-- Wrong vector spaces are non Hilbertian
Han de Bruijn
vectors in V. If such a maximum number does not exist, then we say that
the dimension of V is _infinite_. In a vector space of finite dimension
a set of N independent so called _base_ vectors can be found, such that
any vector v in V can be written as:
v = v_1.e_1 + v_2.e_2 + .. + v_N.e_N
Here { e_k } is the set of base vectors and { v_k } are the components
or coordinates of the vector v. If the vector space is provided with an
inner product (,), then the base vectors can be chosen orthogonal, i.e.
(e_i,e_j) = 0 for i <> j and (e_i,e_j) = 1 for i = j
Any orthogonal set of vectors is independent. Any vectorspace of finite
dimension has an orthogonal base (if the inner product is defined).
Let's proceed now with vector spaces of infinite dimension. Within such
an infinite vector space, we try to build up a base and seek to express
any vector as a linear combination of base vectors, and we try to do it
"as good as possible", by minimizing the following expression:
| x - [ sum_(i=1..N) a_i.e_i ] |
Here is: x = (complex) vector , { e_i } = set of N orthogonal vectors ,
a_i = vector components (complex numbers) with respect to { e_i } .
Theorem: the squared sum | x - [ sum_(i=1..N) a_i.e_i ] |^2 is minimal
with respect to the numbers a_i iff a_i = (x,e_i) = x_i (: define).
And the minimum is: (x,x) - [ sum_(i=1..N) x_i^2 ] >= 0 .
Theorem: the squared sum can be minimized further by increasing N, the
number of vectors in the would-be base of V. Therefore a vector x in V
is approximated best by:
| x - [ sum_(i=1..oo) x_i.e_i ] | = minimum
With no further information available, Bessel's inequality is provable:
(x,x) <= [ sum_(i=1..oo) x_i^2 ] .
Definition: a Hilbert space is an infinite vector space where the base
vectors are a _complete_ set, meaning that:
x = sum_(i=1..oo) x_i.e_i and (x,x) = sum_(i=1..oo) x_i^2 .
The latter equality is known as Parceval's theorem, effectively meaning
that there is an analogue of Pythagoras' theorem in any vector space of
finite or infinite dimension, the latter if Hilbertian.
Hilbert spaces may be nice, but there is NO mathematical necessity for
an infinite vector space to be a Hilbert space.
It's here where 'sci.physics' comes in ..
Miraculous Fact: in Quantum Mechanics, any infinite dimensional vector
space is a Hilbert space. How would come that orthogonal bases, in QM,
are _always_ complete? What would be the consequence if they were not?
Important note: even within _very large_ vector spaces, there is always
a complete base of orthogonal vectors. It's only within TRULY INFINITE
vector spaces that incompleteness of an orthogonal bases is possible in
theory, mathematically, but .. not physically.
The Miraculous Fact is explained by the Hypothesis that truly infinite
does not exist and that, physically speaking, there is only VERY LARGE.
Why not have the same Hypothesis in mathematics then?
Han de Bruijn
Tonico
wrote:
Quick answer: because we mathematicians don't want to and how do you
like that?
Long answer: because we mathematicans don't give a damn what physics
or other kind of a-mathematical mammals want/need/desire, and also
because we mathematicians don't want to and how do you like that?
By the way: the first part, about explaining vectors spaces, bases,
linear independency and stuff: tsk,tsk,tsk...lousy, Han...VERY lousy.
No wonder you usually talk so huge nonsense in this forum, Han: your
mathematical background, even when you cut and past, is rather
pitiful.
Regards
Tonio
> Han de Bruijn
Han de Bruijn
> Quick answer: because we mathematicians don't want to and how do you
> like that?
>
> Long answer: because we mathematicans don't give a damn what physics
> or other kind of a-mathematical mammals want/need/desire, and also
> because we mathematicians don't want to and how do you like that?
I'd rather expect some _content_ instead of your usual ranting. I have
given you the opportunity to answer with something _substantial_, like
e.g. "What you're saying is not true, because .."
> By the way: the first part, about explaining vectors spaces, bases,
> linear independency and stuff: tsk,tsk,tsk...lousy, Han...VERY lousy.
It's some essentials in a nutshell, what else do you expect? Do I have
to cut and paste a whole textbook on linear algebra, to make my point
clear?
> No wonder you usually talk so huge nonsense in this forum, Han: your
> mathematical background, even when you cut and past, is rather
> pitiful.
Did it from the top of my head, actually. But never mind. This posting
expresses my _wondering_ about the fact that infinite vector spaces in
Quantum Mechanics always have a complete base, despite of the fact that
such is not necessarily so in mathematics. What's wrong with wondering
about such things? Suppose you never suffer from such feelings, do you?
Han de Bruijn
Tonico
wrote:
> Tonico wrote:
> > Quick answer: because we mathematicians don't want to and how do you
> > like that?
>
> > Long answer: because we mathematicans don't give a damn what physics
> > or other kind of a-mathematical mammals want/need/desire, and also
> > because we mathematicians don't want to and how do you like that?
>
> I'd rather expect some _content_ instead of your usual ranting. I have
> given you the opportunity to answer with something _substantial_, like
> e.g. "What you're saying is not true, because .."
>
*****************************************************************
So now I am the one ranting, uh Han? Oooo-kay! And about the
substantial something: several people already tried that with you in
those last two threads and it was pointless.
*****************************************************************
> > By the way: the first part, about explaining vectors spaces, bases,
> > linear independency and stuff: tsk,tsk,tsk...lousy, Han...VERY lousy.
>
> It's some essentials in a nutshell, what else do you expect? Do I have
> to cut and paste a whole textbook on linear algebra, to make my point
> clear?
>
> > No wonder you usually talk so huge nonsense in this forum, Han: your
> > mathematical background, even when you cut and past, is rather
> > pitiful.
>
> Did it from the top of my head, actually. But never mind. This posting
> expresses my _wondering_ about the fact that infinite vector spaces in
> Quantum Mechanics always have a complete base, despite of the fact that
> such is not necessarily so in mathematics. What's wrong with wondering
> about such things? Suppose you never suffer from such feelings, do you?
>
***************************************************************
No, I don't...at least not with some more or less basic stuff. But I
guess that's just because I actually studied maths.
You see, Han: sometimes it's kind of a progressive process. When one
begins studying calculus, sometimes it seems obvious any "decent"
function is derivable iff it is continue...until one checks the
absolute value func. Then it looks like there can't be a continuous
function which isn't derivable in too many points, until one learns of
non-derivable everywhere but continuous everywhere func's, etc.
You say all the inf. vec. spaces appearing in QM have always complete
bases, something which I don't know if it is true, but it never minds.
I could tell you that in mathematics we sometimes talk of inf.
dimensional vec. spaces without even mentioning completeness, inner
product and stuff...why? Because there are spaces without those
things.
It seems now like you're pissed off about mathematics having infinite
vector spaces [sic] and not being Hilbert spaces, unlike (according to
you) QM...but again, that is YOUR problem.
You talk about " TRULY INFINITE vector spaces", "The Miraculous Fact
is explained by the Hypothesis that truly infinite does not exist and
that, physically speaking, there is only VERY LARGE.", etc.
It's hard to take you seriously when you talk semi-esoteric and mistic
nonsense like this, Han, and ONCE again: we mathematicians don't give
a damn, physically speaking, what physics and other non-mathematical
identities believe that mathematics SHOULD deal with.
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
>The dimension of a vectorspace V is the maximum number N of independent
>vectors in V. If such a maximum number does not exist, then we say that
>the dimension of V is _infinite_. In a vector space of finite dimension
>a set of N independent so called _base_ vectors can be found, such that
>any vector v in V can be written as:
>
> v = v_1.e_1 + v_2.e_2 + .. + v_N.e_N
>
>Here { e_k } is the set of base vectors and { v_k } are the components
>or coordinates of the vector v. If the vector space is provided with an
>inner product (,), then the base vectors can be chosen orthogonal, i.e.
>
> (e_i,e_j) = 0 for i <> j and (e_i,e_j) = 1 for i = j
>
>Any orthogonal set of vectors is independent. Any vectorspace of finite
>dimension has an orthogonal base (if the inner product is defined).
>
>Let's proceed now with vector spaces of infinite dimension.
Everything below applies only to inner-product spaces,
not general vector spaces.
>Within such
>an infinite vector space, we try to build up a base and seek to express
>any vector as a linear combination of base vectors, and we try to do it
>"as good as possible", by minimizing the following expression:
>
> | x - [ sum_(i=1..N) a_i.e_i ] |
>
>Here is: x = (complex) vector , { e_i } = set of N orthogonal vectors ,
>a_i = vector components (complex numbers) with respect to { e_i } .
>
>Theorem: the squared sum | x - [ sum_(i=1..N) a_i.e_i ] |^2 is minimal
>with respect to the numbers a_i iff a_i = (x,e_i) = x_i (: define).
>And the minimum is: (x,x) - [ sum_(i=1..N) x_i^2 ] >= 0 .
>
>Theorem: the squared sum can be minimized further by increasing N, the
>number of vectors in the would-be base of V. Therefore a vector x in V
>is approximated best by:
> | x - [ sum_(i=1..oo) x_i.e_i ] | = minimum
>
>With no further information available, Bessel's inequality is provable:
>
> (x,x) <= [ sum_(i=1..oo) x_i^2 ] .
>
>Definition: a Hilbert space is an infinite vector space where the base
>vectors are a _complete_ set, meaning that:
>
> x = sum_(i=1..oo) x_i.e_i and (x,x) = sum_(i=1..oo) x_i^2 .
No, that's not the definition of "Hilbert space". It is a somewhat
correct sloppy version of the definition of "complete orthonormal
basis". But the definition of "Hilbert space" is "complete inner-
product space". The word "complete" in the two definitions
means different things.
>The latter equality is known as Parceval's theorem, effectively meaning
>that there is an analogue of Pythagoras' theorem in any vector space of
>finite or infinite dimension, the latter if Hilbertian.
>
>Hilbert spaces may be nice, but there is NO mathematical necessity for
>an infinite vector space to be a Hilbert space.
>
>It's here where 'sci.physics' comes in ..
>
>Miraculous Fact: in Quantum Mechanics, any infinite dimensional vector
>space is a Hilbert space.
Erm, this is nonsense. It _could_ be true that Hilbert spaces
are the only spaces that come up in QM - that's not the same
thing.
>How would come that orthogonal bases, in QM,
>are _always_ complete? What would be the consequence if they were not?
>
>Important note: even within _very large_ vector spaces, there is always
>a complete base of orthogonal vectors. It's only within TRULY INFINITE
>vector spaces that incompleteness of an orthogonal bases is possible in
>theory, mathematically, but .. not physically.
Not that it matters, but this is nonsense as well. {(1,0)} is an
incomplete orthonormal set in the finite-dimensional space R^2.
>The Miraculous Fact is explained by the Hypothesis that truly infinite
>does not exist and that, physically speaking, there is only VERY LARGE.
>
>Why not have the same Hypothesis in mathematics then?
Because there's no good reason for it, and it would mean
we couldn't do a lot of the math we want to do.
I mean duh, that wasn't hard.
