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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 of these two terms can be again divided into two terms because the differential (which arises from $\vec{\nabla}_i$ ) must operate on two functions ( $\vec{\Psi}_2$ is a function of set #1) and we need the differential of a product (essentially $\frac{d}{dx}A(x)B(x)=B(x)\frac{d}{dx}A(x)+A(x)\frac{d}{dx}B(x)$ ). Writing out the first term from above explicitly, we have

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

Notice that I have commuted $\vec{\Psi}_2$ to the left. I have done this because I want to avoid the ambiguity in representation which would arise if I left the order the same. Now, let us look at the first term of the latest expression. The sum is being taken over set #1 and $\vec{\Psi}_1$ is not a function of set #2. It follows that these terms have exactly the same value for all possible selections from set #2. This means that those terms may be factored out of the integral. Thus, the first term of this latest expression (the latest “first term”) becomes:

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

The integral indicated there is quite obviously equal to unity: i.e., it is exactly the total probability of all possibilities for set #2. So, let us lay aside the first term of our finished expression as:

$\left\{\sum_{i=\#1} \vec{\alpha}_i \cdot \vec{\nabla}_i \right\}\vec{\Psi}_1.$

Mark this as term #1.

We now need to work backwards and examine the remaining terms in detail. In order to be sure we have not omitted a term, let us next look at the second term of that “latest expression”.

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

The only thing we can do with this term is to point out that $\vec{\Psi}_1$ is not a function of set #2 (thus it can be factored from the integral) and point out that the integral of a sum is identical to the sum of the integrals. The resulting expression is:

$\vec{\Psi}_1\int\vec{\Psi}_2^\dagger \cdot \left\{\sum_{i=\#1}\vec{\alpha}_i \cdot \vec{\nabla}_i \right\} \vec{\Psi}_2dV_2.$

We will lay this term aside to be picked up later. You can mark this as Term A.

Now, let us go back to the next previous term. That would be the remaining term with a differential operator (the second term of the original pair containing the $\vec{\nabla}$ operator):

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

Again, the only thing we can do here is to point out is that $\vec{\Psi}_1$ is not a function of set #2. Thus the $\vec{\nabla}$ operator (which the sum has constrained to set #2) does not yield any differentials of $\vec{\Psi}_1$ (the product rule does not apply because there is no differentiable product). Once again $\vec{\Psi}_1$ can be factored from the integral and the sum. The result can then be written:

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

Since the only difference between this term and term A expressed earlier is the range of the sum and the fact that $\vec{\Psi}_1$ function is on opposite sides of the expression, if one uses parenthesis to indicate that whatever is inside the parenthesis is evaluated first (thus, even when the sum over the $\vec{\nabla}$ operator includes set #1, it still does not operate on $\vec{\Psi}_1$ as that function is outside the parenthesis), we can add these two terms together and obtain the expression:

$\left\{\int\vec{\Psi}_2^\dagger \cdot \sum_i\vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_2 dV_2\right\}\vec{\Psi}_1.$

Mark this as term #2. As I have said several times, this representation is slightly ambiguous but the intention can be taken from context as clear because of the integral (what is contained within the parenthesis takes precedence over any other operations). The differential operator does not operate on $\vec{\Psi}_1$ because what is in the parenthesis evaluates to an ordinary algebraic expression and does not end up being a differential operator.

That brings us down to the portion of our original equation which involves the Dirac delta functions. One thing is nice and that is the fact that commutation brings on no difficulties here. The beta operators only appear once and they commute with all the other functions which are ordinary mathematical functions. The only issue of importance is the range of the sums on i and j. We have only three possibilities: i and j are chosen from set #1, i and j are chosen from set #2 and i and j are chosen from opposite sets. Let us examine the opposite sets case first. The expression to be evaluated is,

$\int\vec{\Psi}_2^\dagger \cdot \left\{ \sum_{i=\# 1 j=\#2}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) +\sum_{i=\# 2 j=\#1}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) \right\}\vec{\Psi}_1 \vec{\Psi}_2 dV_2$

Since the Dirac delta function is defined to yield a positive result only when its argument vanishes, the two sums shown above are exactly equal (i and j are merely inverted) and the result is exactly twice the result of evaluating either sum by itself. Put this together with the fact that $\vec{\Psi}_1$ can be factored from the integral and it should be clear that this term can be written:

$\left\{2 \sum_{i=\#1 j=\#2}\int \vec{\Psi}_2^\dagger \cdot \beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1$

which we can Mark as term #3.

The case where i and j are both taken from set #1 evaluates to a rather simple expression as everything, including the Dirac delta function, can be factored from the integral (the only expressions which are a function of set #2 are $\vec{\Psi}_2^\dagger$ and $\vec{\Psi}_2$ ) the result can be written,

$\left\{ \sum_{i \neq j (\#1)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\int \vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1.$

Since the integral obviously evaluates to unity (it is again the sum over the probability of all possibilities for set #2) that expression may be removed and one has,

$\left\{ \sum_{i \neq j (\#1)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j) \right\}\vec{\Psi}_1.$

Add this expression to term #1 and one has:

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

The final expression to evaluate is the case where i and j are both taken from set #2. In that case, all that can be said is that $\vec{\Psi}_1$ may be factored out of both the integral and the sum (it is entirely dependent upon set #1 and does not change for any such set of i and j or any change in the arguments of set #2). Thus we obtain:

$\left\{\int \vec{\Psi}_2^\dagger \cdot \left[ \sum_{i \neq j (\#2)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\right]\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1.$

This term can be added to term #2. The result can be written as

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

This completes the evaluation of the left hand side of our algebraically altered fundamental equation:

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

All which remains is to evaluate the right hand side of the equal sign which was explicitly.

$\int \vec{\Psi}_2^\dagger \cdot K\frac{\partial}{\partial t}\vec{\Psi}_1 \vec{\Psi}_2 dV_2.$

The only problem here is that we must use the product rule to evaluate the differential with respect to t.

$\frac{\partial}{\partial t}\vec{\Psi}_1 \vec{\Psi}_2 = \vec{\Psi}_2 \frac{\partial}{\partial t}\vec{\Psi}_1 +\vec{\Psi}_1\frac{\partial}{\partial t}\vec{\Psi}_2$

Thus, noting again that $\vec{\Psi}_1$ (and its differential with respect to t) can be factored from the integral, the right hand side of the equal sign becomes,

$\left\{\int \vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2 dV_2\right\} K \frac{\partial}{\partial t}\vec{\Psi}_1 +K\left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1$

where again I have used the parenthesis to indicate that everything inside the parenthesis is to be evaluated first: i.e., the second term does not imply any differentiation of $\vec{\Psi}_1$ . Of course the integral in the left hand term once again evaluates to unity for the same reasons given earlier. The final result for the right side of the equal sign is,

$K \frac{\partial}{\partial t}\vec{\Psi}_1 +K\left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1$

The final result is exactly what is shown in the first post of this thread.

Quote:

Originally Posted by Doctordick

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

I hope you don't find that too tricky to figure out.

Quote:

Originally Posted by AnssiH

Glad we are not in particular hurry

So am I. I also hope you appreciate the work it took to write and debug all that LaTex code I just composed.

Have fun -- Dick

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