Patrick Van Esch
Aug29-04, 02:39 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>I have a few problems with the Many World interpretation of quantum\nmechanics.\n\nI think I\'ve hit upon a major problem with MW, but I might be wrong.\nSo I will post my reasoning below, and await the comments of the\nknowledgeable.\n\nThe problem is the following: what prevents us from giving "equal\nprobabilities" to each of the branches of the universe, which will\nthen differ from the Bohr probabilities.\nDeWitt\'s proof that branches where we have histories that do not\nfollow Bohr probabilities have Hilbert norm going to 0 is not\nsatisfying, because Many World proponents want at all price to avoid\nan a priori connection between the Hilbert norm and a probability,\nbecause they want to derive the Bohr rule. So a non-zero norm of a\nbranch, even if it is very small, should a priori not have anything to\nsay about the probability of being in that branch.\n\nDeutsch\'s theorem is said to solve the issue if we accept a few extra\naxioms which should be "reasonable". It is explained in the following\npaper:\n\nhttp://philsci-archive.pitt.edu/archive/00001030/00/decshortarx.pdf\n\nIn short, Deutsch\'s theorem tells us that an observer in a branch,\nbefore splitting, has only one rational way of assigning probabilities\nto future branches, and these are exactly the Bohr probabilities.\n\nHowever, I think this is sneaked into the "reasonable axioms".\nThe trick is made clear in stage 3 of the proof, on page 15.\n\nLet us reduce a bit the possibilities, for I don\'t need all the\ngenerality. Let us assume that |psi> lives in an n-dimensional\nhilbertspace, and that X is an observable with n distinct eigenvalues\nx_i. Let P be a set of n real numbers, c_i.\nWhat Deutsch tries to prove is that given a triple: {psi, X, P}, we\ncan define one and only one function V, that satisfies the properties\ncited (dominance, substitutivity etc...), and then V = |<psi|x_i>|^2 x\nc_i.\nFrom the properties V has to satisfy, it is clear that any set of\n"probabilities" one could reasonably assign to the set of eigenvalues\nshould be able to build up a V = p_i x c_i.. So indeed, taking\nDeutsch\'s theorem at first sight, if he can prove that the ONLY\npossible values of p_i are those given by the Bohr rule, leaves no\nroom for discussion: the only consistent way of assigning\nprobabilities to a state psi and an observable X is the Bohr rule.\nSo the Bohr rule has been derived from the other postulates, instead\nof being explicitly given.\nBut when looking at the original idea of MW, one is inclined to give\neach branch an equal probability. How come that we cannot assign\n"equal" probabilities to each possibility of outcome? Why can\'t we set\np_i = 1/n ? That is against Deutsch\'s theorem. So what gives ?\nWell, if you look at the proof in step 3, I think he sneaks in the\nhilbert length of the components by using the unitary equivalence\n(ME). How does he do this ?\nIn the proof of step 3, our n is set equal to 2. He builds a unitary\ncorrespondence between the 2-dim original hilbert space, and a much\nbigger hilbert space, where he assigns to the image of lambda-1 as\nmany dimensions as needed to have the length of a1 and to lambda-2 as\nmany dimensions as needed to have length a2. So it is as if we could\nsay that a Everett world of length a1 can be considered as split in a1\n"equivalent" worlds and one of length a2 identically split in a2\nworlds ; so after assigning "equal probabilities" to these split\nworlds we can recover our Born rule.\n\nSo Deutsch\'s theorem is a direct consequence of ME (measurement\nequivalence) where it is required to have equal probabilities if the\nquantum systems are related by unitary transformations which map\nbetween hilbert spaces of different dimensions. Everybody who has\nbeen raised by the Bohr rule will of course recognize the special\nstatus of unitary transformations, and accept that statement. They\nleave the inproduct, and hence the Bohr rule, invariant. So it\ndoesn\'t seem shocking to REQUIRE equivalence of probabilities under\nunitary transformations. But in MW, we do away with this a priori\nrule and we try to prove that the Bohr probabilities come out of it.\nSo we shouldn\'t include a priori equivalence of probabilities under\nunitary transformations, because that AMOUNTS TO POSTULATING THE BOHR\nRULE. In fact, Deutsch\'s proof works only if we also require ME under\ndifferent dimensionalities of Hilbert spaces. If we do not require\nthat, then the p_i = 1/n rule satisfies all of Deutsch\'s requirements.\n\nSo my conclusion is that Deutsch can only deduce the Bohr rule from MW\nafter adding it as a postulate.\n\nNow my question is: is this a known problem (with a known answer?), or\ndid I miss something or what ?\n\ncheers,\nPatrick.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>I have a few problems with the Many World interpretation of quantum
mechanics.
