The Mathematical Substructure Underlying Quantum Field Theory, and Why I Like Anthony Zee's Book so Much
Jay R. Yablon
I have often made reference to the textbook by Anthony Zee, Quantum
Field Theory in a Nutshell. One of the things I find very attractive
about this book is the way in which he deals very simply with the
mathematical substructure which underlies quantum field theory. I
specifically have in mind Appendix 1 in Chapter 1.2 and his overall
Appendix A. At
http://jayryablon.files.wordpress.com/2009/11/zee-qft-substructure.pdf,
I have posted these two excerpts from Zee's book, as a backdrop for some
further rooting through the mathematical substructure of quantum field
theory, which I have also posted at:
In particular, I have offered some thoughts about resolving the "measure
problem" that shows up in efforts to quantize gravitation, and about how
using the Dirac delta as the D-->0 limit of a Gaussian integral provides
a possible connection between the mathematical substructures of gauge
theory and gravitational theory.
That is what I so much like about Zee's book: he does an excellent job
in laying out the mathematical substructure of QFT, rather than try to
calculate out everything under the sun, and provides an extremely
penetrating way to look at how everything ultimately hangs together.
I look forward to discussing all of this.
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Oh No
As I have commented before, I find this very odd indeed. I have done
courses at Cambridge, a PhD at London, I have read several books of
quantum field theory, perhaps half a dozen in varying degrees of detail,
and I have given a first principles development showing the real
mathematical substructure on my website. I find no overlap whatsoever
with what Zee does in these excerpts.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Jay R. Yablon
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
news:nu+FKHCt...@charlesfrancis.wanadoo.co.uk...
Well, maybe you should have also studied Zee's book. ;-)
If you are saying that Zee is wrong (as opposed to merely giving an
approach with which you are not familiar), you should try to articulate
why.
Jay.
brad
> Dear Friends:
>
> I have often made reference to the textbook by Anthony Zee, Quantum
> Field Theory in a Nutshell. �One of the things I find very attractive
> about this book is the way in which he deals very simply with the
> mathematical substructure which underlies quantum field theory. �I
> specifically have in mind Appendix 1 in Chapter 1.2 and his overall
> I have posted these two excerpts from Zee's book, as a backdrop for some
> further rooting through the mathematical substructure of quantum field
> theory, which I have also posted at:
>
>
> In particular, I have offered some thoughts about resolving the "measure
> problem" that shows up in efforts to quantize gravitation, and about how
> using the Dirac delta as the D-->0 limit of a Gaussian integral provides
> a possible connection between the mathematical substructures of gauge
> theory and gravitational theory.
>
> That is what I so much like about Zee's book: he does an excellent job
> in laying out the mathematical substructure of QFT, rather than try to
> calculate out everything under the sun, and provides an extremely
> penetrating way to look at how everything ultimately hangs together.
Is (15) not a stationary 4 dimensional wave? If so I like it!
But this is the mathematical description of inertia (in my theory).
Matter in motion would continuously create such a disturbance
and the created disturbance merely defines a shorter spacetime path.
By least action principle matter follows *easiest* path.
A minimum of 2 bodies will create such a disturbance that will result
in g.
That type of formula says 16pi. At least, that's the path I'm on.
Brad
Jay R. Yablon
"brad" <lbjohn...@yahoo.com> wrote in message
news:ad7c74dd-4f31-4562...@p33g2000vbn.googlegroups.com...
Can you be specific, mathematically, how it is that you look at (15) and
see a stationary wave?
> But this is the mathematical description of inertia (in my theory).
> Matter in motion would continuously create such a disturbance
> and the created disturbance merely defines a shorter spacetime path.
> By least action principle matter follows *easiest* path.
>
> A minimum of 2 bodies will create such a disturbance that will result
> in g.
> That type of formula says 16pi. At least, that's the path I'm on.
What do you mean by "formula says 16pi"?
Jay
>
> Brad
>
Jay R. Yablon
of the last week can be put into a one-equation, one page synopsis,
linked below:
http://jayryablon.files.wordpress.com/2009/11/one-equation-synopsis.pdf
I would like to get agreement that this one equation, which I would
maintain is a *mathematical identity*, is correct as a mathematical
matter.
Then, we can talk about the possible physics applications.
For those of you who have followed me "in labor" this past week, that
may not have been a pretty thing. Here, I think / hope, is the "baby."
Oh No
>> courses at Cambridge, a PhD at London, I have read several books of
>> quantum field theory, perhaps half a dozen in varying degrees of
>>detail,
>> and I have given a first principles development showing the real
>> mathematical substructure on my website. I find no overlap whatsoever
>> with what Zee does in these excerpts.
>
>Well, maybe you should have also studied Zee's book. ;-)
Perhaps I did not express myself strongly enough. The extracts you have
presented are enough to convince me that under no circumstances should
anyone study Zee's book. Zee himself should probably be taken out and
shot. For example what he gives as Wick's theorem isn't actually a
statement of Wick's theorem at all, but is followed with a vague
rambling not entirely unrelated to Wick's theorem, and the straight
replacement of J by iJ and a with -ia in 12 & 13 is quite unjustified,
and gives a quite misleading notion of what one can do in maths.
>
>If you are saying that Zee is wrong (as opposed to merely giving an
>approach with which you are not familiar), you should try to articulate
>why.
>
I think it is wrong to call something "the central identity of quantum
field theory" when it actually has no particular role in quantum field
theory. In fact Gaussians have no role in qft, or in qm generally, in
which wave functions are complex, not real like Gaussians.
The only way in which this identity even has an application or use is to
consider the gaussian as a nascent delta function. Then of course you
will be able to get the same results as by using any other delta
function. However this is not the nascent delta function which occurs
naturally from the use of Hilbert space in qm, and its use is be quite
artificial and contrived. Calling it a
brad
> "brad" <lbjohnson1...@yahoo.com> wrote in message
Actually, what I see looks like the frequency of a 1 dimensional, time
dependent wave in a homogeneous medium.
I'm unsure how to translate that into 4 D spacetime.
frequency=angular frequency/2pi.
The right side of 15 is 4times the unit disturbance per +++- No?
Lagrangian_EH/2pi is *unit*. correct?
Maybe I'm mistaken . But, isn't Lagrangian_EH a defined spacetime?
In essence a class of disturbance? And 2pi the geometric resolution?
To my mind if it's a type of disturbance then the equation is similar
to yours.
If it's generated by matter in motion then the disturbance will be
dependent on the instantaneous
speed of that matter. It can'tpropagate faster than it is created .
Essentially a standing wave for moving matter. I don't think I'm
disagreeing with Einstein here.
The remainder of my post is even more speculative. I should not have
posted it!
Brad
neuropulp
town square. It seems that Prince Charles has issued an execution
order
upon the much-respected wise man Anthony Galileo de Zee for heresy,
after
seeing by a tiny extract of the wise man's tome...
Prince Charles decrees:
> The extracts you have presented are enough to convince me
> that under no circumstances should anyone study Zee's book.
> Zee himself should probably be taken out and shot. [...]
> the straight replacement of J by iJ and a with -ia in 12 &
> 13 is quite unjustified, and gives a quite misleading notion
> of what one can do in maths.
But my dear Prince, 'tis but an application of elementary
analytic continuation and no heresy at all.
> it is wrong to call something "the central identity of
> quantum field theory" when it actually has no particular
> role in quantum field theory. In fact Gaussians have no role
> in qft, or in qm generally, in which wave functions are
> complex, not real like Gaussians.
Gaussian integrals quickly become crucial in the path integral
approach
to QFT, both in the sense that one needs Gaussian (Wiener) measure to
make sense of the path integral (aka "generating functional), and also
that perturbative calculations by this method rely on functional
derivatives applied to the generating functional.
(BTW, Gaussian integrals are also useful in qm, in particular quantum
optics based on the coherent state approach. They also occur in more
generalized coherent state theory.)
Please spare this wise man from your wrath, oh Prince.
---
LoL from the Princess as she returns to her rest.
Ken S. Tucker
Hi Princess and all.
I (we) sure hope you will return to full health, get well soon :-).
I have Zee's book and admire our resident genius's Charles
and Jay, though I'll side with Charles (except for the shooting),
from the PoV of a physics student studying Zee's book.
But I'm lurkin' this thread to see how things turn out.
Regards
Ken S. Tucker
Oh No
>>From her hospital bed, the Princess hears a commotion outside in the
>town square. It seems that Prince Charles has issued an execution
>order
>upon the much-respected wise man Anthony Galileo de Zee for heresy,
>after
>seeing by a tiny extract of the wise man's tome...
>
>Prince Charles decrees:
>
> > The extracts you have presented are enough to convince me
> > that under no circumstances should anyone study Zee's book.
> > Zee himself should probably be taken out and shot. [...]
> > the straight replacement of J by iJ and a with -ia in 12 &
> > 13 is quite unjustified, and gives a quite misleading notion
> > of what one can do in maths.
>
>But my dear Prince, 'tis but an application of elementary
>analytic continuation and no heresy at all.
Allow me to take a slightly simpler instance in which all may see that
there is, at least potentially, a problem.
I = integral_0^oo dx exp(-ax)
Clearly for Real a>0 we have
I = 1/a
Now put a -> -a Do we have I = -1/a? What about substituting a -> ia ?
What I am saying is that even if Zee's result is correct, his
justification is inadequate.
> > it is wrong to call something "the central identity of
> > quantum field theory" when it actually has no particular
> > role in quantum field theory. In fact Gaussians have no role
> > in qft, or in qm generally, in which wave functions are
> > complex, not real like Gaussians.
>
>Gaussian integrals quickly become crucial in the path integral
>approach
>to QFT, both in the sense that one needs Gaussian (Wiener) measure to
>make sense of the path integral (aka "generating functional), and also
>that perturbative calculations by this method rely on functional
>derivatives applied to the generating functional.
Forgive me if I am wrong, beloved princess, but I think Zee said the
Central identity of Quantum Field Theory, not the Central Identity of
the path integral approach to quantum field theory.
>(BTW, Gaussian integrals are also useful in qm, in particular quantum
>optics based on the coherent state approach. They also occur in more
>generalized coherent state theory.)
Making an appearance in is somewhat different from being central to. Can
you explain how the integrand remains Gaussian, and real, when it
evolves?
>Please spare this wise man from your wrath, oh Prince.
>
Come, Princess, thou knowest that my wrath is just a pose, for the
entertainment of the audience, and has no intent behind it.
>LoL from the Princess as she returns to her rest.
>
Mike
> Gaussian integrals quickly become crucial in the path integral
> approach
> to QFT, both in the sense that one needs Gaussian (Wiener) measure to
> make sense of the path integral (aka "generating functional), and also
> that perturbative calculations by this method rely on functional
> derivatives applied to the generating functional.
>
The path integral can be derived by inserting the identity an infinite
number of time and then substituting the gaussian form of the dirac
delta for each of the inner products <xi|xj>. Typically the gaussian
form is justified through a convoluted reasoning using the schrodinger
equation. But I wonder if one could more directly infer the use of a
gaussian form for the <xi|xj>. Then the path integral would be derived
entirely on principle alone.
Is there any merit to this reasoning for the use of the gaussian form
of the dirac delta function for each of the <xi|xj>? Since the path
integral includes every product, <xi|xj>, with equal weight, then
singling out any one for special consideration must have been an
arbitrary and thus totally random choice. Does this mean that the
inner product <xi|xj>, which is a distribution, must be represented by
a random process itself, by a gaussian distribution? What else could
<xi|xj> mean if not the probability of obtaining it? That's how we are
accustomed to thinking about <xi|xj> in QM. But more generally, is
that how we should consider and inner product, as a amplitude or
probability of obtaining it? It seems to me that <xi|xj> has no
meaning apart from the context of all the other vectors in the space
that |xi> and |xj> are a part of. In fact it seems necessary to
consider all the rest of the |xn> in order to even calculate <xi|xj>.
For we have that <xi|xj> = 0 if i not equal j and <xi|xj>=1 if i=j. So
it seems that the evaluation of <xi|xj> depends on the context of
choicing it from all the other possible <xi|xj>. Any further thoughts
out there? Maybe I'm touching on some mathematics that I'm not
familiar with
brad
>>
> Actually, what I see looks like the frequency of a 1 dimensional, time
> dependent wave in a homogeneous medium.
> I'm unsure how to translate that into 4 D spacetime.
I wish to elaborate on this point. The formula for this disturbance
has
2pi in the denominator. If I expand it to 3 dimensions then the
denominator
will go to 8pi. If two objects interact then the space of their
interaction
will be some combination of two disturbances; so the denominator of
the
formula describing that interaction will include the term - 16pi.
However, the time parameter is not included in this formulation. If
the first
formula (with 8pi) is considered to be the wave associated with
inertial
motion then that motion is described as velocity . How do I describe
Time?
Or is it already spacetime?
If two objects are interacting then the motions will be described as
accelerations,
and the time parameter is squared. If a new *force* is applied then an
acceleration
would also result in some time dependent spacial distortion. At least
until a new
velocity is attained. In light of the previous paragragh should my
wave equations have
16pi or 32pi in the denominator?
I am uncertain as to whether time may be an inate property of motion;
and so, should not
be factored into the original dimensional figuration. Or is it
integral to space. I know
what GR says. I'm just not sure if it's emergent.
This hypothesis demands that a more severe disturbance would result
from greater
velocities; and so proper time must vary with velocity as per SR. A
more complex
disturbance results from the interaction of 2 objects and the
resulting complexity
gives rise to accelerated motion including gravity.
Because the disturbance is velocity dependent, it becomes increasingly
difficult to alter
the trajectory. But, this is the definition of mass- resistance to an
acceleration.
So inertia, mass, and gravity are all different aspects of the same
relationship between matter and space.
