Re: Paradoxes
- From: Tom Roberts <tjroberts137@xxxxxxxxxxxxx>
- Date: Mon, 24 Mar 2008 10:05:00 -0500
Juan R. González-Álvarez wrote:
Tom Roberts wrote on Sat, 22 Mar 2008 10:58:29 -0500:But in quantum physics, such determinism fails, as many phenomena are
inherently random. Knowledge of the configuration at one instant in time
is insufficient to determine its configuration at other times; at best
all one can do is compute probabilities for the different possible
configurations.
You are repeating a common misunderstanding.
No, I am not. You are attempting to change the subject (to non-relativistic quantum theory). Answer this:
There is a muon at rest at time t=0. What state will it be in at time t=1 microsecond -- will it be a muon or will it have decayed?
[This is just one of MANY examples I could have used; it is
the simplest.]
Quantum mechanical evolution (Schrödinger) is perfectly deterministic.
Certainly, in NON-RELATIVISTIC quantum mechanics. But it does not permit one to answer the above question about the muon, because muon decay is outside its domain. In QFT, evolution is not at all deterministic. The latter is the established theory of "quantum physics" (my original phrase), while the former is an approximation to the latter.
Also: Hidden in your discussion is the fact that "state" changed meaning -- you introduced an unacknowledged PUN that probably confuses many readers. In the original discussion, "state" was implicitly classical; when you introduced the Schroedinger equation, you introduced a non-rel. quantum meaning -- the "state" represented by the wave function, which evolves deterministically via the Sch. eqn., is merely probabilistic in the sense of the initial (classical) meaning of "state".
The Quantum Field Theory meaning of "state" I implicitly used FOR THIS EXAMPLE is closer to the classical meaning than the non-rel QM meaning you used. In particular, it is completely unambiguous whether or not the muon has decayed at t=1 us, and this aspect of the state of the system is equivalent to a classical meaning of "state". So this serves just fine as an example of non-determinism, because "state" did not radically change meaning, and yet QFT can only compute probabilities for this aspect of the state.
Nondeterminism arises only during the measurement process.
So why can't you give a "deterministic" answer to the above question about the muon's state? There is no "nondeterminism" whatsoever in the measurement of whether it is a muon or an electron and 2 neutrinos.
The answer, of course, is that non-rel QM and (relativistic) QFT are DIFFERENT -- the former is merely an APPROXIMATION to the latter. A major aspect of that approximation is the assumption that no objects change type (e.g. nothing ever decays); in both QFT and the real world, many quantum objects decay, invalidating non-rel. QM.
Just like Newtonian mechanics is useful within its domain,
so too is non-rel. QM useful within its domain. Both are
useless outside their respective domains, and both are merely
approximations to better theories with wider domains.
IOW: I said "quantum physics" and you read "non-relativistic quantum mechanics" when that is not at all what I actually said (or had in mind). Please do not confuse an approximation to a theory (non-rel. QM) with the theory itself (QFT).
> [...further elaboration completely missing my point]
Tom Roberts
.
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