Deriving Schrödinger's Equation from my Fundamental Equation

Doctordick · 07-13-2008

This post is essentially a quote of post #194 on the thread “What can we know of reality?”

I want to make it clear to anyone who reads this that the issue is not really a solution of my fundamental equation but rather an examination of possible solutions. By definition, $\vec{\Psi}$ is a mathematical representation of our expectations. Those expectations are the result of a flaw free explanation of reality. The explanation itself is a epistemological construct which provides a consistent and flaw free explanation of the past. As such, I have no real interest in the actual solution or how it was achieved; my only interest is in the fact that such a solution exists: i.e., you do in fact have expectations.

There are two facts extant here: first, a function (a method of obtaining one's expectations from a given set of known elements: i.e., $\vec{\Psi}$ ) exists and that function must be a solution to my fundamental equation. Furthermore, if I understand that flaw-free explanation, the method of obtaining the appropriate expectations is known to me. It is very important here to remember that $\vec{\Psi}$ is a mathematical representation of our expectations and is not necessarily a correct representation of the future. What I am trying to point out is that our expectations are never necessarily correct (see Kriminal99's post on induction); what is being enforced is that the known past is consistent with those expectations,not the future. The future is a totally unknown issue. Our only defense of our expectations is that the volume of information which goes to make up the past is far far in excess of the next “present” (from our perspective): i.e., it would be rather ridiculous to conclude that anything in the next “present” would be sufficiently significant to be a major alteration to the net past (that would be “all the information we are trying to make sense of”).

With that in mind, the equation of interest is

$\left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) \right\}\vec{\Psi} = K\frac{\partial}{\partial t}\vec{\Psi}.$

This expression is quite analogous to a differential equation describing the evolution of a many body system which, as anyone competent in physics knows, is not an easy thing to solve. What we would like to do is to reduce the number of arguments to something which can be handled: i.e., we want to know the nature of the equations which must be obeyed by a subset of those variables. In an interest towards accomplishing that result, my first step is to divide the problem into two sets of variables: set number one will be the set referring to our “valid” ontological elements (together with the associated tau indices) and set number two will refer to all the remaining arguments. I will refer to these sets as #1 and #2 respectively. (You should comprehend that #1 must be finite and that #2 can possibly be infinite.) Now, when we started this whole thing, I defined the probability of specific expectations to be given by the squared magnitude of $\vec{\Psi}$ under the argument that such a notation (that abstract vector) can represent absolutely any method of getting from one set of numbers to another: i.e., there exists no operation capable of yielding one's expectations which cannot be represented by such a structure.

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 $\vec{x}_i$ ),

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

Substituting this result into our fundamental equation, what we obtain can be written

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

At this point, it is important to realize that set #2 consists of invalid ontological elements created for the purpose of constraining set #1 to what they actually were. I often used to ask the question, “how does one tell the difference between an electron and a Volkswagen?” No one except Anssi seemed to ever grasp the essence of that question. The answer is of course: “context”. In my original proof, arbitrary invalid ontological elements were added until one achieved the state where knowing the specific indices of any n-1 elements associated with a given t index would guarantee that the index of the missing element could be determined. Under this picture, set #2 is certainly context as since they are invalid ontological elements, they can be anything so long as they are consistent with the explanation: i.e., the only requirement here is that they need to obey the fundamental equation. Thus it is that I will take the position that, if we know a flaw-free explanation, we know the method of obtaining our expectations for set #2: i.e., we know $\vec{\Psi}_2$ . If we left multiply the above equation by $\vec{\Psi}_2^\dagger$ (forming the inner or dot product with the algebraically modified $\vec{\Psi}_2$ ) and integrate over the entire set of arguments referred to as set #2, we will obtain the following result:

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

Notice that $\int \vec{\Psi}_2^\dagger \cdot\vec{\Psi}_2dV_2$ equals unity by definition of normalization. Furthermore, the tau axis was introduced for the sole purpose of assuring that two identical indices associated with valid ontological elements existing in the same B(t) ( now being represented by an $(x,\tau)_t$ point in the $x,\tau$ plane) would not be represented by the same point. We came to the conclusion that this could only be guaranteed in the continuous limit by requiring $\vec{\Psi}_1$ to be asymmetric with regard to exchange of arguments. 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.

