Science > Physics > Variance of Quantum Operators is Incorrectly Defined
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Science > Physics |
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"OsherD" |
| Date: |
21 Aug 2005 01:52:22 AM |
| Object: |
Variance of Quantum Operators is Incorrectly Defined |
From Osher Doctorow
COPYRIGHT NOTICE
Variance of Quantum Operators is Incorrectly Defined
Copyright By Owner Osher Doctorow Ph.D.
First Published 2005
The variance of quantum operators (in practice used for self-adjoint
quantum operators on Hilbert space) is incorrectly defined.
If I...dx is the integral with regard to some variable x, and A is an
operator corresponding to an observable O, then the definition of
quantum expectation (which serves as a model for quantum variance) is:
1) E(A) = I(w*Aw)dx
Since Born interprets w*w = ww* for wavefunction w to be a probability
density f (let's say fX(x)), it might be thought that (1) generalizes
the well-known mathematical probability-statistics expectation of a
random variable X:
2) E(X) = I(xfX(x))dx
with integration over the real line. Actually, more generally E(X) of
(2) has to be replaced by a multiple/iterated integral because in
general we don't just have fX(x) depending on a single real value X but
rather a joint probability density function fX1,X2,...,Xn(x1, x2, ...,
xn) depending on n > = 1 arguments.
The question then becomes: does w*w = ww* represent a probability
density function depending on one argument or more than one argument?
If it depends on more than one argument, then we have the remarkable
Theorem:
Theorem. If w*w = ww* = f(x1,x2,...,xn) for n > 1, then X1, X2, ...,
Xn (taking vlaues x1, x2, ..., xn) cannot be "conjugate" as the
Heisenberg Uncertainty Principle (HUP) since their joint probability
density function f(x1,x2,...,xn) is defined in its entire non-null
domain and hence the random variables are simultaneously defined.
Proof. Direct from the definitions. Q.E.D.
But the Schrodinger equation involves partial derivatives of w with
regard to more than one argument except in the one-dimensional wave
case, so this knocks out all of quantum theory in higher than one
dimension any time that two allegedly conjugate variables are involved.
If there is only one variable on which w depends, then not only do the
previous objects of my recent threads hold, but the definition (1)
doesn't make sense since it doesn't distinguish between w depending on
different single variables!
In mathematical probability-statistics, we distinguish between pdfs for
different random variables X1, X2 by writing them respectively as:
3) fX1(x1)
4) fX2(x2)
That isn't done for the wave function w. The wavefunction is
unqualified and presumably doesn't change as random variables to which
it refers change! Hence it is the same for any two random variables,
hence there are no "conjugate" random variables!
Of course, quantum theorists may want to change the one-variable wave
function w and start referring to it as wX1, wX2, etc., but it is
rather late in the day for that, not to mention the horrible
multivariate pdf situation like w(x1,x2,...,xn) for n > 1.
Next question :>)
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Variance of Quantum Operators is Incorrectly Defined |
21 Aug 2005 02:10:05 AM |
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From Osher Doctorow
I meant to type "previous objections", not "previous objects".
By the way, there is a rather curious fact about generalizing
mathematics developed in one field or domain to a different domain.
Namely, if you generalize into different phases (phase in the sense of
basically different physical state like gas, liquid, solid, plasma,
superfluid, superconductor, arguably black hole, liquid crystal, etc.),
you may make very serious mistakes.
It so happens that when you generalize from the real to the arbitrary
complex domain or even to the imaginary domain in complex variables,
you are simultaneously moving into a different scale or several scales,
namely the real scale and the imaginary scale and their paired or added
representations. If you can do it for quantum theory expectations and
variances, why can't do you it for superluminal motion in special
relativity, which technically/formally arises from sqrt(1 - v^2/c^2) if
v > c in which latter case the square root becomes imaginary! The
argument that you can't see superluminal motion is roughly as useless
as the argument that you can't see the quantum domain. As for whether
there are any known indirect effects (which there seem to be for
quantum theory) for superluminal motion, how do we know since almost
nobody has been looking for them? As a matter of fact, though,
superluminal group velocities have been confirmed, and inflation at
least requires that universe as a whole expand superluminally but not
its parts if relativity holds!