>Han de Bruijn
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Han de Bruijn
> On Mon, 15 Sep 2008 09:41:46 +0200, Han de Bruijn
> <Han.de...@DTO.TUDelft.NL> wrote:
>
>>The dimension of a vectorspace V is the maximum number N of independent
>>vectors in V. If such a maximum number does not exist, then we say that
>>the dimension of V is _infinite_. In a vector space of finite dimension
>>a set of N independent so called _base_ vectors can be found, such that
>>any vector v in V can be written as:
>>
>> v = v_1.e_1 + v_2.e_2 + .. + v_N.e_N
>>
>>Here { e_k } is the set of base vectors and { v_k } are the components
>>or coordinates of the vector v. If the vector space is provided with an
>>inner product (,), then the base vectors can be chosen orthogonal, i.e.
>>
>> (e_i,e_j) = 0 for i <> j and (e_i,e_j) = 1 for i = j
>>
>>Any orthogonal set of vectors is independent. Any vectorspace of finite
>>dimension has an orthogonal base (if the inner product is defined).
>>
>>Let's proceed now with vector spaces of infinite dimension.
>
> Everything below applies only to inner-product spaces,
> not general vector spaces.
You're quite right.
>>Within such
>>an infinite vector space, we try to build up a base and seek to express
>>any vector as a linear combination of base vectors, and we try to do it
>>"as good as possible", by minimizing the following expression:
>>
>> | x - [ sum_(i=1..N) a_i.e_i ] |
>>
>>Here is: x = (complex) vector , { e_i } = set of N orthogonal vectors ,
>>a_i = vector components (complex numbers) with respect to { e_i } .
>>
>>Theorem: the squared sum | x - [ sum_(i=1..N) a_i.e_i ] |^2 is minimal
>>with respect to the numbers a_i iff a_i = (x,e_i) = x_i (: define).
>>And the minimum is: (x,x) - [ sum_(i=1..N) x_i^2 ] >= 0 .
>>
>>Theorem: the squared sum can be minimized further by increasing N, the
>>number of vectors in the would-be base of V. Therefore a vector x in V
>>is approximated best by:
>> | x - [ sum_(i=1..oo) x_i.e_i ] | = minimum
>>
>>With no further information available, Bessel's inequality is provable:
>>
>> (x,x) <= [ sum_(i=1..oo) x_i^2 ] .
>>
>>Definition: a Hilbert space is an infinite vector space where the base
>>vectors are a _complete_ set, meaning that:
>>
>> x = sum_(i=1..oo) x_i.e_i and (x,x) = sum_(i=1..oo) x_i^2 .
>
> No, that's not the definition of "Hilbert space". It is a somewhat
> correct sloppy version of the definition of "complete orthonormal
> basis". But the definition of "Hilbert space" is "complete inner-
> product space". The word "complete" in the two definitions
> means different things.
See Tonico? That's what I mean by "answer with something _substantial_".
I may not always agree with Ullrich, but _he_ certainly knows something
about the _details_ in mathematics while _you_ are only pretending that
you do. Thanks David !
>>The latter equality is known as Parceval's theorem, effectively meaning
>>that there is an analogue of Pythagoras' theorem in any vector space of
>>finite or infinite dimension, the latter if Hilbertian.
>>
>>Hilbert spaces may be nice, but there is NO mathematical necessity for
>>an infinite vector space to be a Hilbert space.
>>
>>It's here where 'sci.physics' comes in ..
>>
>>Miraculous Fact: in Quantum Mechanics, any infinite dimensional vector
>>space is a Hilbert space.
>
> Erm, this is nonsense. It _could_ be true that Hilbert spaces
> are the only spaces that come up in QM - that's not the same
> thing.
Correction. I think you're quite right. So let's change the conjecture
into this: Hilbert spaces are the only infinite spaces that come up in
Quantum Mechanics.
>>How would come that orthogonal bases, in QM,
>>are _always_ complete? What would be the consequence if they were not?
>>
>>Important note: even within _very large_ vector spaces, there is always
>>a complete base of orthogonal vectors. It's only within TRULY INFINITE
>>vector spaces that incompleteness of an orthogonal bases is possible in
>>theory, mathematically, but .. not physically.
>
> Not that it matters, but this is nonsense as well. {(1,0)} is an
> incomplete orthonormal set in the finite-dimensional space R^2.
That's not what I mean and I think you know it. In finite vector spaces
(inner product spaces) there _always_ exists a complete orthogonal base,
like {(1,0),(0,1)} in R^2.
This is _not so_ with infinite dimensional inner product spaces. Right ?
WG
Division by zero. [infinities]
Renormalization;
While it works, some mathematicians and physicists feel uncomfortable about using it to get rid of infinities. Richard Feynmann calls it a dippy process without much solid mathematical basis.
Infinites arise of course by division by 0, which leads to an undefined answer.
But wait a second; we live in a mathematical universe, math simply works and is explanatory to the highest degree of precision. Every aspect of the universe grinds to the precision of math. Why does math work so well for everything .but lets us down for this one anomaly which is division by zero..
Have we got something wrong here, should we be looking at the concept of zero a little differently?
This is where I'm going to step out on a limb and suggest something radical.
In the quantum world there is no such thing as zero, only minimum quantities [i.e. Plank quantities].
Let me explain.
In pure math A-A=0 of course, but we do not live in a pure math world, we live in a quantum world where heisenburgs uncertainty principle [HUP] comes into play. In our real world we deal with applied math (not pure), all quantities are measured quantities and have units associated with them. Because of HUP any measurement has a degree of uncertainty and can never be known with complete accuracy at some level.
Take for example a measurement of distance on a line graph;
1 meter - 1 meter = 0 correct???
Wrong. In quantum mechanics one can never know on the line graph where the meter starts closer than one Plank length [1.6160 x 10-35 m]. One can never know where that 1 meter measurement ends to within 1 Plank length. Subtraction results in the same conundrum. The answer 1M-1M can never be less than the unknown we started with i.e. [1 Plank length].
This argument applies to all of measurement since there is a Plank time, Plank mass and a Plank value for every measurement one can do.
Another possible way to obtain a zero in math is to subtract objects not measurements. For example say one has an object sitting on a table and take away that object. 1 apple - 1 apple = 0 apples, correct?
This is certainly true in Newtonian space, but in the quantum universe HUP introduces a degree of uncertainty to even an apple on a table. HUP states position and Momentum [and conversely, time and energy] cannot be know to exact precision at some level.
Also in Einsteins Space time one has to consider non locality, entanglement and Bells theorem. While one can remove the apple in Newtonian space, one can never remove all traces of the apple in Einsteins space time. At some level the apple has left its signature on the universe from that point [position and time] and continue to effect the universe from that point on. There has to be a minimum effect [i.e. Plank quantity].
Everett's "Sum Over Paths (Universal Wave Function)" and Feynman's Path Integrals approach seems to fit here as well. In this approach one deals with probabilities, and the probability of the path (or wave function) of any apple in the universe crossing the table can never be zero, it can only be a minimum quantity, ie. a plank quantity.
It seems something of a coincidence that string theory which has been so successful at eliminating infinities, has done so by constructing strings which have dimensions roughly equal to Plank dimensions.
Minimum dimensions appear to have done the trick here!
Addendum;
Pure Math for the most part deals with the manipulation of unitless pure numbers and symbols, and it is here that we may frequently experience infinities [division by zero].
But in the real quantum universe [a grainy not continuous universe] we observe no infinities so it should follow that there can be no divisions by zero in applied math only divisions by minimum quantities.
Since infinities can't be dealt with, the problem must reside in our use of zero.
It appears Quantum mechanics and HUP may provide a solution to this dilemma.
"Han de Bruijn" <Han.de...@DTO.TUDelft.NL> wrote in message news:65a94$48ce5818$82a1e228$69...@news1.tudelft.nl...
Virgil
Because mathematics is not limited by any physical analogies.
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
That's why I said "not that it matters".
Your main point, or what it seems to me was your main
point, is below. It's curious that you had no comment
on my comment on your main point.
Or maybe it's not curious - your main point was
simply stupid, and there's really not much more to
be said about it.
>In finite vector spaces
>(inner product spaces) there _always_ exists a complete orthogonal base,
>like {(1,0),(0,1)} in R^2.
>
>This is _not so_ with infinite dimensional inner product spaces. Right ?
>
>>>The Miraculous Fact is explained by the Hypothesis that truly infinite
>>>does not exist and that, physically speaking, there is only VERY LARGE.
>>>
>>>Why not have the same Hypothesis in mathematics then?
>>
>> Because there's no good reason for it, and it would mean
>> we couldn't do a lot of the math we want to do.
>>
>> I mean duh, that wasn't hard.
>
>Han de Bruijn
David C. Ullrich
Han de Bruijn
In retrospect, it might be that we have a chicken and egg problem here.
Once it is discovered that the base of an infinite inner product space V
is incomplete, then we can move as many of the (non-base) vectors to the
outer space of V (i.e. remove them from V) as is needed to end up with a
inner product space that HAS a complete base for the remaining vectors.
Perhaps this, trivially, is the whole reason why inner product spaces in
QM are always complete. The orthogonal base vectors are the eigenvectors
belonging to the eigenvalue problem of a Hermitian operator. Any vector
in the Hilbert space is only admissible if it can be expressed into the
base vectors.
But, whatever the reason is, I want to _understand_ WHY oh WHY any base
in a QM (Hilbert) space is always complete. Seems a legitimate quest ..
Han de Bruijn
victor_me...@yahoo.co.uk
> But, whatever the reason is, I want to _understand_ WHY oh WHY any base
> in a QM (Hilbert) space is always complete. Seems a legitimate quest ..
The ask a quantum mechanic, not a mathematician.
Victor Meldrew
"I don't believe it!"
sorl...@hotmail.com
<snip>
> Let's proceed now with vector spaces of infinite dimension. Within such
> an infinite vector space, we try to build up a base and seek to express
> any vector as a linear combination of base vectors, and we try to do it
> "as good as possible", by minimizing the following expression:
>
> | x - [ sum_(i=1..N) a_i.e_i ] |
>
> Here is: x = (complex) vector , { e_i } = set of N orthogonal vectors ,
> a_i = vector components (complex numbers) with respect to { e_i } .
>
> Theorem: the squared sum | x - [ sum_(i=1..N) a_i.e_i ] |^2 is minimal
> with respect to the numbers a_i iff a_i = (x,e_i) = x_i (: define).
> And the minimum is: (x,x) - [ sum_(i=1..N) x_i^2 ] >= 0 .
>
> Theorem: the squared sum can be minimized further by increasing N, the
> number of vectors in the would-be base of V. Therefore a vector x in V
> is approximated best by:
> | x - [ sum_(i=1..oo) x_i.e_i ] | = minimum
>
> With no further information available, Bessel's inequality is provable:
>
> (x,x) <= [ sum_(i=1..oo) x_i^2 ] .
>
> Definition: a Hilbert space is an infinite vector space where the base
> vectors are a _complete_ set, meaning that:
>
Wrong definition. A Hilbert space is a linear space equipped with an
inner product that is complete in the sense that every Cauchy sequence
converges. It is a *theorem* that every Hilbert space has an
orthonormal base that spans the whole space (is complete in your
terminology). The theorem is proved as an easy application of Zorn's
lemma, but there may be, I am not sure, constructively valid versions
of it if you impose some restrictions like separability (good enough
for the applications to QM). In the Hilbert spaces that appear in
applications (e.g. QM) you can many times explicitely construct such
an orthonormal base.
> x = sum_(i=1..oo) x_i.e_i and (x,x) = sum_(i=1..oo) x_i^2 .
>
> The latter equality is known as Parceval's theorem, effectively meaning
> that there is an analogue of Pythagoras' theorem in any vector space of
> finite or infinite dimension, the latter if Hilbertian.
>
> Hilbert spaces may be nice, but there is NO mathematical necessity for
> an infinite vector space to be a Hilbert space.
>
Sure. There are linear spaces, there are inner product spaces and then
there are Hilbert spaces. Simple as that.
> It's here where 'sci.physics' comes in ..
>
Here comes the crank again.
> Miraculous Fact: in Quantum Mechanics, any infinite dimensional vector
> space is a Hilbert space. How would come that orthogonal bases, in QM,
> are _always_ complete? What would be the consequence if they were not?
>
As always, you got things upside down. This is not a miraculous fact,
it is actually a straightforward consequence of a mathematical fact.
Theorem A: Every inner product space can be canonically completed.
So working strictly with Hilbert spaces, instead of general inner
product spaces, is no restriction at all and that is why in the rules
of quantum mechanics you simply assume that the state spaces are
Hilbert spaces. There is nothing miraculous at work here, except
perhaps your own obtuseness.
> Important note: even within _very large_ vector spaces, there is always
> a complete base of orthogonal vectors. It's only within TRULY INFINITE
> vector spaces that incompleteness of an orthogonal bases is possible in
> theory, mathematically, but .. not physically.
>
What do you mean by _very large_? Finite dimensional but with
dimension n "very large" whatever that means? There are two different
phenomena here, that you conflate because you are ignorant of even
basic mathematics.
1. Every finite dimensional inner product space is complete.
2. Every finite dimensional inner product space has an orthonormal
base. This is hardly surprizing since saying that a linear space
finite dimensional is *by definition* the assertion that the space has
a *finite* base and Gram-Schmidt-ification provides us with the
necessary orthonormal base.
> The Miraculous Fact is explained by the Hypothesis that truly infinite
> does not exist and that, physically speaking, there is only VERY LARGE.
>
> Why not have the same Hypothesis in mathematics then?
Because, first we don't need the hypothesis, second because the
hypothesis is BAD BAD BAD. Why it is bad? Because in many
constructions, what comes out is an inner product space that we *then*
complete by applying theorem A above. This happens in mathematical
physics *also*. Off the top of my head, ever heard of the GNS
construction? Geometric quantization? Of course not, because not only
you are ignorant of mathematics but you have a narrow and parochial
view of physics.
Regards,
G. Rodrigues
David C. Ullrich
This is total nonsense.
> Perhaps this, trivially, is the whole reason why inner product spaces in
> QM are always complete. The orthogonal base vectors are the eigenvectors
> belonging to the eigenvalue problem of a Hermitian operator. Any vector
> in the Hilbert space is only admissible if it can be expressed into the
> base vectors.
>
> But, whatever the reason is, I want to _understand_ WHY oh WHY any base
> in a QM (Hilbert) space is always complete. Seems a legitimate quest ..
This is an utterly stupid "quest". First, saying "any base is complete"
is nonsense. Hard to explain exactly why, since it's hard to know
exactly what you mean. In my terminology a basis for a Hilbert
space is a complete orthonormal set, by _definition_, so if you're
talking about that then a base is always complete for exactly the
same reason as an even number is always divisible by 2.
_possibly_ by "base" you mean "orthonormal set", whether complete
or not. If so then it's simply not true that a base is always
complete - it's simply the case that complete bases are the
ones that get _used_.
> Han de Bruijn
--
David C. Ullrich
JEMebius
There is one mathematical fact that apparently did not play any role in this still ongoing
discussion on "wrong vector spaces are not Hilbertian":
There exist non-separable Hilbert spaces, for instance Harald Bohr's classical vector
space of complex-valued almost-periodic (AP) functions on the reals.
Main theorem: the complex-valued AP functions on the reals are precisely the finite
goniometric polynomials
Sum from k=1 to k=M of Ak * exp (2.pi.Nk.x) where the frequencies Nk are arbitrary reals
and their limits according to the norm derived from the inner product
(F, G) = lim for T to infinity of [1/2T * integral from -T to +T of F(x)*G(x)].
Like in separable Hilbert space, Parseval's inequality and equality are valid.
But unlike in separable Hilbert space, no single countable orthonormal basis exists to
represent =all= possible finite and infinite goniometric series. Any countable orthonormal
basis yields a separable Hilbert proper subspace of the space of all AP functions.
For each frequency N there is a basis function B(N): x -> exp (2.pi.N.x). The B(N)s form
an uncountable orthonormal basis for the Hilbert space of complex-valued AP functions on
the reals. To each separable Hilbert subspace of AP functions there belongs a countable
set of basis functions B(N), and conversely.
I guess this will help settling the debate around the issues of "large" versus "very
large" and of "moving around sets of base and non-base vectors".
Now about the question "why only separable Hilbert spaces in Quantum Mechanics?":
IMO there are at least the following two good reasons:
(A) Separable Hilbert space appears to be an adequate tool for describing certain broad
aspects of QM;
(B) The theory of AP functions was brand-new in the 1920s and virtually unknown at that
time in the areas of QM and theoretical physics in general.
As far as I know AP functions were never used in QM; neither in the Old QM
(pre-Copenhagen) nor in the beginning and later years of the New QM (post-Copenhagen).
AP function space as a mathematical model for QM may very well be overdone.
Good luck: Johan E. Mebius
Rotwang
>
> [...]
>
> But, whatever the reason is, I want to _understand_ WHY oh WHY any base
> in a QM (Hilbert) space is always complete. Seems a legitimate quest ..
Actually, in practice physicists often work with pre-Hilbert spaces
(i.e. complex vector spaces with a positive definite inner-product
which are not necessarily complete) which we refer to as "Hilbert
spaces" even though they aren't. But if one is attempting to do
rigorous mathematical physics then there are reasons why we want our
state spaces to be complete; somebody who knows a bit about rigorous
QFT could tell you more about what those reasons are than I, but an
example, I think, is that one needs the completeness property to prove
the existence of a projection operator onto a subspace with a given
value for an observable. You might find some of the replies in this
thread helpful:
http://groups.google.co.uk/group/sci.math/browse_thread/thread/ace94cf8241dd629/0d1a5f118aa13005
Regarding your OP, it's not true that "in Quantum Mechanics, any
infinite dimensional vector space is a Hilbert space". On the
contrary, many of the vector spaces which arise naturally are not
complete (for example the intermediate spaces in the construction of
Fock space as an infinite direct sum of N-particle subspaces), and so
we need to take their completion. So the fact that the spaces used in
QM are Hilbert spaces is not a "Miraculous Fact" but rather an
artifice added by hand to make the theory work. Mathematics is not so
constrained, and the reason that mathematicians study infinite
dimensional pre-Hilbert spaces which are incomplete is because such
things arise naturally among the spaces that mathematicians consider,
whether they like it or not.
Rotwang
> And from a different slant;
>
> Division by zero. [infinities]
>
> Renormalization;
>
> While it works, some mathematicians and physicists feel
> uncomfortable about using it to get rid of infinities.
> Richard Feynmann calls it a dippy process without much
> solid mathematical basis.
>
> Infinites arise of course by division by 0, which leads
> to an undefined answer.
Actually the infinities that necessitate renormalisation don't arise
from division by zero. They arise from integrals of divergent
functions over the whole of |R or |R+. And many physically meaningful
quantities calculated from the theory become ill-defined not because
they involve division by zero but because they involve infinity
divided by infinity.
Virgil
Perhaps the physical limitations imposed by QM are not the same as the
mathematical limitations of infinite dimensional real or complex vector
spaces.
After all, the set of all real valued functions on an arbitrary set form
a real vector space of 'dimension' equal to the cardinality of their
common domain.
Rotwang
>
> Perhaps the physical limitations imposed by QM are not the same as the
> mathematical limitations of infinite dimensional real or complex vector
> spaces.
>
> After all, the set of all real valued functions on an arbitrary set form
> a real vector space of 'dimension' equal to the cardinality of their
> common domain.
I don't think this is true unless we consider only functions which are
zero on a set whose complement is finite, is it? For example, isn't
the cardinality of any Hamel basis for the space of sequences of real
numbers uncountable?
Virgil
<7dd17ab8-e5e8-408c...@k13g2000hse.googlegroups.com>,
Rotwang <sg...@hotmail.co.uk> wrote:
I stand corrected.
Han de Bruijn
> _possibly_ by "base" you mean "orthonormal set", whether complete
> or not. If so then it's simply not true that a base is always
> complete - it's simply the case that complete bases are the
> ones that get _used_.
Exactly ! And I'm _only_ questioning WHY oh WHY "complete bases are the
ones that get _used_". Repeat: this is a quite legitimate quest and it's
IMO _not_ a stupid one.
Han de Bruijn
Han de Bruijn
> On 16 Sep, 12:40, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
>
>>[...]
>>
>>But, whatever the reason is, I want to _understand_ WHY oh WHY any base
>>in a QM (Hilbert) space is always complete. Seems a legitimate quest ..
>
> Actually, in practice physicists often work with pre-Hilbert spaces
> (i.e. complex vector spaces with a positive definite inner-product
> which are not necessarily complete) which we refer to as "Hilbert
> spaces" even though they aren't. But if one is attempting to do
> rigorous mathematical physics then there are reasons why we want our
> state spaces to be complete; somebody who knows a bit about rigorous
> QFT could tell you more about what those reasons are than I, but an
> example, I think, is that one needs the completeness property to prove
> the existence of a projection operator onto a subspace with a given
> value for an observable. You might find some of the replies in this
> thread helpful:
>
> http://groups.google.co.uk/group/sci.math/browse_thread/thread/ace94cf8241dd629/0d1a5f118aa13005
Yes! Now we are getting somewhere, finally .. Thanks !
> Regarding your OP, it's not true that "in Quantum Mechanics, any
> infinite dimensional vector space is a Hilbert space". On the
> contrary, many of the vector spaces which arise naturally are not
> complete (for example the intermediate spaces in the construction of
> Fock space as an infinite direct sum of N-particle subspaces), and so
> we need to take their completion. So the fact that the spaces used in
> QM are Hilbert spaces is not a "Miraculous Fact" but rather an
> artifice added by hand to make the theory work. Mathematics is not so
> constrained, and the reason that mathematicians study infinite
> dimensional pre-Hilbert spaces which are incomplete is because such
> things arise naturally among the spaces that mathematicians consider,
> whether they like it or not.
Han de Bruijn
Han de Bruijn
> On 15 Set, 08:41, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
>
> <snip>
>
>>Let's proceed now with vector spaces of infinite dimension. Within such
>>an infinite vector space, we try to build up a base and seek to express
>>any vector as a linear combination of base vectors, and we try to do it
>>"as good as possible", by minimizing the following expression:
>>
>> | x - [ sum_(i=1..N) a_i.e_i ] |
>>
>>Here is: x = (complex) vector , { e_i } = set of N orthogonal vectors ,
>>a_i = vector components (complex numbers) with respect to { e_i } .
>>
>>Theorem: the squared sum | x - [ sum_(i=1..N) a_i.e_i ] |^2 is minimal
>>with respect to the numbers a_i iff a_i = (x,e_i) = x_i (: define).
>>And the minimum is: (x,x) - [ sum_(i=1..N) x_i^2 ] >= 0 .
