Might be a basic question

  • #1
Hello!

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

[tex] \int d^3 p \mid p \rangle \langle p \mid = 1[/tex]

Knowing this, how can you calculate

[tex] \int d^3 p \mid p \rangle p \langle p \mid [/tex]

?

Thanks a lot!
 
Last edited:

Answers and Replies

  • #2
tom.stoer
Science Advisor
5,766
161
What do you want to calculate? It's an operator you can apply to bras and kets

I prefer

[tex]\int d^3p |p\rangle\langle p| p [/tex]
 
Last edited:
  • #3
113
0
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.
 
  • #4
tom.stoer
Science Advisor
5,766
161
That's why I prefer

[tex]\int d^3p |p\rangle\langle p| \hat{p}^\dagger = \int d^3p |p\rangle\langle p| p [/tex]
 
  • #5
505
0
operator times operator
 
  • #6
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.
 
  • #7
tom.stoer
Science Advisor
5,766
161
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
 

Related Threads on Might be a basic question

Replies
13
Views
6K
Replies
29
Views
6K
Top