Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
A silly question about quantum mechanics?
Let ψ₁, ψ₂, ... , ψᵥ be an orthonormal basis set.
And,
ψ = c₁ψ₁ + c₂ψ₂ + ... + cᵥψᵥ
It is given in the lecture note I am provided with that since 〈ψᵤ|ψᵥ〉 = δᵤᵥ we can find
cᵣ = 〈ψᵣ|ψ〉
The proof is given and I understand that. The problem is, how can one apply this to find cᵣ without really knowing c₁, c₂, ..., cᵣ, ..., cᵥ? (ψ = c₁ψ₁ + c₂ψ₂ + ... + cᵥψᵥ)
@PleaseSt...Yes the result might sound trivial and that's exactly why I don't understand what the note says. Let me put it in this way..
If I define ψ as a superposition of an orthonormal basis ψ₁, ψ₂, ... , ψᵥ .... then I already know the mixing coefficients. Then why on earth should I bother "finding" them using 〈ψᵣ|ψ〉?
2 Answers
- 1 decade agoFavorite Answer
This is silly. I do not even understand your problem here. What do you mean without really knowing c₁, c₂, ..., cᵣ, ..., cᵥ ?
You know c₁, c₂, cᵥ because you defined ψ already as a superposition of an orthonormal basis ψ₁, ψ₂, ... , ψᵥ ....
ψ = c₁ψ₁ + c₂ψ₂ + ... + cᵥψᵥ
So by 〈ψᵤ|ψᵥ〉 = δᵤᵥ any 〈ψᵣ|ψ〉 = cᵣ
ie.. 〈ψ₁|ψ〉 = c₁
EDIT:
After I was done I figured that was what you meant.
Even though you can easily see that is true, you still have to justify that is true. Thats how you do it. Simple but necessary.
This is important in the generalized statistical interpretation. Say ψᵣ is an eigenfunction corresponding to a discrete eigenvalue qn of a hermitian operator,
then |cᵣ|^2 is the probability that measurement on ψ will yield the eigenvalue qn.
You probably already know that.. and that seems trivial, but you still have to define the formalism of the theory properly using mathematics.
Also how do you make all those nice symbols?
- Anonymous1 decade ago
yeah, that's just silly.