Deriving Schrödinger's Equation from my Fundamental Equation

Doctordick · 08-16-2008

Hi Anssi, sorry you were ill; I hope you are feeling better now. I would have answered your post more quickly but I wanted to be as clear as possible so that my comments would not be misunderstood.

Quote:

Originally Posted by AnssiH

Well not exactly. It means Schrödinger's equation is a tautology from "not making undefendable assumptions about the meaning of any raw data".

This I would have to agree with; my deduction is very much based upon “not making undefendable assumptions”.

Quote:

Originally Posted by AnssiH

I.e. you remember those symmetries that were discussed at the early stages of the deduction? Shift symmetry to the assignment of labels etc? Those symmetries are the source of Schrödinger's equation being valid. Your worldview doesn't actually need to make those symmetry assumptions, but in exchange there would always exist some undefendable assumptions (much like assuming an ontological "center point" for the universe). Modern physical models do make those symmetry assumptions (for the most part anyway), and that is why they look alike with DD's deductions.

Here I wouldn't quite put it the way you did; I think you have things a little backward (of course I could be a little prejudice there

). I would instead say that my work indicates the relationships found by modern physicists have to be true (at least as a good approximation) and it shouldn't be too surprising that after three hundred years of comparing their expectations based on their epistemological constructs to actual results (their experiments) they should come up with some of these same relationships. And that would include their discovery of symmetry arguments; however, I have never heard of anyone even considering those symmetry relationships to be a more fundamental starting point then time and/or dimensionality itself.

I would comment that both Rade and LaurieAG are, to a great extent, trolls offering little consideration to what is actually being said. LaurieAG is new to me but there are others who agree. Rade has been on my back for many years already and I am tired of dealing with his contributions,

I will comment on his failure to include the single most important aspect of my comment in his quote above. I have included the important part in square brackets here

Quote:

Originally Posted by Rade

So, here then I think we find understanding why physicists claim the DD equation is philosophy [and outside their interest], philosophers claim it is mathematics [and outside their interest], and mathematicians claim it is physics [and outside their interest]--all are correct.

It is indeed, outside their interest and, I might say, outside their expertise.

That brings me to something I just read in the August 16 issue of “Science News”

Quote:

Originally Posted by Douglas Osheroff

Those discoveries that most change the way we think about nature cannot be anticipated... Beware of subtle unexplained behavior; don't dismiss it. Frequently nature does not knock with a very loud sound but rather a very soft whisper, and you have to be aware of subtle behavior which may in fact be a sign that there is interesting physics to be had.

Too much noise from professionals battening down the hatches against all attacks on established authority can easily drown out those subtle observations; like the fact that “clocks don't measure time” but rather measure Einstein's invariant interval.

I think Rade and LaurieAG add random noise for the sole purpose of disrupting discussions over their heads. I have a strong suspicion that Rade thinks the sole purpose of knowledge is to cover up stupidly. I have had the following as a sign on the wall of my office for many years.

Knowledge is Power
And the most popular abuse of that
power is to use it to hide stupidity.

I used to have that as part of my signiture.

Have fun -- Dick

AnssiH · 08-17-2008

Quote:

Originally Posted by Doctordick

Hi Anssi, sorry you were ill; I hope you are feeling better now.

Slowly getting there... Still been having some sort of flu aftermath and my head's been all sore, haven't had the energy to do much of anything... I kind of feel like I'm getting old with strange aches in my head... I press the top of my head and feel it at the right side of my upper jaw, what the hell?

Yeah I couldn't really figure out what Rade is saying... Anyway, I probably sound like a broken record player but I'll be trying to get to the topic soon

-Anssi

LaurieAG · 08-19-2008

Quote:

Originally Posted by Doctordick

I would comment that both Rade and LaurieAG are, to a great extent, trolls offering little consideration to what is actually being said. LaurieAG is new to me but there are others who agree.

When I was studying calculus and advanced maths at high school (we had to wait for proofs from first principles in first year uni) I came to the conclusion that if, during the working out of a solution, I proved that 1=1 or 0=0 then I realised that I did not have a solution for the stated problem due to errors in my process and would have to start again if I wanted to get the real answer.

Haven't you realised this yet?

Rade · 08-20-2008

LOL, poor DD, does my name cause such a negative reaction that you cannot even comprehend a compliment when placed on the tip of your nose ?--clearly your 'sign-on-the-wall' is well placed.

