The Spectral Theorem for Unbounded Operators in Quantum Mechanics

  • Thread starter Thread starter mrandersdk
  • Start date Start date
  • Tags Tags
    Qm Theorem
Click For Summary
The discussion focuses on the rigorous formulation of the spectral theorem for unbounded operators in quantum mechanics, particularly in relation to the resolution of the identity. It highlights the transition from discrete spectral measures to continuous ones, emphasizing the integral representation I = ∫σ(I) λ dE(λ) and its connection to inner products in Hilbert spaces. The conversation explores how the Stieltjes integral arises in this context and its relationship to Lebesgue measure through spectral projections. Participants seek references that provide proofs of this form of the spectral theorem and explanations of spectral projections for various operators. The need for clarity on the mathematical foundations and generalizations of these concepts is evident.
mrandersdk
Messages
243
Reaction score
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

I = \int_{\sigma (I)} \lambda d E(\lambda)

makes sence. I guess somehow taking the inner product with a bra and a ket should get me something like

<\phi|T|\psi> = \int <\phi|x><x|T|\psi> dx

so comparing this with

<\phi|T|\psi> = <\phi| \int_{\sigma (I)} \lambda d E(\lambda) T |\psi>

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?
 
Physics news on Phys.org
You might find this reference handy. In particular, appendix B section 6.5 defines this integral to be computed 'pointwise':

THEOREM: A selfadjoint operator A in a Hilbert space \mathcal{H} possesses a unique spectral resolution \{ E_\alpha \} such that
A = \int_{-\infty}^{+\infty} \alpha dE_\alpha,​
meaning that for each \psi in the domain \mathcal{D}_A one has the convergent Stieltjes integral
(\psi, A\psi) = \int_{-\infty}^{+\infty} \alpha d(\psi, E_\alpha\psi).​
 
This seems very interesting. They state a new form of the theorem, saying that you actually get a Stieltjes-Lebesgue integral when taking the iner product.

And as a example the make the spectral projections for X, so it makes sence. Have I understood it right.

Do you any refferences where they proof this form of the spectral theorem, and where they explaine why the spectral projection fx of X is

(E_{x_0}\psi)(x) = \psi(x)

and why this leads to

d(\psi,E_{x}\psi)= ||\psi(x)||^2 dx

and can this be made general to

d(\phi,E_{x}\psi)= \phi^*(x) \psi(x) dx.

And maybe how the spectral projection is calculated for other operators.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
3
Views
5K