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:

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

The above is an example of the kind of function the indices on our valid ontological elements must obey; however, it is still in the form of a many body equation and is of little use to us if we cannot solve it. In the interest of learning the kinds of constraints the equation implies, let us take the above procedure one step farther and search for the form of equation a single index must obey (remember the fact that we added invalid ontological elements until the index on any given element could be recovered if we had all n-1 other indices). We may immediately write $P_1$ (set #1) = $P_0(\vec{x}_1,t)P_r$ (remainder of set #1 given $\vec{x}_1$ ,t). Note that $\vec{x}_1$ can refer to any index of interest as order is of no significance. Once again, we can deduce that there exist algorithms capable of producing $P_0$ and $P_r$ ; I will call these functions $\vec{\Psi}_0$ and $\vec{\Psi}_r$ respectively. It follows that $\vec{\Psi}_1$ may be written as follows:

$\vec{\Psi}_1(\vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n, t)= \vec{\Psi}_0(\vec{x}_1,t)\vec{\Psi}_r(\vec{x}_1,\vec{x}_2, \cdots, \vec{x}_n, t).$

If I make this substitution in the earlier equation for $\vec{\Psi}_1$ , I will obtain the following relationship:

$\left\{\sum_{i=1}^n \vec{\alpha}_i \cdot \vec{\nabla}_i +f(\vec{x}_1,\vec{x}_2, \cdots,\vec{x}_n,t)\right\}\vec{\Psi}_0\vec{\Psi}_r = K\frac{\partial}{\partial t}(\vec{\Psi}_0\vec{\Psi}_r).$

Once again I point out that $\vec{\Psi}_r$ constitutes the context for $\vec{\Psi}_0(\vec{x}_1,t)$ . Once again, I will take the position that, if we know dthe flaw-free explanation represented by $\vec{\Psi}$ , we know our expectations for the set of indices two through n, set “r”,: i.e., we know $\vec{\Psi}_r$ (the context). As before, if we now left multiply the above equation by $\vec{\Psi}_r^\dagger$ (forming the inner or dot product with the algebraically modified $\vec{\Psi}_r$ ) and integrate over the entire set of arguments referred to as set “r” (the remainder after $\vec{x}_1$ has been specified), we will obtain the following result:

$\vec{\alpha}_1\cdot \vec{\nabla}_1\vec{\Psi}_0 + \left\{\int \vec{\Psi}_r^\dagger\cdot \left[ \sum_{i=1}^n \vec{\alpha}_i \cdot \vec{\nabla}_i +f(\vec{x}_1,\vec{x}_2, \cdots,\vec{x}_n,t)\right] \vec{\Psi}_r dV_r\right\}\vec{\Psi}_0 = K\frac{\partial}{\partial t}\vec{\Psi}_0\ + K\left\{\int \vec{\Psi}_r^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_r dV_r \right\}\vec{\Psi}_0.$

Notice once again that $\int \vec{\Psi}_r^\dagger \cdot\vec{\Psi}_rdV_2$ equals unity by definition of normalization. Notice also that the term $\vec{\alpha}_1\cdot \vec{\nabla}_1$ appears both standing alone and inside the integral over the indices represented by the set “r”; this occurs because $\vec{\Psi}_r$ is a function of $\vec{x}_1$ and the chain rule applies to differential operation on the product function $\vec{\Psi}_0\vec{\Psi}_r$ .

Now, this resultant may be a linear differential equation in one variable but it is not exactly in a form one would call “transparent”. In the interest of seeing the actual form of possible solutions allow me to discuss an approximate solution discovered by setting three very specific constraints to be approximately valid. The first of these three is that the data point of interest, $\vec{x}_1$ , is insignificant to the rest of the universe: i.e., $P_r$ is, for practical purposes, not much effected by any change in the actual form of $\vec{\Psi}_0$ : i.e., feed back from the rest of the universe due to changes in $\vec{\Psi}_0$ can be neglected. The second constraint will be that the probability distribution describing the rest of the universe is stationary in time: that would be that $P_r$ is, for practical purposes, not a function of t. If that is the case, the only form of the time dependence of $\vec{\Psi}_r$ which satisfies temporal shift symmetry is $e^{iS_rt}$ .

