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Old 07-31-2008   #8 (permalink)
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AnssiH
Understanding


 



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Re: Deriving Schrödinger's Equation from my Fundamental Equation

Doh, sorry it took me a while to reply. I was not subscribed to this thread and instead kept checking your post history to see when you post your reply... I expected to see it posted after your note to the old thread... ...and didn't notice at all you'd already posted it before

Quote:
Originally Posted by Doctordick View Post
I also hope you appreciate the work it took to write and debug all that LaTex code I just composed.
Certainly. I find it incredibly time consuming to type in any LaTex code since there's no good way to preview it. It would probably be possible to implement the post editing so that it would display the resulting LaTex render somewhere in real time as you are typing the code... ...but I guess that would make life too easy.

Anyhow, it was all very helpful, and I just spent the time to walk through it carefully with little baby steps, and was able to follow just about all of it. I would still call my understanding of all that algebra very superficial as I would not be able to perform those steps myself, although they seem to make perfect sense following your representation.

Just couple of questions;

Quote:

\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.
I am a bit uncertain what you mean by "what is in the parenthesis evaluates to an ordinary algebraic expression and does not end up being a differential operator". Do you just mean that if one was to expand that sum into its explicit components, the differential operator would not be there to create any ambiguity? (Just that when doing that, one would have to explicitly know, from the context, that the differential operator is not meant to operate on \vec{\Psi}_1, otherwise the expanded expression would turn out wrong?)

Also I am a bit shaky with the term dV_2 appearing in places. I remember asking about its meaning before, and it had something to do with having to consider how large an area of possibilities we are considering... Still I am somewhat puzzled about what is it doing there, where ever I see it appearing...

Anyway, thank you for the detailed explanation, I can continue from here soon...

-Anssi
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