Reference on spectral theory for selfadjont non-compact operators I cant find any book which treats spectral theory for selfadjont non-compact operators and in particular existance of eigenvectors.
I had a look in Krein and Gohberg's book from -69 but I cant find a desierd result.
 A: As soon as you loose compactness the spectral theorem doesn't just talk about eigenvalues, but the entire spectrum, which may now contain a continuous part. So there is a leap from the simple compact operator spectral theorem to the spectral theorem for bounded operators.
I personally like the treatment in say M. Reed & B. Simon "Methods of modern mathematical physics - functional analysis" (pg. ~225). Another good book is W. Rudin "Functional Analysis" (see pg. 321).
Also have a look here for the spectal theorem for bounded oparators.
Addition:
Seems like you are interested in eigenvalues. A word of caution must be given here. A bounded self adjoint operator may have no eigenvalues. Consider for instance $M \colon L^2([0,1]) \to L^2([0,1])$ given by $(Mf)(x) = xf(x)$ has no eigenvalues. 
If you are interested in Schrödinger type operators i suggest (as FreeziiS.) that you look at volume IV of Reed & Simons book (it is presented as an area known as perturbation theory). Also T. Kato's book "Perturbation Theory" is worth looking at. 
A: IMHO, good references for this subject are:
"J. Blank, P. Exner, M. Havlíček, Hilbert Space Operators in Quantum Physics"
"Brian C. Hall, Quantum Theory for Mathematicians"
You may also have a look here: References
A: This was mentioned in the comments, but you didn't answer those comments, so I'm still not sure if you are aware that "eigenvalue" and "eigenvector" are almost useless notions when dealing with non-compact operators. Below are some examples. 


*

*Some selfadjoint operators have a spectral decomposition in terms of eigenvalues. For instance,  $P:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$, given by 
$$
P(a_1,a_2,a_3,\ldots)=(a_1,0,a_3,0,a_5,0,\ldots)
$$
is selfadjoint, has eigenvalues $0$ and $1$, and it has a spectral decomposiion 
$$
P=1\,P+0\,(I-P).
$$
So the eigenspaces for $P$ are 
$$
\{a_1,0,a_2,0,a_3,0,\ldots):\ a\in\ell^2\},\ \ \text{ and } \{(0,a_1,0,a_2,0,\ldots):\ a\in\ell^2\}
$$
and it is possible to construct a basis of eigenvectors. 

*Consider, for $f\in L^\infty[0,1]$, the operator $T_f:L^2[0,1]\to L^2[0,1]$ given by $T_fg=fg$. It is not hard to show that $M_f$ is bounded with norm $\|f\|_\infty$, and it is selfadjoint. Its spectrum is the essential range of $f$. For $T_f$ to have an eigenvalue $\lambda$, we would need the equality 
$$
\lambda g=T_fg=fg
$$
for some $g\in L^2[0,1]$.
It follows that $f=\lambda$ wherever $g\ne0$, and one can deduce that $f$ would have to be of the form $\lambda\,1_E+h\,1_{E^c}$. So, for example, we could have 


*

*$T=M_f$ with $f=1_{[0,1/2]}$, and then $f$ has eigenvalues $0,1$ and it admits a basis of eigenvectors (this is exactly the same example as in 1.). 

*$T=M_f$ with $f=1_{[0,1/2]}+x\,1_{1/2,1]}$. Here $1$ is an eigenvalue, and any function supported in $[0,1/2]$ is an eigenvector. But there are no other eigenvalues, and $T$ does not admit a basis of eigenvectors. 

*As already mentioned in the comments, if $T=M_f$ with $f$ the identity function $f(x)=x$, then $T$ has no eigenvalues at all. 
A: I found a result that coverd my needs, in the last chapter of Conways "A course in F.A".
If $N$ is normal and $\lambda$ is in the complement of the essential spectra then it is a eigenvalue of finite multiplicity.
Which is what I needed. I wasnt awere of that this is essentially the "compact" part of the operator. The general case is very involved.
