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Old Feb 14th 2016, 11:09 AM   #21
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Originally Posted by studiot View Post
Your statement [A, B] is a standard inner product notation.
That is incorrect. [A, B] is called the commutator of A and B and is defined as follows

[A, B] = AB - BA

Originally Posted by studiot View Post
If it is meant to be an inner product how is

the inner product [Force, Distance] incompatible?
A and B are observables which means that they're operators. There is no operator for force and force plays no role in quantum mechanics.
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Old Feb 14th 2016, 12:28 PM   #22
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With the greatest respect it is not incorrect.

I have seen it used for the inner product.

Kreider, page 257 equation 2 for example.

I have also seen other notations used, there are many unfortunately.

That is why it is encumbent upon the promoter to specify what he means by his notation.

You did not.

I am sorry I misuderstood but notation does not detract from the correctness of my assertions.

Incidentally my reference discusses the mathematical relationship between the Schwartz inequality, Cauchy inequality, Parsevals theorem and others quite simply.

Kreider, Kuller, Ostberg and Perkins

An Introduction to Linear Analysis

Also unfortunately another popular inner product notation < > is also used for Bra and Ket in Qunatum Mechanics so is best avoided.

The dot notation is good if you also embolden the vectors.

The use of square brackets in relation to commutators derives from group theory

[a, b] = a^-1b^-1ab for ab in the group.

as you correctly say in operator theory the commutatior of two operators AB is AB - BA.

Oh and by the way, does not AB - BA conform to the mathematical definition of an inner product for the vectors A and B?
(An inner product is a map from a vector space to the underlying field, in this case that of the complex numbers, although physicists only want the real part)

Oh and by the by the way bra and ket is also a form of inner product.

They get everywhere don't they?

The only Schrodinger equations for a semi infinite metallically bonded we can solve are the Konig Penny equations.
These lead to another one of these inequalities theat mathematicainas solve in the complex domain and physicists solve by saying the only real solutions are the ones that lead to bonding and anti bonding orbitals in the solid state.

That is halfway between QM and classical mechanics, but I was asked for examples in classical mechanics, so here is another example fom the same stable.

Of course the triangle inequality goes back to Euclid, but the inner product inequality leads to Nyquist's theorem.

Last edited by studiot; Feb 14th 2016 at 03:02 PM.
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Old Feb 22nd 2016, 02:52 AM   #23
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Originally Posted by studiot View Post
With the greatest respect it is not incorrect.
I still disagree, of course. In this post I'll explain why. oint. I'll take the most well known quantum mechanics texts, which I have, and post the definition of commutator as these texts use them. I'll then post the exact definition as given by that text and post the reference to the text.

First I'll start with Science World - Wolfram

First take a look at the definition and symbol for inner product at:
http://mathworld.wolfram.com/InnerProduct.html

The symbol for it is <a, b>. It's a generalization of the dot product.

Next we'll look at commutators:
http://scienceworlScienceWorld - Wolfram
[[b]A,B] = AB - BA
http://scienceworld.wolfram.com/phys...mmutators.html
In quantum mechanics, two quantities can be measured to any degree of precision only if their operators commute, in which case they have simultaneous eigenfunctions
Now we'll look at Wolfram - Mathworld
https://en.wikibooks.org/wiki/Quantu...nd_Commutators
Note the definition from the Contents under Mathematical Definition of Commutator which is located at the following link:
https://en.wikibooks.org/wiki/Quantu..._of_Commutator
It says (I can't do carets in ascii so I'm using boldface instead).
[A, B] = AB - BA
From Commutators
http://quantummechanics.ucsd.edu/ph1...s/node109.html
[p, x] = px - xp
The following are from my collection of quantum mechanics textbooks;

Quantum Mechanics - 3rd Ed. by Eugen Merzbacher. Commutator is defined on page 37 as [F, H] = FH - HF (Eq. 3.45).

Introductory Quantum Mechanics - 3rd Ed. by Richard L. Liboff. Commutator is defined on page 134 as [A, B] = AB - BA (Eq. 5.49).

Quantum Mechanics in a Nutshell by Gerald D. Mahan. Commutator is defined on page 37 as [A, B] = AB - [B]B[b]A (Eq. 2.167).

Quantum Mechanics: The Theoretical Minimum; What you need to know to start doing physics by Leonard Susskind & Art Friedman. Commutator is defined on page 111 as [L, M] = LM- ML.

Introduction to Quantum Mechanics - 2nd Ed. by David J. Griffiths. Commutator is defined on page 43 as [A, B] = AB - BA (Eq. 2.48).

Lectures in Quantum Mechanics by Steven Weinberg. Commutator is defined on page 25 as [A, B] = AB - BA

Originally Posted by studiot View Post
I have seen it used for the inner product.
When using a term in a particular field, like here where we're talking about quantum mechanics, the definition is as defined in that field if there is no other way that it's used as is the case in QM. Therefore since the symbol [A, B] never has any other meaning other than [A, B] = AB - BA there is no need to define it unless its used when discussing QM.

Originally Posted by studiot View Post
Kreider, page 257 equation 2 for example.
If that is a text what is the name of it?

Originally Posted by studiot View Post
I have also seen other notations used, there are many unfortunately.
I've never seen it used any other way. Please provide a few examples.

Originally Posted by studiot View Post
That is why it is encumbent upon the promoter to specify what he means by his notation.

You did not.
I disagree. The commutator is never used in any other way in QM so there is no need to do that since the meaning is implied. I didn't do so because there was no reason for it.
This entire thread is on QM. In particular its about the uncertainty principle so its meaning is implied. In fact if you look up the uncertainty principle in Wikipedia you would have seen this. That's what I do almost all the time. I recommend that you do the same thing too.

