- #1
Sina
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Hello,
I was reading von neumann's mathematical foundations of quantum mechanics book and I am confused at a certain part regarding finding the resolutions of position and momentum operator. I will ask my first question (I will think more about the second one before posting :p)
For anyone who has the book the page is 129. He says that if the operator is the position operator defined as solving the eigenvalue equation Af(q) = qf(q) = λfλ(q) (using 1 variable for simplicity), then the eigenvectors to this function can only be almost everywhere zero functions with the exception for the point q=λ. He says that while these functions are zero (from a lebesue measure perspective) we may take linear combinations for some (λ1...)to form non-zero functions (the reason for taking the linear combionations although irrelevant to my question is to in a heurestic way build resolutions for this operator by looking at the closed linear manifolds spanned by these combionations). This set has to be essentialy uncountable ofcourse or else it is again a zero function (from Lebesgue measure point of view).
Ʃλ [itex]\in[/itex] I fλ
where I is an uncountable set with say a supremum λ0. How is this process formalized? I am asking this because, I think he also uses that such a summation can be used to represent any function that is only non-zero before λ0 and this is the part that confuses me (as I have not really dealt with concrete examples of sums over uncountable index sets).
Then ofcourse if this is possible E[λ0] = Ʃλ<λ0P[fλ] (where [fλ] is the closed linear subspace spanned by this /or these eigenvectors for the eigenvalue λ) goes to identity as λ0 goes to infinity.
So my question is basically can you represent any function f(x) as the uncountable sum of the functions fλ(x) s.t fλ(λ) = f(λ) and is zero else where. I have never seen such a thing. If I have misunderstood his construction, please correct me. If I can understand my mistake (or lack of knowledge) here I think I can get over the second question about momentum operators.
Best wishes
Sina
I was reading von neumann's mathematical foundations of quantum mechanics book and I am confused at a certain part regarding finding the resolutions of position and momentum operator. I will ask my first question (I will think more about the second one before posting :p)
For anyone who has the book the page is 129. He says that if the operator is the position operator defined as solving the eigenvalue equation Af(q) = qf(q) = λfλ(q) (using 1 variable for simplicity), then the eigenvectors to this function can only be almost everywhere zero functions with the exception for the point q=λ. He says that while these functions are zero (from a lebesue measure perspective) we may take linear combinations for some (λ1...)to form non-zero functions (the reason for taking the linear combionations although irrelevant to my question is to in a heurestic way build resolutions for this operator by looking at the closed linear manifolds spanned by these combionations). This set has to be essentialy uncountable ofcourse or else it is again a zero function (from Lebesgue measure point of view).
Ʃλ [itex]\in[/itex] I fλ
where I is an uncountable set with say a supremum λ0. How is this process formalized? I am asking this because, I think he also uses that such a summation can be used to represent any function that is only non-zero before λ0 and this is the part that confuses me (as I have not really dealt with concrete examples of sums over uncountable index sets).
Then ofcourse if this is possible E[λ0] = Ʃλ<λ0P[fλ] (where [fλ] is the closed linear subspace spanned by this /or these eigenvectors for the eigenvalue λ) goes to identity as λ0 goes to infinity.
So my question is basically can you represent any function f(x) as the uncountable sum of the functions fλ(x) s.t fλ(λ) = f(λ) and is zero else where. I have never seen such a thing. If I have misunderstood his construction, please correct me. If I can understand my mistake (or lack of knowledge) here I think I can get over the second question about momentum operators.
Best wishes
Sina
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