_x`, y` respectively..
     
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