- #1

- 2

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter csutton1
- Start date

- #1

- 2

- 0

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

- #2

siddharth

Homework Helper

Gold Member

- 1,130

- 0

Could you be more specific? Proof of what?

- #3

- 2

- 0

- #4

siddharth

Homework Helper

Gold Member

- 1,130

- 0

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

Science Advisor

Homework Helper

- 4,795

- 22

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

gives the momentum of a state?

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

- #6

JK423

Gold Member

- 394

- 7

Assume that this relation for the mean values is valid:

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

Then substitute <x>=(Ψ,

<P>=(Ψ,

- #7

malawi_glenn

Science Advisor

Homework Helper

- 4,795

- 22

- #8

JK423

Gold Member

- 394

- 7

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

Science Advisor

Homework Helper

- 4,795

- 22

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.

Share: