- #1
mrandersdk
- 246
- 1
On wiki http://en.wikipedia.org/wiki/Borel_functional_calculus in the paragraf 'Resolution of the identity' there is said
'In physics literature, using the above as heuristic, one passes to the case when the spectral measure is no longer discrete and write the resolution of identity as ... '
How is this made rigorous. I had had a course in C*-algebras and proven the spectral theorem for bounded operators, I know most of physical are unbounded, but there must be a connection?
How is it constructed such that
[tex] I = \int_{\sigma (I)} \lambda d E(\lambda) [/tex]
makes sence. I guess somehow taking the inner product with a bra and a ket should get me something like
[tex] <\phi|T|\psi> = \int <\phi|x><x|T|\psi> dx [/tex]
so comparing this with
[tex] <\phi|T|\psi> = <\phi| \int_{\sigma (I)} \lambda d E(\lambda) T |\psi>[/tex]
how do I revieve the lebesgue measure, from the resolution of identity, and see that this is the same, if it even is. I hope it is clear what my problem is?
'In physics literature, using the above as heuristic, one passes to the case when the spectral measure is no longer discrete and write the resolution of identity as ... '
How is this made rigorous. I had had a course in C*-algebras and proven the spectral theorem for bounded operators, I know most of physical are unbounded, but there must be a connection?
How is it constructed such that
[tex] I = \int_{\sigma (I)} \lambda d E(\lambda) [/tex]
makes sence. I guess somehow taking the inner product with a bra and a ket should get me something like
[tex] <\phi|T|\psi> = \int <\phi|x><x|T|\psi> dx [/tex]
so comparing this with
[tex] <\phi|T|\psi> = <\phi| \int_{\sigma (I)} \lambda d E(\lambda) T |\psi>[/tex]
how do I revieve the lebesgue measure, from the resolution of identity, and see that this is the same, if it even is. I hope it is clear what my problem is?