- #1
Doctordick
- 634
- 0
Dear Sirs,
I am posting here for one reason and one reason only. There are apparently a great number of powerful people on the main forum who would like me to go away. If they were the only people on the forum, I would certainly do so; however, I think leaving would do a great disservice to the readers on the forum. Physics is a very important subject with far reaching implications and, as such I think careful thought about its foundations is worthwhile to any student (or professor) of the field.
I originally posted to the quantum mechanics thread because I thought the probability of reaching cognizant minds with an interest in the foundations of their field was higher there than anywhere else. My original thread was immediately moved to "Theory Development" which I accepted as a prerogative of the powers; however, I have now been locked out of posting to that sub forum.
I have tried to present a very simple point to which absolutely no one on the physics forum has yet responded. The point can be broken into two issues. In the interest keeping things simple, would someone competent please respond to my first issue which is the extent of the applicability of the expression
[tex]
P(\vec{x},t) = \vec{\Psi}^{\dagger}(\vec{x},t)\cdot\vec{\Psi}(\vec{x},t)dv
[/tex]
It is clearly an expression of far reaching consequences in quantum mechanics; however, it is my position that the expression is of far deeper significance than is ordinarily attributed to it. I hold that there exists no algorithm which will yield (as a result of that algorithm) a real number between zero and one which cannot be represented by that expression.
The proof of that statement is quite straight forward.
1) Anything which can be referenced can be represented by a set of numbers.
2) An algorithm is defined to be a procedure which transforms something into something else: i.e. from the above, this can be represented by one set of numbers being transformed into a second set of numbers.
3) Both [itex](\vec{x},t)[/itex] and [itex]\vec\Psi[/itex] can be used to represent an arbitrary set of numbers.
4) Given [itex]\Psi[/itex], it is always possible to define [itex]\vec{\Psi}^{\dagger}[/itex] such that the inner product, [itex]
\vec{\Psi}^{\dagger}(\vec{x},t)\cdot\vec{\Psi}(\vec{x},t)
[/itex], is a real number greater than or equal to zero.
5.) Probability is defined to be a real number between zero and one.
It follows that there exists no statement of probability of an occurrence which cannot be written in the form given above.
Either what I have just said is true or false: i.e., the proof is valid or it is not.
As an aside (issue #2), it follows that the correct answer to any question concievable resolves down to finding the algorithm [itex]\vec{\Psi}[/itex] which will yield the probability of each and every possible answer (represented by the expression for the argument of [itex]\vec{\Psi}[/itex])
Given the possible importance of that fundamental representation, I would appreciate it if anyone who sees an error in my proof, would please point it out to me. If you believe my proof holds water, I refer you to the locked thread:
https://www.physicsforums.com/showthread.php?t=39508
In particular, that thread was locked before I could comment on some of the posts made there which I would like very much to answer.
Russell, referring to
https://www.physicsforums.com/showthread.php?p=291013#post291013
You seem to missing a very important issue when you bring up the validity of my representation of probability. That statement is a mathematical statement, not a physics statement. If you can follow my proof, the central issue of that particular statement (and the reason I presented it) is that there exists no algorithm for determining any expectation of anything which cannot be put in that form.
The simple answer is, of course, yes it does! We have a great number of specific problems whose answer is expressed exactly in that form. Since Tom sees the issue in terms of the problems he has learned to solve, his "intuitive" position on the validity of the expression is: "I have to know the problem before I can answer the question of its validity!". When he does that, he misses the entire point of my presentation.
There's a line in "Harry Potter and the Order of the Phoenix" which just seems to fit this situation exactly. Hagrid, speaking of the giants, says, "overload them with information an' they'll kill yeh jus' to simplify things".
I think Tom finds my presentation overloads his ability to think things out. I conclude that because he supports BaffledMatt's statement, "Why else does he hide the logic of his arguments by making his written theory so incomprehensible?" with the comment, "He did make a point, and he is right".
Check it all out and see if you can comprehend what I am getting at.
By the way, I have made no presentation of a theory in any way.
Have fun -- Dick
I am posting here for one reason and one reason only. There are apparently a great number of powerful people on the main forum who would like me to go away. If they were the only people on the forum, I would certainly do so; however, I think leaving would do a great disservice to the readers on the forum. Physics is a very important subject with far reaching implications and, as such I think careful thought about its foundations is worthwhile to any student (or professor) of the field.
