# Thread: Has anyone ever quantized newtons laws?

1. ## Has anyone ever quantized newtons laws?

Has anyone ever attempted to quantize Newtons equations, and indeed if they have, what are the equations used for?

I had a stab at quantizing just for fun. I began with this classical Newtonian equation

$-kx = \frac{\partial P}{\partial t}$

To begin, I simply re-wrote this using the respective quantum operators,

$-i \hbar \frac{\partial}{\partial x} (\frac{\partial}{\partial t}) = -kx$

I hit it with a wave function given as $\Psi = \psi(x)\phi(t)$ which produces

$-i\hbar \frac{\partial}{\partial x}(\frac{\partial \phi(t)}{\partial t} \psi(x)) = -kx \psi(x) \phi(t)$

Dividing through by $\Psi$ gives

$-i \hbar \frac{\partial}{\partial x}(\frac{\partial \phi (t)}{\partial t}\frac{1}{\phi(t)}) = -kx$

using the method of separation of variables and dealing with the time-dependant case we end up with

$-i \hbar(\frac{\partial \phi (t)}{\partial t}) = \lambda \phi(t)$

where $\lambda$ is just some constant. Rearranging the equation above yields

$(\frac{\partial \phi (t)}{\partial t}) = -\frac{i \lambda}{\hbar} \partial t$

Integration yields

$\int \frac{\partial \phi(t)}{\phi t} = \int -\frac{i \lambda}{\hbar} \partial t \rightarrow -\frac{i \lambda}{\hbar} t + C$

The solution then is

$C e^{-\frac{i \lambda}{\hbar} t}$

2. Pseudoscience , please. There are so many errors, it is not even worth commenting.

3. Originally Posted by Aethelwulf
Where? I have never quantized Newtons laws before, so if there are errors, be a man and step up and show.
Reiku,

What you are doing is called "numerology", has nothing to do with quantization.

4. Originally Posted by Aethelwulf
Has anyone ever attempted to quantize Newtons equations, and indeed if they have, what are the equations used for?

I had a stab at quantizing just for fun. I began with this classical Newtonian equation

$-kx = \frac{\partial P}{\partial t}$
Err, no. The correct form is $-kx=\frac{dp}{dt}$. Do you understand the difference?

To begin, I simply re-wrote this using the respective quantum operators,

$-i \hbar \frac{\partial}{\partial x} (\frac{\partial}{\partial t}) = -kx$
The above is a "nothing", definitely not an equation. Do you understand why? Do you want to continue, Reiku?

5. Originally Posted by Aethelwulf
$\frac{\partial P}{\partial t}$ ignoring what derivatives I am using for now. Just gonna stick with what I have used so far. Altogether with the quantum operators in place, the equation simply becomes

$-i \hbar \frac{\partial}{\partial x} (\frac{\partial}{\partial t}) = -kx$

Now, why wouldn't that be an equation?
Give it a rest, Reiku.

6. Quantum mechanics follows from classical mechanics by converting Poisson brackets into operator commutation relations. Much of the mathematical structure in one is in the other, you just have to know how to phrase it properly.

As Tach has already commented, this isn't going about things in a careful, proper manner, it's just Reiku spewing out more of his nonsense. I guess he thinks if he scatter guns enough **** onto the walls something is going to stick.

Aethelwulf/Reiku, given everyone knows your past posting behaviour and even in your current incarnation you've shown a complete lack of understanding for the mathematics you post and the underlying physics threads like this are not to be made in the maths/physics subforum. Until such time as you can demonstrate a working grasp of the relevant mathematical physics your 'musings' are to be put in the pseudo or alternative theories forums. Which is where this is going. You want to play physicist by spewing out stuff you think is complicated and will convince people you're not wasting your existence. We're not here to feed your delusions of mediocrity.

7. Typically you don't site the sources you copy from. This is sort of a break from tradition!

8. Originally Posted by Aethelwulf
$-i \hbar \frac{\partial}{\partial x} (\frac{\partial}{\partial t}) = -kx$

Now, why wouldn't that be an equation?
Because you don't equate differential operators like that. If, for example, $\partial_{t}\phi = \partial_{x}\phi$ that doesn't mean you can write $\partial_{t} = \partial_{x}$, such an expression is meaningless. $\partial_{t}\phi = \partial_{x}\phi$ says that there is some field $\phi$ where at every point in space and time it's partial derivative wrt time is equal to its partial derivative wrt space. $\partial_{t} = \partial_{x}$ is saying "These two operators are identical, they not only have the same action on some function, they have the same action on all functions".

Yes, there's the relation $\hat{p} \leftrightarrow -i\hbar \partial_{x}$ used in quantum mechanics but this is relating two operators which are identical. You arrived at this dubious conclusion because you assumed all you had to do to go from classical to quantum is just change notation, ie assume all classical p's become quantum p's etc and then just mess with things using quantum mechanical expressions.