And I wonder if Time is an emergent dimension dependent on the
relationship between matter and space.
Brad
Oh No
Prince Charles is pacing the throne room, muttering to himself. He has
sorely missed the attentions of Princess Spank-a-lot, and had been most
excited by her sudden reappearance, but now she has again vanished.
Perhaps a different tack is required,
Thus spake Oh No <No...@charlesfrancis.wanadoo.co.uk>
>Thus spake neuropulp <neur...@yahoo.com.au>
>>>From her hospital bed, the Princess hears a commotion outside in the
>>town square. It seems that Prince Charles has issued an execution
>>order
>>upon the much-respected wise man Anthony Galileo de Zee for heresy,
>>after
>>seeing by a tiny extract of the wise man's tome...
>>
>What I am saying is that even if Zee's result is correct, his
>justification is inadequate.
I should have said "even though... "
The fair Portia has spoken eloquently in Antonio's defence, but it does
not excuse that no mention has been made that absolute convergence is
necessary to the proof. Antonio should have made mention, and if proving
absolute convergence is too tedious (as mayhap be the case) then he
should have said so, and not have presented the result as though
derived.
>>Please spare this wise man from your wrath, oh Prince.
Ah, now I see the problem. Of course Zee is not worthy of special
treatment. What we should actually do is line up all the authors of text
books, as well as professors of qft, who do not pay attention to such
issues as absolute convergence, and the order of taking limits, and have
the lot of them shot. The days when one could say "its a material field,
so it must give convergence" are past. There is no justification for the
claim that quantum fields are somehow material, beyond the crazy ideas
of Pascual Jordan, who was a Nazi and probably even believed in having
people shot, so we could start with him if he wasn't already dead.
By Royal Decree
Prince Charles
neuropulp
> Thus didn't spake the masked quantum damsel [...]
Ha! Prince Charles, you've succeeded in becoming only the
second person on spf to make _me_ laugh out loud. (And I
thank the inventor of monocryl surgical dressings that I
didn't get taken down to the O.R. again as a result.)
> Prince Charles is pacing the throne room, muttering to
> himself. He has sorely missed the attentions of Princess
> Spank-a-lot, and had been most excited by her sudden
> reappearance, but now she has again vanished.
She is too tired for anything more than a short
conversation. Certainly not for enthusiastic spanking.
Basically, I didn't really disagree with anything you said.
But since you haven't read Zee (iiuc), it takes some effort
to explain that apparent contradiction.
Since you've made my morning in this sad place a little less
sad, I'll make the effort...
I think Zee's book achieves it stated aims, which are to
give a broad brush overview of many areas of application of
QFT, aimed at the physics student who knows some QM, but is
not a maths major, and who is rearing in the harness to get
into QFT somehow.
Can one achieve proficiency in a range of QFT math areas and
a respectably wide range of useful QFT calculations just by
studying Zee's book? No, not even close. But that's not what
the book was ever intended for.
Used wisely, Zee's book should help inspire students to get
into more thorough and difficult material. Used unwisely, it
creates a false sense that the superficial presentation of
the path integrals (and indeed the canonical formalism)
is mathematically adequate for all purposes.
> What we should actually do is line up all the authors of
> text books, as well as professors of qft, who do not pay
> attention to such issues as absolute convergence, and the
> order of taking limits, and have the lot of them shot.
IMHO, we should line up all the naughty boys who don't read
publications thoroughly before passing judgment, and have
the lot them severely spanked. But... hmmm... no. In your
case that would just encourage the bad behaviour.
---
LoL from the Princess!
P.S: Sir Mike, maybe I didn't really understand what you were
saying or what question you were trying to ask. But if
you're not familiar with "rigged Hilbert spaces" and their
role in quantum theory, that's probably something you
should look at if you want to know the rigorous meaning
of <xi|xj> in QM. Try quant-ph/0502053.
Approximating delta fns by a sequence of Gaussians is all
well and good, but it's not unique because sequences of other
so-called "nascent" delta fns also exist. (Look up the Wikipedia
entry on the Dirac delta.) What matters is the limit. But for
that, one needs to understand distribution theory well.
Oh No
>
> > Thus didn't spake the masked quantum damsel [...]
>
>Ha! Prince Charles, you've succeeded in becoming only the
>second person on spf to make _me_ laugh out loud. (And I
>thank the inventor of monocryl surgical dressings that I
>didn't get taken down to the O.R. again as a result.)
It is indeed good to know that there has not been permanent damage to
the Princess's sense of humour. There were serious fears at court when
it was learned that she was not at all amused by the image of the Prince
charging around the countryside issuing challenges to automobiles in
vengeance for her condition. Btw, he couldn't catch the moving ones, and
those that were parked remained silent, refusing to respond to his
challenge. So, he was forced to declare them all cowards and come home
without doing any damage at all.
>I think Zee's book achieves it stated aims, which are to
>give a broad brush overview of many areas of application of
>QFT, aimed at the physics student who knows some QM, but is
>not a maths major, and who is rearing in the harness to get
>into QFT somehow.
>
>Can one achieve proficiency in a range of QFT math areas and
>a respectably wide range of useful QFT calculations just by
>studying Zee's book? No, not even close. But that's not what
>the book was ever intended for.
>
>Used wisely, Zee's book should help inspire students to get
>into more thorough and difficult material. Used unwisely, it
>creates a false sense that the superficial presentation of
>the path integrals (and indeed the canonical formalism)
>is mathematically adequate for all purposes.
This sounds like excellent advice, and I hope is appreciated by Sir Jay.
> > What we should actually do is line up all the authors of
> > text books, as well as professors of qft, who do not pay
> > attention to such issues as absolute convergence, and the
> > order of taking limits, and have the lot of them shot.
>
>IMHO, we should line up all the naughty boys who don't read
>publications thoroughly before passing judgment, and have
>the lot them severely spanked. But... hmmm... no. In your
>case that would just encourage the bad behaviour.
but then if I were better behaved and said nothing, both I and Sir Jay
would not have benefited from the Princess's correction.
Is there any chance she might be able to find the energy for
http://papers.rqgravity.net/RQGQED.pdf ?
If you remember she came to a halt because, owing to a sudden attack of
dysentery, something very unpleasant had been left on the page. The mess
has now been cleared up. The equations are even numbered, whether I
reference them or not.
>---
>LoL from the Princess!
>
>P.S: Sir Mike, maybe I didn't really understand what you were
>saying or what question you were trying to ask. But if
>you're not familiar with "rigged Hilbert spaces" and their
>role in quantum theory, that's probably something you
>should look at if you want to know the rigorous meaning
>of <xi|xj> in QM. Try quant-ph/0502053.
>
>Approximating delta fns by a sequence of Gaussians is all
>well and good, but it's not unique because sequences of other
>so-called "nascent" delta fns also exist. (Look up the Wikipedia
>entry on the Dirac delta.) What matters is the limit. But for
>that, one needs to understand distribution theory well.
>
I second this. Regrettably the table of nascent delta functions which
was on this page has been moved to a button on the talk page. I would be
interested to know if anyone agrees with me that it should be restored.
I actually think a better approach is the one I use in my papers and
website using finite dimensional Hilbert space. This leads to a natural
choice of nascent delta function
http://rqgravity.net/SomeBitsofMathematics#RepresentationsOfTheDeltaFunc
tion
Though I confess the relationship between what I have done and rigged
Hilbert space is by no means as straightforward as I had anticipated.
Ken S. Tucker
On Nov 26, 9:39 am, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
...
> I have Zee's book and admire our resident genius's Charles
> and Jay, though I'll side with Charles (except for the shooting),
> from the PoV of a physics student studying Zee's book.
> But I'm lurkin' this thread to see how things turn out.
> Regards
> Ken S. Tucker
Thanks to Neuropulp for this ref,
http://en.wikipedia.org/wiki/Dirac_delta
It appeals to an electronic teky, like me, after staring at
meters, silly scopes and frequency analyzers for years.
Jay, I checked the index and contents of Zee's for a clear
statement on Conservation of Energy (CoE) as he understands it,
and had a problem finding one, it could be that is too basic
to be included.
Power (W), such as outputted by a 60 Watt bulb, is demo'd from
the photoelectric effect to be quantized.
Would you generalize that power is quantized?
Let's assume so.
Also "Current" (I=charges (e) / sec) is assumed quantized,
then Voltage = W/I is quantized.
Let's generalize Voltage to the EM potential 4-vector A_u
then quantize by excluding continuous variation using the
covariant derivative,
A_u;v = 0, but the partial A_u,v =/=0 .
We can snug that up by placing a charge q subject to the
EM potential (and the voltage) using,
q*A_u;v= 0 = P_u;v expressed as an energy.
Using that technique, we obtain CoE, by quantizing the
energy density,
T_uv;w = 0 , though T_uv,w =/=0.
We see at first,
P_u;v = 0 = P_u,v - {s,uv} P_s , (1)
P_u,v = {s,uv} P_s (2)
with { } being a Christoffel of the 2nd kind,
and "s" is a summy dummy index.
Changing Energy is power, so I suppose we can call
that a "Christoffel Powered Universe", (kinda pretty eh),
We still have the EM field tensor via the asymmetric,
q*F_uv = a*(A_u,v - A_v,u) from above,
and we're quantized.
I think Eq.(2) embodies a quantized CoE, however
it appears to need more constraint.
Regards
Ken S. Tucker
Mike
> >saying or what question you were trying to ask. But if
> >you're not familiar with "rigged Hilbert spaces" and their
> >role in quantum theory, that's probably something you
> >should look at if you want to know the rigorous meaning
> >of <xi|xj> in QM. Try quant-ph/0502053.
>
> >Approximating delta fns by a sequence of Gaussians is all
> >well and good, but it's not unique because sequences of other
> >so-called "nascent" delta fns also exist. (Look up the Wikipedia
> >entry on the Dirac delta.) What matters is the limit. But for
> >that, one needs to understand distribution theory well.
>
>
> I actually think a better approach is the one I use in my papers and
> website using finite dimensional Hilbert space. This leads to a natural
> choice of nascent delta functionhttp://rqgravity.net/SomeBitsofMathematics#RepresentationsOfTheDeltaFunc
> tion
>
> Though I confess the relationship between what I have done and rigged
> Hilbert space is by no means as straightforward as I had anticipated.
>
Path integrals, delta functions, and Hilbert spaces, Oh my!
These seem to touch on issues that concern me at present. It seems a
Feynman type path integral can be obtained by using the recursive
property of the Dirac delta function and then substituting the
gaussian form of the delta function for the infinite number of delta
functions inside the infinite number of integrations of the path
integral. It's a pretty easy procedure. See details at:
http://hook.sirus.com/users/mjake/delta_physics.htm
This give what appears to be at least the path integral with a
lagrangian for a free particle. But what about the potential term of
the lagrangian, and what about lagrangians with other kinds of fields
like those used in QFT? So I wonder if the next step is to replace the
velocity in the lagrangian with any function. The only restriction on
the type of functions that can be used is that they are square
integrable so that the exponential still exists. And I understand that
the space of square integrable functions forms a Hilbert space. Maybe
this is why Hilbert spaces are relevant in QM. One of the questions I
have with this idea is what it does to the gaussian delta function if
one replaces the (x-x_0) term with an arbitrary function. Is it still
a gaussian? Is it still a delta function which integrates to 1? I
wonder how much functional analysis I'll have to learn in order to
address these issues? Your comments would be appreciated. Thanks.
Jay R. Yablon
"neuropulp" <neur...@yahoo.com.au> wrote in message
news:f1d6e246-f451-45de...@o9g2000prg.googlegroups.com...
>
> > Thus didn't spake the masked quantum damsel [...]
>
> Ha! Prince Charles, you've succeeded in becoming only the
> second person on spf to make _me_ laugh out loud. (And I
> thank the inventor of monocryl surgical dressings that I
> didn't get taken down to the O.R. again as a result.)
Well Princess, you probably know as well as I do, so as not to be
outdone, that Prince Charles has been trying really hard to do so for
many months. ;-)
In all seriousness, we are all very glad to see you back in action, and
pray that your recovery is going well and will continue to do so. You
are a delightful amelioration in tone to all of the alpha males who tend
to inhabit SPF and all of the physics newsgroups, and your substantive
commentary it always top grade! And the princess pleading for leniency
for wise man Zee following the edicts of the Court brings classical,
Shakespearean drama to the proceedings of our group.
>
> > Prince Charles is pacing the throne room, muttering to
> > himself. He has sorely missed the attentions of Princess
> > Spank-a-lot, and had been most excited by her sudden
> > reappearance,
I am sure he has . . . someone please dispatch a firehose with some cold
water. ;-)
> but now she has again vanished.
>
> She is too tired for anything more than a short
> conversation. Certainly not for enthusiastic spanking.
>
> Basically, I didn't really disagree with anything you said.
> But since you haven't read Zee (iiuc), it takes some effort
> to explain that apparent contradiction.
>
> Since you've made my morning in this sad place a little less
> sad, I'll make the effort...
>
> I think Zee's book achieves it stated aims, which are to
> give a broad brush overview of many areas of application of
> QFT, aimed at the physics student who knows some QM, but is
> not a maths major, and who is rearing in the harness to get
> into QFT somehow.
>
> Can one achieve proficiency in a range of QFT math areas and
> a respectably wide range of useful QFT calculations just by
> studying Zee's book? No, not even close. But that's not what
> the book was ever intended for.
>
> Used wisely, Zee's book should help inspire students to get
> into more thorough and difficult material. Used unwisely, it
> creates a false sense that the superficial presentation of
> the path integrals (and indeed the canonical formalism)
> is mathematically adequate for all purposes.
>
I would agree completely with Princess on this, and as the provocateur
of this discussion (Sir Charles in not the only one here who may
overstate a point for dramatic effect), am glad that you have restored
balance on both ends of the spectrum.