I think I've hit upon a major problem with MW, but I might be wrong.
So I will post my reasoning below, and await the comments of the
knowledgeable.
The problem is the following: what prevents us from giving "equal
probabilities" to each of the branches of the universe, which will
then differ from the Bohr probabilities.
DeWitt's proof that branches where we have histories that do not
follow Bohr probabilities have Hilbert norm going to is not
satisfying, because Many World proponents want at all price to avoid
an a priori connection between the Hilbert norm and a probability,
because they want to derive the Bohr rule. So a non-zero norm of a
branch, even if it is very small, should a priori not have anything to
say about the probability of being in that branch.
Deutsch's theorem is said to solve the issue if we accept a few extra
axioms which should be "reasonable". It is explained in the following
paper:
http://philsci-archive.pitt.edu/archive/00001030/00/decshortarx.pdf
In short, Deutsch's theorem tells us that an observer in a branch,
before splitting, has only one rational way of assigning probabilities
to future branches, and these are exactly the Bohr probabilities.
However, I think this is sneaked into the "reasonable axioms".
The trick is made clear in stage 3 of the proof, on page 15.
Let us reduce a bit the possibilities, for I don't need all the
generality. Let us assume that |\psi> lives in an n-dimensional
hilbertspace, and that X is an observable with n distinct eigenvalues
x_i. Let P be a set of n real numbers, c_i.
What Deutsch tries to prove is that given a triple: {\psi, X, P}, we
can define one and only one function V, that satisfies the properties
cited (dominance, substitutivity etc...), and then V = |<\psi|x_i>|^2 xc_i.
From the properties V has to satisfy, it is clear that any set of
"probabilities" one could reasonably assign to the set of eigenvalues
should be able to build up a V = p_i x c_i.. So indeed, taking
Deutsch's theorem at first sight, if he can prove that the ONLY
possible values of p_i are those given by the Bohr rule, leaves no
room for discussion: the only consistent way of assigning
probabilities to a state \psi and an observable X is the Bohr rule.
So the Bohr rule has been derived from the other postulates, instead
of being explicitly given.
But when looking at the original idea of MW, one is inclined to give
each branch an equal probability. How come that we cannot assign
"equal" probabilities to each possibility of outcome? Why can't we set
p_i = 1/n ? That is against Deutsch's theorem. So what gives ?
Well, if you look at the proof in step 3, I think he sneaks in the
hilbert length of the components by using the unitary equivalence
(ME). How does he do this ?
In the proof of step 3, our n is set equal to 2. He builds a unitary
correspondence between the 2-dim original hilbert space, and a much
bigger hilbert space, where he assigns to the image of \lambda-1 as
many dimensions as needed to have the length of a1 and to \lambda-2 as
many dimensions as needed to have length a2. So it is as if we could
say that a Everett world of length a1 can be considered as split in a1
"equivalent" worlds and one of length a2 identically split in a2
worlds ; so after assigning "equal probabilities" to these split
worlds we can recover our Born rule.
So Deutsch's theorem is a direct consequence of ME (measurement
equivalence) where it is required to have equal probabilities if the
quantum systems are related by unitary transformations which map
between hilbert spaces of different dimensions. Everybody who has
been raised by the Bohr rule will of course recognize the special
status of unitary transformations, and accept that statement. They
leave the inproduct, and hence the Bohr rule, invariant. So it
doesn't seem shocking to REQUIRE equivalence of probabilities under
unitary transformations. But in MW, we do away with this a priori
rule and we try to prove that the Bohr probabilities come out of it.
So we shouldn't include a priori equivalence of probabilities under
unitary transformations, because that AMOUNTS TO POSTULATING THE BOHR
RULE. In fact, Deutsch's proof works only if we also require ME under
different dimensionalities of Hilbert spaces. If we do not require
that, then the p_i = 1/n rule satisfies all of Deutsch's requirements.
