Deriving Schrödinger's Equation from my Fundamental Equation

Rade · 05-08-2009

Quote:

Originally Posted by AnssiH

Rade, I posted a reply to:

http://hypography.com/forums/philoso...tml#post264413

AnssiH....THANK YOU....I will provide further comments-questions at this new link and not post on this thread anymore....and to your question....
..."I can only hope this post is helpful to you, and I hope you appreciate the effort"...
.....the answer is a very appreciated yes.

AnssiH · 05-10-2009

Quote:

Originally Posted by Doctordick

The term:

$\left\{ 2K^2\frac{\partial^2}{\partial t^2} + q^2 \right\}$

(which is correct) does not posses that important negative sign and thus (as written) does not factor. That is why it was changed to

$\left\{ 2K^2\frac{\partial^2}{\partial t^2} -(i q)^2 \right\}$

Ah, right.

My non-sophisticated mathematical mind thinks it would have been little bit clearer if the $-(i q)^2$ had not been factored to $q^2$ in the previous step, only to be turned back to $-(i q)^2$ in the next turn... :I But then, I guess it makes no difference to anyone who actually knows their math.

Quote:

Originally Posted by Doctordick

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.

I don't understand what that approximation means conceptually, maybe you can elaborate on that. I.e., what is ultimately the justification for the approximation.

Since I don't understand what it means, I'll just take it on faith to get onwards:

Quote:

In this case, the first term on the right may be replaced by -2iq and, after devision by 2q, we have

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

Okay, that looks all valid to me.

And I guess here you have just chosen to not substitute the $\sqrt{2}K \frac{\partial}{\partial t}$ with $-iq$ in the second term, even though (or because?) it would just remove the whole term... ?

Quote:

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.

Well, I wrote down:

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

And I've been staring at that for little while now but I just don't know which way to move with it... :I

Help!

-Anssi

Doctordick · 05-10-2009

Quote:

Originally Posted by AnssiH

My non-sophisticated mathematical mind thinks it would have been little bit clearer if the $-(i q)^2$ had not been factored to $q^2$ in the previous step, only to be turned back to $-(i q)^2$ in the next turn... :I But then, I guess it makes no difference to anyone who actually knows their math.

You are, of course, correct. Mathematicians often simplify things all the time without considering the possible confusion it might create. It never even dawned on me that I had done that. I should not have as it served no purpose.

Quote:

Originally Posted by AnssiH

I don't understand what that approximation means conceptually, maybe you can elaborate on that. I.e., what is ultimately the justification for the approximation.

The justification is a couple of lines later in that same post. I was shown that factorization by the same professor who taught me introductory quantum mechanics (the same one to whom I explained my childish solution to relativity way back in 1964). He showed it to me in terms of the Klein–Gordon equation.

$\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\Psi -\nabla^2\Psi +\frac{m^2c^2}{\hbar^2}\Psi =0$

Which is normally referred to as the relativistic expression of Schrödinger's equation. By convention, all effects of the interactions is usually presumed to be embedded in “m”. If that is not “presumed”, and, instead, the last term is written

$\frac{m_0^2c^2}{\hbar^2}\Psi +V\Psi$

where “ $m_0$ ” is taken to be “the rest mass”, the term

$\frac{1}{c^2}\frac{\partial^2}{\partial t^2} +\frac{m_0^2c^2}{\hbar^2}$

can be factored in exactly the way I have just factored it, in which case, the Kline-Gordon equation (for non relativistic situations, becomes Schrödinger's equation with only a minor alteration in the definition of V: a shift by a constant). Years ago, (when I first used that factorization to deduce Schrödinger's equation from my equation) I went to tell him about it. I was told that he was suffering from Alzheimer's and not available to talk to anyone. Isn't that the way life is?

Quote:

Originally Posted by AnssiH

And I guess here you have just chosen to not substitute the $\sqrt{2}K \frac{\partial}{\partial t}$ with $-iq$ in the second term, even though (or because?) it would just remove the whole term... ?

The second term reduces to almost zero (it can't actually be zero because this is an approximation). The first term is roughly 2q which is not anywhere near zero so it can be considered very similar to 2q (only off by a small percentage). So, yes; it would unjustifiably remove the whole term and essentially remove the time derivative (which will be defined to be energy) from the equation.

Quote:

Originally Posted by AnssiH

Well, I wrote down:

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

And I've been staring at that for little while now but I just don't know which way to move with it... :I

Just work with the right hand side. First, multiply it out and get the following

$-i\left\{\sqrt{2}K \frac{\partial}{\partial t} + iq \right\} \vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}= -i\sqrt{2}K\frac{\partial}{\partial t} \vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}+q \vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}$

Now, for the moment, look at that first term,

$-i\sqrt{2}K\frac{\partial}{\partial t}\vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}$ .

