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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.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?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, sorryIsn'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. (?)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?
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.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.This is valid for self-adjoint operators with a purely discrete spectrum only. [...]