AnssiH · 08-25-2008

I ran through post #7 again to refresh my memory and it still seemed to make sense. So to pick up from where I'm at with the OP...

Quote:

Originally Posted by Doctordick

Notice that $\int \vec{\Psi}_2^\dagger \cdot\vec{\Psi}_2dV_2$ equals unity by definition of normalization. Furthermore, since the tau axis was introduced for the sole purpose of assuring that two identical indices associated with valid ontological elements existing in the same $(x,\tau)_t$ would not be represented by the same point,

I suppose that should read "...assuring that two identical ontological elements, being referred to by the same "X" index, would not exist in the same point..." or something along those lines.

Quote:

we came to the conclusion that $\vec{\Psi}_1$ must be asymmetric with regard to exchange of arguments.

So, I dug this up and I suppose it is the issue explained in post #180 of What can we know of reality:
What can we know of reality?

Quote:

Originally Posted by Doctordick

When we actually let the number of possibilities go to infinity and include all possibilities, we run into the circumstance where the difference between two indices can go to zero. Now, if we have two indices who differ by exactly zero, is it not true that they are the same? If they are the same then the two points which were to represent different noumena become a single point and the purpose for which the tau axis was created is no longer effective. I got around this difficulty by requiring $vec{Psi}$ to be asymmetric with respect to exchange; by driving the probability density to exactly zero, this will guarantee the difficulty never arises.

Seems to make sense.

Quote:

If that is indeed the case (as it must be) then the second term in the above equation will vanish identically as $\vec{x}_i$ can never equal $\vec{x}_j$ for any i and j both chosen from set #1.

By "second term" you must refer to:
$\sum_{i \neq j (\#1)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j)$

Seems to make sense that it would vanish.

So this is where we stand;

$\left\{\sum_{\#1} \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j (\#1)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\right\}\vec{\Psi}_1 + \left\{2 \sum_{i=\#1 j=\#2}\int \vec{\Psi}_2^\dagger \cdot \beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 \right. +$
$\left.\int \vec{\Psi}_2^\dagger \cdot \left[\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j (\#2)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j) \right]\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1+K \left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1$

Quote:

If the actual function $\vec{\Psi}_2$ were known (i.e., a way of obtaining our expectations for set #2 is known), the above integrals could be explicitly done and we would obtain an equation of the form:

$\left\{\sum_{i=1}^n \vec{\alpha}_i \cdot \vec{\nabla}_i +f(\vec{x}_1,\vec{x}_2, \cdots,\vec{x}_n,t)\right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1.$

The function f must be a linear weighted sum of alpha and beta operators plus one single term which does not contain such an operator. That single term arises from the final integral of the time derivative of $\vec{\Psi}_2$ on the right side of the original representation of the result of integration:

$\int \vec{\Psi}_2^\dagger\cdot\frac{\partial}{\partial t}\vec{\Psi}_2dV_2.$

So here I need help. I do not quite understand how things unfold into that equation, and what or how the "function f" appears there, or what it means that a function is a "linear weighted sum of...", and how does that single term arise from the integral of the time derivative of $\vec{\Psi}_2$ ...

I suppose I shouldn't plow onwards until I understand that step.

-Anssi

Doctordick · 08-25-2008

Quote:

Originally Posted by AnssiH

I suppose I shouldn't plow onwards until I understand that step.

Yes sir; please stop the moment anything is not clear.

Quote:

Originally Posted by AnssiH

I suppose that should read "...assuring that two identical ontological elements, being referred to by the same "X" index, would not exist in the same point..." or something along those lines.

Actually, I kind of find the phrase, “exist in the same point” applying to the actual ontological elements a little bothersome. On rereading the original line, I suspect the real problem is the length of the sentence. I'll think about it and edit the paragraph to something better.

I take your comment, “Seems to make sense” to mean you understand the consequence of requiring asymmetry; your presumed reference was right on the money.

Quote:

Originally Posted by AnssiH

By "second term" you must refer to:
$\sum_{i\neq j (\#1)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j)$

Seems to make sense that it would vanish.

That is correct.

