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- Xref: sparky sci.physics:21805 sci.math:17447
- Newsgroups: sci.physics,sci.math
- Path: sparky!uunet!psinntp!scylla!daryl
- From: daryl@oracorp.com (Daryl McCullough)
- Subject: Re: Bayes' theorem and QM
- Message-ID: <1992Dec28.023234.18394@oracorp.com>
- Organization: ORA Corporation
- Date: Mon, 28 Dec 1992 02:32:34 GMT
- Lines: 48
-
- jbaez@riesz.mit.edu (John C. Baez) writes:
-
- >>When you say that quantum mechanics is fundamentally probabilistic, do
- >>you mean (A) QM is a probabilistic theory with no known deterministic
- >>completion, or (B) QM is a probabilistic theory that is known *not* to
- >>have a deterministic completion?
-
- >Neither, since I'm not interested in so-called "completions" of
- >quantum mechanics, which is already complete enough for me.
-
- I don't care whether you are *interested* in deterministic completions,
- to say that quantum mechanics is fundamentally probabilistic to me means
- that there is no deterministic completion of quantum mechanics.
-
- >What I mean is this. In classical mechanics, all pure states are
- >dispersion-free. In quantum mechanics this is not so.
-
- >Less tersely: in classical mechanics, in a pure state one can calculate
- >a numerical value for every observable; an ideal measurement of this
- >observable should give this number as an answer. Probability
- >distributions for the value of an observable are only different from
- >delta functions in the case of mixed states (i.e., states in which one
- >doesn't know a maximal amount of information about what's going on, as
- >are used in statistical mechanics.) We can say that probability theory
- >is only needed classically if you have some ignorance about what the
- >system is up to. In quantum mechanics, even in a pure state one can
- >only calculate a probability distribution for the value of an
- >observable, and generically this is not a delta function, but has
- >nonzero "dispersion" or standard deviation.
-
- If this is supposed to explain why quantum mechanics is fundamentally
- a probabilistic theory, it fails. The terminology "pure state"
- *assumes* that there is no further deterministic completion of quantum
- mechanics, that there is no more to know once one knows the pure state.
-
- If there is a deterministic completion of quantum mechanics, then that
- will simply mean that the terminology "pure state" is a misnomer.
-
- Daryl McCullough
- ORA Corp.
- Ithaca, NY
-
- P.S. By a "deterministic completion of quantum mechanics", I mean a
- deterministic theory predicting (in principle) the values of all
- observables as a function of past values of observables (plus possibly
- "hidden variables") such that all the probabilistic predictions of
- quantum mechanics are explained in terms of ignorance of initial
- conditions.
-