# Might be a basic question

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!

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## Answers and Replies

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tom.stoer
Science Advisor
What do you want to calculate? It's an operator you can apply to bras and kets

I prefer

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

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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
Science Advisor
That's why I prefer

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

operator times operator

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
Science Advisor
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