Quote:

Originally Posted by AnssiH

So this is where we stand;

Not quite. I wouldn?t have included the term we just decided must vanish. I would have said, this is where we stand:

$\sum_{\#1} \vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1 + \left\{2 \sum_{i=\#1 j=\#2}\int \vec{\Psi}_2^\dagger \cdot \beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 \right. +$
$\left.\int \vec{\Psi}_2^\dagger \cdot \left[\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j (\#2)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j) \right]\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1+K \left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1$

Quote:

Originally Posted by AnssiH

So here I need help. I do not quite understand how things unfold into that equation, and what or how the "function f" appears there, or what it means that a function is a "linear weighted sum of...", and how does that single term arise from the integral of the time derivative of $\vec{\Psi}_2$ ...

Maybe it will be a little clearer if I make a minor rewrite of the above equation. What is important is that the alpha and beta operators can be factored from the integrals, they operate on the terms of the sums over i and j, not the arguments of those terms.

$\sum_{\#1} \vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1 + \left\{2 \sum_{i=\#1 j=\#2}\beta_{ij}\int \vec{\Psi}_2^\dagger \cdot \delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 +\sum_i \vec{\alpha}_i\cdot \int\vec{\Psi}_2^\dagger \cdot \vec{\nabla}_i \vec{\Psi}_2dV_2 \right. +$
$\left.\sum_{i \neq j (\#2)}\beta_{ij}\int \vec{\Psi}_2^\dagger \cdot \delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1+K \left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1$

Now, all those integrals essentially yield numerical results which are functions of the arguments from set #1. All the arguments from set #2 have been integrated over (essentially those integrals can be seen as functional contributions arising from the probability distribution of indices taken from set #2 summed over all possibilities (remember that the integral is defined to be the result of a sum taken in the limit where the size of the elements goes to zero as the number of elements goes to infinity; $\vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2$ is the density of the elements in that sum (as a function of the arguments) and dV_2 is the differential element which drives the net element to zero in the limit.

A “weighted sum” is a sum of terms where each term has a weight assigned to it: five of these, four of those, fiftytwo of a third thing, etc, etc. The sum above (after the integrals are done) is a simple weighted sum of alpha and beta operators where the integrals provide the weights (these weights will end up being functions of the arguments from set #1 (those from set #2 are integrated out). The adjective “linear” simply means that every term contains only one of those elements: there are no terms which contain a product of two such elements. The function “f” is no more than a symbol which stands for that sum: i.e., “the function f must be a linear weighted sum of alpha and beta operators” is exactly what I have just said.

And lastly, the term $\int \vec{\Psi}_2^\dagger \cdot\frac{\partial}{\partial t}\vec{\Psi}_2 dV_2$ is the only term arising from the integration operation which does not contain either a alpha or a beta operator. Thus we have “one single term which does not contain such an operator”. Thus it is that we know that the resulting equation above can be written in the form:

$\left\{\sum_{i=1}^n \vec{\alpha}_i \cdot \vec{\nabla}_i +f(\vec{x}_1,\vec{x}_2, \cdots,\vec{x}_n,t)\right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1.$

The actual form of f is determined entirely by the means by which we obtain our expectations for the arguments of set #2 which, as I said earlier, are the fabricated references we created to insure that the arguments of set #1 are consistent with our experiences. So, what should we do? We should perform experiments to determine the form of f. Actually, that would be pretty much a waste of time because the equation we have is still a many body equation and there exists no method of solving it. That is why I proceed to the next step.

I hope that makes a little sense to you.

Have fun -- Dick

AnssiH · 09-06-2008

Hi, sorry I'm being slow again...

One question that just occurred to me. Since set #1 and set #2 obey the same flaw free explanation, aren't $\vec{\Psi}_1$ and $\vec{\Psi}_2$ the same function? Or is the issue rather that, if one could actually tell the elements of #1 & #2 apart, they would obey different $\vec{\Psi}$ functions?

Quote:

Originally Posted by Doctordick

I hope that makes a little sense to you.

Well it does make little sense to me... ...but I was hoping it'd make a lot of sense

Heh, seriously though, it was helpful, even though I could not really understand everything you said;

Quote:

Maybe it will be a little clearer if I make a minor rewrite of the above equation. What is important is that the alpha and beta operators can be factored from the integrals, they operate on the terms of the sums over i and j, not the arguments of those terms.

$\sum_{\#1} \vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1 + \left\{2 \sum_{i=\#1 j=\#2}\beta_{ij}\int \vec{\Psi}_2^\dagger \cdot \delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 +\sum_i \vec{\alpha}_i\cdot \int\vec{\Psi}_2^\dagger \cdot \vec{\nabla}_i \vec{\Psi}_2dV_2 \right. +$
$\left.\sum_{i \neq j (\#2)}\beta_{ij}\int \vec{\Psi}_2^\dagger \cdot \delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1+K \left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1$

I see the alpha and beta operators have been moved outside of integrals, but I am not really sure what it means that they operate on the terms of the sums, but not the arguments of those terms, and I have no idea why that means that they can be factored from the integrals.