I turns out -- per some parallel threads in sci.physics.research, but it
takes them too darned long to post things over there -- that the
calculations I used at the outset of this post contained an extremely
careless error overlooked in haste, and is not of lasting value other
than in the lessons learned and insofar as it got me thinking a great
deal about delta functions and nascent deltas and their possible role in
physics and whether the Gaussians of QED are to be regarded from some
perspective as nascent deltas and what this would mean if so and how
this might be exploited.
Zee is in many ways a good book for someone attempting creative work in
physics, because while he is not exhaustive by his own clear admission,
he does focus on particular aspects of physics which he feels show the
underlying elegance of QFT and which his own intuition tells him are the
pieces of QFT which are most likely to be centerpieces in further
advances in the subject. He may or may not be right, and in some
instances is likely right and in others may be off base. But, as
important as it is to have people like Weinberg writing the exhaustive
encyclopedia, it is equally important to have a person of stature such
as Zee writing a book emphasizing what he thinks will turn out to be the
lasting pieces in a "to be developed" unifying theory. Of course that
is a subjective judgment, but when you put things in balance, both
approaches are necessary and neither is to be scorned.
For a thread which started with something containing a careless error
and a bit of mischievous overstatement by me which Sir Charles swallowed
with the hook and the line and the sinker, this has actually been very
productive in illuminating certain core questions about QFT which I
submit are at the forefront of physics, controversial, and by no means
settled. One of these, is that of how one balances the path integral
formulation of QFT, with the canonical formulation. Clearly, the
teaching pedagogy (and the history) of physics greatly leans toward the
canonical formulation, and Zee make a deliberate effort to weight things
toward the path integral, perhaps to restore some balance, and perhaps
because as x-phy said over in spr, the path integral formulation is the
"artist's" view of QFT. And, history has shown that in the end, it is
the "artistic" views, e.g., Maxwell and Einstein, which survive over
time in a transcendent and lasting way. Certainly, I personally find
the path integral formulation much more palatable, because one can
assimilate this on an intuitive ground, whereas the canonical
formulation seems too much like a collection of rules which are they way
that are because that's the way they are. Or, as someone once said,
just shut up and study!
The question that this raises for me, is this: Zee states at one point
in his first few sections that the path integral and the canonical
formulations are "complementary" approaches to quantum field theory. He
also points out that in many cases one approach illuminates some things
more clearly than other, and in other cases, vice versa. My question
now, is this: where are those point of contact where one can see most
clearly, the interrelation between these two complementary views /
approaches. Put in other words, is there some "correspondence" theorem
that clearly shows as directly as possible, where these two formulations
meet up and are seen as the same side of one coin? Where is the place
of unity between these two approaches?
In the hope that I will motivate the Princes to start spanking someone
again because the Doctor says this would be an important step forward in
her recovery ;-) let me at least lay out what my intuition tells me
about all of this. Heisenberg uncertainty, which is the first step one
gets to from the canonical commutation relationships, is fundamentally
connected with the mathematics of Fourier transforms. My 24-year old
son Joshua who I am proud to say is starting a doctoral program in
quantum optics and is perhaps headed to a dissertation about gravity
waves, says it very well: "If you get into a fight with a bully, ask him
to step outside into momentum space." When one sees things things in
Fourier terms, then the fact that a tight Gaussian in one space is a
broad Gaussian in a complementary space is extremely natural and does
not require one to think in terms of some ad hoc set of canonical rules.
Things like position / momentum uncertainties seem not odd, but a
natural consequence of Fourier math. This is just the way that Fourier
transforms of Gaussian functions inherently operate, and it seems
perfectly natural.
If one this thinks of the canonical formulation as one that centers
around the complementary spreads of Gaussians in spacetime and momentum
(Fourier) space, then that meets up very nicely with the path integrals
which are the focus of Zee, because at least when one is looking at
gauge theory path integrals (and I'll set aside the Einstein-Hilbert
action for now because having chased that one for the last couple of
weeks all I can say is that that is a strange animal in comparison), the
mathematical core of the generating functional is a Gaussian, and the
path integral transforms a Gaussian in the space of "spacetime and
fields" (via the term "expi(A.D.A)" into a complementary Gaussian in the
space of "momentum and sources" (via "expi(J.D^-1.J)").
So, at some basic level, it seems to me that the two main formulations
of QFT find their common ground around Gaussian distributions and
Fourier transformations. I say this as an "intuitive" reaction to all
that I see, with the intent of simplifying rather than complexifying. I
would be interested in seeing who has put some precision to the question
of where these two formulations meet up, and how they do so.
Jay
>
> > What we should actually do is line up all the authors of
> > text books, as well as professors of qft, who do not pay
> > attention to such issues as absolute convergence, and the
> > order of taking limits, and have the lot of them shot.
>
> IMHO, we should line up all the naughty boys who don't read
> publications thoroughly before passing judgment, and have
> the lot them severely spanked. But... hmmm... no. In your
> case that would just encourage the bad behaviour.
>
> ---
> LoL from the Princess!
I know it will be a heresy to Prince Charles, but rigor must be balanced
with creativity and even some artistic license, if physics is to be
advanced. And that has to all be done recognizing that there will be
bumps and bruises and mistakes and even spankings along the way. If one
is looking to pass exams, then by all means, be rigorous at every step
and do not ignore anything. But creativity is a different animal --
even in a tightly constrained pursuit such as physics.
Princess, again, welcome back, and wishing you much good health and
speedy recovery.
Jay
>
> P.S: Sir Mike, maybe I didn't really understand what you were
> saying or what question you were trying to ask. But if
> you're not familiar with "rigged Hilbert spaces" and their
> role in quantum theory, that's probably something you
> should look at if you want to know the rigorous meaning
> of <xi|xj> in QM. Try quant-ph/0502053.
>
> Approximating delta fns by a sequence of Gaussians is all
> well and good, but it's not unique because sequences of other
> so-called "nascent" delta fns also exist. (Look up the Wikipedia
> entry on the Dirac delta.) What matters is the limit. But for
> that, one needs to understand distribution theory well.
>
Mike recently posted to clarify his inquiry.
Jay R. Yablon
"Mike" <mj...@sirus.com> wrote in message
news:3c328d7c-011f-472c...@9g2000yqa.googlegroups.com...
We are in parallel thinking patterns. Actually, however, reverse the
direction of your thinking. Starting with a delta, once cannot
exclusively state what the nascent function is. But, starting with a
Gaussian, it only goes one way -- into a delta.
So, start with the Gaussians of path integral QED. Treat these as
nascent deltas. DON'T use the +i epsilon prescription to avoid the
pole, but fearlessly run into the poles and accept delta functions as
the consequence. As these Gaussians approach deltas, the momentum space
propagator denominator p^2-m^2 approaches zero. And, it seems, the x-y
in the spacetime propagator D(x-y) approaches 0.
What this all means, I am still not sure. I had thought that this
limiting case would be quantum gravitation, but the math that made me
think this was in error. Yet, my intuition tells me that this should be
the end result, i.e., that quantum gravitation should be the limiting
case of QED in which Gaussians, treated as nascent deltas, turn to true
deltas. How to show this, if it is so, I do not know.
Jay
neuropulp
> Path integrals, delta functions, and Hilbert spaces, Oh my!
>
> These seem to touch on issues that concern me at present. It
> seems a Feynman type path integral can be obtained by using
> the recursive property of the Dirac delta function and then
> substituting the gaussian form of the delta function for the
> infinite number of delta functions inside the infinite
> number of integrations of the path integral. It's a pretty
> easy procedure.
It's also pretty easy to indulge in "dishonest mathematics"
(to borrow a colleague's phrase). The infinite-dimensional
measure "dx1 dx2 dx3 ....." does not exist. See
http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure
> See details at:
>
> http://hook.sirus.com/users/mjake/delta_physics.htm
Uh oh! That site looks intensely theological. I don't know what a
seriously sinful little girl like me can add.
(Hmmm. Maybe I should be calling you "Pastor Mike"?)
> [...] I understand that the space of square integrable
> functions forms a Hilbert space. Maybe this is why Hilbert
> spaces are relevant in QM.
Sounds like you need to study a textbook on QM. Try this one:
Ballentine, "Quantum Mechanics - A Modern Development".
> I wonder how much functional analysis I'll have to learn in
> order to address these issues?
Quite a lot, by the sounds of things. Functional analysis is needed
even to just understand the mathematics of nonrelativistic QM in an
honest thorough way.
neuropulp
> he couldn't catch the moving ones, and those that were
> parked remained silent, refusing to respond to his
> challenge. So, he was forced to declare them all cowards
> and come home without doing any damage at all.
Then Sir Basil of Fawlty is a better man than you. His vehicle
was also silent but he gave it a good thrashing anyway.
> Is there any chance she might be able to find the energy for
> http://papers.rqgravity.net/RQGQED.pdf ?
In the drawer beside my hospital bed are 6 unread textbooks,
and not a few papers (and soon to be 8 unread textbooks if
my parents indulge my request for an early Xmas present and
buy me those 2 expensive volumes of Zeidler that recently got
favorable mention by the Fearsome Sage of Vienna on spr).
But if there's something specific you want to me look at in
RQGQED.pdf, maybe I can fit it in between physio sessions,
provided it doesn't require me to study the whole thing
thoroughly.
> I actually think a better approach is the one I use in my
> papers and website using finite dimensional Hilbert space.
> This leads to a natural choice of nascent delta function
> http://rqgravity.net/SomeBitsofMathematics#RepresentationsOf
> TheDeltaFunction
On that webpage you write:
> Standard quantum theory uses continuous transforms,
> introducing mathematical difficulties concerning the
> definability of these states.
That's not really correct. The states are well-defined in
the context of rigged Hilbert space. It's the equal-point
multiplication of operator-valued distributions that
confounds everyone.
> Though I confess the relationship between what I have done
> and rigged Hilbert space is by no means as straightforward
> as I had anticipated.
Rigged Hilbert space is almost certainly what you get in the
limit as the lattice spacing goes to zero. In a Gelfand triple
Omega \subset H \subset Omega^* (take Omega as the Schwartz space
and Omega^* as its dual, the space of tempered distributions),
a key point is that H is dense in Omega^* which is where the
Dirac bras and kets live. More explicitly, the gaussians or the
narrow spikes (or whatever) live in a Hilbert space H (being
square-integrable), but the limit does not -- it lives in
Omega^*.
neuropulp
> we are all very glad to see you back in action,
No, I'm nowhere near that yet. Definitely not capable of any
serious "action" even though I'm starting to get seriously
HOR-OR-OR-NY in this (non-private) hospital bed.
> [...] got me thinking a great deal about delta functions and
> nascent deltas and their possible role in physics and
> whether the Gaussians of QED are to be regarded from some
> perspective as nascent deltas and what this would mean if so
> and how this might be exploited.
The Gaussian stuff you see in path integrals is a bit of an
accident, arising from the fact that free Hamiltonians tend
to be quadratic in the fields. The latter (or rather the
group representation theory behind it) is more fundamental.
> [...] how one balances the path integral formulation of QFT,
> with the canonical formulation. [...] And, history has shown
> that in the end, it is the "artistic" views, e.g., Maxwell
> and Einstein, which survive over time in a transcendent and
> lasting way.
Unless the artistic view turns out to be rubbish physically.
(Higher-dimensional string theory can be very artistic.)
> Certainly, I personally find the path integral formulation
> much more palatable, because one can assimilate this on an
> intuitive ground, whereas the canonical formulation seems
> too much like a collection of rules which are they way that
> are because that's the way they are.
Then maybe you don't yet appreciate the foundations of the
canonical formulation well enough. Both formulations rely on
a Lagrangian with merely-postulated interaction term.
> Or, as someone once said, just shut up and study!
I prefer to think of that phrase more as
"stop wanking and start studying".
> where are those point of contact where one can see most
> clearly, the interrelation between these two complementary
> views / approaches.
The path integral is just a solution (albeit ill-defined) of
the Schrodinger equation for the case of infinite degrees of
freedom. Both approaches are just trying to construct
unitary irreducible representations of a dynamical algebra
of observables (which includes a Hamiltonian).
(And grasshopper, when you can decode that last sentence
correctly, you will have learned. :-)
Did you ever acquire Ballentine's QM textbook? The way he
derives QM basics from (nonrelativistic) symmetries is an
excellent starting point to understand what "unitary
irreducible representations of a dynamical algebra of
observables" means in detail.
> Heisenberg uncertainty, which is the first step one gets
> to from the canonical commutation relationships, is
> fundamentally connected with the mathematics of Fourier
> transforms.
You'll get further after you understand that the maths of
Fourier transforms (and the associated *distribution*
theory) is simply what one gets when trying to construct a
unitary representation of the canonical commutation
relations. The CCRs can't be represented in a finite-dim
Hilbert space (theorem), but in an infinite-dim Hilbert
space at least one of the P,Q operators is unbounded
(theorem), hence can't be defined on every state in the
Hilbert space (theorem). That pushes us to Dirac bra/ket
formalism which is made rigorous by rigged Hilbert space
theory (aka Gelfand triples). Fourier transforms are just a
change of basis in the Omega^* space that I mentioned above
in an answer to Prince Charles.
> My 24-year old son Joshua who I am proud to say is starting
> a doctoral program in quantum optics
Does he have Mandel & Wolf? Might be a good end-of-year prezzie.
> and is perhaps headed to a dissertation about gravity waves,
> says it very well: "If you get into a fight with a bully,
> ask him to step outside into momentum space."
He's obviously never had to do that for real. Otherwise,
he'd have found out the hard way that such a change of basis
can't affect any physical results.
> Things like position / momentum uncertainties seem not odd,
> but a natural consequence of Fourier math.
Those uncertainties are really just a consequence of the
CCRs and unitary representation thereof. The Fourier math
stuff is just a particular representation.
