FOM: non-computable waves - QM case
Jeffrey Ketland
ketland at ketland.fsnet.co.uk
Wed Aug 23 13:05:20 EDT 2000
Joe Shipman concluded his interesting and informative message with:
*************
At this point it is no longer useful to talk about "making predictions to
test the theory". If theory has passed all the tests so far, it makes more
sense to assume the theory as a new axiom and start deriving new mathematics
by resorting to experiment! This shows that mathematics is not logically
prior to physics;
>for another result in that direction see the last section of my thesis [J.
Shipman, "Cardinal Conditions for Strong Fubini Theorems", Trans. Am. Math.
Soc. 321 (1990), pp.465-481].
************
Thanks for that message, Joe. That's the sort of thing I'm interested in -
especially the (provocative) idea that physics might be "logically prior to
mathematics". I agree that Pour-El/Richards uncomputability results don't
seem to give "harnessable physical uncomputability" (since they require
infinite precision in measurements), whereas the Chaitin approach you
describe (which I don't fully grasp) seems to give a form of uncomputability
which can be measured.
I thought a bit about applying the Pour-El/Richards ideas to the
(non-relativistic) Quantum Mechanics case
1. Quantum mechanics time evolution
In the last post, I mentioned the problem of applying Pour-El/Richards 1st
Main Theorem (bounded operators preserve computability, unbounded ones
don't). to the QM case (i.e. Schroedinger equation). The Schroedinger
equation is
(1) H f(x,t) = ih (d/dt) f(x,t)
where f is a wave-function in the Hilbert space (usually L^2[R]) and H is
the Hamiltonian operator on the Hilbert space.
I am now guessing that applying the Pour-El/Richards theorem to this
equation is quite simple, in virtue of the fact that the solutions of (1)
can be written,
(2) f(x, t) = U(t) f(x, 0)
Where U(t) is the time-evolution operator,
(3) U(t) = exp{-i Ht /h}.
H is usually unbounded, because it contains derivatives, but because U(t) is
a phase factor of modulus 1, it follows that the norm of state U(t) f(x, 0)
is invariant. I.e.,
(4) || U(t) f(x, 0) || = || f(x, 0) ||
In particular, this implies that the quantum mechanics time-evolution
operator U(t) is *bounded*, since by definition, for any Hilbert space
vector f in L^2[R], || f || is bounded.
The time-evolution operator is a unitary operator. In Quantum Field theory,
this is called "unitarity" and is important in axiomatic presentations of
QFT.
(Apparently, there are situations in perturbative QFT, where you use Feynman
integrals to calculate scattering amplitudes, where you have violations of
unitary - this is usually considered a problem with the theory, or at least
with the mathematical techniques for figuring out physical predictions).
If this is right - if the QM time evolution operator is unitary and thus
bounded - then using Pour-El/Richards 1st Main Theorem, one gets:
Theorem: If a quantum mechanical system is set up with a computable
wavefunction f(x,0), then future Schroedinger time-evolved state f(x, t) is
computable also, for all t.
(Can it be as straightforward as that? - perhaps I've goofed up somewhere).
2. "Collapse" of the wavefunction.
To complete the treatment for quantum mechanics, one also has to consider
the "collapse prescription" also - i.e., what happens to the wavefunction
when you measure a system.
If a physical system is in quantum mechanical state described by
wavefunction f(x, t), then measuring an observable O causes the system to
collapse into some eigenvector v(x) (with eigenvalue o) of O (with
probability determined by modulus squared of the co-efficients of the
expansion of f(x, t) in terms of the eigenvectors of O).
(Of course, this is tremendously controversial - there are "no collapse"
interpretations of QM as well).
In Chapter 4, Pour-El/Richards give:
Second Main Theorem (p. 128): "under mild side conditions, a self-adjoint
operator has computable eigenvalues, but the sequence of eigenvalues need
not be computable" (summary, p. 2).
and:
The Eigenvector Theorem (p. 133): "there is an effectively determined
bounded self-adjoint operator T* such that 0 is an eigenvalue, but none of
the eigenvectors with this eigenvalue are computable" (summary, p. 2).
It is thus conceivable that that there is a (rather spooky) observable O
(i.e., some self-adjoint operator O, and perhaps something we could in
principle set up an experiment to measure), which has non-computable
eigenvectors. If that's right, we could have a quantum mechanical system in
a computable state f(x), and "push" it into a non-computable state g(x) by
measuring this observable O* (where g(x) would be some non-computable
eigenvector of this operator).
Regards - Jeff
Jeffrey Ketland
Dept of Philosophy, University of Nottingham
Nottingham NG7 2RD United Kingdom
Tel: 0115 951 5843
Home: 0115 922 3978
E-mail: jeffrey.ketland at nottingham.ac.uk
Home: ketland at ketland.fsnet.co.uk
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