Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ladder operators and the momentum and position commutator

  1. Jul 3, 2014 #1
    When using Fourier's trick for determining the allowable energies for stationary states, Griffiths introduces the a+- operators. When factoring the Hamiltonian, the imaginary part is assigned to the momentum operator versus the position operator. Is there a reason for this? If :

    a-+ = k(ip + mwx)(-ip + mwx), and the commutator is (xp-px), Is

    a+- = k(-ip = mwx)(ip + mwx) ?

    If so, is the commutator (px-xp)?

    Thanks in advance for your input.
  2. jcsd
  3. Jul 4, 2014 #2
    Hi kmchugh,

    The Fourier transform of the nabla (or del) operator and partial time derivative operator is [itex] \ \ \ \ \hat F(\nabla) = ik \ \ \ \ \ [/itex] [itex]\hat F(\frac{\partial}{\partial t}) = -i\omega[/itex]

    Where spatial variables [itex]x, y, z[/itex] are transformed into the wavenumber vector [itex]k[/itex] and the time variable is transformed into the angular frequency scalar [itex]\omega[/itex]

    The first order nabla operator is associated with momentum and its transform contains [itex]i[/itex], meaning that it is an imaginary value when compared to the phase of position variables [itex]x, y, z[/itex].
    Last edited: Jul 4, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook