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Understanding Ladder Operators

  1. Mar 12, 2010 #1
    Calculate [Lz,L+]

    By defintion ladder operators are:

    Important Relations:
    LxLy = i[tex]\hbar[/tex]Lz, LyLz = i[tex]\hbar[/tex]Lx, LzLx = i[tex]\hbar[/tex]Ly

    Lx = ypz - zpy, Ly = xpz - zpx, Lz = xpy - ypx

    To start solving;
    Lz - (Lx + iLy) = 0
    Multiply through by [tex]\hbar[/tex]:
    [tex]\hbar[/tex]Lz - [tex]\hbar[/tex]Lx + i[tex]\hbar[/tex]Ly

    The i[tex]\hbar[/tex]Ly is equal to LzLx. From this point
    I've tried varying approaches in attempt to cancel variable out, but have failed. I have a feeling this problem can be solved easier. Should I try to use spherical coordinates instead of Cartesian? From trying to figure this out I have stumbled upon the answer but I would like to know how to produce the answer.

    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 12, 2010 #2


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    This is not even close to correct ... where did you get that zero from? Where are the products of the bracketed operators?

    The above is not correct ... the correct relation is:

    [tex]i\hbar L_{y}=[L_{z},L_{x}][/tex]

    where the bracketed term on the right is the commutator of Lz and Lx. (the initial thing you are solving for is also a commutator).

    Do you know how to expand a commutator? Do you understand it's significance in QM? That is the place to start on this problem.
    Last edited: Mar 12, 2010
  4. Mar 12, 2010 #3
    Didn't even recognize that, thanks.

    Starting over using the correct commutator rules: AB - BA = 0
    LzL+-L+Lz=0 replacing L+ with Lx+iLy and multiplying the Lz through

    => LzLx+iLzLy-LxLz-iLyLz

    Could you show me where you found the relation you used. I found a very similar one with the hbar in it.
  5. Mar 12, 2010 #4


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    Yeah ... mine should have an hbar in it too ... sorry about that .. it is fixed now.
  6. Mar 12, 2010 #5


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    AB-BA does not equal 0 in general. It only equals zero if A and B commute. Otherwise AB-BA=[A,B] that you calculated earlier.
  7. Mar 12, 2010 #6
    Does LzLx = LxLz ? I want to know if the relations hold true for both of them or if I have to add a negative sign (or something of that nature). Don't angular momentums commute?
  8. Mar 13, 2010 #7
    Thanks for the clarification. It's a measurement of the extent to which they do NOT commute, correct? I do not see where a non-zero commutator would be useful though.
  9. Mar 13, 2010 #8


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    No, the different components of angular momentum along the cartesian axes most certainly do not commute. You can verify this for your self using the definition of angular momentum in terms of the cross product of the position and momentum vectors. Just replace the classical momenta with the corresponding QM operators and you will see that in general:

    [tex][L_{a},L_{b}]=\pm i\hbar L_{c}[/tex]

    where a, b and c are all different Cartesian axes. The sign will change based on the ordering of the of the axes ... if you have them ordered as xyzxyzxyz... it will be positive, if you have them ordered as zyxzyxzyx... it will be negative. (This can be stated very simply using a symbol called the Levi-Civita symbol, but that is a side issue .. google it if you are interested)

    Good gracious! Haven't you studied the Heisenberg uncertainty principle yet? Quite simply, much of the weirdness of QM can be traced back to the fact that the properties of quantum states must inevitably give rise to non-commuting operators. The standard HUP formulation in terms of position and momentum arises because those two operators don't commute.
  10. Mar 13, 2010 #9
    So that's where that comes from ;) I'm having trouble identifying all the equations from very similar ones. You seem to have a keen grasp of QM, where did you learn from? Some kind of organization in this maelstrom of QM ideas and variables would be nice...

    I've seen similar ideas but never a name for it. Thanks for that.

    Ah sorry, I didn't realize HUP was a commutation. I simply memorized the formula.
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