Of course, all these comments hold for Clifford algebra/Spacetime
Calculus (a la Hestenes of Arizona State U.) including generalizations
to quaternions and octonians, etc. But, hey, nobody said it would be
easy.
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Variance of Quantum Operators is Incorrectly Defined |
21 Aug 2005 03:06:38 AM |
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From Osher Doctorow
Real variables constitute an Achilles Heel for the HUP, while complex
variables are sufficiently obscure even to theorists in an intuitive
sense that they obscure a multitude of "hand-waving".
For example, we know that there isn't anything remotely similar to the
HUP in real-valued mathematical probability-statistics, but this is
"explained" by many quantum theorists by the claim that in:
1) U(x)U(p) > = h/4pi
where U( ) is uncertainty and x is position, p is momentum, the
quantity h "goes to 0" classically so that for real variables (1) just
says U(x)U(p) > = 0 which for variances and standard deviations is
always true.
It is true that h has a "mysterious" role across different quantum vs
classical scenarios, but most of the mystery can be traced to the one
or two original equations in the quantum theory, which is not a mystery
at all.
What really would be convincing would be some indication that U(x) and
U(p) get "continuously deformed" somewhere, perhaps to a constant, as
one approaches the real domain, or that they either approach or diverge
from some "natural boundary" related to them physically. There isn't
any such thing.
As soon as the wave function replaces the imaginary part of the complex
expression for w by 0, so that it becomes real, HUP fails.
Equally surprising, there are inequalities involving moments and
expectations and variances in mathematical probability-statistics of
the real-valued type, but none of them tend toward 0 in a way that
would cross-validate the HUP.
For example, the Schwarz inequality (Cauchy-Schwartz) tells us in
real-valued mathematical probability/statistics:
2) [E(XY)]^2 < = (EX^2)(EY^2) if EX^2 and EY^2 are finite
which in turn implies:
3) [Cov(X, Y)]^2 < = Var(X)Var(Y)
which in turn implies:
4) rho(X, Y) < = 1 (rho the correlation coefficient between X and Y)
But rho(X, Y) doesn't go anywhere "parallel to h" or parallel to
anything else, and neither does the covariance Cov(X, Y) or E(XY),
which depend very much on exactly what X and Y are and how they're
related and which in fact cover the "gamut" of relationships no matter
what "else" is held constant.
So the left hand side of the HUP behaves entirely differently from its
right hand side "as h --> 0". This is the type of cross-validation
that should have been done long, long ago, say a few billion dollars
earlier :>)
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Variance of Quantum Operators is Incorrectly Defined |
21 Aug 2005 03:24:28 AM |
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From Osher Doctorow
About a year or so ago, I did a book review for Isis, the History of
Science Society journal, on a microfiche of one of Chebyshev's works.
Chebyshev was a mathematical probability-statistics pioneer, and one of
his most interesting inequalities is called Chebyshev's inequality:
1) P(/X - E(X)/ > = t) < = Var(X)/t^2 if E(X) and Var(X) are finite
This has a rather interesting consequence much used in statistics:
2) P(/X - E(X)/ > = kVar(X)) < = 1/k^2, k = 1, 2, 3, ... under the
conditions of (1)
which tells you the approximate proportion of probability distributions
within one, two, three, etc. standard deviations on either side of the
population mean.
Chebyshev, also written Tchebysheff, Tchebyshev, etc., was before
Heisenberg, and no, there wasn't any trend in variances with "h --> 0".
And equation (2) is used far more often than the HUP and with
considerable success. If you multiplied the difference of the above
expressions by the same difference with X replaced by some other random
variable Y, you'd get Var(X)Var(Y), but it isn't going anywhere with or
without Heisenberg or h.
Osher Doctorow
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