That is a product of two different functions of t. The product rule of differentiation says that such a differential yields two terms which are: the differential of the first term times the second term plus the first term times the differential of the second term. Explicitly, that would be

$-i\sqrt{2}K\frac{\partial}{\partial t} \vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}= -i\sqrt{2}K\left\{\frac{\partial}{\partial t} \vec{\phi}(x,t)\right\}e^{\frac{-iqt}{K\sqrt{2}}}-i\sqrt{2}K\vec{\phi}(x,t)\frac{\partial}{\partial t}e^{\frac{-iqt}{K\sqrt{2}}}=$

$-i\sqrt{2}K\left\{\frac{\partial}{\partial t}\vec{\phi}(x,t)\right\}e^{\frac{-iqt}{K\sqrt{2}}} -i \sqrt{2}K\vec{\phi}(x,t)\frac{-iq }{K\sqrt{2}}e^{\frac{-iqt}{K\sqrt{2}}}=-i\sqrt{2}K\left\{\frac{\partial}{\partial t}\vec{\phi}(x,t)\right\}e^{\frac{-iqt}{K\sqrt{2}}} -q\vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}$

Now, go back and substitute that for the first term (when we multiplied the right hand side out). The right hand side will now be

$-i\sqrt{2}K\left\{\frac{\partial}{\partial t}\vec{\phi}(x,t)\right\}e^{\frac{-iqt}{K\sqrt{2}}} -q\vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}+q \vec{\phi}(x,t)e^{\frac{-iqt}{K\sqrt{2}}}$

And the last two terms exactly cancel out. Going back to the equation we started with (as it now appears, without those two last terms), we have

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

where the exponential function explicitly appears in every term but is not to be differentiated anywhere and it may be factored out (we can divide it out or we can multiply the whole equation from the right by $e^{\frac{-iqt}{K\sqrt{2}}}$ , it all results in exactly the same effect). We are left with the equation

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

That mathematical maneuver is well known to anyone who does classical quantum mechanics. The right hand side of that equation is a conserved quantity (which will soon be identified with energy). The mathematical maneuver I just introduced you to simply changes that conserved quantity by a constant. In classical physics, the zero reference for energy can always be changed by a constant and has no physical consequences at all. In modern physics people tend to believe this characteristic of physics is no longer true and they come up with a thing they call “zero point energy” which I suspect is actually a figment of their imagination; of course, I tend to believe the fact that reality obeys physics is a figment of their imagination.

Only Anssi will understand the meaning of that comment so don't everybody else jump off the deep end!

Have fun -- Dick

AnssiH · 05-11-2009

Ah, okay, I was able to walk through all of that succesfully, after some head scratching and using google calculater to figure out what -i*-i...

I didn't stop and think about that commentary about the Klein-Gordon equation too deeplythough and don't understand it completely. I suppose it was just additional commentary here anyway.

...to be continued soon...

-Anssi

AnssiH · 05-16-2009

Quote:

Originally Posted by Doctordick

We are left with the equation

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

Quote:

Originally Posted by Doctordick

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)

My attempt to write that step down is:

$\left\{\frac{-\hbar c}{2q} \frac{\partial^2}{\partial x^2} + \frac{-\hbar c}{2q} G(x)\right\} \vec{\phi}(x,t) = -\hbar c (-i)\sqrt{2}K \frac{\partial}{\partial t}\vec{\phi}(x,t)$

I can only hope it is correct... :I

Assuming it is, I tried to follow the next step:

Quote:

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.

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

Well, I figure the first definition $m=\frac{q\hbar}{c}$ (mass?) comes to play in the first term, and after a lot of head scratching, I was finally able to figure out the steps, when I did it in reverse. At least I hope everything I do here is valid (just had to trust my logic but I know it's easy for me to make errors):

So I started with your end result, and substituted the $m$ with $\frac{q\hbar}{c}$

$-\left(\frac{\hbar^2}{2m}\right) = -\left( \frac{\hbar^2} {2 \left( \frac{q\hbar}{c} \right)} \right)$

I suppose that can be written as:

$-\left(\frac{\hbar^2}{2} \cdot \frac{c}{q\hbar} \right)$

Which seems to factor to what I had:
$\frac{-\hbar c}{2q}$

So if that's all valid, I think understand the left hand side of...

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

...since the second term, $V(x)$ , seems to be just a straightforward substitution (except I don't know what it represents, the wiki page for Schrödinger equation talks about "time independent potential energy of particle", but I don't understand what that means either)

Finally, I'm focusing on the right hand side, substituting $c$ with $\frac{1}{K\sqrt{2}}$ :

$-\hbar c \left(-i\sqrt{2}K\right) \frac{\partial}{\partial t}\vec{\phi}(x,t) = -\hbar \frac{1}{K\sqrt{2}} (-i)\sqrt{2}K \frac{\partial}{\partial t}\vec{\phi}(x,t)$

I suppose $\sqrt{2}K$ factors right out, leaving me with:

$-\hbar (-i) \frac{\partial}{\partial t}\vec{\phi}(x,t)$

Since your end result is:

$i\hbar\frac{\partial}{\partial t}\vec{\phi}(x,t)$

I suppose the negatives just factor out normally, even with the imaginary number involved. If that's a correct assumption, I think I understand this part too...