At this point, we must carefully analyze the development of the function f created when we integrated over set #2 in our earlier example. As mentioned at the time, f was a linear weighted sum of alpha and beta operators except for one strange term introduced by the time derivative of $\vec{\Psi}_2$ . Please note that, if $P_r$ is insensitive to $\vec{\Psi}_0$ and stationary in time then so is $P_2$ . This follows directly from the fact that $P_2$ is the probability distribution of the “invalid” ontological elements required to constrain the “valid” ontological elements to what is to be explained. There is certainly no required time dependence if the set to be explained has no time dependence, nor can there be any dependence upon $\vec{\Psi}_0$ if the set “r” can be seen as uninfluenced by $\vec{\Psi}_0$ . This leads to the conclusion that

$K\left\{\int \vec{\Psi}_2^\dagger \frac{\partial}{\partial t}\vec{\Psi}_2dV_2\right\}\vec{\Psi}_1=iKS_2\vec{\Psi}_1$

and that the function “f” may be written $f=f_0 -iKS_2$ where [imath]f_0 is entirely made up of a linear weighted sum of alpha and beta operators. So long as the above constraints are approximately valid, our differential equation for $\vec{\Psi}_0(\vec{x}_1,t)$ may be written in the following form.

$\vec{\alpha}_1\cdot \vec{\nabla}\vec{\Psi}_0 + \left\{\int \vec{\Psi}_r^\dagger\cdot \left[ \sum_{i=1}^n \vec{\alpha}_i \cdot \vec{\nabla}_i +f_0(\vec{x}_1,\vec{x}_2, \cdots,\vec{x}_n,t)\right] \vec{\Psi}_r dV_r\right\}\vec{\Psi}_0 = K\frac{\partial}{\partial t}\vec{\Psi}_0\ + iK\left(S_2+S_r\right)\vec{\Psi}_0.$

For the simple convenience of solving this differential equation, this result clearly suggests that one redefine $\vec{\Psi}_0$ via the definition $\vec{\Psi}_0 = e^{-iK(S_2+S_r)t}\vec{\Phi}$ . If one further defines the integral within the curly braces to be $g(\vec{x}_1)$ , $\vec{x}_1$ being the only variable not integrated over, the equation we need to solve can be written in an extremely concise form:

$\left\{\vec{\alpha}\cdot \vec{\nabla} + g(\vec{x})\right\}\vec{\Phi} = K\frac{\partial}{\partial t}\vec{\Phi},$

which implies the following operational identity:

$\vec{\alpha}\cdot \vec{\nabla} + g(\vec{x}) = K\frac{\partial}{\partial t}.$

That is, as long as these operators are operating on the appropriate $\vec{\Phi}$ they must yield identical results. If we now multiply the original equation by the respective sides of this identity, recognizing that the multiplication of the alpha and beta operators yields either one half (for all the direct terms) or zero (for all the cross terms) and defining the resultant of $g(\vec{x})g(\vec{x})$ to be $\frac{1}{2}G(\vec{x})$ (note that all alpha and beta operators have vanished), we can write the differential equation to be solved as

$\nabla^2\vec{\Phi}(\vec{x},t) + G(\vec{x})\vec{\Phi}(\vec{x},t)= 2K^2\frac{\partial^2}{\partial t^2}\vec{\Phi}(\vec{x},t).$

At this point we must turn to analysis of the impact of our $\tau$ axis, a pure creation of our own imagination and not a characteristic of the actual data defining the collection of referenced elements we need to explain. Since we are interested in the implied probability distribution of x, we must (in the final analysis) integrate over the probability distribution of tau. Since tau is a complete fabrication of our imagination, the final $P(x.\tau,t)$ certainly cannot depend upon tau. It follows directly from this observation that the dependence of $\vec{\Phi}$ on tau must (at worst) be of the form $e^{iq\tau}$ . It follows directly from this observation that the differential equation can be written.

