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Momentum operator in Schrodinger equation

  1. Apr 16, 2008 #1
    Why momentum is replaced by momentum operator in Schrodinger equation ?
  2. jcsd
  3. Apr 16, 2008 #2


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  4. Apr 16, 2008 #3


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    A key basic idea in QM is that the commutator of p(momentum) and q(position),

    qp - pq = i (The tradition in high energy physics is to use units in which c=1, and h bar=1)

    For his equation to hold, both q and p must be operators. Note also that this commutator is responsible for the Heisenberg Uncertainty Principle.

    In fact this is an assumption, but a good one, and as basic as it gets. You would benefit from a bit of homework, as suggested by malawi_glenn above.
    Reilly Atkinson
  5. Apr 17, 2008 #4
    So we have a set of quantum rule or postulate that can not be derived. That will be a bit strange, because then where did Schrodinger get his equation?
  6. Apr 17, 2008 #5
    He took one part de Broglie and one part Einstein and mixed it together... on a serious note, [tex]p \mapsto -i \hbar \tfrac{\partial}{\partial x}[/tex] follows from de Broglie's work and is based on analogies between classical mechanics of particles and waves.
    Last edited: Apr 17, 2008
  7. Apr 18, 2008 #6


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    That is not strange, every field has its postulates (axioms) that can't be derived nor be prooved.

    However, you may postulate something, then from those axioms derive formulas and relations which can be verified and thus one can tell if you can build up a coherent system (a paradigm) or not.

    Here you can read Schrodingers first publication about his new equation:

    Hanve fun :-)
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