vanhees71 said:
How can ##\rho## (assuming it's what's called the statistical operator in the standard interpretation) be a "beable", if it depends on the picture of time evolution chosen? The same holds for operators representing observables.
I meant to say, in any
fixed picture, ##\rho## is a beable. In the thread where you posted this, we were silently using the Schrödinger picture. Picture change are like coordinate changes.
The situation is the same as when considering a position vector as a beable of a classical system, by fixing the coordinate system.
vanhees71 said:
What's a physical quantity (...) are
$$P(t,a|\rho)=\sum_{\beta} \langle t, a,\beta|\rho(t)|t,a,\beta \rangle,$$
where ##|t,a,\beta## and ##\rho(t)## are the eigenvectors of ##\hat{A}## and ##\rho(t)## the statistical operator, evolving in time according to the chosen picture of time evolution. In the standard minimal interpretation ##P(t,a|\rho)## is the probability for obtaining the value ##a## when measuring the observable ##A## precisely at time ##t##.
Yes, this physical quantity is the q-observable ##P(t,a|\rho)=\langle B\rangle##, where
$$B=\sum_{\beta} |t,a,\beta \rangle\langle t, a,\beta|$$
Thus you agree that at least certain q-observables are physical quantities, and we are getting closer.
vanhees71 said:
Now, before one discuss or even prove anything concerning an interpretation, one must define, what's the meaning of this expression in the interpretation. I still didn't get, as what this quantity is interpreted in the thermal interpretation, because you forbid it to be interpreted as probabilities.
I do not forbid it; I only remove it from the foundations, and allow q-expectations (rather than eigenvalues) to be interpreted as the true properties (beables). See the previous post #590.
In cases
where someone actually performs many microscopic measurements of ##A## in the textbook sense, this q-expectation has indeed the statistical meaning you describe.
But there are other q-observables associated to macroscopic objects (all properties considered in thermodynamics; e.g., the mass of a particular brick of iron) which can be measured by
actually performing only a single measurement, and measurement statistics over single events (or over unperformed measurements) is meaningless. The thermal interpretation still applies since it is independent of statistics.
vanhees71 said:
On the other hand, I think it's pretty safe to say the universe, on a large space-time scale, is close to local thermal equilibrium, as defined in standard coordinates of the FLRM metric, where the CMBR is in local thermal equilibrium up to tiny fluctuations of the relative order of ##10^{-5}##.
I agree. This implies that the
exact state of the universe is ##\rho=e^-S/k_B##, where the entropy operator ##S## of the universe is
approximately given by an integral over the energy density operator and particle density operators, with suitable weights (intensive fields). The coarse-graining inherent in the neglect of field products in an expansion of ##S## into fields makes ##S## exactly equal to such an expression and defines
exact local equilibrium as an
approximate state of the universe.
Thus the state of the universe is fairly well, but not in all details, specified by our current knowledge.