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

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Discussion Overview

The discussion revolves around the momentum operator for a free particle that possesses definite momentum and energy. Participants explore whether this operator is equivalent to the general momentum operator and seek clarification on its application to wave functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions if the momentum operator for a free particle with definite momentum and energy is the same as the general momentum operator, suggesting it might be represented as -ih/2PI()*partial/partial_x.
  • This participant justifies their claim by stating that with definite momentum, the uncertainty in momentum (delta p) is zero, while the uncertainty in position (delta x) approaches infinity.
  • Another participant asserts that the operator itself does not depend on the wave function but will yield different results when applied to different wave functions.
  • A request for clarification is made regarding examples of wave functions that exhibit definite momentum and energy.
  • A specific wave function, \Psi(x,t)=A \exp [i(px - Et) / \hbar], is provided as an example to apply the momentum operator.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the momentum operator and its application, indicating that multiple competing views remain on the topic.

Contextual Notes

There are unresolved aspects concerning the application of the momentum operator to specific wave functions and the implications of definite momentum and energy on the operator's behavior.

Who May Find This Useful

This discussion may be useful for students and researchers interested in quantum mechanics, particularly those exploring the properties of operators and wave functions in the context of momentum and energy.

trelek2
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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:)
 
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The operator does not depend on the wave function, but will give a different result on different wave functions.
 
I don't understand what do you mean, can you give an example eg. the function with a definite momentum and energy?
 
[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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