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Four momentum

  1. Aug 17, 2007 #1
    can the 4 momentum (since it is a vector) be considered an operator?
     
  2. jcsd
  3. Aug 17, 2007 #2

    jtbell

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    Staff: Mentor

    Please explain your question. Why do you think there might be a connection with it being a vector? In non-relativistic QM, there are operators for both vector quantities (e.g. momentum and position) and scalar quantities (e.g. energy). In relativistic QM, it's similar, except that now we talk about four-vectors instead of three-vectors.
     
    Last edited: Aug 17, 2007
  4. Aug 17, 2007 #3
    I'm usually not fanatic about precise meaning of terminology, but isn't the term "scalar" fixed to mean invariant quantities quite unanimously?
     
  5. Aug 17, 2007 #4

    jtbell

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    In a relativistic context, yes. I was thinking of the non-relativistic context in which energy is a scalar, and thereby mixing relativistic and non-relativistic contexts. Oops.

    (and of course the 4-vector momentum operator in effect includes energy as the zero-th component)
     
  6. Aug 17, 2007 #5
    Yes, the 4-vector of energy-momentum has components [itex] H, P_x, P_y, P_z [/itex]. In quantum mechanics they are represented by operators of energy (or Hamiltonian) H and momentum [itex] \mathbf{P}[/itex].
     
  7. Aug 17, 2007 #6

    jtbell

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    I now realize that there are two interpretations of your question:

    (a) Can the 4-momentum be considered an operator because it is a vector? That is, does its being a vector imply that it is also an operator?

    or

    (b) Can the 4-momentum be considered an operator even though it is a vector? That is, does its being a vector prevent it from also being an operator?

    This kind of ambiguity is why it's important not to be too terse in your questions. It's especially important if English is not your native language, because it's difficult even for native speakers to be both terse and precise at the same time! :frown:
     
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