> [...] rigor must be balanced with creativity and even some
> artistic license, if physics is to be advanced.
Here's a prayer for you to say at bedtime:
"God please grant me time for study, time for creating, time
for wanking, and the wisdom to know the difference."
Oh No
>news:f1d6e246-f451-45de...@o9g2000prg.googlegroups.com..
>.
>>
>> > Thus didn't spake the masked quantum damsel [...]
>>
>> Ha! Prince Charles, you've succeeded in becoming only the
>> second person on spf to make _me_ laugh out loud. (And I
>> thank the inventor of monocryl surgical dressings that I
>> didn't get taken down to the O.R. again as a result.)
>
>Well Princess, you probably know as well as I do, so as not to be
>outdone, that Prince Charles has been trying really hard to do so for
>many months. ;-)
Careful, Sir Jay, for I am sure that, like mine sword, mine pen can be
far more cutting than thine. It is in fact the second time I have been
the second person on spf to make the Princess laugh, which means I am
still the only one to have done it on purpose.
> One of these, is that of how one balances the path integral
>formulation of QFT, with the canonical formulation.
The real challenge is to construct QFT from Fock space, which is what I
would like the Princess to look at.
> Clearly, the teaching pedagogy (and the history) of physics greatly
>leans toward the canonical formulation, and Zee make a deliberate
>effort to weight things toward the path integral, perhaps to restore
>some balance, and perhaps because as x-phy said over in spr, the path
>integral formulation is the "artist's" view of QFT. And, history has
>shown that in the end, it is the "artistic" views, e.g., Maxwell and
>Einstein, which survive over time in a transcendent and lasting way.
If one identifies "artist's" with mathematical. More strictly of course
the modern form of Maxwell's equations is due to Heaviside, as Maxwell
had 20 equations some of which were redundant. The reason these
formulations survive, like Newton's laws, is that they show how the
whole subject is contained in few postulates. The fact that they do so
elegantly (which I think is what you mean by "artist") is also a sign of
good mathematics.
>Certainly, I personally find the path integral formulation much more
>palatable, because one can assimilate this on an intuitive ground,
>whereas the canonical formulation seems too much like a collection of
>rules which are they way that are because that's the way they are. Or,
>as someone once said, just shut up and study!
I agree. The usual production of the canonical formulation is the
antithesis of maths imv. Imv, one must start with interpretation, then
produce mathematics.
>The question that this raises for me, is this: Zee states at one point
>in his first few sections that the path integral and the canonical
>formulations are "complementary" approaches to quantum field theory.
>He also points out that in many cases one approach illuminates some
>things more clearly than other, and in other cases, vice versa. My
>question now, is this: where are those point of contact where one can
>see most clearly, the interrelation between these two complementary
>views / approaches. Put in other words, is there some "correspondence"
>theorem that clearly shows as directly as possible, where these two
>formulations meet up and are seen as the same side of one coin? Where
>is the place of unity between these two approaches?
As I say, starting from interpretation one finds both approaches. As you
say, the canonical formulation centres on the changes of bases implicit
in the FT. The path integral, or “sum over all paths” has as natural
interpretation that the sum over paths is a logical OR between the
possible paths that might be detected if an experiment could be done to
trace the path
>> > What we should actually do is line up all the authors of
>> > text books, as well as professors of qft, who do not pay
>> > attention to such issues as absolute convergence, and the
>> > order of taking limits, and have the lot of them shot.
>>
>> IMHO, we should line up all the naughty boys who don't read
>> publications thoroughly before passing judgment, and have
>> the lot them severely spanked. But... hmmm... no. In your
>> case that would just encourage the bad behaviour.
>>
>> ---
>> LoL from the Princess!
>
>I know it will be a heresy to Prince Charles, but rigor must be
>balanced with creativity and even some artistic license, if physics is
>to be advanced.
That is only one stage of the advance of physics. The other is that if
rigor is not then imposed, what has be done becomes meaningless and
ultimately valueless. The very absence of rigor has lead to things like
goldstone bosons, Higgs particles, string theory CDM, dark energy,
inflation to name a few. If science advances at all it will advance to
the point where advocates of these theories are as much a laughing stock
as alchemists, or the advocates of phlogiston.
Oh No
>Prince Charles wrote:
>
> > he couldn't catch the moving ones, and those that were
> > parked remained silent, refusing to respond to his
> > challenge. So, he was forced to declare them all cowards
> > and come home without doing any damage at all.
>
>Then Sir Basil of Fawlty is a better man than you. His vehicle
>was also silent but he gave it a good thrashing anyway.
You don't say!! The cad!!. He shall be cast into the tower if he is
truly guilty of this breach of the knight's code.
> > Is there any chance she might be able to find the energy for
> > http://papers.rqgravity.net/RQGQED.pdf ?
>
>In the drawer beside my hospital bed are 6 unread textbooks,
>and not a few papers (and soon to be 8 unread textbooks if
>my parents indulge my request for an early Xmas present and
>buy me those 2 expensive volumes of Zeidler that recently got
>favorable mention by the Fearsome Sage of Vienna on spr).
>
>But if there's something specific you want to me look at in
>RQGQED.pdf, maybe I can fit it in between physio sessions,
>provided it doesn't require me to study the whole thing
>thoroughly.
Well the appendices just contain standard calculations. There shouldn't
be much point in studying them, so it is only 20, fairly small, pages.
The real questions are whether, as I believe, I have constructed field
operators for qed in a satisfactory manner, whether I have given an
adequate derivation of Maxwell's equations, and, I guess, whether the
treatment of the Landau pole suffices.
> > I actually think a better approach is the one I use in my
> > papers and website using finite dimensional Hilbert space.
> > This leads to a natural choice of nascent delta function
> > http://rqgravity.net/SomeBitsofMathematics#RepresentationsOf
> > TheDeltaFunction
>
>On that webpage you write:
>
> > Standard quantum theory uses continuous transforms,
> > introducing mathematical difficulties concerning the
> > definability of these states.
>
>That's not really correct. The states are well-defined in
>the context of rigged Hilbert space. It's the equal-point
>multiplication of operator-valued distributions that
>confounds everyone.
yes, that is true. otoh as I am intending to write a less mathematical
and more introductory type of account, would you not think it valid to
describe the need for rigged Hilbert space as "mathematical
difficulties", albeit difficulties that are actually resolved?
>
> > Though I confess the relationship between what I have done
> > and rigged Hilbert space is by no means as straightforward
> > as I had anticipated.
>
>Rigged Hilbert space is almost certainly what you get in the
>limit as the lattice spacing goes to zero.
That is what we both thought, as I recall. but when I actually came to
define the limit, I found it nowhere near as straightforward as I had
supposed. I continue to think it likely that I am just really rusty, and
perhaps I am missing something straightforward, but unless we can
actually specify the relationship, we cannot claim it.
>In a Gelfand triple
>Omega \subset H \subset Omega^* (take Omega as the Schwartz space
>and Omega^* as its dual, the space of tempered distributions),
>a key point is that H is dense in Omega^* which is where the
>Dirac bras and kets live. More explicitly, the gaussians or the
>narrow spikes (or whatever) live in a Hilbert space H (being
>square-integrable), but the limit does not -- it lives in
>Omega^*.
>
>---
>LoL from the Princess!
>
Oh No
>
>Path integrals, delta functions, and Hilbert spaces, Oh my!
>
>These seem to touch on issues that concern me at present. It seems a
>Feynman type path integral can be obtained by using the recursive
>property of the Dirac delta function and then substituting the
>gaussian form of the delta function for the infinite number of delta
>functions inside the infinite number of integrations of the path
>integral. It's a pretty easy procedure. See details at:
>
>http://hook.sirus.com/users/mjake/delta_physics.htm
>
>This give what appears to be at least the path integral with a
>lagrangian for a free particle. But what about the potential term of
>the lagrangian, and what about lagrangians with other kinds of fields
>like those used in QFT? So I wonder if the next step is to replace the
>velocity in the lagrangian with any function. The only restriction on
>the type of functions that can be used is that they are square
>integrable so that the exponential still exists.
Certainly not any function. There are further restrictions due to
conservation of probability and covariance.
>And I understand that
>the space of square integrable functions forms a Hilbert space. Maybe
>this is why Hilbert spaces are relevant in QM. One of the questions I
>have with this idea is what it does to the gaussian delta function if
>one replaces the (x-x_0) term with an arbitrary function.
The answer to this question is the next topic of the webpage I gave you.
http://rqgravity.net/SomeBitsofMathematics#DeltaFunctionOfAFunction
(the generalised scaling property)
>Is it still
>a gaussian? Is it still a delta function which integrates to 1? I
>wonder how much functional analysis I'll have to learn in order to
>address these issues? Your comments would be appreciated. Thanks.
>
I confess I am no expert in path integrals, but I think you have got to
a point where it would be worth your while to invest some effort into
understanding more about foundations of qm and also qed. Obviously I
will recommend you to my own account, which is, I think, much more
succinct and to the point than anything you will find in text books, but
it would be wise not to restrict yourself to that.
Mike
Actually, I consider the delta function as the mathematical starting
point because I first derived the path integral from purely logical
considerations in which it turns out that the delta function
represents the logical connective of material implication. See details
at:
http://hook.sirus.com/users/mjake/QMfromlogic.htm
Here the path integral is derived as an implication of the conjunction
of all facts. It is an implication of the conjunction, not an
equality. So we will never be able to go backwards to prove the
conjunction because the implication does not go backwards. But a
conjunction of all facts seems like a fair assumption to start with
about reality.
In my development, the conjunction implies a disjunction of all
possible conjunctions of all possible implications. Then, when
multiplication is assigned to conjunction and addition assigned to
disjunction, as is usually done in probability spaces, the path
integral results. Of course, you could, I suppose work backwards
starting with the path integral. Break up the action integral into
little pieces, assign a delta to each piece, replace implication for
deltas, replace conjunction for multiplication, replace disjunction
for addition. And you end up with the logic that is implied by a
conjunction of all things.
>
> What this all means, I am still not sure. �I had thought that this
> limiting case would be quantum gravitation, but the math that made me
> think this was in error. �Yet, my intuition tells me that this should be
> the end result, i.e., that quantum gravitation should be the limiting
> case of QED in which Gaussians, treated as nascent deltas, turn to true
> deltas. �How to show this, if it is so, I do not know.
>
I haven't gotten that far yet, but I'm thinking that if we start with
the delta as fundamental, quantum gravity (which hopefully includes
the Hilbert-Einstein action) will result by expressing the delta
function and the path integral in terms of curved spacetime and
proceding as before.
Mike
> �Thus spake Mike <mj...@sirus.com>
> >And I understand that
> >the space of square integrable functions forms a Hilbert space. Maybe
> >this is why Hilbert spaces are relevant in QM. One of the questions I
> >have with this idea is what it does to the gaussian delta function if
> >one replaces the (x-x_0) term with an arbitrary function.
>
> The answer to this question is the next topic of the webpage I gave you.
>
> http://rqgravity.net/SomeBitsofMathematics#DeltaFunctionOfAFunction
>
> (the generalised scaling property)
I'm not sure this works. The second line of the proof simply assumes
that delta(u) is a delta function that integrates to 1. That does not
tell us for which kinds of functions u=g(x) the delta still integrates
to 1. Or more specifically, if the gaussian exponential delta function
were to replace the (x-x_0) term with some function f(x), does it
still remain a gaussian distribution or even a delta function which
integrates to 1?
>
> >Is it still
> >a gaussian? Is it still a delta function which integrates to 1? I
> >wonder how much functional analysis I'll have to learn in order to
> >address these issues? Your comments would be appreciated. Thanks.
>
> I confess I am no expert in path integrals, but I think you have got to
> a point where it would be worth your while to invest some effort into
> understanding more about foundations of qm and also qed.
QM is usually given in an axiomatic form which suggests a foundation.
But no reasons are ever given for these axioms; they are given as is
to either accept or reject. I have already looked. I am not going to
find a book that explains where QM comes from, except that
historically it works, which is the shut up and calculate mentality.
But now I feel that I have a real chance of explaining the origin of
QM, and possibly even quantum gravity in the sequel.
Oh No
>On Nov 30, 12:17�pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> �Thus spake Mike <mj...@sirus.com>
>
>> >And I understand that
>> >the space of square integrable functions forms a Hilbert space. Maybe
>> >this is why Hilbert spaces are relevant in QM. One of the questions I
>> >have with this idea is what it does to the gaussian delta function if
>> >one replaces the (x-x_0) term with an arbitrary function.
>>
>> The answer to this question is the next topic of the webpage I gave you.
>>
>> http://rqgravity.net/SomeBitsofMathematics#DeltaFunctionOfAFunction
>>
>> (the generalised scaling property)
>
>I'm not sure this works. The second line of the proof simply assumes
>that delta(u) is a delta function that integrates to 1.
Its a given that delta is a delta function. That's why its called delta.
>That does not
>tell us for which kinds of functions u=g(x) the delta still integrates
>to 1.
As the integration is with respect to u, this has no bearing.
>> I confess I am no expert in path integrals, but I think you have got to
>> a point where it would be worth your while to invest some effort into
>> understanding more about foundations of qm and also qed.
>
>QM is usually given in an axiomatic form which suggests a foundation.
>But no reasons are ever given for these axioms; they are given as is
>to either accept or reject. I have already looked. I am not going to
>find a book that explains where QM comes from,
This is addressed at
http://rqgravity.net/FoundationsOfQuantumTheory
and in subsequent pages
Oh No
>>tell us for which kinds of functions u=g(x) the delta still integrates
>>to 1.
>
>As the integration is with respect to u, this has no bearing.
>
differentiable.
Mike
> Thus spake Mike <mj...@sirus.com>
> >QM is usually given in an axiomatic form which suggests a foundation.