$\left\{\frac{\partial^2}{\partial x^2} - q^2 + G(x)\right\}\vec{\Phi}(x,t)= 2K^2\frac{\partial^2}{\partial t^2}\vec{\Phi}(x,t).$

Notice that, if the term $q^2$ is moved to the right side of the equal sign, we may factor that side and obtain,

$\left\{\frac{\partial^2}{\partial x^2} + G(x)\right\}\vec{\Phi}(x,t)=\left\{\sqrt{2}K\frac{\partial}{\partial t}- iq\right\}\left\{\sqrt{2}K\frac{\partial}{\partial t}+iq\right\}\vec{\Phi}(x,t).$

At this point, I will invoke a third approximation. I will concern myself only with cases where $K\sqrt{2}\frac{\partial}{\partial t}\vec{\Phi} \approx -iq\vec{\Phi}$ to a high degree of accuracy. In this case, the first term on the right may be replaced by -2iq and, after devision by 2q, we have

$\left\{\frac{1}{2q}\frac{\partial^2}{\partial x^2}+\frac{1}{2q}G(x)\right\}\vec{\Phi}(x,t)= -i\left\{\sqrt{2}K \frac{\partial}{\partial t} + iq \right\}\vec{\Phi}(x,t).$

Once again, the form of the equation suggests we redefine $\vec{\Phi}$ via an exponential adjustment $\vec{\Phi}(x,t)=\vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}$ , thus simplifying the differential equation by removing the final iq term. To anyone familiar with modern physics, the equation should be beginning to look very familiar. In fact, if we multiply through by $-\hbar c$ (which clearly has utterly no impact on the solution as it multiplies every term) and make the following definitions directly related to constants already defined,

$m=\frac{q\hbar}{c}$ , $c=\frac{1}{K\sqrt{2}}$ and $V(x)= -\frac{\hbar c}{2q}G(x)$

it turns out that the equation of interest (without the introduction of a single free parameter: please note that no parameters not defined in the derivation of the equation have been introduced) is exactly one of the most fundamental equations of modern physics.

$\left\{-\left(\frac{\hbar^2}{2m}\right)\frac{\partial^2}{\partial x^2}+ V(x)\right\}\vec{\phi}(x,t)=i\hbar\frac{\partial}{\partial t}\vec{\phi}(x,t)$

This is, in fact, exactly Schroedinger's equation in one dimension.

This is a truly astounding conclusion. The fact that the probability of seeing a particular number in a stream of totally undefined numbers can be deduced to be found via Schroedinger's equation, no matter what the rule behind those numbers might be, is totally counter intuitive. It is extremely important that we check the meaning of the three constraints I placed on the problem in terms of the conclusion reached.

The first two are quite obvious. Recapping, they consisted of demanding that the data point under consideration had negligible impact on the rest of the universe and that the pattern representing the rest of the universe was approximately constant in time. These are both common approximations made when one goes to apply Schroedinger's equation: that is, we should not be surprised that these approximations made life convenient. What is important is that Schroedinger's equation is still applicable to physical situations where these constraints are considerably relaxed. In other words, the constraints are not required by Schroedinger's equation itself.

The serious question then is, what happens to my derivation when those constraints are relaxed. If one examines that derivation carefully, one will discover that the only result of these constraints was to remove the time dependent term from the linear weighted sum expressed by g(x). If this term is left in, G(x) will be complicated in three ways: first, the general representation must allow for time dependence; second, the representation must allow for terms proportional to $\frac{\partial}{\partial x}$ and, finally, the resultant V(x) will be a linear sum of the alpha and beta operators.

The time dependence creates no real problems: V(x) merely becomes V(x,t). The terms proportional to $\frac{\partial}{\partial x}$ correspond to velocity dependent terms in V and, finally, retention of the alpha and beta operators essentially forces our deductive result to be a set of equation, each with its own V(x,t). All of these results are entirely consistent with Schroedinger's equation, they simply require interactions not commonly seen on the introductory level. Inclusion of these complications would only have served to obscure the fact that what was deduced was, in fact, Schroedinger's equation.