The correct way to express the classical equation you're looking at would not to change the classical p into a quantum operator p but instead into the expectation of that operator. This is well known to anyone who has actually done quantum mechanics and not just copied it from Google hits, in that classical quantities relate to measurements of operators. Want to see how it's done, seeing as you obviously don't know how?

We've got some wave function $|\psi(t)\rangle$, which satisfies the Schrodinger equation, $-i\partial_{t}|\psi(t) = \hat{H}|\psi(t)\rangle$. This is trivially solved to $|\psi(t)\rangle = e^{i\hat{H}t}|\psi(0)\rangle$. We want to measure some quantity A, which is associated to a quantum mechanical operator $\hat{A}$. This operator is time independent, it has an infinite dimensional matrix representation where all entries are constant. Therefore we get $A(t) = \langle \psi(t)|\hat{A}|\psi(t)\rangle$. Putting in the solution for the wave function we can collect all of the time dependence into a single term, $\langle \psi(t)|\hat{A}|\psi(t)\rangle = \langle \psi(0)| e^{-i\hat{H}t}\hat{A}e^{i\hat{H}t}|\psi(0)\rangle \equiv \langle \psi(0)| \hat{A}(t)|\psi(0)\rangle$. This is how you get different 'pictures' in quantum mechanics. The Schrodinger picture considers the operators time independent and the states time dependent, while the Heisenberg picture is the other way around. Either way we might want to ask what the time derivative of this observable is, $\partial_{t}A(t)$. Well we just use the chain rule and the fact $[f(\hat{O}),\hat{O}] = 0$ always, $\partial_{t}A(t) = \langle \psi(0) | -\hat{H}\hat{A}(t) + \hat{A}(t)\, \hat{H} |\psi(0)\rangle = \langle \psi(0) | [\hat{H},\hat{A}(t)] | \psi(0)\rangle \equiv [H,A(t)]$. In the Heisenberg picture the states are time independent so we get $0 = \langle \psi(0)| \Big( \partial_{t}\hat{A}(t) - [\hat{A}(t),\hat{H}] \Big) | \psi(0)\rangle$. This is the Heisenberg Equation of Motion for the time dependent operators and their observables.

So let's put in $\hat{A} = \hat{p}$. Well we know that $\hat{H} = \frac{1}{2}\hat{p}^{2} + U(\hat{x})$ and the first term goes and we're left with $\partial_{t}\hat{p}(t) = \langle [\hat{p},U(\hat{x})] \rangle$. It's easy to show that $[\hat{p},U(\hat{x})] |\psi(0)\rangle = -(\partial_{x}U)(\hat{x})|\psi(0)\rangle$ and thus we have not an operator equation but an equation for the observables of operators as $\dot{p}(t) = -\nabla U(x(t))$, exactly as the Newtonian mechanics would say if you started with the classical Hamiltonian $\frac{1}{2}m\dot{x}^{2} + U(x)$.

The fact you don't know this shows how you've never done any real quantum mechanics, you would have come across this in an introductory course or textbook on quantum mechanics, as it is a nice example to illustrate quantum mathematical methods while returning a result the students should be familiar with. Yet another piece of evidence you're trying to talk about things you don't understand.

Originally Posted by Aethelwulf
Actually, I consider what I have done relatively easy.
It's generally the most ignorant who have the most inflated, unjustified view of their abilities.

Originally Posted by Aethelwulf
All I did was swap some classical physics for the canonical quantization method.
No, you read what the classical expresions where, you looked up a bunch of formulae from quantum mechanics and you indiscriminantly applied them on the assumption such usages were valid. As I just explained, they are not.

Originally Posted by Aethelwulf
So easy for you to say I am ''playing physics'' when I have a genuine interest in it like the next person here.
No, you don't. You have an interest but not a genuine one. A genuine one wouldn't start thread after thread throwing around expressions he doesn't understand, pretending to be doing actual physics when it is neither actual physics nor demonstrating any understanding of actual physics (quite the opposite in fact). It is very very clear you do not have a working understanding of this area of physics, both from this thread and your Planck thread, never mind all your previous accounts. There's plenty of people here who don't understand quantum mechanics and most of them, the half rational ones, don't start thread after thread spewing out algebra they can't do properly, claiming its about physics they don't understand. You don't have a genuine interest, you have an interest in appearing to be genuine and informed when in fact you are neither. I get it, you want to put on the proverbial white coat, say something which sounds complicated and delude yourself you've achieved something. Unfortunately it results in precisely the opposite.

Originally Posted by Aethelwulf
you're not exactly saying my post is wrong in any way.
You regularly try this line of flawed reasoning, ie if I don't explicitly say "This is wrong" then you assume I'm accepting it is right. The fact I might neither have the time nor the inclination to go through a post line by line doesn't mean I accept it correct. You should have learnt that by now, given all the times you've tried this, only to have me turn around and nail your post to the wall in extreme detail. Besides, even if there were no one here explaining your mistakes you'd still be practically innumerate and extremely bad at quantum mechanics and relativity. Even if this forum were filled with thousands of people posting "Wow, what a genius you are!" and "Surely you are the most insightful physicist ever!" and giving you offers of marriage or free beer and blow jobs, it wouldn't accomplish anything. No one with any mathematics or physics education would believe you, it wouldn't get you a research job or a PhD or a publication. It would accomplish sweet F.A..

