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

1. Feb 21, 2009

### 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:)

2. Feb 21, 2009

### clem

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

3. Feb 21, 2009

### trelek2

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

4. Feb 21, 2009

### Staff: Mentor

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

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

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