A question about resolutions of operators

  • Thread starter Sina
  • Start date
  • Tags
    Operators
In summary, operators are mathematical functions or transformations that are crucial in resolution methods. They allow for manipulation and transformation of equations, and are commonly used in fields such as physics and engineering. By providing a systematic approach to solving complex problems, resolutions of operators are relevant in scientific research and can be applied to real-world problems in various fields.
  • #1
Sina
120
0
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
 
Last edited:
Physics news on Phys.org
  • #2


Dear Sina,

Thank you for your question regarding von Neumann's mathematical foundations of quantum mechanics. I can understand your confusion with the construction of the resolution of the position operator. Let me try to explain it in a simpler way.

Firstly, the position operator is defined as solving the eigenvalue equation Af(q) = qf(q) = λfλ(q). This means that the operator A, when acting on a function f(q), gives back the eigenvalue λ multiplied by the same function f(q). In other words, the function f(q) is an eigenvector of the operator A with the eigenvalue λ.

Now, von Neumann goes on to say that these eigenvectors can only be almost everywhere zero functions, with the exception of the point q=λ. This means that the function f(q) is zero everywhere except at the point q=λ. However, we can take linear combinations of these functions, denoted as Ʃλ \in I fλ, where I is an uncountable set of eigenvalues. This means that we can take linear combinations of all the eigenvectors, with different eigenvalues, to form non-zero functions. This set of functions is essentially uncountable, as you correctly pointed out, and is spanned by the closed linear manifolds of these linear combinations.

To answer your question, yes, any function f(x) can be represented as the uncountable sum of the functions fλ(x) with the condition that fλ(λ) = f(λ) and is zero elsewhere. This is because the set of functions fλ(x) is essentially uncountable and spans all possible functions that can be constructed from linear combinations of eigenvectors with different eigenvalues.

I hope this helps to clarify the concept for you. If you have any further questions or need more clarification, please do not hesitate to ask. Best of luck with your studies!
 

1. What is meant by the term "operator" in the context of resolutions?

An operator in the context of resolutions refers to a mathematical function or transformation that acts on a vector space to produce another vector or scalar value.

2. How do operators play a role in resolution methods?

Operators play a crucial role in resolution methods by allowing us to manipulate and transform equations or systems of equations in order to find solutions or prove their existence.

3. What are some common examples of operators used in resolution methods?

Some common examples of operators used in resolution methods include differentiation, integration, matrix operations, and differential equations.

4. How are resolutions of operators relevant in scientific research?

Resolutions of operators are relevant in scientific research as they provide a powerful tool for solving complex problems and equations, particularly in fields such as physics, engineering, and mathematics.

5. Can resolutions of operators be applied to real-world problems?

Yes, resolutions of operators can be applied to real-world problems in various fields, such as data analysis, signal processing, and image recognition, to name a few. They provide a systematic and efficient approach to finding solutions and understanding complex systems.

Similar threads

  • Calculus
Replies
17
Views
1K
Replies
3
Views
803
  • Linear and Abstract Algebra
Replies
6
Views
932
Replies
2
Views
816
  • Linear and Abstract Algebra
Replies
3
Views
928
  • Linear and Abstract Algebra
Replies
8
Views
959
Replies
9
Views
1K
  • Quantum Physics
Replies
31
Views
2K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Quantum Physics
Replies
14
Views
811
Back
Top