- #1

csutton1

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Does anyone have a momentum operator proof? The book I'm using is skipping a lot of steps .

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- Thread starter csutton1
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- #1

csutton1

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Does anyone have a momentum operator proof? The book I'm using is skipping a lot of steps .

- #2

siddharth

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Could you be more specific? Proof of what?

- #3

csutton1

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- #4

siddharth

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Yes, what about it? I don't understand what proof you have in mind. Maybe, if you post the exact problem statement, and where you are stuck, it'll be easier to help.

- #5

malawi_glenn

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[tex] -i\hbar \nabla [/tex]

gives the momentum of a state?

Be specific, and write down for us where you get stucked.

- #6

JK423

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Assume that this relation for the mean values is valid:

<P>=m*d<x>/dt.

Then substitute <x>=(Ψ,

<P>=(Ψ,

- #7

malawi_glenn

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- #8

JK423

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I mentioned a way to calculate momentum`s operator in {x} representation from a known classical equation. What`s wrong about that?

What do you mean by

"You want to show that the p operator is the generator of translations and that it fulfills certain commutation relations" ?

You can show these these things without knowing what operator p equals to?

- #9

malawi_glenn

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but is not a proof, as you said.

but I know what p equals, so I can proove that it is the generator of translation and fulfills certain commutator relations. I did not say that i was about to derive the momentum operator, which is done by doing the steps backwards. I.e I demand translations symmetry and that it fulfills certain commutator relations.

It is quite simple.

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