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Quick question about raising and lowering operators (ladder operators)

  1. Mar 14, 2013 #1
    Reading through my QM text, I came across this short piece on ladder operators that is giving me trouble (see picture). What I am struggling with is how to get to equations 2 and 3 from equation 1.

    Can someone point me in the right direction? Where does the i infront of the x go?
     

    Attached Files:

  2. jcsd
  3. Mar 14, 2013 #2

    tiny-tim

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    Hi PhysicsGirl90! :smile:
    It disappeared when they multiplied the whole thing by the constant i/√(ωh) :wink:
     
  4. Mar 15, 2013 #3
    Hey tiny-tim,

    Thanks for your suggestion. I tried it but i get stuck trying to get the same equation as the text. I have included what i got so far in the picture. Can you give it a look and tell me what I am doing wrong?
     

    Attached Files:

  5. Mar 15, 2013 #4

    tiny-tim

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    Hey PhysicsGirl90! :smile:

    (just got up :zzz:)

    the RHS of what you got is the same as in the text :wink:

    (apart from a constant, and you can always multiply an eigenvector by any constant!)
     
  6. Mar 15, 2013 #5

    DrClaude

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    This is a very sloppy "derivation" of the ladder operators. First of all
    is not correct, since [itex]\hat{x}[/itex] and [itex]\hat{p}[/itex] are operators. (By the way, it should be [itex]d/dx[/itex], not [itex]d/d\hat{x}[/itex].) You can get inspiration from what algebraic factoring would give, if these were ordinary variables, in order to investigate operators that look like the roots. But [itex]a_+[/itex] and [itex]a_-[/itex] are not "derived" this way.

    You can also notice that [tex]\hat{X} = \sqrt{\frac{m \omega}{\hbar}} \hat{x}[/tex] and [tex]\hat{P} = \frac{1}{\sqrt{m \hbar \omega}} \hat{p}[/tex] are dimensionless.
     
  7. Mar 15, 2013 #6
    Thanks again tiny-tim and thank you DrClaude for your help
     
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