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Ladder operators to find Hamiltonian of harmonic oscillator

  1. Mar 4, 2015 #1

    Maylis

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    Gold Member

    Hello, I was just watching a youtube video deriving the equation for the Hamiltonian for the harmonic oscillator, and I am also following Griffiths explanation. I just got stuck at a part here, and was wondering if I could get some help understanding the next step (both the video and book glanced over this part)

    Video Link

    The derivation starts at about 8:00

    First defining
    $$\hat {a}_{\pm} = \frac {1}{\sqrt{2 \hbar m \omega}}( \mp i \hat {p} + m \omega \hat {x})$$

    $$ \hat {a}_{-} \hat {a}_{+} = \frac {1}{2 \hbar m \omega}(i \hat {p} + m \omega \hat {x})(-i \hat {p} + m \omega \hat {x}) $$
    $$ = \frac {1}{2 \hbar m \omega}( \hat {p}^{2} + i m \omega \hat {p} \hat {x} - i m \omega \hat {x} \hat {p} + m^{2} \omega^{2} \hat {x}^{2}) $$
    $$ = \frac {1}{2 \hbar m \omega}( \hat {p}^{2} - i m \omega (\hat {x} \hat {p} - \hat {p} \hat {x}) + m^{2} \omega^{2} \hat {x}^{2})$$
    And I know the commutator ##[\hat {x}, \hat {p}] = i \hbar##
    $$ = \frac {1}{2 \hbar m \omega}( \hat {p}^{2} - i m \omega ( i \hbar) + m^{2} \omega^{2} \hat {x}^{2} )$$
    At this point both the video and Griffiths stop, although each do something different
    I have no idea how Griffiths goes from this
    upload_2015-3-5_0-11-14.png
    To this
    upload_2015-3-5_0-14-12.png
    I will continue with what I was doing,
    $$ = \frac {1}{2 \hbar m \omega}( \hat {p}^{2} + m \omega \hbar + m^{2} \omega^{2} \hat {x}^{2})$$
    $$ = \frac {1}{2 \hbar m \omega}\hat {p}^{2} + \frac {1}{2 \hbar^{2} m^{2} \omega^{2}} + \frac {m \omega \hat {x}^{2}}{2 \hbar} $$
    And from here I am not sure how both the video and Griffiths conclude
    upload_2015-3-5_0-14-34.png
    How do I get to the answer from my steps? Since I am not skipping..
     

    Attached Files:

    Last edited: Mar 4, 2015
  2. jcsd
  3. Mar 4, 2015 #2
    There is an i too much in the term in the middle


    The term in the middle is again wrong...



    Basically the last two equations you wrote before the Griffiths insert are wrong.
    Once you have them right, you know that the Hamiltonian is
    [tex]H=\dfrac{p^2}{2 m} + \dfrac{1}{2} m \omega^2 x^2[/tex]
     
  4. Mar 4, 2015 #3

    Maylis

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    Gold Member

    Woops, I caught my error and now I am able to see how it was done. Thanks.
     
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