Interpretation of the Heisenberg picture in QM

[vanhees71]

Well, but #29 just before your posting shows the confusion coming out of the idea to (wrongly) associate abstract mathematical entities of the theory with the real things in the lab. As a theoretician myself, I feel, it's quite important to look at a real lab from time to time and how "quantum things" are prepared and observed.

 

[atyy]

Well, but #29 just before your posting shows the confusion coming out of the idea to (wrongly) associate abstract mathematical entities of the theory with the real things in the lab. As a theoretician myself, I feel, it's quite important to look at a real lab from time to time and how "quantum things" are prepared and observed.

But that is in the Heisenberg picture, I think the OP does understand how to apply the Schroedinger and Heisenberg pictures to experiment, and also how to "interpret" the Schroedinger picture in a naive but helpful way in which the state is real FAPP.

I think he is only asking whether the Heisenberg picture also has a naive interpretation which is also helpful to the intuition. (As far as I know, it doesn't. If it does it is something like the system is unchanged, but what you consider a position measurement changes - but I think most people will not consider that helpful. Mathematically, of course the time evolution is just a rotation, and it doesn't matter if the system rotates and you keep still, or you rotate and the system keeps still.)

 

[vanhees71]

There is no difference in interpretation depending on the picture of time evolution used. So how can it make sense to associate the one or other mathematical element of the theory with the one or other physics piece of reality?

 

[atyy]

There is no difference in interpretation depending on the picture of time evolution used. So how can it make sense to associate the one or other mathematical element of the theory with the one or other physics piece of reality?

It depends on what one means by interpretation. If in Copenhagen, one is working at the level of true reality, then there is no difference in interpretation. But if in Copenhagen, one is working at the level of FAPP reality, then the state is the state in the sense of classical physics - it is the complete specification of of an individual system within the theory. And if one is trying to solve the measurement problem, then so far the Schroedinger picture seems privileged, since Bohmian Mechanics and Many-Worlds both privilege the Schroedinger picture.

It is a bit like in classical mechanics it doesn't matter whether we use Newtonian, Lagrangian or Hamiltonian formulations. However, in quantum mechanics, the Hamiltonian formulation is privileged. Even if one were to argue for the Lagrangian formulation via path integrals, the interpretation is dramatically changed, since the idea that the particle takes an extremal path is no longer true.

 

[Jilang]

@Jilang

O.k., I'm happy with that.
What I'm unhappy with is that in the standard phrasing the word "state" is used although the state does not fully describe the system at any point in time (except at the initial point). As far as I'm aware, everywhere else in physics (classical mechanics, thermodynamics), a "state" of a system describes the system at a certain time in point.
So I find it more clear to say that the "state" meant in the Heisenberg picture is the "initial state" (or initial condition), that the observables (operators) evolve (and the nice thing about the Heisenberg picture is that they evolve independently of the initial state, i.e., I can calculate the evolution of the ooperators once and for all and then apply them to different initial states) and that to actually calculate any measurable quantity I of course need to apply the operator to the initial state. Would you say that this interpretation/phrasing is incorrect?
PS: I do agree that the association of the observable with the measurement apparatus only was highly problematic (if not so say, wrong...)

I think you are getting there (although you are still perhaps subconsciously identifying the state with the particle!), however it is misleading to identify observables with operators in any picture (as Vanhees pointed out). Obervables arise as a function of both the operator and the state. If the interpretation works for you, stick with it. It's only maths.

 

[atyy]

I can calculate the evolution of the ooperators once and for all and then apply them to different initial states) and that to actually calculate any measurable quantity I of course need to apply the operator to the initial state. Would you say that this interpretation/phrasing is incorrect?

If I understand what you mean by "once and for all", this isn't right. After a measurement, you still have to collapse either the wave function or the observables.

If you want a very shut up and calculate formula, you can try Eq 37 in Laloe's http://arxiv.org/abs/quant-ph/0209123. But the classical/quantum cut is fundamental in the traditional formulation of quantum mechanics, and it is natural to identify the state with the quantum side, and the operators with the classical side. But as with all observations, the physics is relative, and it is only the relative that is absolute (for example, the universe is absolutely expanding relative to a family of observers) , so one can imagine that the system is evolving relative to apparatus, or that the apparatus is evolving relative to the system.

More in this spirit is given in section 1.3 of Wiseman and Milburn's https://books.google.com/books?id=ZNjvHaH8qA4C&printsec=frontcover#v=onepage&q&f=false.

 

[vanhees71]

where "collapse" simply means to adapt your state to the gained knowledge from the measurement. It's rather a preparation than a measurement procedure.

 

[atyy]

where "collapse" simply means to adapt your state to the gained knowledge from the measurement. It's rather a preparation than a measurement procedure.

I would rather say that measurement is a means of state preparation.

 

[Sonderval]

@Jilang
Thanks for the encouragement.

@atty
The collapse itself is not part of either the Schrödinger or the Heisenberg time evolution, so I don't really think this argument applies (It's what Penrose calls the R-operation and it's not unitary). For calculating expectation values etc. no collapse is needed.

 

[atyy]

@atty
The collapse itself is not part of either the Schrödinger or the Heisenberg time evolution, so I don't really think this argument applies (It's what Penrose calls the R-operation and it's not unitary). For calculating expectation values etc. no collapse is needed.

The collapse is needed in order to calculate correlations (which are expectation values) between sequential measurement outcomes. For example, it is needed in the Bell test experiments.

 

[vanhees71]

Where do you need collapse in Bell-test experiments? To the contrary, you must make sure that no "collapse" (i.e., decoherence) occurs to the entangled properties of your state. That's why such experiments are usually difficult and usually made with biphotons which are easy to prepare and not too sensitive concerning decoherence.