1. Aug 9, 2004

### humanino

An observable in QM is to be represented by a self-adjoint/hermitean operator, so that the eigenvalues are real numbers.

I could not find the answer to : are all self-adjoint/hermitean operators actually observables ?

Thank you for any help !

2. Aug 9, 2004

### marlon

yes !!! Why else would we need them in QM ?

All info is in the observables

regards
marlon

3. Aug 9, 2004

### humanino

What you are saying is :
if I can exhibit ANY self-adjoint operator, then you can provide an experiment to find its spectrum ?!

4. Aug 9, 2004

### humanino

that only normal operators are needed :$$[A,A^\dagger] = 0$$.
And it is even argued that we use self-adjoint operators because they are easy to handle !

I am currently confused :uhh:

Last edited: Aug 9, 2004
5. Aug 9, 2004

### marlon

What I am saying is that we need hermitian operators for two reasons. First the eigenvalues are real, which corresponds to fysical results. So yes any such operator produces a spectrum which has to correspond to some kind of real fysical situation that is to be verified in experiments. Second : we need them because they can be diagonalized with the eigenvalus on the diagonal. This makes the math a lot easier. Besides this is what we are always looking for when working with tensors in GTR for example. The more symmetry, the easier the matrix representations of the opertors get. The more zero's the better !!! that is the rule

regards
marlon

6. Aug 9, 2004

### humanino

helped me to understand that one needs a complete set of commuting operators, spanning the entire space, and that it was realized by normal operators.

Then one can argue that, whatever the basis, one can change it for another one. The position basis for instance. Or the momentum basis. And any result in any basis can be expressed in the preffered basis, thus being observable.

This is rather disappointing to me.

7. Aug 9, 2004

### humanino

Hey Marlon by the way, I remember we are born almost the same week ! Thank you again for your help ! Please add any comment to my last post !

8. Aug 9, 2004

### marlon

Attention, though we have QM-operators for which the above commutation-relation does not count. Such operators have no mutual eigenvalues. This is not good, because when you measure the first one (i.e. project it out on some chosen basis) you automatically influence the eigenvalues (and thus the measurement-results) of the second one. When the commutator is zero the operators can be diagonalized at the same time and their quantumnumbers entirely describe the wave-function of a system. E.G. : l (s-p-d-f-... energylevel) and m (magnetic number like spin up or down) quantumnumbers of an atom.

9. Aug 9, 2004

### marlon

oui je sais, ceux de janvier sont les meilleurs...(je suis belge, voilà la raison que je parle le français)

10. Aug 9, 2004

### marlon

why ??? you should be happy with that

11. Aug 9, 2004

### humanino

Well, not quite. My concern is : if one discovers a new operator, not commuting with the usual ones, but having some relevance and exhibiting self-adjointness. One can claim it is an observable, but does it necessarilly have its own physical interpretation ?

This is what happened with the spin ! The angular momentum was not conserved without spin, so one concluded it is an observable and particles do have intrisic angular momentum.

Now we know there is not any hidden variable, so we can safely hope we did not forget anything to describe entirely a particle, and there is no new operator to discover. I'm not quite confident about this last argument.

12. Aug 9, 2004

### marlon

The way spin was introduced is a very nice example of how QM and even more QFT work. WE are always looking for conserved quantities because they correspond to certain symmetries of the math (here i am again with symmetry). The concept with total spin as a conserved quantity of which the operator is a Casimir-operator gives a very accurate description of the properties of atoms and so on...

You can be certain that for all atomic energy-levels we have the exact amount of necessary quantum-numbers that label each level, corresponding to experiments. More energy-levels and thus more operators could only occur when we change the fundamental properties of the elementary particles, but then we would be changing tyhe basis of QFT. Why do that, keeping in mind it gives us the best model up till now when it comes to atomic-phenomena : i.e. the standard model

regards
marlon

13. Aug 9, 2004

### humanino

I realize I might just as well question the validity of quantum mechanics ! Well, I hope every physicists needs doubts now and then. :uhh: :zzz:

What are you working on Marlon ?

14. Aug 9, 2004

### marlon

just finished university in Ghent, master in theoretical fysics ; MY THESIS WAS ON THE QFT AND QUARKCONFINEMENT. In september i start to study nanotechnology and photonics at the university of Brussels. I hear that Paris is very high leveled on these subjects right ???

15. Aug 9, 2004

### humanino

They hope so, I can't really compare
Well, I went three month at McGill University last year to study photonics (Multi-tunable-wavelenghts pulsed laser, in fiber) They had pretty much the same level as back in France, but more money

sounds hot topic (to me at least) I understand we could hear each other. But be carefull : I'm an experimentalist. Please talk slowly :tongue2:

16. Aug 9, 2004

### humanino

Read the Polyakov's (father) "Gauge fields and strings" yet ?
I tried. :uhh: :surprise: :rofl:

17. Aug 9, 2004

### marlon

be happy, normally them experimentalists have a much bigger bank-account. I want to become one of them too, hence the reason for this nanotechnology thing. Oh, yes, i also find it interesting...

marlon the poor theory-guy

18. Aug 9, 2004

### marlon

no, is it available on site ??? I am interested...

19. Aug 9, 2004

### humanino

It is one of the greatest book for you to read I guess. It is kind of his diary, out of which he wrote a book. He goes beyond established theorems to discuss his views on the future developpement. For instance, he speaks about the link classical statistics/QFT, and stresses the importance of the Wilson (loop) operator. Everything has gauge confinement as general background.

I apologize to the people who really understand Polyakov's book. :shy:

20. Aug 9, 2004

### marlon

I WOULD LIKE TO HAVE THIS BOOK? WHERE CAN I GET IT???