Re: 10 questions on QM postulates

Igor Khavkine wrote:
> On 2005-05-15, Hendrik van Hees <hees@xxxxxxxxxxxxx> wrote:
>
>
>>>1) How can the mathematical concept of hermitian operators be
>>>translated into physical terms? What are all the conditions which
>>>ultimately restrict operators such that they can be only hermitian as
>>>opposed to any other type?
>>
>>That's a very deep question, and I also asked it to my professor when he
>>introduced the concept. His answer was: "The operators, describing
>>observables, must have only real eigenvalues, and Hermitean operators
>>have only real eigenvalues."
>>
>>In my opinion, that's a short but not really satisfactory answer. After
>>thinking about the foundations of quantum theory from time to time (I
>>apply quantum theory every day, but one does not think about its
>>foundations too often in this kind of work, so it's more a hobby of
>>mine to think about it from time to time, especially when I hear
>>something about modern quantum-optics experiments, which help a lot to
>>understand the foundations of quantum theory), I think the real
>>mathematical reason that the observables can be described without any
>>inconsistencies by Hermitean (or to be more precise by essentially
>>self-adjoint) operators is the spectral theorem, i.e., each quantum
>>state can be described by a superposition of a complete set of
>>generalized eigenvectors of an self-adjoint operator.
>>
>>I do not claim that an operator must be necessarily self-adjoint to meet
>>this property. I'm not sure wheter it is or not, it might be sufficient
>>that an operator commutes with its adjoint, which is usually called a
>>"normal operator". Perhaps somebody more familiar with the mathematical
>>details may answer this question.
>
>
> One way to look at self-adjointness is that self adjoint operators are
> the only ones that can be full recovered just from their expectation
> values.
>
> Let A be a self-adjoint operator, then we can define a quadratic form
>
> a(psi) = <psi|A|psi>.
>
> In turn, if we know a(psi) for all |psi>, not just the eigenstates of A,
> then if a is real valued and satisfies the parallelogram law
>
> a(psi+phi) + a(psi-phi) = 2(a(psi) + a(phi)),
>
> Then the matrix elements of A can be recovered via the so-called
> polarization formula
>
> <psi|A|phi> = (1/4) sum_{n=0..3} i^n a(psi + i^n phi).

All this holds for arbitrary Hermitian operators, and has nothing
to do with self-adjointness or spectral conditions.


> Once we have a self-adjoint operator, the spectral theorem gives us a
> decomposition of the Hilbert space into orthogonal eigenspaces. Thus
> this property can be derived just from some properties of observables,
> i.e. expectation values.

No. There are Hermitian operators that are not self-adjoint,
and that may have complex spectrum!

This is discussed, e.g., in the Mathematical Physics treatises by
Reed and Simon, or by Thirring.


Arnold Neumaier

.

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