> >But no reasons are ever given for these axioms; they are given as is
> >to either accept or reject. I have already looked. I am not going to
> >find a book that explains where QM comes from,
>
> This is addressed at
>
> http://rqgravity.net/FoundationsOfQuantumTheory
>
> and in subsequent pages
>
It's interesting that we are both talking about how implication (the
If,Then type of statements) are represented by Dirac delta functions.
Is this part of the first description of Birkhoff and von Neumann?
I've also seen this "implication" language (or hypotheticals, if you
perfer) from Ariel Caticha's work. Are you aware of his work in this
regard? Your development, like Caticha's incorporates "implicaton" (or
hypotheticals) as a practical matter without justification. My
development derives implication from the first principles of sample
spaces. Also, you simply declare your space of hypotheticals to be a
vector space. This seems ad hoc, done only because it is useful, with
no real proof or justification. But this is exactly what I am asking
for, CAN a space of dirac delta functions (represented by gaussian
exponential functions) form a Hilbert space. And since you simply
appeal to the use of hypotheticals and vector space out of shear
convenience, I doubt very much that you are prepared to address the
actual justification of their use. But maybe I have tweeked your
interest.
neuropulp
> It is in fact the second time I have been the second person
> on spf to make the Princess laugh,
Oh dear. I don't remember the 1st time you were the 2nd person.
Maybe some of my neurons are still pulped.
Looks like I'll need another brain MRI.
> which means I am still
> the only one to have done it on purpose.
Oh, Sir Jay did it on purpose - since I meant "make me laugh
in a good way".
>> But if there's something specific you want to me look at in
>> RQGQED.pdf, maybe I can fit it in between physio sessions,
>> provided it doesn't require me to study the whole thing
>> thoroughly.
> Well the appendices just contain standard calculations.
> There shouldn't be much point in studying them, so it is
> only 20, fairly small, pages. The real questions are
> whether, as I believe, I have constructed field operators
> for qed in a satisfactory manner, whether I have given an
> adequate derivation of Maxwell's equations, and, I guess,
> whether the treatment of the Landau pole suffices.
Ummm,... Er... those are seriously difficult questions if you want
rigorous
answers. TBH, I'm not sure I'm adequately qualified for such a task
(which is why I'm trying to get through several advanced textbooks
while I'm stuck here).
But I've put your paper in my pile. Don't hold your breath, though.
>> Rigged Hilbert space is almost certainly what you get in the
>> limit as the lattice spacing goes to zero.
> That is what we both thought, as I recall. but when I
> actually came to define the limit, I found it nowhere near
> as straightforward as I had supposed. I continue to think it
> likely that I am just really rusty, and perhaps I am missing
> something straightforward,
Probably not. Rigorous distribution theory is difficult. You've got to
extend your operators into the dual/antidual spaces over nuclear
subspaces of your Hilbert space, then analyze everything in weak
topology.
I read recently about how Sobolev tried various stuff involving duals,
but it is Schwartz who got credit for establishing the theory -
because
he used (sequences of) nuclear spaces.
---
LoL from the Princess.
Oh No
>Prince Charles compares the size of his sword with Sir Jay's...
>
>
>>> RQGQED.pdf, maybe I can fit it in between physio sessions,
>>> provided it doesn't require me to study the whole thing
>>> thoroughly.
>
>> Well the appendices just contain standard calculations.
>> There shouldn't be much point in studying them, so it is
>> only 20, fairly small, pages. The real questions are
>> whether, as I believe, I have constructed field operators
>> for qed in a satisfactory manner, whether I have given an
>> adequate derivation of Maxwell's equations, and, I guess,
>> whether the treatment of the Landau pole suffices.
>
>Ummm,... Er... those are seriously difficult questions if you want
>rigorous
>answers.
yeah, but considering the questions, I think I have found seriously easy
answers. I think there has been an implicit assumption that the answers
must be difficult, and that very assumption has meant that the real
answer has been overlooked.
>TBH, I'm not sure I'm adequately qualified for such a task
>(which is why I'm trying to get through several advanced textbooks
>while I'm stuck here).
Actually I am reasonably sure you are. You are certainly more qualified
than a typical reviewer or editor at the IOP, and I doubt you are less
qualified than me to write it. Incidentally is your background maths or
physics? You certainly know more math than a typical physician and more
physic than a typical mathmo.
>
>But I've put your paper in my pile. Don't hold your breath, though.
Ok, I'm holding my breath.
>
>
>>> Rigged Hilbert space is almost certainly what you get in the
>>> limit as the lattice spacing goes to zero.
>
>> That is what we both thought, as I recall. but when I
>> actually came to define the limit, I found it nowhere near
>> as straightforward as I had supposed. I continue to think it
>> likely that I am just really rusty, and perhaps I am missing
>> something straightforward,
>
>Probably not. Rigorous distribution theory is difficult.
Is this an acceptance that when I said "mathematical difficulties
concerning the definability of these states" it was probably ok, or do
you think I still need to re-express myself? (havent thought how yet, as
my brain works at snail's pace these days).
>You've got to
>extend your operators into the dual/antidual spaces over nuclear
>subspaces of your Hilbert space, then analyze everything in weak
>topology.
>I read recently about how Sobolev tried various stuff involving duals,
>but it is Schwartz who got credit for establishing the theory -
>because
>he used (sequences of) nuclear spaces.
>
Possibly if I was really familiar with Schwartz work I would see how to
set up a direct correspondence with his sequences, but I'm not. As it
happens it is of academic interest only, as I don't ever need to go into
distribution theory proper.
Oh No
>On Nov 30, 3:51�pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Mike <mj...@sirus.com>
>
>
>If,Then type of statements) are represented by Dirac delta functions.
>Is this part of the first description of Birkhoff and von Neumann?
Birkhoff and von Neumann asserted on general grounds that qm has the
formal structure of a language, but did not interpret that language.
>I've also seen this "implication" language (or hypotheticals, if you
>perfer) from Ariel Caticha's work. Are you aware of his work in this
>regard?
I haven't been, but it seems worth a look.
>Your development, like Caticha's incorporates "implicaton" (or
>hypotheticals) as a practical matter without justification.
I don't agree with that. I think I do justify the structure by
translating into sentences in English
>My
>development derives implication from the first principles of sample
>spaces.
This makes it difficult to see what it means, imv.
>Also, you simply declare your space of hypotheticals to be a
>vector space. This seems ad hoc, done only because it is useful, with
>no real proof or justification.
Again, I don't agree. I formally show why disjunction gives us a vector
space. Again, that is something which I think you require but do not
make clear in simple terms.
>But this is exactly what I am asking
>for, CAN a space of dirac delta functions (represented by gaussian
>exponential functions) form a Hilbert space.
Obviously not, since a gaussian + a gaussian is not necessarily a
gaussian. But it can be a basis for a Hilbert space. I'm afraid that if
you need to ask this question, you do need to learn some linear algebra.
>And since you simply
>appeal to the use of hypotheticals and vector space out of shear
>convenience, I doubt very much that you are prepared to address the
>actual justification of their use. But maybe I have tweeked your
>interest.
Well I hope to have tweeked your interest, but I think you should reread
if you have not seen the justification here, since justification is the
main feature of my treatment.
Mike
> Thus spake Mike <mj...@sirus.com>
>
> >hypotheticals) as a practical matter without justification.
>
> I don't agree with that. I think I do justify the structure by
> translating into sentences in English
You're "translating into sentences in English" the same old quantum
mechanical concepts which begs further explanation. This is not a
justification from a mathematical point of view. One is still left
asking why things are as they are.
>
> >My
> >development derives implication from the first principles of sample
> >spaces.
>
> This makes it difficult to see what it means, imv.
If all facts being consistent implies a Feynman type path integral,
then what's the problem? I'm not sure I understand your objection.
>
> >Also, you simply declare your space of hypotheticals to be a
> >vector space. This seems ad hoc, done only because it is useful, with
> >no real proof or justification.
>
> Again, I don't agree. I formally show why disjunction gives us a vector
> space. Again, that is something which I think you require but do not
> make clear in simple terms.
I just reread your thesis on Quantum Foundations. In one sentence you
declare that alternative states require a disjunction symbolized by
the + sign. In a few sentences later you simply use this + sign for
arithmatic addition. This jump of logic could be explained a bit
better.
It seems clear that you are simply restating the axioms of QM as
presented in every other book I've read. But it seems a bit
intellectually dishonest to say that a presentation is mathematically
and philosophically justified just because the word axioms is used in
your founding premises.
>
> >But this is exactly what I am asking
> >for, CAN a space of dirac delta functions (represented by gaussian
> >exponential functions) form a Hilbert space.
>
> Obviously not, since a gaussian + a gaussian is not necessarily a
> gaussian. But it can be a basis for a Hilbert space. I'm afraid that if
> you need to ask this question, you do need to learn some linear algebra.
OK, I think I can accept the generalized scaling property of the Dirac
delta as you present. Since the parameter approaches zero in the limit
around the zeros of the g(x), the slope of g(x) becomes linear and
works like the constant scaling property of the delta function. This
breaks up the delta into a number of deltas each centered at the zeros
of g(x). So it is not equivalent to a single delta. I'm not sure what
to do with that. Maybe the next step is to see how this also breaks up
the path integral. Then maybe the various portions can be equated to
each other, or something like that.
neuropulp
> You are certainly more qualified
> than a typical reviewer or editor at the IOP, and I doubt you are less
> qualified than me to write it. Incidentally is your background maths or
> physics? [...]
Alas, this masked quantum damsel must remain coy, merely smiling
sweetly at such questions, lest her mask be dislodged incrementally.
>> Rigorous distribution theory is difficult.
> Is this an acceptance that when I said "mathematical difficulties
> concerning the definability of these states" it was probably ok, or do
> you think I still need to re-express myself?
Perhaps a brief mention that rigged Hilbert space is necessary for
rigorous definition. Although... if you never used such things
thereafter, and stay in a discrete space, maybe that just confuses
the reader.
If you avoid implying anywhere (directly or indirectly) that their
non-definability in ordinary Hilbert space forces one to discrete
spaces as the only alternative, that would do.
---
LoL from the rarely-stony-faced Princess.
Oh No
> On Dec 1, 1:28�pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
> > Thus spake Mike <mj...@sirus.com>
>
> > >Your development, like Caticha's incorporates "implicaton" (or
> > >hypotheticals) as a practical matter without justification.
>
> > I don't agree with that. I think I do justify the structure by
> > translating into sentences in English
>
> You're "translating into sentences in English" the same old quantum
> mechanical concepts which begs further explanation. This is not a
> justification from a mathematical point of view. One is still left
> asking why things are as they are.
The concepts are just classical concepts of doing measurement. The
mathematical structure is exposed in the structure of statements in
the subjunctive mood.
>
> > >My
> > >development derives implication from the first principles of sample
> > >spaces.
>
> > This makes it difficult to see what it means, imv.
>
> If all facts being consistent implies a Feynman type path integral,
> then what's the problem? I'm not sure I understand your objection.
There are a number of problems, starting from the nebulous nature of
the space of all facts, which I find quite unsuitable as a basis on
which to define mathematical structure. For example it immediately
invokes Russel's paradox. Is the following statement a fact?
"This statement is not a fact."
One is left thinking that you mean something by fact, but have not
worked out, and certainly have not expressed, exactly what it is.
Then there is the problem that you cannot even write down an integral
unless you have first defined a vector space, the very thing you
object to me doing as a basic mathematical necessity.
> > >Also, you simply declare your space of hypotheticals to be a
> > >vector space. This seems ad hoc, done only because it is useful, with
> > >no real proof or justification.
>
> > Again, I don't agree. I formally show why disjunction gives us a vector
> > space. Again, that is something which I think you require but do not
> > make clear in simple terms.
>
> I just reread your thesis on Quantum Foundations. In one sentence you
> declare that alternative states require a disjunction symbolized by
> the + sign. In a few sentences later you simply use this + sign for
> arithmatic addition. This jump of logic could be explained a bit
> better.
Again I think this shows your lack of familiarity with abstract
algebra, and with linear algebra and vector spaces in particular. I
refer you back to
http://rqgravity.net/IntroductionToVectorSpace
but I think for your requirements you really ought to get a book or
two.
> It seems clear that you are simply restating the axioms of QM as
> presented in every other book I've read. But it seems a bit
> intellectually dishonest to say that a presentation is mathematically
> and philosophically justified just because the word axioms is used in
> your founding premises.
>
That is not the reason. The reason is that I have exhibited something
concrete, viz formal conditional clauses, which obey those axioms.
> > >But this is exactly what I am asking
> > >for, CAN a space of dirac delta functions (represented by gaussian
> > >exponential functions) form a Hilbert space.
>
> > Obviously not, since a gaussian + a gaussian is not necessarily a
> > gaussian. But it can be a basis for a Hilbert space. I'm afraid that if
> > you need to ask this question, you do need to learn some linear algebra.
actually the use of gaussians as nascent delta funcions is a serious
disadvantage here, as you don't naturally get an orthogonal basis. You
would be better to follow the development in my paper
http://papers.rqgravity.net/RQGFoundations.pdf
in which I show the equivalence between the step function and the one
described on the website
>
> OK, I think I can accept the generalized scaling property of the Dirac
> delta as you present. Since the parameter approaches zero in the limit
> around the zeros of the g(x), the slope of g(x) becomes linear and
> works like the constant scaling property of the delta function. This
> breaks up the delta into a number of deltas each centered at the zeros
> of g(x). So it is not equivalent to a single delta. I'm not sure what
> to do with that. Maybe the next step is to see how this also breaks up
> the path integral. Then maybe the various portions can be equated to
> each other, or something like that.
