is a position an observable ?

audioloop

are observables only those quantities which commute with system's Hamiltonian ?

bhobba

No. If it commutes with the Hamiltonian it simply means you can measure both it and energy simultaneously. It also means it probably will be conserved if the Hamiltonian has no specific time dependence - actually there is no probably about it - it will be conserved.

Thanks
Bill

Marioeden

As to your original question, position is an observable. Obviously from Heisenberg's Uncertainty Principle, the uncertainty in the momentum would then be infinite, so you have the classic statement of how if you know where a particle is, you have no idea where it's going or how fast. The converse statement is also true, you can know its momentum but consequently have no idea where it is.

OhYoungLions

As far as I know, in principle ANY Hermitian operator is associated with an observable, which are simply the eigenvalues of that operator. For the vast majority of such operators, the observable is not very interesting. For example, the Hamiltonian operator is associated with the observable "Energy". If the operators associated with two observables commute, then your system can be in a state where both observables can be simultaneously well defined.

Remember: just because a quantity is called an observable in QM doesn't mean we know how to build a machine to measure it, or that we even care about it.

jfy4

also, it can (and I've seen it done) be argued that position is the only observable.

bhobba

Yes I have seen it argued as well - but its based on the silly idea the outcome of any observation is the position of a pointer or something like that. People like that are stuck in a time warp IMHO and are not in the computer age. Observations can be captured digitally not having anything to do with position at all.

Thanks
Bill

vanhees71

It depends on how you define "position" and which system to look at whether there is such a thing as a position observable.

In non-relativistic physics for (necessarily massive) particles of any spin, there always exists a position observable defined via the representation theory of the Poincare group, which gives you the 10 conserved quantities and their commutation relations (energy, momentum, angular momentum, center-of-momentum coordinates). Then you can define the position variable as that not explicitly time dependent observable which together with the momentum coordinates fulfills the Heisenberg algebra
[tex][\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk}.[/tex]
In relativistic physics also for all massive particles there exists a position observable of this kind, for massless particles that's the case only for particles with spin 0 and spin 1/2. For all massless particles of higher spin, especially also for photons, there is no position observable in the strict sense. For details, see Arnold Neumaier's Theoretical Physics FAQ:

http://www.mat.univie.ac.at/~neum/ph.../position.html

audioloop

coordinates.

lugita15

The issue is this: position is the primary way humans interact with the world. We see where objects are, we hear them, etc. Even if the information is stored digitally, we still have to acquire that information somehow, in a digital display for instance, and that involves sight. So the argument is that we only directly deal with position, and everything else we conclude about the world comes indirectly, from interpretation of the positional data of the senses.

A. Neumaier

Sight is not position but reception of the electromagnetic field. Position is reconstructed from what we see by a nontrivial process.

bhobba

Exactly - scratching my head why anyone would think otherwise.

And reading information on a computer screen has nothing to do with the position of the screen or the position of whatever you use to present the information.

Thanks
Bill