Expectation value in momentum space

AI Thread Summary
The discussion centers on proving that the expectation value of momentum, <p>, can be expressed as <p>= ∫φ(p)*pφ(p)dp using the Fourier transform relationship between wave functions in position and momentum space. It is confirmed that this relationship holds for any operator O(p) due to the commutation with the exponential function e^(±ipr/ℏ). Participants discuss the challenges of sharing detailed proofs using LaTeX and the need to demonstrate the commutation of O(p) with the exponential operator. A hint is provided to consider the Maclaurin series for further understanding. The conversation emphasizes the applicability of the proof across different operators in quantum mechanics.
VVS2000
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Homework Statement
The problem was to show that given φ(p), and wave function Ψ(r), prove that <p>= ∫φ(p)*pφ(p)dp and <p²>=∫φ(p)*p²φ(p)dp
Relevant Equations
Ψ(r)=1/2πℏ ∫ φ(p)exp(ipr/ℏ)dp
<p>=∫Ψ(r)*pΨ(r)dr
so from Fourier transform we know that
Ψ(r)=1/2πℏ∫φ(p)exp(ipr/ℏ)dp
I proved that <p>= ∫φ(p)*pφ(p)dp from <p>=∫Ψ(r)*pΨ(r)dr
so will the same hold any operator??
 
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VVS2000 said:
Homework Statement:: The problem was to show that given φ(p), and wave function Ψ(r), prove that <p>= ∫φ(p)*pφ(p)dp and <p²>=∫φ(p)*p²φ(p)dp
Relevant Equations:: Ψ(r)=1/2πℏ ∫ φ(p)exp(ipr/ℏ)dp
<p>=∫Ψ(r)*pΨ(r)dr

so from Fourier transform we know that
Ψ(r)=1/2πℏ∫φ(p)exp(ipr/ℏ)dp
I proved that <p>= ∫φ(p)*pφ(p)dp from <p>=∫Ψ(r)*pΨ(r)dr
so will the same hold any operator??
Without seeing your work I can't really say if what you did would hold. But, yes, we can do this with any operator O(p). The usual method involves noting that ##e^{\pm ipr/ \hbar}## commutes with p. Any operator O(p) will commute with the exponentials, so just put that into your proof instead of p. (You may need to prove that you can do this. Think of how you prove that the exponential operator ##e^{\pm ipr/ \hbar}## commutes with p.)

-Dan
 
topsquark said:
Without seeing your work I can't really say if what you did would hold. But, yes, we can do this with any operator O(p). The usual method involves noting that ##e^{\pm ipr/ \hbar}## commutes with p. Any operator O(p) will commute with the exponentials, so just put that into your proof instead of p. (You may need to prove that you can do this. Think of how you prove that the exponential operator ##e^{\pm ipr/ \hbar}## commutes with p.)

-Dan
yeah, thing is I am still learning Latex to use it to type equations here, so It would be difficult to get in the whole proof. I tried uploading the pic of my work but there was some issue
yeah I think will try to prove why O(p) will commute
 
VVS2000 said:
yeah, thing is I am still learning Latex to use it to type equations here, so It would be difficult to get in the whole proof. I tried uploading the pic of my work but there was some issue
yeah I think will try to prove why O(p) will commute
Hint: Think Maclaurin series.

-Dan
 
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