Going from system & observable to hilbert space & operator?

1. Sep 11, 2011

nonequilibrium

Hello,

Given a system and an observable that one wants to measure in it, how does on get the (or a?) relevant hilbert space and the suitable operator in it? The examples I've come across so far seem to rely on... well, I'd call it "vague reasoning", but the word 'reasoning' seems too much. It seems like the idea is "well we're just guessing and then predict some things and it turns out that they're experimentally verified", so I was wondering if there's a more elegant way to determine them?

On a related note, what constitutes a measurement as corresponding to an operator? Apparently something called "weak measurement" is not allowed, which is basically (for example) inference about momentum based on a position measurement. That's okay, I suppose, neglecting that kind of measurement, but then I expect, in return, a clear definition of what is considered as a "valid" measurement. Or can it not be given and is it left intentionally vague? To try and start the concrete specification of a "valid" measurement, I might suggest the following: "it" has to happen on a specific time t (?).

2. Sep 12, 2011

dextercioby

You;re touching the most sensitive point of QM, namely the interpretational one which contains the measurement issue. I can only address your first bolded question. The Hilbert space is dictated by the hypothesis of the problem which determine the observables needed to 'label' the physical states. For example, a spinless massive Galilean particle free to move in all space has a number of observables like position, energy, momentum, electric charge... among which a number of algebraic relations exist. One can build with these observables an algebra whose GNS representation provides the suitable Hilbert space, in this case L^2(R^3, dx).

3. Sep 12, 2011

nonequilibrium

Ah, so if I understand correctly (in regard to your answer to my first question) the current quantum theory does not predict a quantity like spin, but tells you what happens if it were to exist?

But to what extent must we supply information about the concept "spin"? How much physical information regarding "spin" must we inject ourselves before we can deduce the rest of its properties out of the quantum formalism? (I hope this question is clear, if not, let me know, I'll try to formulate it better)

4. Sep 12, 2011

dextercioby

Incidentally, spin is a derived concept in a formulation of the theory in which observables are generators of symmetries. So one starts with the possible symmetries of the physical system and derives the observables (spin, energy, momentum, position, etc.). One could, of course, proceed the other way around, i.e. postulate the observables and derive from them the symmetries, but this is not really physical (though common practice in many textbooks).

So QT can not only predict the existence of an observable, but also fully describe what we can do with it.
The system has Galilean invariance (the full Galilean group of symmetries of space-time is an invariance group for the particle/system of particles).

5. Sep 12, 2011

nonequilibrium

That sounds really interesting! I've torrented (shh) Ballentine's book on QM; in case you know it: is that a good source to get that kind of method from? (I think he does it in the correct order, as you state) For my class of QM we'll be using Jean-Louis Basdevant & Jean Dalibard, but you probably won't know that one, but anyway it looks less promising.

So if I take your two posts together, am I justified in saying that Galilean invariance also indicates your Hilbert space? The reason for this: the Galilean invariance gives you your observables (your 2nd post) and your observables give you your Hilbert space (your 1st post).

6. Sep 12, 2011

dextercioby

I own Ballentine and I'm 1000% satisfied with his presentation (based on work by TF Jordan in the 70's). Yes, symmetries determine everything. Observables and states, i.e. the Hilbert space. Read Ballentine.

7. Sep 12, 2011

nonequilibrium

ordered on amazon :)