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

[trelek2]

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:)

 

[Meir Achuz]

The operator does not depend on the wave function, but will give a different result on different wave functions.

 

[trelek2]

I don't understand what do you mean, can you give an example eg. the function with a definite momentum and energy?

 

[jtbell]

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

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