Quote:

Now, all those integrals essentially yield numerical results which are functions of the arguments from set #1.

Meaning, the results of the integrals have a dependency on the arguments from set #1?

Even that final integral from the left side of the equation, which does not seem to contain any arguments from #1? I.e. $\sum_{i \neq j (\#2)}\beta_{ij}\int \vec{\Psi}_2^\dagger \cdot \delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2$

Quote:

All the arguments from set #2 have been integrated over (essentially those integrals can be seen as functional contributions arising from the probability distribution of indices taken from set #2 summed over all possibilities (remember that the integral is defined to be the result of a sum taken in the limit where the size of the elements goes to zero as the number of elements goes to infinity; $\vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2$ is the density of the elements in that sum (as a function of the arguments) and dV_2 is the differential element which drives the net element to zero in the limit.

I'm afraid I understand almost nothing of the above

Can you explain it with more detail?

Quote:

A “weighted sum” is a sum of terms where each term has a weight assigned to it: five of these, four of those, fiftytwo of a third thing, etc, etc. The sum above (after the integrals are done) is a simple weighted sum of alpha and beta operators where the integrals provide the weights (these weights will end up being functions of the arguments from set #1 (those from set #2 are integrated out). The adjective “linear” simply means that every term contains only one of those elements: there are no terms which contain a product of two such elements. The function “f” is no more than a symbol which stands for that sum: i.e., “the function f must be a linear weighted sum of alpha and beta operators” is exactly what I have just said.

And lastly, the term $\int \vec{\Psi}_2^\dagger \cdot\frac{\partial}{\partial t}\vec{\Psi}_2 dV_2$ is the only term arising from the integration operation which does not contain either a alpha or a beta operator. Thus we have “one single term which does not contain such an operator”. Thus it is that we know that the resulting equation above can be written in the form:

$\left\{\sum_{i=1}^n \vec{\alpha}_i \cdot \vec{\nabla}_i +f(\vec{x}_1,\vec{x}_2, \cdots,\vec{x}_n,t)\right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1.$

So that "one single term which does not contain alpha nor beta operator" is simply one of the arguments of $f$ ? (because I was wondering why I can't see it anywhere

I now have an incredibly vague idea about what the above means (which is a lot more than I earlier had), and how the equation could be expressed that way after the integrals are done. Very, uncomfortably vague. I could take it on faith for now and proceed to next step, unless you think this might bite me in the ass sooner or later?

-Anssi

ps. Wow, the LaTex in quotes works now! Excellent, thanks to whoever fixed that.

Doctordick · 09-07-2008

Hi Anssi! Don't worry about being slow.

Quote:

Originally Posted by AnssiH

One question that just occurred to me. Since set #1 and set #2 obey the same flaw free explanation, aren't $\vec{\Psi}_1$ and $\vec{\Psi}_2$ the same function? Or is the issue rather that, if one could actually tell the elements of #1 & #2 apart, they would obey different $\vec{\Psi}$ functions?

The elements of the entire universe obey exactly the same flaw free explanation does not mean that the probability distribution (what is being calculated via $\vec{\Psi}$ ) is the same for both sets. The probability distributions for some arguments are a function of the probability distributions of other events. The probability that you will have no problems driving to work is dependent upon the probability there is air in your tires. The whole finished result, $\vec{\Psi}$ , is a coherent expression. When everything is included, $\vec{\Psi}$ must obey my fundamental equation. I have divided the entire set of arguments into two sets: set #1 and set #2. Having done that, I point out the following;

Quote:

Originally Posted by Doctordick

Having divided the arguments into two sets, a competent understanding of probability should lead to acceptance of the following relationship: the probability of #1 and #2 (i.e., the expectation that these two specific sets occur together) is given by the product of two specific probabilities: $P_1$ (#1), the probability of set number one, times $P_2$ (#2 given #1), the probability of set number two given set number one exists. The existence of set #1 in the second probability is necessary as the probability of set #2 can very much depend upon that existence. At this point, exactly the same argument used to defend $\vec{\Psi}$ as embodying a method of obtaining expectations (the probability distribution) for the entire collection of arguments can be used to assert that there must exist abstract vector functions $\vec{\Psi}_1$ and $\vec{\Psi}_2$ which will yield, respectively $P_1$ and $P_2$ .