Originally Posted by studiot View Post
I am sorry I misuderstood but notation does not detract from the correctness of my assertions.
What assertions are you referring to?

Originally Posted by studiot View Post
Also unfortunately another popular inner product notation < > is also used for Bra and Ket in Qunatum Mechanics so is best avoided.
In almost all quantum mechanics texts both the commutator [A, B] and the inner product <a, b> are used. As I showed you using the Wolfram dictionaries they have one and only one meaning. And that's what I assumed to be true to the best of my knowledge. If its used any other way then I've never seen it and that would be the source of your disagreement with me.

Originally Posted by studiot View Post
Oh and by the way, does not AB - BA conform to the mathematical definition of an inner product for the vectors A and B?
The former applies only to the operators while the later only applies to dot product.

Originally Posted by studiot View Post
Oh and by the by the way bra and ket is also a form of inner product.
Yup. I'm quite aware of that, than you.

Originally Posted by studiot View Post
They get everywhere don't they?
Yeah. They sure do!

Originally Posted by studiot View Post
The use of square brackets in relation to commutators derives from group theory

[a, b] = a^-1b^-1ab for ab in the group.
My abstract algebra text does not use the symbol [ , ]. And the commutator used in abstract algebra is entirely different then the commutator in general. In fact it only refers to the "commutator of the group."

Last edited by Pmb; Feb 22nd 2016 at 09:16 PM.
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Old Sep 2nd 2016, 07:33 AM   #24
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Delay is the meaning of uncertainty
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Old Sep 2nd 2016, 08:20 AM   #25
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Originally Posted by Instantonly View Post
Delay is the meaning of uncertainty
Let me be the first to say "What the (ahem)?!" Could you expand upon that?

-Dan
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Old Sep 2nd 2016, 08:28 AM   #26
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Originally Posted by topsquark View Post
Let me be the first to say "What the (ahem)?!" Could you expand upon that?

-Dan
A photon polarises in the double slit experiment because it has length defined by it's polarisation/vacuum engagement. If the leading edge of the photon goes in one hole as up the trailing edge has to go through the other as down. There does have to be a radial limit to how far between holes provides a result doesn't there? A particle's spin is confined by it's time/vacuum delay. Confine the particle's vacuum delay and you can measure it's position and velocity precisely.

Uncertainty of outcome causes delay in human decision making.

One statement has to be correct at least or is that enough explanation?

Last edited by Instantonly; Sep 2nd 2016 at 09:01 AM.
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Old Sep 2nd 2016, 12:04 PM   #27
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Originally Posted by Instantonly View Post
A photon polarises in the double slit experiment because it has length defined by it's polarisation/vacuum engagement. If the leading edge of the photon goes in one hole as up the trailing edge has to go through the other as down. There does have to be a radial limit to how far between holes provides a result doesn't there? A particle's spin is confined by it's time/vacuum delay. Confine the particle's vacuum delay and you can measure it's position and velocity precisely.

Uncertainty of outcome causes delay in human decision making.

One statement has to be correct at least or is that enough explanation?
First off: This thread was started by studiot who wanted to have a discussion on the Uncertainty Principle. I highly doubt he was looking for anything beyond the Standard Model. Please keep this in mind in the future: don't do this in threads that are looking for answers in the Standard Model.

I would say that studiot's questions have been answered fully enough so I'm going to address your post now.

1. Photons do not have a length.
2. What do you mean by "engagement?"
3. Time/vacuum delay?
4. No, you can't measure a particle's position and momentum precisely at the same time. That's the whole meaning of the Uncertainty Principle...The whole point of this thread! There has been no sign of it ever having been violated. It has nothing at all to do with "human decision making."

You really need a review of Quantum Mechanics.

-Dan
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Old Sep 2nd 2016, 03:47 PM   #28
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Originally Posted by topsquark View Post
First off: This thread was started by studiot who wanted to have a discussion on the Uncertainty Principle. I highly doubt he was looking for anything beyond the Standard Model. Please keep this in mind in the future: don't do this in threads that are looking for answers in the Standard Model.

I would say that studiot's questions have been answered fully enough so I'm going to address your post now.

1. Photons do not have a length.
2. What do you mean by "engagement?"
3. Time/vacuum delay?
4. No, you can't measure a particle's position and momentum precisely at the same time. That's the whole meaning of the Uncertainty Principle...The whole point of this thread! There has been no sign of it ever having been violated. It has nothing at all to do with "human decision making."

You really need a review of Quantum Mechanics.

-Dan
a) Photon's have a length defined by polarisation dissociation developed by it's total vacuum exposure.
b)It has nothing to do with observation or decision making(don't know how you brought that into the topic).
c)I'm not certain you have read the revised OP to the thread I posted.

I can only apologise to studiot for hijacking his thread with thinking.
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Old Sep 2nd 2016, 04:15 PM   #29
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Originally Posted by Instantonly View Post
a) Photon's have a length defined by polarisation dissociation developed by it's total vacuum exposure.
b)It has nothing to do with observation or decision making(don't know how you brought that into the topic).
c)I'm not certain you have read the revised OP to the thread I posted.

I can only apologise to studiot for hijacking his thread with thinking.
a) Photons are the quanta of the electromagnetic field. They are point particles so far as we know.

b)
Uncertainty of outcome causes delay in human decision making.
c) Nope. I didn't. There's no point. You have so many conceptual problems here it isn't worth my time.

Learn some Physics before you try to shoot it down.

Unless studiot wants to re-open this, thread closed.

-Dan
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