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Momentum operator for a free particle with a definite momentum and Energy

  1. Feb 21, 2009 #1
    Hi!

    I need an explanation:
    Is the momentum operator for a free particle with a definite momentum and energy the same as what we know as the momentum operator in general?

    Is it just -ih/2PI()*partial/partial_x?

    With the justification that since the momentum is definite, delta p is 0, but delta x must go to infinity so differentiating wrt. x gives

    (momentum_operator)*wave_func.=momentum*wave func?

    If anyone could clarify I'd be grateful:)
     
  2. jcsd
  3. Feb 21, 2009 #2

    clem

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    Science Advisor

    The operator does not depend on the wave function, but will give a different result on different wave functions.
     
  4. Feb 21, 2009 #3
    I don't understand what do you mean, can you give an example eg. the function with a definite momentum and energy?
     
  5. Feb 21, 2009 #4

    jtbell

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    Staff: Mentor

    [tex]\Psi(x,t)=A \exp [i(px - Et) / \hbar][/tex]

    Apply the momentum operator to it and see what you get. :smile:
     
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