Expectation value in momentum space

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SUMMARY

The discussion centers on proving the expectation values in momentum space using Fourier transforms. The equation Ψ(r)=1/2πℏ∫φ(p)exp(ipr/ℏ)dp is established, leading to the proof that

=∫φ(p)*pφ(p)dp and =∫φ(p)*p²φ(p)dp. Participants confirm that this proof can be generalized to any operator O(p) by utilizing the commutation property of the exponential operator e^{±ipr/ℏ} with momentum p. The discussion emphasizes the importance of proving the commutation relation for a comprehensive understanding.

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  • Fourier Transform in Quantum Mechanics
  • Expectation Values in Quantum Mechanics
  • Commutation Relations of Operators
  • Basic LaTeX for Equation Formatting
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  • Study the properties of Fourier Transforms in quantum mechanics
  • Learn about the commutation relations of quantum operators
  • Explore the proof of expectation values for various operators
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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
 
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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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