That brings us down to the final constraint, $K\sqrt{2}\frac{\partial}{\partial t}\vec{\Phi}\approx -iq\vec{\Phi}$ . If we multiply this relationship through by $i\hbar$ and divide by $K\sqrt{2}$ the definitions given for m and c above imply the constraint can be written

$i\hbar\frac{\partial}{\partial t}\vec{\Phi}\approx q\hbar c \vec{\Phi}= \left( \frac{q\hbar}{c}\right) c^2\vec{\Phi} = mc^2\vec{\Phi}.$

The term $mc^2$ should be familiar to everyone and the left hand side, $i\hbar\frac{\partial}{\partial t}$ , should be recognized as the energy operator from the standard Schroedinger representation of quantum mechanics. Putting these two facts together, it is clear that the redefinition of $\vec{\Phi}$ to $\vec{\phi}$ in the above deduction was completely analogous to adjusting the zero energy point to non-relativistic energies. This step is certainly necessary as Schroedinger's equation is well known to be a non-relativistic approximation: i.e., Schroedinger's equation is known to be false if this approximation is not valid.

A very strange thing has happened: that the above approximation is necessary is not surprising; that it arose the way it did is rather astonishing as we have arrived at the expression $E=mc^2$ without even mentioning the concept of relativity. This certainly implies that at least some aspects of relativity seem to be embedded in the paradigm I am presenting. That will turn out to be exactly correct and will become overtly evident a few posts from here.

Meanwhile, the fact that the Schroedinger equation is an approximate solution to my equation leads me to put forth a few more definitions. Note to Buffy: there is no presumption of reality in these definitions; they are no more than definitions of abstract relationships embedded in the mathematical constraint of interest to us. That is, these definitions are entirely in terms of the mathematical representation and are thus defined for any collection of indices which constitute references to the elements the function $\vec{\Psi}$ was defined to explain.

First, I will define ”the Energy Operator” as $i\hbar\frac{\partial}{\partial t}$ (and thus, the conserved quantity required by the fact of shift symmetry in the t index becomes “energy”: i.e., energy is conserved by definition). A second definition totally consistent with what has already been presented is to define the expectation value of “energy” to be given by

$E=i\hbar\int\vec{\Psi}^\dagger\cdot\frac{\partial}{\partial t}\vec{\Psi}dV.$

I am putting this forward as a definition of the expectation value of energy for the sole reason that the concept is then applicable to the various functions I have proceeded through in deducing the Schroedinger equation above. What is important here is that the energy so defined is not conserved in the approximations used above (when the individual individual reference indices of ontological elements are examined) but rather that, when the entire collection of indices referring to these elements is represented by the appropriate function, total energy so defined will be conserved.

In addition, the comparison with Schroedinger's equation also suggests the definition of another mathematical operator which can, via exactly the same analogy, be called "the Momentum Operator" as $-i\hbar\frac{\partial}{\partial x}$ (and thus, the conserved quantity required by the fact of shift symmetry in the “x” index becomes “momentum”: i.e., the total momentum of the entire collection of referrences to our ontological elements will be conserved via the constraint $\sum\frac{\partial}{\partial x_i}\vec{\Psi}=0$ ). Once again, a second definition total consistent with what has already been presented is to define the expectation value of “momentum” to be given by

$P=-i\hbar\int\vec{\Psi}^\dagger\cdot\frac{\partial}{\partial x}\vec{\Psi}dV.$

Once again, this says nothing about the conservation of an individual indices “momentum”. The momentum of an individual index is a function of actual $\vec{\phi}$ describing the expectation of the element referranced by that index. Nevertheless, it does imply that the total momentum of all the referrence indices will be conserved.