This is a step forward, at least
Mike
> On 1 Dec, 23:11, Mike <mj...@sirus.com> wrote:
>
> > If all facts being consistent implies a Feynman type path integral,
> > then what's the problem? I'm not sure I understand your objection.
>
> There are a number of problems, starting from the nebulous nature of
> the space of all facts, which I find quite unsuitable as a basis on
> which to define mathematical structure. For example it immediately
> invokes Russel's paradox. Is the following statement a fact?
>
> � � �"This statement is not a fact."
>
> One is left thinking that you mean something by fact, but have not
> worked out, and certainly have not �expressed, exactly what it is.
No so. I even give an example. I say "just looking around us we see
that this thing exists and that thing exists and this thing over there
exists, etc". Obviously the facts I'm referring to are physical
entities. And I propose that this simple conjunction of all (physical)
facts is trustworthy down to the smallest detail. So this is where I
start my thesis and from there derive a logical version of the path
integral and convert it to a mathematical version of the path
integral.
>
> Then there is the problem that you cannot even write down an integral
> unless you have first defined a vector space, the very thing you
> object to me doing as a basic mathematical necessity.
If you are saying that integration necessarily requires or even
defines a vector space that would be interesting. I've not seen that
requirement anywhere.
> > I just reread your thesis on Quantum Foundations. In one sentence you
> > declare that alternative states require a disjunction symbolized by
> > the + sign. In a few sentences later you simply use this + sign for
> > arithmatic addition. This jump of logic could be explained a bit
> > better.
>
> Again I think this shows your lack of familiarity with abstract
> algebra, and with linear algebra and vector spaces in particular. I
> refer you back to
>
> http://rqgravity.net/IntroductionToVectorSpace
>
> but I think for your requirements you really ought to get a book or
> two.
>
I think I have had a thorough introduction to vector spaces. Although
it's been a while since I was intimate with the details, I don't
remember ever seeing how vector math justifies simply turning a
disjunction into addition, or conjunction into multiplication. Is
there an absolute necessary conversion from logic connectives to
vector spaces? I've not ever seen this before. The only place I've
ever seen a justification for converting logic connective to
arithmatic operation is in measure theory. Yes, I also may need to go
back and fill in this gap in my development as well. But I think I'm
closer since I'm starting from a sample space to begin with on which
measures are defined, and I'm already using a delta function which can
be considered a very basic type of measure.
> > It seems clear that you are simply restating the axioms of QM as
> > presented in every other book I've read. But it seems a bit
> > intellectually dishonest to say that a presentation is mathematically
> > and philosophically justified just because the word axioms is used in
> > your founding premises.
>
> That is not the reason. The reason is that I have exhibited something
> concrete, viz formal conditional clauses, which obey those axioms.
>
You do not realize what you're saying. When you even mention the word
conditional, you're acknowledging the existence of individual
propositions, those used for the premise and those used for the
conclusion of which a conditional statement is made. So even here you
are agreeing with me that propositions in a general sense is a
starting place to develop physics. Next, implicitly you are labeling
proposition with coordinates just as I do. For the separate
proposition of your conditionals are denoted with bras and kets which
are distiguished with spacetime labels. You write |x> for a premise
and your write <y| for your conclusion to form a conditional which you
write as <y|x>. So I would appreciate it if you not fault me for my
labeling of arbitrary proposition with coordinates because that is
exactly what you are doing.
> > > >But this is exactly what I am asking
> > > >for, CAN a space of dirac delta functions (represented by gaussian
> > > >exponential functions) form a Hilbert space.
>
> > > Obviously not, since a gaussian + a gaussian is not necessarily a
> > > gaussian. But it can be a basis for a Hilbert space. I'm afraid that if
> > > you need to ask this question, you do need to learn some linear algebra.
>
> actually the use of gaussians as nascent delta funcions is a serious
> disadvantage here, as you don't naturally get an orthogonal basis. You
> would be better to follow the development in my paper
>
Yes, I'm still in the process of trying to figure out how all this
fits together. And I suppose it's possible that I've failed to
consider something easy and simple. That's why I'm on this new group
trying to tap other people's experience. I've seen how a delta
function can be represented by an inner product. But I don't yet see
how that justifies jumping to the conclusion that this IS equaivalent
to a vector space. Just because an inner production can be seen as a
delta function does not mean that the delta function justifies all the
extra structure of a vector space. Or does it?
But if I'm going to use your development as foundational, then you're
going to have to justify the use of a vector space to begin with. I
don't see how you get from disjunction to addition so readily, and I
don't see how adding states allows you to automatically assume a
vector space.
Oh No
>
>http://hook.sirus.com/users/mjake/delta_physics.htm
line 3 is incorrect. You cannot in general take the product of
distributions.
Oh No
>On Dec 2, 9:10�am, Oh No <n...@charlesfrancis.wanadoo.co.uk> wrote:
>> On 1 Dec, 23:11, Mike <mj...@sirus.com> wrote:
>>
>> > If all facts being consistent implies a Feynman type path integral,
>> > then what's the problem? I'm not sure I understand your objection.
>>
>> There are a number of problems, starting from the nebulous nature of
>> the space of all facts, which I find quite unsuitable as a basis on
>> which to define mathematical structure. For example it immediately
>> invokes Russel's paradox. Is the following statement a fact?
>>
>> � � �"This statement is not a fact."
>>
>> One is left thinking that you mean something by fact, but have not
>> worked out, and certainly have not �expressed, exactly what it is.
>
>No so. I even give an example. I say "just looking around us we see
>that this thing exists and that thing exists and this thing over there
>exists, etc". Obviously
That is my objection. What is obvious to you is not at all obvious, and
is actually not well specified.
>the facts I'm referring to are physical
>entities.
But a fact is not a physical entity. It may be a statement about a
physical entity. But physical entity is not well defined. Is a statement
not a physical entity, if it is written in black and white? And am I
really to consider facts like "the cat sat on the mat" together with
statements about electrons.
>And I propose that this simple conjunction of all (physical)
>facts is trustworthy down to the smallest detail.
But it most certainly isn't. The issue with qm is that we cannot easily
state what the facts are.
>So this is where I
>start my thesis and from there derive a logical version of the path
>integral and convert it to a mathematical version of the path
>integral.
>
>>
>> Then there is the problem that you cannot even write down an integral
>> unless you have first defined a vector space, the very thing you
>> object to me doing as a basic mathematical necessity.
>
>If you are saying that integration necessarily requires or even
>defines a vector space that would be interesting. I've not seen that
>requirement anywhere.
Have you looked at a formal definition of a delta function?
http://en.wikipedia.org/wiki/Distribution_(mathematics)
>> > I just reread your thesis on Quantum Foundations. In one sentence you
>> > declare that alternative states require a disjunction symbolized by
>> > the + sign. In a few sentences later you simply use this + sign for
>> > arithmatic addition. This jump of logic could be explained a bit
>> > better.
>>
>> Again I think this shows your lack of familiarity with abstract
>> algebra, and with linear algebra and vector spaces in particular. I
>> refer you back to
>>
>> http://rqgravity.net/IntroductionToVectorSpace
>>
>> but I think for your requirements you really ought to get a book or
>> two.
>>
>
>I think I have had a thorough introduction to vector spaces. Although
>it's been a while since I was intimate with the details, I don't
>remember ever seeing how vector math justifies simply turning a
>disjunction into addition,
It is only necessary to be able to write down formal rules which obey
the axioms of vector space.
> or conjunction into multiplication.
This is a basic requirement of probability theory, on which my treatment
is based.
> Is
>there an absolute necessary conversion from logic connectives to
>vector spaces? I've not ever seen this before. The only place I've
>ever seen a justification for converting logic connective to
>arithmatic operation is in measure theory. Yes, I also may need to go
>back and fill in this gap in my development as well. But I think I'm
>closer since I'm starting from a sample space to begin with on which
>measures are defined, and I'm already using a delta function which can
>be considered a very basic type of measure.
that is starting at the deep end :-)
>> > It seems clear that you are simply restating the axioms of QM as
>> > presented in every other book I've read. But it seems a bit
>> > intellectually dishonest to say that a presentation is mathematically
>> > and philosophically justified just because the word axioms is used in
>> > your founding premises.
>>
>> That is not the reason. The reason is that I have exhibited something
>> concrete, viz formal conditional clauses, which obey those axioms.
>>
>
>You do not realize what you're saying. When you even mention the word
>conditional, you're acknowledging the existence of individual
>propositions, those used for the premise and those used for the
>conclusion of which a conditional statement is made.
of course.
>So even here you
>are agreeing with me that propositions in a general sense is a
>starting place to develop physics.
Or you are agreeing with me. Certainly there is an overlap of ideas,
which is what makes discussion worthwhile. We do have to thrash out the
detail, however.
> Next, implicitly you are labeling
>proposition with coordinates just as I do.
I would say that I am doing that explicitly. You appear to be doing it
implicitly, which is again one of my objections. You need to make this
kind of thing absolutely explicit and clear.
>
>> > > >But this is exactly what I am asking
>> > > >for, CAN a space of dirac delta functions (represented by gaussian
>> > > >exponential functions) form a Hilbert space.
>>
>> > > Obviously not, since a gaussian + a gaussian is not necessarily a
>> > > gaussian. But it can be a basis for a Hilbert space. I'm afraid that if
>> > > you need to ask this question, you do need to learn some linear algebra.
>>
>> actually the use of gaussians as nascent delta funcions is a serious
>> disadvantage here, as you don't naturally get an orthogonal basis. You
>> would be better to follow the development in my paper
>>
>
>Yes, I'm still in the process of trying to figure out how all this
>fits together. And I suppose it's possible that I've failed to
>consider something easy and simple. That's why I'm on this new group
>trying to tap other people's experience. I've seen how a delta
>function can be represented by an inner product. But I don't yet see
>how that justifies jumping to the conclusion that this IS equaivalent
>to a vector space.
you can't have an inner product unless you *already* have a vector
space.
>But if I'm going to use your development as foundational, then you're
>going to have to justify the use of a vector space to begin with. I
>don't see how you get from disjunction to addition so readily, and I
>don't see how adding states allows you to automatically assume a
>vector space.
>
In essence, the ability to add states and multiply by scalars is the
definition of vector space. Again, if you don't already know this you
must study real mathematical treatments of linear algebra, not just rely
on elementary accounts of vectors.
Mike
The delta function can also be viewed as a measure. See
http://en.wikipedia.org/wiki/Dirac_delta_function#As_a_measure
And as such it can be integrated against any "continuous compactly
supported functions ƒ". And a little latter the article states that
the delta function is a continuous compactly supported function.
A typical development of the path integral is to insert the identity
an infinite number of time where the identity is
$|x1><x1|dx1
And when inserted an infinite number of times inside a typical inner
product we get
<x|x0>=$$...$<x|x1><x1|x2><x2|x3>...<xn-21xn><xn|x0>dx1dx2...dxn
Then as you say on your website that the inner product is treated as a
dirac delta, then here you also would have products of distributions
as well. For you are the one saying that the |xi> is a vector space so
that the <xi|xj> is an inner production represented by a delta
function. So it seems you have the same issue.
Oh No
>Prince Charles wrote:
>>Incidentally is your background maths or
>> physics? [...]
>
>Alas, this masked quantum damsel must remain coy, merely smiling
>sweetly at such questions, lest her mask be dislodged incrementally.
>LoL from the rarely-stony-faced Princess.
>
Ah, well that's the give away. Undergrad physics is intensely boring, so
you must be a matmo.
Oh No
>On Dec 3, 6:17 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Mike <mj...@sirus.com>
>>
>>
>>
>> >http://hook.sirus.com/users/mjake/delta_physics.htm
>>
>> line 3 is incorrect. You cannot in general take the product of
>> distributions.
>>
>
>
>http://en.wikipedia.org/wiki/Dirac_delta_function#As_a_measure
>
>And as such it can be integrated against any "continuous compactly
>supported functions ƒ". And a little latter the article states that
>the delta function is a continuous compactly supported function.
I cannot find anywhere where it says that. In any case it is not true.
The article does say "The measure delta is not absolutely continuous
with respect to the Lebesgue measure — in fact, it is a singular
measure." Moreover the delta function is not a function at all.
>A typical development of the path integral is to insert the identity
>an infinite number of time where the identity is
>
>$|x1><x1|dx1
>
>And when inserted an infinite number of times inside a typical inner
>product we get
>
><x|x0>=$$...$<x|x1><x1|x2><x2|x3>...<xn-21xn><xn|x0>dx1dx2...dxn
That looks like inserting it a finite number, n, times. Don't I recall
that the Princess recently pointed out to you that this integral is not
defined for infinite n?
>
>Then as you say on your website that the inner product is treated as a
>dirac delta, then here you also would have products of distributions
>as well. For you are the one saying that the |xi> is a vector space so
>that the <xi|xj> is an inner production represented by a delta
>function. So it seems you have the same issue.
What I do on the website is not strictly orthodox, because I work with
finite dimensional vector space. This means I have products of
functions, not products of distributions. I do this because a great deal
of care must be taken with limits in qft in general, and qed in
particular. I do it rather more rigorously in the papers.
Mike
> Thus spake Mike <mj...@sirus.com>
>
> >And when inserted an infinite number of times inside a typical inner
> >product we get
>
> > <x|x0>=$$...$<x|x1><x1|x2><x2|x3>...<xn-21xn><xn|x0>dx1dx2...dxn
>
> That looks like inserting it a finite number, n, times. Don't I recall
> that the Princess recently pointed out to you that this integral is not
> defined for infinite n?
>
Yes, that Dirac delta function seems a bit crazy. I'm sure I was
speaking too loosely when I said insert the identity an infinite
number of times. I should have said insert it a finite number of times
just short of an infinite number of times. I think the problem with
the delta is that you are dealing with more than one limiting process
and which limit is considered first might cause confusion. There's the
limiting process associated with the normalizing integral of the delta
function that equals 1. And then there's the parameter inside the
delta function that goes to zero so that the delta function approaches
infinite at its zero argument. But I think it's clear that the
integrating limiting process is always considered first before the
parameter limiting process. For you have to have the integral of the
delta equal 1 for any value of the inside parameter.