All looks valid?

-Anssi

Doctordick · 05-16-2009

Hi Anssi, I am sorry it took me so long to respond to your post. I left it to last because I wanted to give you serious attention: i.e., I wanted to get the trivial things (the other posts) out of the way first but they just kept posting and I couldn't get finished. I finally gave up on that agenda and went to your post.

Quote:

Originally Posted by AnssiH

...since the second term, $V(x)$ , seems to be just a straightforward substitution (except I don't know what it represents, the wiki page for Schrödinger equation talks about "time independent potential energy of particle", but I don't understand what that means either)

Sorry I made the thing so difficult for you. Instead of writing the first term in the form m=, I should divided both sides by qm and obtained the expression 1/q = and then it also would have been a simple substitution.

$\frac{1}{q}= \frac{\hbar}{mc}$

But you figured it out anyway. You did fine.

Quote:

Originally Posted by AnssiH

All looks valid?

Yes, everything you did was valid (a little round about but absolutely valid).

Regarding the second term, as per wiki, V(x) is the time independent potential energy of the particle (per Schrödinger). In my equation it is the the change in the energy of the universe due the change in position of the element of interest. Remember, we removed that component of the energy of the rest of the universe which didn't change (that S_r term) a long time ago. So this is an algebraic function which tells you how the potential energy changes with respect to position of the element of interest. Mine is a consequence of the direct calculation of energy of the whole universe while Schrödinger's is a phenomenological function determined by experiment. They both refer to exactly the same phenomena.

Thus my equation is exactly Schrödinger's equation without introducing any constants of any kind. You should be aware of the fact that K establishes a fixed velocity which I call v_? in that thread An “analytical-metaphysical” take on Special Relativity! You need to understand that thread in order to understand where c comes from (that is, the fact that it is not really a free parameter).

Before we go on to Dirac's equation, we probably ought to go through the Special Relativity thread first. Other than that, I think we are ready to go on.

Have fun -- Dick

AnssiH · 05-17-2009

Quote:

Originally Posted by Doctordick

Yes, everything you did was valid (a little round about but absolutely valid).

Heh, so I suspected.... Great.

Quote:

Regarding the second term, as per wiki, V(x) is the time independent potential energy of the particle (per Schrödinger). In my equation it is the the change in the energy of the universe due the change in position of the element of interest. Remember, we removed that component of the energy of the rest of the universe which didn't change (that S_r term) a long time ago. So this is an algebraic function which tells you how the potential energy changes with respect to position of the element of interest. Mine is a consequence of the direct calculation of energy of the whole universe while Schrödinger's is a phenomenological function determined by experiment. They both refer to exactly the same phenomena.

Well, my understanding of Schrödinger's Equation is quite shallow, (I don't know how it is conventionally looked at) and I would like to understand it a bit better... Don't really know what potential energy means in quantum mechanical context :I

Quote:

Thus my equation is exactly Schrödinger's equation without introducing any constants of any kind. You should be aware of the fact that K establishes a fixed velocity which I call v_? in that thread An “analytical-metaphysical” take on Special Relativity! You need to understand that thread in order to understand where c comes from (that is, the fact that it is not really a free parameter).

Before we go on to Dirac's equation, we probably ought to go through the Special Relativity thread first. Other than that, I think we are ready to go on.

Well, I read the rest of the OP, and I can understand what you are saying there little bit. Not everything, but I understand that you are drawing out the relationship $E = mc^2$ - at this point seems completely plausible to me that that relationship was embedded to the original constraints. And you defined Energy, Momentum and Mass as conserved quantities that, I imagine, correspond very much to how they are defined in modern physics.

If that's enough then yes, let's concentrate on the special relativity for a while...

-Anssi

Doctordick · 05-17-2009

Quote:

Originally Posted by AnssiH

Well, my understanding of Schrödinger's Equation is quite shallow, (I don't know how it is conventionally looked at) and I would like to understand it a bit better... Don't really know what potential energy means in quantum mechanical context :I

Fundamentally, it is little more than a statement of energy conservation: kinetic energy (the energy of motion) plus potential energy (the energy of position) equals total energy. The kinetic energy is related to the momentum, a spacial differential and the total energy is related to the time derivative. These differential relationships actually arise through mathematical analysis of procedures designed to solve Newton's equations. They also yield our understanding of electromagnetic phenomena as expressed by Maxwell's equations: they connect the momentum and energy of a photon to Newton's idea of momentum and energy. I wouldn't really worry about it if I were you; just take it as the definition of energy and momentum (and, in my case, mass). It turns out that it fits exactly the phenomena Newton defined to be momentum, energy and mass (via connections to experiments: i.e., expectations given by $\psi^\dagger(\vec{x},t)O\psi(\vec{x},t) dx$ where O stands for an operator). Except for a little uncertainty.

Quote:

Originally Posted by AnssiH

If that's enough then yes, let's concentrate on the special relativity for a while...

Wonderful; I will consider this thread to be essentially closed.

See you on the other thread -- Dick