Finally, I would like to introduce a third operator defended by exactly the same analysis provided above. This third operator is completely fictional as it arises from shift symmetry in the fictional axis tau. I will call this operator "the Mass Operator" and define it as $-i\frac{\hbar}{c}\frac{\partial}{\partial \tau}$ . Likewise, this leads to a second definition: the expectation value of “mass” to be given by

$m=-i\frac{\hbar}{c}\int\vec{\Psi}^\dagger\cdot\frac{\partial}{\partial \tau}\vec{\Psi}dV.$

Once again, I have managed to define a term (a mathematical operator) applicable to each and every referrence index to every element in the entire collection. The relationship between referrence indices implied here is a little more involved than energy and momentum. The fact that tau is a totally fictional axis requires not only shift symmetry (which yields conservation of mass when summed over the entire collection) but also yields conservation of mass on the referrence index level as nothing can actually be a function of tau in the final analysis. That is, not only do we have shift symmetry (which yields total mass as a conserved quantity) but we also have the fact that no details of the final result cannot possibly be a function of tau. This leads to the conclusion that the “mass” of individual referrences to valid ontological elements cannot be a function of tau.

I'll see what kinds of objections that presentation leads to before I will go on. As a comment to Buffy, this is still a completely abstract paradigm and there is utterly no implied relationship to reality. All I have done is show that there always exists a paradigm designed to yield expectations from a set of numbers which can see those numbers as elements approximately obeying Schroedinger's equation: i.e., time, position, mass, momentum and energy are all terms which can be defined for any collection of numerical indices to be analyzed. Once upon a time (back in the mid eighties) an economics professor asked me what what I was doing had to do with economics and I composed a paper for him showing exactly how all the above concepts could be mapped directly into economic theory. Not only that, but most all the economists already knew most of it; they already use terms like “energy” and “momentum” in their own discussions of trends and what kinds of changes one should expect. These are quite well defined universal concepts applicable to any numerical analysis.

Have fun -- Dick

LaurieAG · 07-14-2008

Quote:

Originally Posted by Doctordick

Since we are interested in the implied probability distribution of x, we must (in the final analysis) integrate over the probability distribution of tau. Since tau is a complete fabrication of our imagination, the final certainly cannot depend upon tau.

Hi Doctordick,

SchrÃ¶dinger equation - Wikipedia, the free encyclopedia

Quote:

So that not only is the probability of finding the particle the same everywhere, but the probability flux is as expected from an object moving at the the classical velocity p / m.

The reason that the Schrodinger equation admits a probability flux is because all the hopping is local and forward in time.

Doctordick · 07-15-2008

Quote:

Originally Posted by LaurieAG

Hi Doctordick,

SchrÃ¶dinger equation - Wikipedia, the free encyclopedia

I do not understand the purpose of your post. What are you trying to say?

Dick

LaurieAG · 07-15-2008

Quote:

Originally Posted by Doctordick

I do not understand the purpose of your post. What are you trying to say? Dick

Hi Doctordick,

If the probability distribution of the imaginary tau is constant, just like the Schrödinger probability (without the classic velocity flux), then the core of your proof derives from this relationship.

Doctordick · 07-17-2008

Quote:

Originally Posted by LaurieAG

... then the core of your proof derives from this relationship.

I would deny that. I don't think you understand the essence of my presentation.

Sorry -- Dick

Rade · 07-17-2008

Quote:

Originally Posted by Doctordick

I would deny that. I don't think you understand the essence of my presentation.Sorry -- Dick

Not good enough answer. Please explain why you "deny" LaurieAG presentation.

Doctordick · 07-27-2008

Quote:

Originally Posted by Rade

Not good enough answer. Please explain why you "deny" LaurieAG presentation.

Very simple, there is nothing in my derivation which says anything about “a probability flux” or any “hopping”. The standard defense of Schroedinger's equation bears utterly no resemblance to my analytical derivation. LaurieAG's comment is simply not true.