Now, if we try to have a product of two deltas inside the integral,
each delta has it's own separate parameter that approaches zero
independently. But, of course, the integrating process takes
precidence of it. That mean, I think, that the second delta is just a
gaussian function as far as the integration process is concerned. It
can be treated as a function for the purposes of integration. And the
first delta integration forces the integral to equal the second delta
with the value of the first delta zero. Does this sound right?
I'm trying to directly prove that the integration of the two delta
functions equals a third delta by direct algebraic manipulation of the
gaussian from of the two deltas being integrated. But I'm getting
cross products in the exponent that I don't know what to do with.
Maybe you have some insight there.
Oh No
>On Dec 4, 3:51�pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Mike <mj...@sirus.com>
>
>
>each delta has it's own separate parameter that approaches zero
>independently. But, of course, the integrating process takes
>precidence of it. That mean, I think, that the second delta is just a
>gaussian function as far as the integration process is concerned. It
>can be treated as a function for the purposes of integration. And the
>first delta integration forces the integral to equal the second delta
>with the value of the first delta zero. Does this sound right?
No. I have already told you. You can't do this. It is completely wrong.
I would appreciate if you do not repeat the same query which has already
been answered, as it is against the charter and we will be forced to
reject such posts.
>
>I'm trying to directly prove that the integration of the two delta
>functions equals a third delta by direct algebraic manipulation of the
>gaussian from of the two deltas being integrated. But I'm getting
>cross products in the exponent that I don't know what to do with.
>Maybe you have some insight there.
Yes. I have told you. It cannot be defined. This has been proved in the
general case
http://en.wikipedia.org/wiki/Distribution_(mathematics)#Problem_of_multi
plication
Note that although special cases are mentioned in which multiplication
can be defined, and one of those bears on the path integral, the
particular case you are dealing with cannot be defined. Regularisation
is, in essence, about the removal of such products.
Also note that distribution theory at this level is extremely technical
and difficult, even by the standards of those with a mathematical
education extending well beyond graduate level at a top math uni. Way
above any standard of math you have had contact with.
Mike
> Thus spake Mike <mj...@sirus.com>
>
> >On Dec 4, 3:51�pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> >> Thus spake Mike <mj...@sirus.com>
>
> >Now, if we try to have a product of two deltas inside the integral,
> >each delta has it's own separate parameter that approaches zero
> >independently. But, of course, the integrating process takes
> >precidence of it. That mean, I think, that the second delta is just a
> >gaussian function as far as the integration process is concerned. It
> >can be treated as a function for the purposes of integration. And the
> >first delta integration forces the integral to equal the second delta
> >with the value of the first delta zero. Does this sound right?
>
> No. I have already told you. You can't do this. It is completely wrong.
>
> I would appreciate if you do not repeat the same query which has already
> been answered, as it is against the charter and we will be forced to
> reject such posts.
>
Take a look at the following pdf file:
http://users.physik.fu-berlin.de/~kleinert/public_html/kleiner_reb3/psfiles/pthic04.pdf
Equation 2.17 shows an n-tuple of products of dirac delta function
which can be derived from the simple recursion relation of a product
of two dirac delta functions. Notice that equation 2.17 equates to
just one dirac delta function.
This pdf file is from the book, "Path Integrals in Quantum Mechanics,
Statistics, Polymer Physics, and Financial Markets", byt Hagen
Kleinert, page 91.
>
>
> >I'm trying to directly prove that the integration of the two delta
> >functions equals a third delta by direct algebraic manipulation of the
> >gaussian from of the two deltas being integrated. But I'm getting
> >cross products in the exponent that I don't know what to do with.
> >Maybe you have some insight there.
>
> Yes. I have told you. It cannot be defined. This has been proved in the
> general case
>
> http://en.wikipedia.org/wiki/Distribution_(mathematics)#Problem_of_multi
> plication
>
I found a book which shows that the integral of the product of two
gaussian function is yet another gaussian function. Add a limit to
these gaussians and you have the integral of two dirac deltas is a
third dirac delta function. The book is called,"The Feynman Integral
and Feynman's Operational Calculus", by Gerald W. Johnson and Michel
L. Lapidus, page 37, eq 3.2.8. And this book spends quite a bit of
pages explaining the problem of the path integral measure undefined
nature.
I've used LaTex at the following Webpage to duplicate their eq 3.2.8
at:
http://www.physicsforums.com/showpost.php?p=2485658&postcount=10
This equation is also confirmed in the book review at:
Oh No
>On Dec 5, 5:14�pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Mike <mj...@sirus.com>
>>
>> >On Dec 4, 3:51�pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> >> Thus spake Mike <mj...@sirus.com>
>>
>> >Now, if we try to have a product of two deltas inside the integral,
>> >each delta has it's own separate parameter that approaches zero
>> >independently. But, of course, the integrating process takes
>> >precidence of it. That mean, I think, that the second delta is just a
>> >gaussian function as far as the integration process is concerned. It
>> >can be treated as a function for the purposes of integration. And the
>> >first delta integration forces the integral to equal the second delta
>> >with the value of the first delta zero. Does this sound right?
>>
>> No. I have already told you. You can't do this. It is completely wrong.
>>
>> I would appreciate if you do not repeat the same query which has already
>> been answered, as it is against the charter and we will be forced to
>> reject such posts.
>>
>
>Take a look at the following pdf file:
>
>http://users.physik.fu-berlin.de/~kleinert/public_html/kleiner_reb3/psf
>iles/pthic04.pdf
>
>Equation 2.17 shows an n-tuple of products of dirac delta function
>which can be derived from the simple recursion relation of a product
>of two dirac delta functions. Notice that equation 2.17 equates to
>just one dirac delta function.
>
write down such an equation without a great deal of specification as to
how limits are to be taken.
>> >I'm trying to directly prove that the integration of the two delta
>> >functions equals a third delta by direct algebraic manipulation of the
>> >gaussian from of the two deltas being integrated. But I'm getting
>> >cross products in the exponent that I don't know what to do with.
>> >Maybe you have some insight there.
>>
>> Yes. I have told you. It cannot be defined. This has been proved in the
>> general case
>>
>> http://en.wikipedia.org/wiki/Distribution_(mathematics)#Problem_of_multi
>> plication
>>
>
>I found a book which shows that the integral of the product of two
>gaussian function is yet another gaussian function.
This may be true.
>Add a limit to
>these gaussians and you have the integral of two dirac deltas is a
>third dirac delta function.
Try using different representations of the delta function (or nascent
delta functions). You should be able to use a step function. Do you get
the same answer? You can't generalise to a product of delta functions
unless you can show the same result for *all* representations.
Mike
> Thus spake Mike <mj...@sirus.com>
>
> >which can be derived from the simple recursion relation of a product
> >of two dirac delta functions. Notice that equation 2.17 equates to
> >just one dirac delta function.
>
> This is all very non-rigorous. As a mathematician I couldn't possible
> write down such an equation without a great deal of specification as to
> how limits are to be taken.
>
I think it's only necessary to worry about how the limits work in the
first integration of a product of two dirac deltas. For after that it
would just be iterating the process n-times. And I think it's fairly
obvious that as long as the integration process is done first, then it
does not matter in which order to take the limits of the two delta
parameters. Since the limits of the deltas are indepentent, one is
momentarily left fixed as the other goes to zero. And the one with the
momentarily fixed parameter can be considered the test function which
is continuous and has compact support. Or you can hold the parameter
of the other delta momentarily fixed as the first approaches zero.
Since you get the same answer either way, the result is well defined.
Now I think the integration process is done first since there would be
no possible way to even make sense of the integral if one or both of
the delta parameters were allowed to approach zero first. Since the
functional, or inner product, or integral is specified, that process
has to be done first for it to even be accounted for.
> >Add a limit to
> >these gaussians and you have the integral of two dirac deltas is a
> >third dirac delta function.
>
> Try using different representations of the delta function (or nascent
> delta functions). You should be able to use a step function. Do you get
> the same answer? You can't generalise to a product of delta functions
> unless you can show the same result for *all* representations.
>
I don't think it's possible to know in advance EVERY possible
representation of the dirac delta function. There may be some that
differ from others only by a variable that can vary continuously; that
in itself gives an infinite number of representations. And then it may
be possible to represent one in terms of the other, like exponential
represented with trig functions, etc. As long as one is only using the
essential characteristics of the delta function (It integrates to one
in spite of the limiting parameter), then proof using some particular
delta is probably sufficient.
So the question is, can there be delta functions that can be
integrated to 1 (when their parameter is not yet zero) but not be
continuous enough to serve as a test function? Can there be delta
functions that cannot serve as test functions when there limiting
parameter is not yet zero? If you can show that, then my argument does
not work in the general, but the particular example of gaussian deltas
would not be disproved.
FrediFizzx
Peter Woit "Not Even Wrong" has announced that a second edition of Zee's
book is coming out next year.
http://press.princeton.edu/titles/9227.html
Best,
Fred Diether
Oh No
>On Dec 13, 5:00 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> >Add a limit to
>> >these gaussians and you have the integral of two dirac deltas is a
>> >third dirac delta function.
>>
>> Try using different representations of the delta function (or nascent
>> delta functions). You should be able to use a step function. Do you get
>> the same answer? You can't generalise to a product of delta functions
>> unless you can show the same result for *all* representations.
>>
>
>I don't think it's possible to know in advance EVERY possible
>representation of the dirac delta function. There may be some that
>differ from others only by a variable that can vary continuously; that
>in itself gives an infinite number of representations. And then it may
>be possible to represent one in terms of the other, like exponential
>represented with trig functions, etc. As long as one is only using the
>essential characteristics of the delta function (It integrates to one
>in spite of the limiting parameter), then proof using some particular
>delta is probably sufficient.
No. To write down formulae like this you would have to show that the
result is the same for every delta. If not you have an inconsistency,
and by choosing particular deltas you can probably get any answer you
want, so nothing is proven.
>
>So the question is, can there be delta functions that can be
>integrated to 1 (when their parameter is not yet zero) but not be
>continuous enough to serve as a test function? Can there be delta
>functions that cannot serve as test functions when there limiting
>parameter is not yet zero? If you can show that, then my argument does
>not work in the general, but the particular example of gaussian deltas
>would not be disproved.
>
There is a way forward. You can work in finite dimensional Hilbert
space. Then the delta is not a gaussian, but is the one I have used on
my website. Of the top of my head, I think the result you want does hold
for this delta. Then you can do your full calculation for n dimensions,
and if you can take the limit n->oo when you have finished, I think you
will have established your result.
Jay R. Yablon
I will keep an eye out for that.
Jay
"FrediFizzx" <fredi...@hotmail.com> wrote in message
news:7ol5beF...@mid.individual.net...
Mike
> Thus spake Mike <mj...@sirus.com>
> >As long as one is only using the
> >essential characteristics of the delta function (It integrates to one
> >in spite of the limiting parameter), then proof using some particular
> >delta is probably sufficient.
>
> No. To write down formulae like this you would have to show that the
> result is the same for every delta. If not you have an inconsistency,
> and by choosing particular deltas you can probably get any answer you
> want, so nothing is proven.
>
When they write $f(x)D(x-x0)dx = f(x0), do they consider every
possible representation of the dirac delta function, D(x-x0)? No, they
are only considering the particular property of the delta, namely, $D
(x-x0)dx = 1. The only restriction on f(x) is that it is continuous
and has compact support. (Notice that in $D(x-x0)dx = 1, this is the
same as $f(x)D(x-x0)dx with f(x)=1. Is f(x)=1 a function of compact
support?) My question is can there be any representations of the dirac
delta which is either not continuous or does not have compact support
such that it can not be used as an f(x) when its limiting parameter is
a fixed, non-zero number? I can think of a delta that is not
continuous at its end points, namely, one that equals "a" for the
inteval between -1/(2a) and +1/(2a), and zero outside this inteval.
This has the property that it integrates to 1 no matter what the value
of "a". But it goes to infinity at the origin as "a" approaches zero.
It's not continuous at its end points, but it is continous at the
point "0" where it approaches infinity. Even here I think this
function qualifies as an f(x) in $f(x)D(x-x0)dx as long as x0 is not
and end point of f(x).
> >So the question is, can there be delta functions that can be
> >integrated to 1 (when their parameter is not yet zero) but not be
> >continuous enough to serve as a test function? Can there be delta
> >functions that cannot serve as test functions when there limiting
> >parameter is not yet zero? If you can show that, then my argument does
> >not work in the general, but the particular example of gaussian deltas
> >would not be disproved.
>
> There is a way forward. You can work in finite dimensional Hilbert
> space. Then the delta is not a gaussian, but is the one I have used on
> my website. Of the top of my head, I think the result you want does hold
> for this delta. Then you can do your full calculation for n dimensions,
> and if you can take the limit n->oo when you have finished, I think you
> will have established your result.
>
I don't believe that I ever intended to say that "n" IS infinite, only
that it approaches infinity in the limit. So maybe I'm already doing
what you have suggested above.
Oh No
>On Dec 15, 5:16�am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Mike <mj...@sirus.com>
>
>> >As long as one is only using the
>> >essential characteristics of the delta function (It integrates to one
>> >in spite of the limiting parameter), then proof using some particular
>> >delta is probably sufficient.
>>
>> No. To write down formulae like this you would have to show that the
>> result is the same for every delta. If not you have an inconsistency,
>> and by choosing particular deltas you can probably get any answer you
>> want, so nothing is proven.
>>
>
>When they write $f(x)D(x-x0)dx = f(x0), do they consider every
>possible representation of the dirac delta function, D(x-x0)?