It should be clear that, under these definitions (representing the argument $(x,\tau)_i$ as $\vec{x}_i$ ),

$\vec{\Psi}(\vec{x}_1,\vec{x}_2,\cdots, t)=\vec{\Psi}_1(\vec{x}_1,\vec{x}_2,\cdots,\vec{x}_n, t)\vec{\Psi}_2(\vec{x}_1,\vec{x}_2,\cdots, t).$

What is being said here is that $\vec{\Psi}$ can be written as a product of two functions, $\vec{\Psi}_1$ $\vec{\Psi}_2$ . It is very important to realize that the actual arguments of $\vec{\Psi}_2$ must include both set #1 and set #2. When you break this into a product of two functions, one of the two must contain the arguments of the other (note the phrase “ the probability of set number two given set number one exists”) You can talk about the probability of one set sans information about the other set but once you establish that your situation demands set #1, the probability of the other set can depend upon what that first set consisted of. That fact has some very important consequences: the way I have laid it out, $\vec{\Psi}_2$ must be a function of both sets.

You could estimate the probability you will be late to work tomorrow and you could estimate the probability you would find your car with flat tires tomorrow. You could also estimate the probability you will be late to work tomorrow given you had flat tires. These are three entirely different probabilities. The probability you would have flat tires and be late to work is not the probability you will have flat tires times the probability you would be late to work but it is the probability you would have flat tires times the probability you would be late to work “given you have flat tires”. See how that other argument gets into the thing?

Quote:

Originally Posted by AnssiH

I see the alpha and beta operators have been moved outside of integrals, but I am not really sure what it means that they operate on the terms of the sums, but not the arguments of those terms, and I have no idea why that means that they can be factored from the integrals.

They are just operators which operate on $\vec{\Psi}$ , not on the arguments of $\vec{\Psi}$ . They have an index upon them which indicates the term in the sum to which they are attached. Just think of them as something that term is multiplying: i.e., they are simple factors by definition.

Quote:

Originally Posted by AnssiH

Meaning, the results of the integrals have a dependency on the arguments from set #1?

The integrations are over all the arguments of set #2 but not over all the arguments of $\vec{\Psi}_2$ (the arguments of $\vec{\Psi}_2$ include the arguments of set #1 and we are not integrating over set #1). The value of the resulting integrals therefore depends upon the values arguments from set #1; the result of the integral is some function of the arguments of set #1. If we knew what the function $\vec{\Psi}_2$ was (i.e., we knew how to estimate our expectations for set #2) we would know how those expectations depended upon set #1 and we would thus know what that function obtained by integration would be.

Quote:

Originally Posted by AnssiH

Even that final integral from the left side of the equation, which does not seem to contain any arguments from #1? I.e. $\sum_{i \neq j (\#2)}\beta_{ij}\int \vec{\Psi}_2^\dagger \cdot \delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2$

As I said, it is the function $\vec{\Psi}_2$ which contains the arguments from set #1; that is a completely different from the index over which that integral is being summed (the sum is over the arguments of set #2 and that sum must be there because a different result is obtained for every such pair; that Dirac delta function makes each of those integrals different).

Quote:

Originally Posted by AnssiH

I'm afraid I understand almost nothing of the above

Can you explain it with more detail?

Essentially, all I am saying is that integration over all arguments of set #2 is equivalent to a sum of the probabilities for set #2 over all possibilities so that the arguments from set #2 disappear from the representation.

Quote:

Originally Posted by AnssiH

So that "one single term which does not contain alpha nor beta operator" is simply one of the arguments of $f$ ? (because I was wondering why I can't see it anywhere

No, it is not one of the arguments of f. Each and every integral we have done in the above expansion results in some function of the arguments from set #1; that would be a whole set of functions. One could call those functions $G_k(\vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)$ where k is a simple index on that collection of functions (which integral it came from). All of such functions which arose from integrations on the left side of the equation would include either an alpha operator or a beta operator (which we could call Op_k as a factor but the one from the integration on the right side of the equation contains no such operator. Thus it is that

$f( \vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)=G_{right side}( \vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)+ \sum_k G_k( \vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)Op_k$

I was saying something quite simple; $f(\vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)$ is a linear weighted sum of alpha and beta operators (those Op_k) plus a single term which does not contain an alpha or beta operator (that $G_{right side}$ ).