To Anssi on his difficulties. Let me take another tack on this issue as it seems I am failing to communicate some rather simple issues. We started with my deduced fundamental equation:

$\left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) \right\}\vec{\Psi} = K\frac{\partial}{\partial t}\vec{\Psi}.$

I then proposed that I could divide the underlying data (the numerical references to the unknown ontological elements) into two sets (set #1 and set #2). I could then assert that the probability of any given distribution, P (set #1 and set #2) would be equal to the probability of set #1 times the probability of set #2 (under the constraint that set #1 was given):

$P(set \; \# 1 \; and \; set \; \# 2) = P(set \; \# 1)P(set \; \# 2, \; given \; set \; \# 1)$

Purely from the definition of probability, I know that these two probabilities can be individually represented by a scalar product of some vector function of the specific arguments. It follows directly from this fact that I can represent $\vec{\Psi}$ with the expression:

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

where the arguments $(x,\tau)_i$ are represented by the expression $\vec{x}_i$ ), If we make this substitution in the original equation above we have

$\left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) \right\}\vec{\Psi}_1 \vec{\Psi}_2 = K\frac{\partial}{\partial t}\vec{\Psi}_1 \vec{\Psi}_2.$

Because we want to find the equation obeyed by the numerical references defined as set #1 for the case where set #2 is allowed to be any and all possibilities, we need this equation summed over all possibilities for set #2. We accomplish that by left multiplying by $\vec{\Psi}_2^\dagger\cdot$ and integrating over all arguments from set #2. We multiply by $\vec{\Psi}_2^\dagger\cdot$ because we know the result of $\vec{\Psi}_2^\dagger\cdot\vec{\Psi}_2$ ( $\vec{\Psi}_2$ is defined by the fact that this dot product is the probability of having set #2) and we integrate over all the arguments from set #2 because that act will remove those arguments from the equation (the sum constitutes the sum of the probability of any specific set #2 over all possibilities).

I need to make a few minor comments about those acts. First, we left multiply for a very simple reason. If we were dealing with simple numbers only, direction would be of no consequence; however, we are, in this case, dealing with abstract mathematical operators. Now most all mathematical operators are defined by what they do to operators to their right, not by what they do to operators to their left (the definitions simply presume they don't operate to the left).

The central issue of algebra is the fact that, if we have an equation and we do exactly the same thing to both sides of the equation, the equation is still valid. This fact does not require that the equation contain only numbers (as high school algebra is presented), it can contain any kind of properly defined operators. The issue being that the equation is still valid after we left multiply by $\vec{\Psi}_2^\dagger\cdot$ . This would not be true if we “right multiplied” by this function as placing the function on the right would mean that the operators defined in the original equation would operate on that function and, as the operators on the opposite sides of the equal sign might not be the same, the result (which is what the equal sign is referring to) might not be the same and we would thus not be doing the same thing to both sides of the equation (as an algebraic act, the step would be invalid).

The next step is to remove the arguments of set #2 from the equation. To accomplish this, we integrate both sides of the equation over the arguments of set #2. The resulting equation is explicitly

$\int\vec{\Psi}_2^\dagger \cdot \left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) \right\}\vec{\Psi}_1 \vec{\Psi}_2 dV_2=\int \vec{\Psi}_2^\dagger \cdot K\frac{\partial}{\partial t}\vec{\Psi}_1 \vec{\Psi}_2.$

I am going to presume that, up to this point, everything is perfectly clear to you. The next step is to simplify that expression into a form which is inherently more meaningful to us; mostly by factoring out terms which need not be under the integral sign so that we can understand what has to be done to accomplish the relevant integrals. The first thing is to recognize that the integral of a sum is exactly equal to the sum of the integrals of the terms of the sum. That suggests that it would be very valuable to expand the expression from the sums over i and j as represented to sums over i and j taken from the two different defined sets (set #1 and set #2). Notice that the first term of the above equation (the term containing the differential operator $\vec{\nabla}$ ) can be written as a sum of two terms:

$\int\vec{\Psi}_2^\dagger \cdot \left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i \right\}\vec{\Psi}_1 \vec{\Psi}_2 dV_2=\int\vec{\Psi}_2^\dagger \cdot \left\{\sum_{i = \# 1} \vec{\alpha}_i \cdot \vec{\nabla}_i \right\}\vec{\Psi}_1 \vec{\Psi}_2 dV_2 +\int\vec{\Psi}_2^\dagger \cdot \left\{\sum_{i=\#2}\vec{\alpha}_i \cdot \vec{\nabla}_i \right\}\vec{\Psi}_1 \vec{\Psi}_2 dV_2.$

The first