Interpretation of the Heisenberg picture in QM

[atyy]

I can calculate the evolution of teh 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.