Dealing with normalized quantum functions

In summary: It is important to note that this proof only works if the wave function is composed of a complete set of states {|n>}. If this is not the case, then the proof would not hold and the equation <∅|∅>=1=Ʃn<∅|n><n|∅> would not be true.In summary, the conversation discusses the proof of the equation <∅|∅>=1=Ʃn<∅|n><n|∅> if <∅| is normalized. It is noted that the proof only works if the wave function is composed of a complete set of states.
  • #1
Jimmy25
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0

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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  • #2
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
 
  • #3
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)?
 
  • #4
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
 

1. What is a normalized quantum function?

A normalized quantum function is a mathematical description of a quantum system that has been adjusted to have a total probability of 1. This means that the sum of the squared amplitudes of all possible outcomes is equal to 1, making it a valid probability distribution.

2. Why is it important to deal with normalized quantum functions?

Normalized quantum functions are important because they accurately represent the probabilities of different outcomes in a quantum system. This is essential for making accurate predictions and calculations in quantum mechanics.

3. How do you normalize a quantum function?

To normalize a quantum function, you divide each value in the function by the square root of the sum of the squared amplitudes of all possible outcomes. This ensures that the total probability is equal to 1.

4. Can quantum functions be normalized for any quantum system?

Yes, quantum functions can be normalized for any quantum system as long as it follows the principles of quantum mechanics. However, the specific method for normalization may vary depending on the system and the type of quantum function being used.

5. Are there any limitations to dealing with normalized quantum functions?

One limitation of dealing with normalized quantum functions is that they can become increasingly complex and difficult to manipulate as the number of particles or variables in the system increases. This can make it challenging to accurately model and analyze more complex quantum systems.

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