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Seeking a phase angle operator for the QHO

  1. Nov 29, 2015 #1
    According to Daniel Gillespie in A Quantum Mechanics Primer (1970),

    " . . . any observable which in classical mechanics is some well behaved function of position and momentum, f(x,p), is represented in quantum mechanics by the operator [itex] f ( \hat{x} , \hat {p} ) [/itex]. That is,


    a = f (x,p) . . . implies . . . \hat{a} = f ( \hat{x} , \hat {p} ) = f ( x , -i \hbar \frac {d}{dx}) ."


    Apparently this works for finding the angular momentum operators, for example in classical mechanics


    L_z = xp_y - yp_x


    and in quantum mechanics


    \hat{L}_z = \hat{x} \hat{p_y}- \hat{y} \hat{p_x} =

    x(-i\hbar \frac {\partial}{\partial y}) - y(-i\hbar \frac {\partial}{\partial x} )= -i\hbar (x \frac {\partial}{\partial y} - y \frac {\partial}{\partial x})


    Now I am wondering if this idea can be applied to the harmonic oscillator. Specifically, since phase angle


    \theta = arc tan ( \frac {p}{ x \sqrt {km} } )


    (1) can I make a quantum phase angle operator by replacing the p and x above with their corresponding quantum operators?


    (2) what are the phase angle eigenvalues for a QHO?

    Last edited: Nov 29, 2015
  2. jcsd
  3. Dec 3, 2015 #2
    a bump and a follow-up:

    Last night I worked out that for the classical harmonic oscillator, if we define


    N=H/ \hbar \omega


    then the Poisson bracket between N and phase angle [itex] \theta [/itex] is [itex] 2 / \omega [/itex].

    Is there any connection between this and the photon number - angle phase uncertainly relation?
  4. Dec 3, 2015 #3
  5. Dec 3, 2015 #4
    Thanks AndresB, looks very interesting. Will try to take it in this weekend when I have a few free minutes.
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