# The spectral theorem and QM

1. Jun 29, 2008

### mrandersdk

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?

2. Jun 29, 2008

### Hurkyl

Staff Emeritus
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
$$(\psi, A\psi) = \int_{-\infty}^{+\infty} \alpha d(\psi, E_\alpha\psi).$$​

3. Jun 29, 2008

### mrandersdk

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

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

$$d(\psi,E_{x}\psi)= ||\psi(x)||^2 dx$$
$$d(\phi,E_{x}\psi)= \phi^*(x) \psi(x) dx$$.