# Can a self adjoint operator have a continous spectrum ?

1. Jul 21, 2007

### Klaus_Hoffmann

Can a self adjoint operator have a continous spectrum ??

If we have a self adjoint operator

$$Ly_{n} = \lambda _{n}$$

can n take arbitrary real values (n >0 ) in the sense that the spectrum will be continous ?

and in that case, what is the orthonormality condition for eigenfunctions

$$<y_{n} |y_{m}>= \delta (n-m)$$

where 'd' is dirac delta, as a generalization of discrete case of Kronecker delta. could someone put an example ? (since all the cases from QM i know the spectrum is discrete) thankx

2. Jul 21, 2007

### matt grime

Just look up spectral analysis, please, Jose. C* algebras, stuff like that. (I seem to remember making this request a lot in the past - have you followed that advice?)

The orthonormality condition is precisely what it ought to be - two things are orthogonal if their (inner) product is 0. I don't know what makes you think that Dirac deltas have anything to do with taking an inner product (which by definition is just a complex number, not a distribution).

This is basic functional analysis, and not differential equations anyway - one of the first things you learn is that there is, I seem to think, a bijection between (arbitrary) compact subsets of C and commutative C* algebras. Perhaps I have missed some hypotheses, but that is a well known result.

Last edited: Jul 21, 2007
3. Jul 21, 2007

### Gokul43201

Staff Emeritus
If this is a physics question (rather than a math question) then yes, physicists indeed use the Dirac delta to represent the inner product between states in a space spanned by continuous eigenkets (but then, physicists do a lot of things that make mathematicians cringe). A reason for this choice is that it provides a working extension to the completeness condition. I've never seen a rigorous development for this though.

If you want to use produce some new math, I strongly advise against starting from physics.

Last edited: Jul 21, 2007