How to Calculate Momentum using QM Completeness Relation?

[luisgml_2000]

Hello!

From the ordinary courses in QM it is known that the momentum kets satisfy the completeness relation

\int d^3 p \mid p \rangle \langle p \mid = 1

Knowing this, how can you calculate

\int d^3 p \mid p \rangle p \langle p \mid

?

Thanks a lot!

 

[tom.stoer]

What do you want to calculate? It's an operator you can apply to bras and kets

I prefer

\int d^3p |p\rangle\langle p| p

 

[nnnm4]

What is the p outside the bra/ket exactly? Do you want it to be a momentum operator? or an eigenvalue for the p-ket? If it's the former I think you have an illegal product so there is no such thing as calculating it. If it's a eigenvalue (ie a real number) then it's just the identity operator multiplied by the scalar p.

 

[tom.stoer]

That's why I prefer

\int d^3p |p\rangle\langle p| \hat{p}^\dagger = \int d^3p |p\rangle\langle p| p

 

[ansgar]

operator times operator

 

[luisgml_2000]

What is the p outside the bra/ket exactly? Do you want it to be a momentum operator?

That p stands for the momentum operator. That's why I'm not sure about how to calculate it. I think the result has something to do with a delta function.

 

[tom.stoer]

your p cannot be an operator as it is in the wrong position, whereas my p (with the hat on top) is an operator; in the second step it is replaced by the eigenvalue as it acts (to the left) on the state vector