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kent davidge
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Suppose a set of basis vectors are eigenvectors of some operator. So they will provide a one dimensional representation of that operator in the vector space?
Then on each 1-dimensional eigenspace one has a 1-dimensional representation.kent davidge said:Suppose a set of basis vectors are eigenvectors of some operator. So they will provide a one dimensional representation of that operator in the vector space?
isn't that the case only if the group is additive?vanhees71 said:If the operator ##\hat{A}## is self-adjoint, then it defines a unitary representation of a one-parameter Lie group via
$$\hat{U}=\exp(-\mathrm{i} \lambda \hat{A}).$$
oh yea, sorryvanhees71 said:Isn't this implied by the definition of "one-parameter Lie group"?
I'm guessing your underlying question is whether the operator can be represented in terms of the eigenvectors. For a self-adjoint operator ##A## on a Hilbert space, with eigenvectors ##|a\rangle##, where ##a## is a eigenvalue of ##A##, one can represent $$A ~=~ \sum_a a\, |a\rangle\langle a| ~.$$ This is a simple version of "the spectral theorem". More general versions are available for more general operators, i.e., operators that are not self-adjoint, but satisfy certain other property(ies). I could sketch more detail if indeed that was what you were really asking. (?)kent davidge said:Suppose a set of basis vectors are eigenvectors of some operator. So they will provide a one dimensional representation of that operator in the vector space?
Oops! Yes -- thank you.vanhees71 said:You mean [...]
This is valid for self-adjoint operators with a purely discrete spectrum only. For Hermitian operators with a continuous spectrum there are no eigenvectors. Or you need to embed the Hilbert space into the dual of a nuclear space and get an analogous representation involving distribution-valued bras and kets and integrals in place of the sum. In the mixed spectrum case you need a combination of sums and integrals.strangerep said:I'm guessing your underlying question is whether the operator can be represented in terms of the eigenvectors. For a self-adjoint operator ##A## on a Hilbert space, with eigenvectors ##|a\rangle##, where ##a## is a eigenvalue of ##A##, one can represent $$A ~=~ \sum_a a\, |a\rangle\langle a| ~.$$ This is a simple version of "the spectral theorem".
Yes, I know. The intent of my (incomplete) post was merely to try and zero in on what the OP was really asking about.A. Neumaier said:This is valid for self-adjoint operators with a purely discrete spectrum only. [...]
An eigenvector is a vector that does not change direction when a linear transformation is applied to it. It only changes in magnitude (scaling factor).
Eigenvectors are often used to represent states or groups in linear algebra. In this context, they represent the "stable" or "fixed" components of a system that do not change under a given transformation.
Eigenvectors with one dimension have only one component and are represented as a single column vector. This means that the vector only changes in magnitude, not direction, when a transformation is applied.
Eigenvectors are calculated by finding the non-zero solutions to a system of linear equations, where the transformation matrix and its eigenvectors are known. This is typically done using matrix algebra and linear algebra techniques.
Eigenvectors have many applications in science, particularly in fields such as physics, engineering, and computer science. They are used to analyze and understand systems that involve linear transformations, such as quantum mechanics and signal processing. They also have practical applications in data analysis and data compression.