Dealing with normalized quantum functions

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SUMMARY

The discussion focuses on the normalization of quantum wave functions, specifically addressing the equation <∅|∅>=1=<∅|n>. Participants clarify that the correct interpretation involves summing over all basis states, leading to the identity matrix representation, Ʃn|n> PREREQUISITES

  • Understanding of quantum mechanics and wave functions
  • Familiarity with normalization conditions in quantum states
  • Knowledge of basis states and completeness relations
  • Basic matrix algebra and identity matrices
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  • Study the concept of completeness in quantum mechanics
  • Explore the role of basis states in quantum wave function expansion
  • Learn about the properties of identity matrices in quantum systems
  • Investigate normalization techniques for non-eigenfunction wave functions
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Jimmy25
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Homework Statement



If <∅| is normalized, show that:

<∅|∅>=1=<∅|n><n|∅>

(where ∅ is a non-eigenfunction wave function composed of Ʃc(n)ψ(n).

Homework Equations





The Attempt at a Solution



I can show that <∅|∅>=Ʃc*(n)c(n) (=1). But the next part of the question asks to use your proof to show Ʃc*(n)c(n)=1 so that's not the way it is supposed to be done. I feel like I'm missing something very simple.
 
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Jimmy25 said:

Homework Statement



If <∅| is normalized, show that:

<∅|∅>=1=<∅|n><n|∅>

(where ∅ is a non-eigenfunction wave function composed of Ʃc(n)ψ(n).

It seems to me that the problem statement must be


<∅|∅>=1=Ʃn<∅|n><n|∅>.

This is just the usual expansion over basis states. What you wrote can't be true for arbitrary n; it only works when you sum over all the basis states.

In matrix language, Ʃn|n><n| is just the identity matrix.

BBB
 
Aha that was what I suspected.

With the correction, is there a way to prove that it is equal to one without having to use that Ʃc*(n)c(n) (=1)?
 
Jimmy25 said:
Aha that was what I suspected.

With the correction, is there a way to prove that it is equal to one without having to use that Ʃc*(n)c(n) (=1)?

I would just say Ʃn|n><n| = I, the identity matrix, because that is the definition of a complete set of states {|n>}. Then

<θ|I|θ> = <θ|θ> =1

because |θ> is normalized. This is not the complicated part of quantum mechanics.

BBB
 

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