Re: 10 questions on QM postulates

Arnold Neumaier wrote:


> Normal operators A have under the above conditions
> a complete spectral resolution of the identity, which means that
> A can be written as a multiplication operator on some function
> space. (If, and only if this function space is discrete,
> there is a complete set of eigenvectors.)
>
> Among the normal operators exactly the Hermitian ones have
> real expectation, and, under the above conditions, real spectrum.
>
> In general, Hermitian operators are always normal and have real
> expectations, but they need not have real spectrum, if the above
> conditions are _not_ satisfied.

I'm not a mathematical physicist, but if I understood the literature
right, usually a Hermitian operator is an operator, whose adjoint has
generally a larger domain than the operator itself.

An operator O, which is defined on domain D \subseteq H, is called
Hermitian, iff

\forall |x>, |y> \in D: <A x|y>=<x|A y>

The adjoint O^\dagger of an operator O with a domain D, which is dense
in H, usually is defined in two steps:

(1) The domain D^\dagger of O^\dagger are all |x>, for which the linear
functional A_O:D->C with

A_O(|y>)=<x|O y>

is continuous.

(2) Due to a well-known theorem in Hilbert-space theory (called Hellmann
theorem, if I remember right), there exists a unique vector |z> \in H,
such that.

A_O(y)=<z|y>. Then the adjoint to O is defined by

\forall |x> \in D^{\dagger}: O^{\dagger} |x>=|z>

If O is Hermitian, it is clear that D \subseteq D^\dagger. Only if
D^{\dagger}=D, O is self-adjoint, and only then we are allowed to write
O=O^{\dagger}, and if I remember right, only then the spectral theorem
holds true.

--
Hendrik van Hees Texas A&M University
Phone: +1 979/845-1411 Cyclotron Institute, MS-3366
Fax: +1 979/845-1899 College Station, TX 77843-3366
http://theory.gsi.de/~vanhees/ mailto:hees@xxxxxxxxxxxxx

.

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