>>
>>Theorem: the squared sum can be minimized further by increasing N, the
>>number of vectors in the would-be base of V. Therefore a vector x in V
>>is approximated best by:
>> | x - [ sum_(i=1..oo) x_i.e_i ] | = minimum
>>
>>With no further information available, Bessel's inequality is provable:
>>
>> (x,x) <= [ sum_(i=1..oo) x_i^2 ] .
>>
>>Definition: a Hilbert space is an infinite vector space where the base
>>vectors are a _complete_ set, meaning that:
>
> Wrong definition. A Hilbert space is a linear space equipped with an
> inner product that is complete in the sense that every Cauchy sequence
> converges.
Yes, I've been sloppy. And I've been corrected already on this by David
C. Ullrich. About Cauchy sequences, Wikipedia says:
http://en.wikipedia.org/wiki/Cauchy_sequence
> It is a *theorem* that every Hilbert space has an
> orthonormal base that spans the whole space (is complete in your
> terminology). The theorem is proved as an easy application of Zorn's
> lemma, but there may be, I am not sure, constructively valid versions
> of it if you impose some restrictions like separability (good enough
> for the applications to QM). In the Hilbert spaces that appear in
> applications (e.g. QM) you can many times explicitely construct such
> an orthonormal base.
http://en.wikipedia.org/wiki/Separable_space
Search for "First examples". Then we find:
<Q> Evidently any topological space which is itself finite or countably
infinite is separable. An important example of an uncountable separable
space is the real line, in which the rational numbers form a countable
dense subset. Similarly the set of all vectors (r_1,...,r_n) in R^n in
which r_i is rational for all i is a countable dense subset of R^n , so
for every n the n-dimensional Euclidean space is separable. [snip] </Q>
So maybe separability is "good enough" for the applications in QM, just
_because_ it is good enough for all _finite_ vector spaces R^n. (We have
analogous constructs in e.g. set theory, where theorems for finite sets
are simply assumed to hold as well, by axiom then, for infinite sets.)
>> x = sum_(i=1..oo) x_i.e_i and (x,x) = sum_(i=1..oo) x_i^2 .
>>
>>The latter equality is known as Parceval's theorem, effectively meaning
>>that there is an analogue of Pythagoras' theorem in any vector space of
>>finite or infinite dimension, the latter if Hilbertian.
>>
>>Hilbert spaces may be nice, but there is NO mathematical necessity for
>>an infinite vector space to be a Hilbert space.
>
> Sure. There are linear spaces, there are inner product spaces and then
> there are Hilbert spaces. Simple as that.
>>It's here where 'sci.physics' comes in ..
>
> Here comes the crank again.
You could have formulated an answer without resorting to hostility, as
others in this thread have managed to do.
>>Miraculous Fact: in Quantum Mechanics, any infinite dimensional vector
>>space is a Hilbert space. How would come that orthogonal bases, in QM,
>>are _always_ complete? What would be the consequence if they were not?
>
> As always, you got things upside down. This is not a miraculous fact,
> it is actually a straightforward consequence of a mathematical fact.
Not "as always", but you _may_ be quite right this time.
> Theorem A: Every inner product space can be canonically completed.
>
> So working strictly with Hilbert spaces, instead of general inner
> product spaces, is no restriction at all and that is why in the rules
> of quantum mechanics you simply assume that the state spaces are
> Hilbert spaces. There is nothing miraculous at work here, except
> perhaps your own obtuseness.
You could have formulated an answer without resorting to hostility, as
others in this thread have managed to do.
>>Important note: even within _very large_ vector spaces, there is always
>>a complete base of orthogonal vectors. It's only within TRULY INFINITE
>>vector spaces that incompleteness of an orthogonal bases is possible in
>>theory, mathematically, but .. not physically.
>
> What do you mean by _very large_? Finite dimensional but with
> dimension n "very large" whatever that means? There are two different
> phenomena here, that you conflate because you are ignorant of even
> basic mathematics.
You could formulate .. oh, well. And, _sigh_, _any_ man or woman in the
street understands what "very large" means. Some people, after they have
learned mathematics, suddenly do not understand anymore common words in
common speech. Or are pretending so, in order to sound more impressive.
To me, those geniouses are impressive who can translate their knowledge
into terms that common people, given some reasonable effort, can grasp.
Some notorious guys fall in this class, such as Einstein, Feynman.
> 1. Every finite dimensional inner product space is complete.
>
> 2. Every finite dimensional inner product space has an orthonormal
> base. This is hardly surprizing since saying that a linear space
> finite dimensional is *by definition* the assertion that the space has
> a *finite* base and Gram-Schmidt-ification provides us with the
> necessary orthonormal base.
>
>>The Miraculous Fact is explained by the Hypothesis that truly infinite
>>does not exist and that, physically speaking, there is only VERY LARGE.
>>
>>Why not have the same Hypothesis in mathematics then?
>
> Because, first we don't need the hypothesis, second because the
> hypothesis is BAD BAD BAD. Why it is bad? Because in many
> constructions, what comes out is an inner product space that we *then*
> complete by applying theorem A above. This happens in mathematical
> physics *also*. Off the top of my head, ever heard of the GNS
> construction? Geometric quantization? Of course not, because not only
> you are ignorant of mathematics but you have a narrow and parochial
> view of physics.
>
> Regards,
> G. Rodrigues
http://en.wikipedia.org/wiki/Gelfand-Naimark-Segal
http://en.wikipedia.org/wiki/Geometric_quantization
What's the big deal of all this ? It truly escapes me.
Han de Bruijn
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
>David C. Ullrich wrote:
>
>> _possibly_ by "base" you mean "orthonormal set", whether complete
>> or not. If so then it's simply not true that a base is always
>> complete - it's simply the case that complete bases are the
>> ones that get _used_.