Yes. It is true in the limit for every possible nascent delta function.
Note that this is not in general true for representations - it is only
true in the limit. There is one exception, described below, when it is
actually true without taking the limit.
>My question is can there be any representations of the dirac
>delta which is either not continuous or does not have compact support
>such that it can not be used as an f(x) when its limiting parameter is
>a fixed, non-zero number?
Yes. No representation is continuous in the limit.
>> >So the question is, can there be delta functions that can be
>> >integrated to 1 (when their parameter is not yet zero) but not be
>> >continuous enough to serve as a test function? Can there be delta
>> >functions that cannot serve as test functions when there limiting
>> >parameter is not yet zero? If you can show that, then my argument does
>> >not work in the general, but the particular example of gaussian deltas
>> >would not be disproved.
>>
>> There is a way forward. You can work in finite dimensional Hilbert
>> space. Then the delta is not a gaussian, but is the one I have used on
>> my website. Of the top of my head, I think the result you want does hold
>> for this delta. Then you can do your full calculation for n dimensions,
>> and if you can take the limit n->oo when you have finished, I think you
>> will have established your result.
>>
>
>I don't believe that I ever intended to say that "n" IS infinite, only
>that it approaches infinity in the limit. So maybe I'm already doing
>what you have suggested above.
In this case you should make it clear, and you should be using the
representation I suggest, not a Gaussian. Interestingly enough, in my
papers I show that the continuum representation I use is actually a
delta function even in finite dimensional Hilbert space (meaning that in
this case a delta function can actually be a function, not a
distribution). I would strongly suggest you follow the development in my
papers in depth. Then you will have something to tie yours to.
Ken S. Tucker
This may be a bit Off Topic, but my 1st serious encounter with that
type of math was in rocketry risk analysis that I sense is related.
The theory was to continuously reduce 'risk' so the integrated
risk was less than one, and a failure event should not occur.
What that did is justify the basis of risk analysis work that I think
is important and reduced failure rate, which leads to a 99.99...%
dependability, as a product of a finite number of components, each
entering into the failure (risk) analysis.
A simple fundamental example is the sum of this famous series,
1/2 + 1/4 + 1/8 ... => 1 , (series 1)
which at infinity provides a single event.
Infinity is a limit, that can be moved to finite by an arbituary
increment within the terms of said series, such as,
(1/2+e1) + (1/4+e2) + (1/8+e3) ... => 1 + f(e)
((There's a calculus for that move, it escapes me just now))
Getting to physics, as I understand the idea, is that a detected
particle is an event, such as Eq.(series 1), and the terms within
that series may be regarded as harmonics in Wave Mechanics.
Regards
Ken S. Tucker
...
Mike
>
> In this case you should make it clear, and you should be using the
> representation I suggest, not a Gaussian. Interestingly enough, in my
> papers I show that the continuum representation I use is actually a
> delta function even in finite dimensional Hilbert space (meaning that in
> this case a delta function can actually be a function, not a
> distribution). I would strongly suggest you follow the development in my
> papers in depth. Then you will have something to tie yours to.
>
> http://rqgravity.net/Papers
>
> Regards
Thank for the advice. I read your paper, "A Smooth Representation of
Finite Dimensional Hilbert Space". Very interesting, but correct me if
I'm wrong. It seems you came up with the dirac delta function from
considerations of Hilbert space. However, in my case that would be
more structure than I wish to assume at this point in my development.
Instead, I'd like to see a Hilbert space emerge from the L^2 space
dictated by the Action integral acquired by use of the dirac delta
function used to construct the path integral. For a came to the Dirac
delta function from purely logical reasons, and it had seems to me to
be more fundamental than Hilbert space.
Oh No
The L2 space is already a Hilbert space, and is much more fundamental
than the delta function, which cannot be defined without a Hilbert
space.
Mike
> The L2 space is already a Hilbert space,
Yes, I know that. Perhap my phrase "emerge from an L2 space" was
impercise. More accurately I meant that the L2 space (thus a Hilbert
space) seems to emerge from the use of the gaussian form of the dirac
delta, which when all the exponents are added up form an integral of a
function (x-xj) squared. If the (x-xj) can be replaced with other
kinds of functions, then the space of all qualifying functions must be
square integrable and hence a Hilbert space. This is why I was asking
about compound functions that are still dirac delta functions.
> and is much more fundamental
> than the delta function, which cannot be defined without a Hilbert
> space.
>
I can certainly see that the existence of a Hilbert space implies a
dirac delta as an inner product. But I'm not seeing it the other way
around. The usual definition of the dirac delta is $D(x-x0)dx=1, with
a limiting parameter that forces an infinity at the zero of the
argument. I don't seems to recognize how that necessarily specifies a
Hilbert space.
Oh No
>On Dec 16, 3:12�pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> The L2 space is already a Hilbert space,
>
>Yes, I know that. Perhap my phrase "emerge from an L2 space" was
>impercise. More accurately I meant that the L2 space (thus a Hilbert
>space) seems to emerge from the use of the gaussian form of the dirac
>delta, which when all the exponents are added up form an integral of a
>function (x-xj) squared. If the (x-xj) can be replaced with other
>kinds of functions, then the space of all qualifying functions must be
>square integrable and hence a Hilbert space. This is why I was asking
>about compound functions that are still dirac delta functions.
>
>> and is much more fundamental
>> than the delta function, which cannot be defined without a Hilbert
>> space.
>>
>
>I can certainly see that the existence of a Hilbert space implies a
>dirac delta as an inner product.
The dirac delta is not an inner product. It is a distribution.
> But I'm not seeing it the other way
>around. The usual definition of the dirac delta is $D(x-x0)dx=1, with
>a limiting parameter that forces an infinity at the zero of the
>argument. I don't seems to recognize how that necessarily specifies a
>Hilbert space.
>
You are talking about functions and integration. How would you not have
a Hilbert space?
glird
>
>It's not continuous at its end points, but it is continuous >at the point "0" where it approaches infinity.
Regardless of what "It" is, and regardless of the value of "the
point", nothing has a value that "approaches infinity" at a point;
because a point has zero extension and no volume. Therefore, although
it may or may not be "continuous at its [TWO] end pointS", it can't
HAVE two end points -- nor a length that approaches infinity or 2
inchs -- at a given point "0"; or at any other point.
glird
Jay R. Yablon
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
news:5gKUYIB4...@charlesfrancis.wanadoo.co.uk...
> Thus spake Mike <mj...@sirus.com>
>>On Dec 16, 3:12 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>>
>>> The L2 space is already a Hilbert space,
>>
>>Yes, I know that. Perhap my phrase "emerge from an L2 space" was
>>impercise. More accurately I meant that the L2 space (thus a Hilbert
>>space) seems to emerge from the use of the gaussian form of the dirac
>>delta, which when all the exponents are added up form an integral of a
>>function (x-xj) squared. If the (x-xj) can be replaced with other
>>kinds of functions, then the space of all qualifying functions must be
>>square integrable and hence a Hilbert space. This is why I was asking
>>about compound functions that are still dirac delta functions.
>>
>>> and is much more fundamental
>>> than the delta function, which cannot be defined without a Hilbert
>>> space.
>>>
>>
>>I can certainly see that the existence of a Hilbert space implies a
>>dirac delta as an inner product.
>
> The dirac delta is not an inner product. It is a distribution.
Correct me if I am wrong, but I think that Mike is referring to the
inner product relation (see Sakurai, Modern Quantum Mechanics,
(1.6.2a):
<a'|a''> = delta(a'-a'') (1)
between any two eigenstates |a'>, |a''> in an infinite-dimensional
Hilbert space, having eigenvalues a', a'' along a continuous spectrum.
Here, delta is the Dirac delta.
>
>> But I'm not seeing it the other way
>>around. The usual definition of the dirac delta is $D(x-x0)dx=1, with
>>a limiting parameter that forces an infinity at the zero of the
>>argument. I don't seems to recognize how that necessarily specifies a
>>Hilbert space.
>>
> You are talking about functions and integration. How would you not
> have
> a Hilbert space?
The infinite-dimensioned Hilbert space certainly implies that there must
be an associated delta as defined above in (1).
The counterpart to (1) in a finite-dimensioned Hilbert space is:
<a'|a''> = delta_a'a'' (2)
Here, delta_ab is the Kronecker delta. What you seem to be circling
here, is the question of whether there is some "nascent delta" that is
associated with (2), such that it becomes (1) for the case of an
infinite-dimensional Hilbert space, and whether such nascent delta, if
it exists, is or is not unique.
It seems that to resolve this question, you want to start with the
Kronecker delta_a'a'' on the one hand, and the Dirac delta(a'-a'') on
the other, and develop a calculus theorem about the limit in which the
former becomes the latter as one approaches an infinite dimensional
space, which theorem uses nascent deltas, if possible.
Jay
Oh No
>> Thus spake Mike <mj...@sirus.com>
>>>On Dec 16, 3:12 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>>>
>>>> The L2 space is already a Hilbert space,
>>>
>>>Yes, I know that. Perhap my phrase "emerge from an L2 space" was
>>>impercise. More accurately I meant that the L2 space (thus a Hilbert
>>>space) seems to emerge from the use of the gaussian form of the dirac
>>>delta, which when all the exponents are added up form an integral of a
>>>function (x-xj) squared. If the (x-xj) can be replaced with other
>>>kinds of functions, then the space of all qualifying functions must be
>>>square integrable and hence a Hilbert space. This is why I was asking
>>>about compound functions that are still dirac delta functions.
>>>
>>>> and is much more fundamental
>>>> than the delta function, which cannot be defined without a Hilbert
>>>> space.
>>>>
>>>
>>>I can certainly see that the existence of a Hilbert space implies a
>>>dirac delta as an inner product.
>>
>> The dirac delta is not an inner product. It is a distribution.
>
>Correct me if I am wrong, but I think that Mike is referring to the
>inner product relation (see Sakurai, Modern Quantum Mechanics,
>(1.6.2a):
>
><a'|a''> = delta(a'-a'') (1)
>
>between any two eigenstates |a'>, |a''> in an infinite-dimensional
>Hilbert space, having eigenvalues a', a'' along a continuous spectrum.
>Here, delta is the Dirac delta.
Perhaps, but one should note several things. First, to talk of this one
already has to have the infinite dimensional Hilbert space. That is the
main thing I was trying to clarify. Second |a'>, |a''> are not actually
vectors in this space - that is where one introduces rigged Hilbert
space, and delta is not a function, so <a'|a''> is not actually an inner
product on Hilbert space. Of course, in non-rigorous treatments it is ok
to pretend that none of these issues matter, but in that case one really
should not claim that anything is proved. I think Sir Mike has been
interested in the much more stringent requirements of a proof.
>
>>
>>> But I'm not seeing it the other way
>>>around. The usual definition of the dirac delta is $D(x-x0)dx=1, with
>>>a limiting parameter that forces an infinity at the zero of the
>>>argument. I don't seems to recognize how that necessarily specifies a
>>>Hilbert space.
>>>
>> You are talking about functions and integration. How would you not
>>have
>> a Hilbert space?
>
>The infinite-dimensioned Hilbert space certainly implies that there
>must be an associated delta as defined above in (1).
>
>The counterpart to (1) in a finite-dimensioned Hilbert space is:
>
><a'|a''> = delta_a'a'' (2)
>
>Here, delta_ab is the Kronecker delta. What you seem to be circling
>here, is the question of whether there is some "nascent delta" that is
>associated with (2), such that it becomes (1) for the case of an
>infinite-dimensional Hilbert space, and whether such nascent delta, if
>it exists, is or is not unique.
I am not circling at all. I have done this explicitly in my papers and
shown that a particular nascent delta is associated with 2. On the other
hand I have not seen a simple way of taking the treatment in all
generality to rigged Hilbert space, although both the Princess and I had
the gut feeling that it should be possible.
>
>It seems that to resolve this question, you want to start with the
>Kronecker delta_a'a'' on the one hand, and the Dirac delta(a'-a'') on
>the other, and develop a calculus theorem about the limit in which the
>former becomes the latter as one approaches an infinite dimensional
>space, which theorem uses nascent deltas, if possible.
>
I have done what I think is needed.
Ilja
> "Oh No" <N...@charlesfrancis.wanadoo.co.uk> wrote in message
> Correct me if I am wrong, but I think that Mike is referring to the
> inner product relation �(see Sakurai, Modern Quantum Mechanics,
> (1.6.2a):
>
> <a'|a''> = delta(a'-a'') � �(1)
>
> between any two eigenstates |a'>, |a''> in an infinite-dimensional
> Hilbert space, having eigenvalues a', a'' along a continuous spectrum.
> Here, delta is the Dirac delta.
> The infinite-dimensioned Hilbert space certainly implies that there must
> be an associated delta as defined above in (1).
No. You can use a discrete basis (like that of eigenstates in a
harmonic oscillator) and use your <a'|a''> = delta_a'a'' also in
infinite-dimensional
Hilbert spaces, without any delta-functions. You need delta-functions
only
if you want to handle the continuous spectrum of some operator in a
similar
way as a discrete basis.
Mike
> Now I think the integration process is done first since there would be
> no possible way to even make sense of the integral if one or both of
> the delta parameters were allowed to approach zero first. Since the
> functional, or inner product, or integral is specified, that process
> has to be done first for it to even be accounted for.
The closest theorem that addresses the order of taking limits is the
Dominate Convergence Theorem, seen here at:
http://en.wikipedia.org/wiki/Dominated_convergence_theorem
But I'm not sure it applies to the Dirac delta function since there
doesn't seem to be a g(x) for which |D(x)|<g(x) for all x, since D(0)
=>00.
This doesn't mean that there is not some other proof that integration
commutes with the parameter's limiting process for the Dirac delta
function.