I originally posted to the quantum mechanics thread because I thought the probability of reaching cognizant minds with an interest in the foundations of their field was higher there than anywhere else. My original thread was immediately moved to "Theory Development" which I accepted as a prerogative of the powers; however, I have now been locked out of posting to that sub forum.
I have tried to present a very simple point to which absolutely no one on the physics forum has yet responded. The point can be broken into two issues. In the interest keeping things simple, would someone competent please respond to my first issue which is the extent of the applicability of the expression
[tex]
P(\vec{x},t) = \vec{\Psi}^{\dagger}(\vec{x},t)\cdot\vec{\Psi}(\vec{x},t)dv
[/tex]
It is clearly an expression of far reaching consequences in quantum mechanics; however, it is my position that the expression is of far deeper significance than is ordinarily attributed to it. I hold that there exists no algorithm which will yield (as a result of that algorithm) a real number between zero and one which cannot be represented by that expression.
The proof of that statement is quite straight forward.
1) Anything which can be referenced can be represented by a set of numbers.
2) An algorithm is defined to be a procedure which transforms something into something else: i.e. from the above, this can be represented by one set of numbers being transformed into a second set of numbers.
3) Both [itex](\vec{x},t)[/itex] and [itex]\vec\Psi[/itex] can be used to represent an arbitrary set of numbers.
4) Given [itex]\Psi[/itex], it is always possible to define [itex]\vec{\Psi}^{\dagger}[/itex] such that the inner product, [itex]
\vec{\Psi}^{\dagger}(\vec{x},t)\cdot\vec{\Psi}(\vec{x},t)
[/itex], is a real number greater than or equal to zero.
5.) Probability is defined to be a real number between zero and one.
It follows that there exists no statement of probability of an occurrence which cannot be written in the form given above.
Either what I have just said is true or false: i.e., the proof is valid or it is not.
As an aside (issue #2), it follows that the correct answer to any question concievable resolves down to finding the algorithm [itex]\vec{\Psi}[/itex] which will yield the probability of each and every possible answer (represented by the expression for the argument of [itex]\vec{\Psi}[/itex])
Given the possible importance of that fundamental representation, I would appreciate it if anyone who sees an error in my proof, would please point it out to me. If you believe my proof holds water, I refer you to the locked thread:
https://www.physicsforums.com/showthread.php?t=39508
In particular, that thread was locked before I could comment on some of the posts made there which I would like very much to answer.
Russell, referring to
https://www.physicsforums.com/showthread.php?p=291013#post291013
You seem to missing a very important issue when you bring up the validity of my representation of probability. That statement is a mathematical statement, not a physics statement. If you can follow my proof, the central issue of that particular statement (and the reason I presented it) is that there exists no algorithm for determining any expectation of anything which cannot be put in that form.
What Tom is missing is the fact that I am not discussing any particular problem here. His statement goes to the issue of the value of the expression with regard to the solution of a particular physics problem. He is absolutely correct: the statement contains utterly no physical information at all. The entire issue of interest to physics is: does it apply to a specific problem of interest?Tom Mattson said:The point is that this statement contains absolutely no physical information. Without knowing the dynamical field equation and boundary conditions that determine [itex]\vec{\Psi}(\vec{x},t)[/itex], there is no way to argue either for or against the equation.
The simple answer is, of course, yes it does! We have a great number of specific problems whose answer is expressed exactly in that form. Since Tom sees the issue in terms of the problems he has learned to solve, his "intuitive" position on the validity of the expression is: "I have to know the problem before I can answer the question of its validity!". When he does that, he misses the entire point of my presentation.
There's a line in "Harry Potter and the Order of the Phoenix" which just seems to fit this situation exactly. Hagrid, speaking of the giants, says, "overload them with information an' they'll kill yeh jus' to simplify things".
I think Tom finds my presentation overloads his ability to think things out. I conclude that because he supports BaffledMatt's statement, "Why else does he hide the logic of his arguments by making his written theory so incomprehensible?" with the comment, "He did make a point, and he is right".
Check it all out and see if you can comprehend what I am getting at.
By the way, I have made no presentation of a theory in any way.
Have fun -- Dick