>
>Exactly ! And I'm _only_ questioning WHY oh WHY "complete bases are the
>ones that get _used_".
You keep changing the question. The answer to why complete bases
are used is that they work.
>Repeat: this is a quite legitimate quest and it's
>IMO _not_ a stupid one.
Right. Just like the guy on sci.carpentry, whose quest is to
discover WHY oh WHY carpenters use hammers to drive
nails instead of using toothpicks.
G. Rodrigues
First, Separability is good enough for QM for physical reasons which I am not going to explain, since you, as a self-proclaimed physicist, know them very well. And finite dimensional spaces are *not* good enough. In other threads, I already gave you several examples of models in QM where infinite dimensional spaces are unavoidable. Really, why do you keep repeating this mantra of yours without giving a single argument or answering to my examples? Are you purposefully blind to the evidence?
Second, the "theorems for finite sets" that hold for infinite sets, are also theorems, that is, they are proved from the axioms (of ZFC, say) via rules of inference, etc. Your statement makes it sound as though they hold because they are introduced as axioms.
<snip>
Yes, the "common man" in the street has an intuitive grasp of the Einstein tensor or the Feynman path integral, sure.
Anyway, counting me as one of those "learned mathematicians" that has forgotten the meaning of common speech, could you please enlighten me on what you mean by "very large"? Because truly, I can only guess at what you mean.
Regards,
G. Rodrigues
Han de Bruijn
> Second, the "theorems for finite sets" that hold for infinite sets, are also theorems, that is, they are proved from the axioms (of ZFC, say) via rules of inference, etc. Your statement makes it sound as though they hold because they are introduced as axioms.
Much less axioms are needed to deal only with finite sets. Try to deny.
> Yes, the "common man" in the street has an intuitive grasp of the Einstein tensor or the Feynman path integral, sure.
One can always distort someone's arguments into something nonsensical.
> Anyway, counting me as one of those "learned mathematicians" that has forgotten the meaning of common speech, could you please enlighten me on what you mean by "very large"? Because truly, I can only guess at what you mean.
Your problem.
Han de Bruijn
G. Rodrigues
>
> > Second, the "theorems for finite sets" that hold
> for infinite sets, are also theorems, that is, they
> are proved from the axioms (of ZFC, say) via rules of
> inference, etc. Your statement makes it sound as
> though they hold because they are introduced as
> axioms.
>
> Much less axioms are needed to deal only with finite
> sets. Try to deny.
>
Hmm... well you do need the axiom of infinity that guarantees the existence of an infinite set. If we are going to "talk" about infinite sets, it seems reasonable, right? What significance exactly do you attach to this triviality?
> > Yes, the "common man" in the street has an
> intuitive grasp of the Einstein tensor or the Feynman
> path integral, sure.
>
> One can always distort someone's arguments into
> something nonsensical.
>
To put it simply, you have no argument. To take the example of QM, there are many things that defy the "common man" intuition and part of the job of education is not only to educate your intuition (admittedly difficult in the case of QM) but also to unlearn some prejudices you have picked up on the way. The "common man" no-nonsense common sense could *not* have come up with something as wild as QM.
> > Anyway, counting me as one of those "learned
> mathematicians" that has forgotten the meaning of
> common speech, could you please enlighten me on what
> you mean by "very large"? Because truly, I can only
> guess at what you mean.
>
> Your problem.
>
Fair enough.
Regards,
G. Rodrigues
Rotwang
>
> <snip>
>
> First, Separability is good enough for QM for physical reasons
> which I am not going to explain, since you, as a self-proclaimed
> physicist, know them very well.
As another self-proclaimed physicist (of sorts) I'd quite like to hear
the answer to this question myself. None of the books I've seen cover
it (actually, in the only book I've read that even mentions
separability, the author says that he will assume that all the Hilbert
spaces he uses are separable and then introduces an uncountable basis
of generalised eigenvectors a few lines later).
Han de Bruijn
>>G. Rodrigues wrote:
>>
>>>Second, the "theorems for finite sets" that hold
>>
>>for infinite sets, are also theorems, that is, they
>>are proved from the axioms (of ZFC, say) via rules of
>>inference, etc. Your statement makes it sound as
>>though they hold because they are introduced as
>>axioms.
>>
>>Much less axioms are needed to deal only with finite
>>sets. Try to deny.
>
> Hmm... well you do need the axiom of infinity that guarantees the existence of an infinite set. If we are going to "talk" about infinite sets, it seems reasonable, right? What significance exactly do you attach to this triviality?
What if we're NOT going to "talk" about infinite sets.
Hard to imagine for you, perhaps ..
>>>Yes, the "common man" in the street has an
>>
>>intuitive grasp of the Einstein tensor or the Feynman
>>path integral, sure.
>>
>>One can always distort someone's arguments into
>>something nonsensical.
>
> To put it simply, you have no argument. To take the example of QM, there are many things that defy the "common man" intuition and part of the job of education is not only to educate your intuition (admittedly difficult in the case of QM) but also to unlearn some prejudices you have picked up on the way. The "common man" no-nonsense common sense could *not* have come up with something as wild as QM.
Feynman has written his "Lecture on Physics, Quantum mechanics", which
quite accessible to newcomers, say undergraduate students. Exactly as
you describe, he adresses the unlearning of prejudices, but he does it
in a manner that's unique: with great care, with great patience. Very
much the same can be said about Einstein, where he explains his theory
in the popular book "The Theory of Relativity" (if I remember well).
>>>Anyway, counting me as one of those "learned
>>
>>mathematicians" that has forgotten the meaning of
>>common speech, could you please enlighten me on what
>>you mean by "very large"? Because truly, I can only
>>guess at what you mean.
>>
>>Your problem.
>
> Fair enough.
>
> Regards,
> G. Rodrigues
Han de Bruijn