I hope that is somewhat clearer.

Have fun -- Dick

AnssiH · 09-13-2008

Quote:

Originally Posted by Doctordick

You could estimate the probability you will be late to work tomorrow and you could estimate the probability you would find your car with flat tires tomorrow. You could also estimate the probability you will be late to work tomorrow given you had flat tires. These are three entirely different probabilities. The probability you would have flat tires and be late to work is not the probability you will have flat tires times the probability you would be late to work but it is the probability you would have flat tires times the probability you would be late to work “given you have flat tires”. See how that other argument gets into the thing?

Ah, right, of course...

Quote:

They are just operators which operate on $\vec{\Psi}$ , not on the arguments of $\vec{\Psi}$ . They have an index upon them which indicates the term in the sum to which they are attached. Just think of them as something that term is multiplying: i.e., they are simple factors by definition.

Hmm, after scratching my head a bit I think I understand what you are saying above... Still - though I may be getting bogged down to details a bit too much - my math knowledge is limited so I'm don't know why that means the alpha & beta operators can be moved outside of the integrals...

Quote:

The integrations are over all the arguments of set #2 but not over all the arguments of $\vec{\Psi}_2$ (the arguments of $\vec{\Psi}_2$ include the arguments of set #1 and we are not integrating over set #1). The value of the resulting integrals therefore depends upon the values arguments from set #1; the result of the integral is some function of the arguments of set #1. If we knew what the function $\vec{\Psi}_2$ was (i.e., we knew how to estimate our expectations for set #2) we would know how those expectations depended upon set #1 and we would thus know what that function obtained by integration would be.

Right, that seems to make sense now.

Quote:

No, it is not one of the arguments of f. Each and every integral we have done in the above expansion results in some function of the arguments from set #1; that would be a whole set of functions. One could call those functions $G_k(\vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)$ where k is a simple index on that collection of functions (which integral it came from). All of such functions which arose from integrations on the left side of the equation would include either an alpha operator or a beta operator (which we could call Op_k as a factor but the one from the integration on the right side of the equation contains no such operator. Thus it is that

$f( \vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)=G_{right side}( \vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)+ \sum_k G_k( \vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)Op_k$

I was saying something quite simple; $f(\vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n,t)$ is a linear weighted sum of alpha and beta operators (those Op_k) plus a single term which does not contain an alpha or beta operator (that $G_{right side}$ ).

Right, okay, I think I understand that now. No time to scratch my head more right now as I need to dash, but I should have time to continue from here tomorrow.

Thanks,
-Anssi

Doctordick · 09-13-2008

Quote:

Originally Posted by AnssiH

Hmm, after scratching my head a bit I think I understand what you are saying above... Still - though I may be getting bogged down to details a bit too much - my math knowledge is limited so I'm don't know why that means the alpha & beta operators can be moved outside of the integrals...

Go back to the definition of an integral. Remember that integral sign started life as a large capital “S” standing for a sum where the number of terms in the sum became infinite. If that sum is to be finite (which is the majority of interesting cases) then the terms being summed must go to zero; that is why the differential factor is there (that dx or dz or, in my particular example, dV₂). The differential term accommodates that characteristic so that we can speak of the function being integrated over as a non-shrinking function.

Or another way to look at it is like this. Suppose we say $z=\int f(x)dx$ where dz is a single element of that supposed sum: i.e., the sum can be written $\int dz =z$ . Then dz=f(x)dx; or, dividing by dx, we have $\frac{dz}{dx}=f(x)$ or f(x) is the differential of z and/or z is the “antidifferential” of f(x).

No matter how it is looked at, it is still essentially a sum over a bunch of terms. The point here being that each and every one of those terms (in any specific integral we are doing here) is multiplied by the same alpha (or beta) operator. Think of the operators as apples, oranges, peaches, grapes, etc.: i.e., what you actually get when you add them together is not defined (if you add 5 apples to 3 oranges, you have 5 apples and 3 oranges).

If you go back to the original deduction of my fundamental equation, you will see that these operators were inserted for a very specific purpose. If we need to go back over that purpose, I will present it again (perhaps more clearly than I did the first time). The central purpose is to represent three very different relationships in a single equation as if they were actually related to one another when they really are not.

Have fun -- Dick

Deriving Schrödinger's Equation from my Fundamental Equation

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