So my conclusion is that Deutsch can only deduce the Bohr rule from MW
after adding it as a postulate.
Now my question is: is this a known problem (with a known answer?), or
did I miss something or what ?
cheers,
Patrick.
mechanics.
I think I've hit upon a major problem with MW, but I might be wrong.
So I will post my reasoning below, and await the comments of the
knowledgeable.
The problem is the following: what prevents us from giving "equal
probabilities" to each of the branches of the universe, which will
then differ from the Bohr probabilities.
DeWitt's proof that branches where we have histories that do not
follow Bohr probabilities have Hilbert norm going to is not
satisfying, because Many World proponents want at all price to avoid
an a priori connection between the Hilbert norm and a probability,
because they want to derive the Bohr rule. So a non-zero norm of a
branch, even if it is very small, should a priori not have anything to
say about the probability of being in that branch.
Deutsch's theorem is said to solve the issue if we accept a few extra
axioms which should be "reasonable". It is explained in the following
paper:
http://philsci-archive.pitt.edu/archive/00001030/00/decshortarx.pdf
In short, Deutsch's theorem tells us that an observer in a branch,
before splitting, has only one rational way of assigning probabilities
to future branches, and these are exactly the Bohr probabilities.
However, I think this is sneaked into the "reasonable axioms".
The trick is made clear in stage 3 of the proof, on page 15.
Let us reduce a bit the possibilities, for I don't need all the
generality. Let us assume that |\psi> lives in an n-dimensional
hilbertspace, and that X is an observable with n distinct eigenvalues
x_i. Let P be a set of n real numbers, c_i.
What Deutsch tries to prove is that given a triple: {\psi, X, P}, we
can define one and only one function V, that satisfies the properties
cited (dominance, substitutivity etc...), and then V = |<\psi|x_i>|^2 xc_i.
From the properties V has to satisfy, it is clear that any set of
"probabilities" one could reasonably assign to the set of eigenvalues
should be able to build up a V = p_i x c_i.. So indeed, taking
Deutsch's theorem at first sight, if he can prove that the ONLY
possible values of p_i are those given by the Bohr rule, leaves no
room for discussion: the only consistent way of assigning
probabilities to a state \psi and an observable X is the Bohr rule.
So the Bohr rule has been derived from the other postulates, instead
of being explicitly given.
But when looking at the original idea of MW, one is inclined to give
each branch an equal probability. How come that we cannot assign
"equal" probabilities to each possibility of outcome? Why can't we set
p_i = 1/n ? That is against Deutsch's theorem. So what gives ?
Well, if you look at the proof in step 3, I think he sneaks in the
hilbert length of the components by using the unitary equivalence
(ME). How does he do this ?
In the proof of step 3, our n is set equal to 2. He builds a unitary
correspondence between the 2-dim original hilbert space, and a much
bigger hilbert space, where he assigns to the image of \lambda-1 as
many dimensions as needed to have the length of a1 and to \lambda-2 as
many dimensions as needed to have length a2. So it is as if we could
say that a Everett world of length a1 can be considered as split in a1
"equivalent" worlds and one of length a2 identically split in a2
worlds ; so after assigning "equal probabilities" to these split
worlds we can recover our Born rule.
So Deutsch's theorem is a direct consequence of ME (measurement
equivalence) where it is required to have equal probabilities if the
quantum systems are related by unitary transformations which map
between hilbert spaces of different dimensions. Everybody who has
been raised by the Bohr rule will of course recognize the special
status of unitary transformations, and accept that statement. They
leave the inproduct, and hence the Bohr rule, invariant. So it
doesn't seem shocking to REQUIRE equivalence of probabilities under
unitary transformations. But in MW, we do away with this a priori
rule and we try to prove that the Bohr probabilities come out of it.
So we shouldn't include a priori equivalence of probabilities under
unitary transformations, because that AMOUNTS TO POSTULATING THE BOHR
RULE. In fact, Deutsch's proof works only if we also require ME under
different dimensionalities of Hilbert spaces. If we do not require
that, then the p_i = 1/n rule satisfies all of Deutsch's requirements.
So my conclusion is that Deutsch can only deduce the Bohr rule from MW
after adding it as a postulate.
Now my question is: is this a known problem (with a known answer?), or
did I miss something or what ?
cheers,
Patrick.