Re: 10 questions on QM postulates

Hendrik van Hees wrote:

> 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.

Yes, for Hermitian operators.
The selfadjoint ones are those with the same domain.

In general, the domain of the adjoint can be larger or smaller.
For example, the adjoint of an annihilation operator has only
the zero vector in its domain.


> 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},

This depends on how one defines O^\dagger. If one is in a C^*-algebra
setting, then A^* is defined independently of any representation in a
Hilbert space, and Hermitian operators are those with A^*=A.
I prefer this notation to the dagger notation.
In a representation in which A is not selfadjoint, it means that
A^dagger is an extension of A^* with a bigger domain.


> and if I remember right, only then the spectral theorem
> holds true.

Yes. This is in agreement with my statement. The existence of a
spectral resolution of A is equivalent to the statement that A
can be written as a multiplication operator on some function space.


Arnold Neumaier

.

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