Your behaviour of sticking your fingers in your ears and declaring I don't respond to certain posts or how you try to pick up on little things like "Is it an equation or an expression!" show you know you've backed yourself into a corner. If you had something good to retort with you'd do it. Instead you're stuck trying to magnify out of all reasonable proportions completely pathetic responses.

Originally Posted by Aethelwulf
Your interest is diverted on me alone, which is lousy moderating.

Originally Posted by Aethelwulf
Maybe then if you agree with him so much, you can start off where he left off and explain why that specific equation isn't actually an equation, secondly, explain why this is psuedoscience! Because to be honest, this only looks like you trying to have a dig at me!
....
You have put this in psuedoscience which I think is way harsh - considering I am not attempting to ''explain any fringe sciences''

9. Originally Posted by Aethelwulf

I hit it with a wave function given as $\Psi = \psi(x)\phi(t)$ which produces

$-i\hbar \frac{\partial}{\partial x}(\frac{\partial \phi(t)}{\partial t} \psi(x)) = -kx \psi(x) \phi(t)$

Dividing through by $\Psi$ gives

$-i \hbar \frac{\partial}{\partial x}(\frac{\partial \phi (t)}{\partial t}\frac{1}{\phi(t)}) = -kx$
Basic calculus says that you don't know what you are talking about. The way you divided the LHS is laughable. Like I said, give it a rest, Reiku.

10. Originally Posted by Aethelwulf
If one actually studies the equation, it's got mixed terms, two spatial derivatives on both sides with one time derivative on one side.

Nothing in mathematics says that this cannot be done. What alphanumeric wrote was

$\partial_x \phi = \partial_t \phi$

which is clearly wrong. What I wrote has something similar to the tune of

$\partial_x \phi = \partial_x \partial_t \phi$

That's allowed. And it certainly wasn't a mathematical discrepancy according to the physics teacher who also speculated this which I linked to. Back to block before I see any posts of his and am tempted to reply.
I didn't say you actually said $\partial_{t} =\partial_{x}$, I was using that as an example of what you did write, which was of the form $\partial_{t}\partial_{x} = \ldots$. I'm wondering whether you're being obtuse deliberately or you're really so thick you didn't understand that. I did explain it in detail, about why you cannot just leave the derivatives hanging like that. Obviously my point went over your head, showing how little you understand this stuff.

Originally Posted by Tach
Basic calculus says that you don't know what you are talking about. The way you divided the LHS is laughable. Like I said, give it a rest, Reiku.
Well spotted, I missed that one having had my attention drawn by other mistakes he made. It's like a magic eye picture, the more you stare at his nonsense, the more things you spot. Except at the end you don't suddenly see an aeroplane or a person, you see a pile of ****.

11. Originally Posted by AlphaNumeric
.

Well spotted, I missed that one having had my attention drawn by other mistakes he made. It's like a magic eye picture, the more you stare at his nonsense, the more things you spot. Except at the end you don't suddenly see an aeroplane or a person, you see a pile of ****.
Doesn't stop Reiku marching on with his BS. It is amusing, really.

12. Originally Posted by Aethelwulf
]

plug in a wave function

$-i\hbar \frac{d \phi(t)}{dt}\psi(x) = \frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} \phi(t) + V(x,t)\psi(x)\phi(t)$

use separation of variables but also by letting the potential depend only on space (a common practice). Before this however, you divide through by the wave function

GIVING

$-i\hbar \frac{d\phi(t)}{dt} \frac{1}{\phi(t)} = -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} \frac{1}{\psi(x)} + V(x,t)$

Neither of you have demonstrated, to me, that you actually have a working grasp of calculus.
Just as bad, reiku, just as bad. Nice try at obfuscating your errors. Thanks for the laughs, keep up the good work, it is entertaining.

13. Originally Posted by Aethelwulf

I hit it with a wave function given as $\Psi = \psi(x)\phi(t)$ which produces

$-i\hbar \frac{\partial}{\partial x}(\frac{\partial \phi(t)}{\partial t} \psi(x)) = -kx \psi(x) \phi(t)$

Dividing through by $\Psi$ gives

$-i \hbar \frac{\partial}{\partial x}(\frac{\partial \phi (t)}{\partial t}\frac{1}{\phi(t)}) = -kx$
Thanks for the laughs, Reiku, thanks for the laughs. Keep it up.

14. Originally Posted by Aethelwulf
I can see why people might say what they say about this site. I heard one not so-polite comment saying it was full retards.
Look in the mirror , Reiku. You suffer from a very rare form of graphomania : you feel compelled to write posts containing mathematical formulas that make absolutely no sense. You should consult a psychiatrist, with the appropriate drugs your affliction may be treatable. Or maybe not, your posts are immensely entertaining.

15. Originally Posted by Aethelwulf
Yup... the place is full of retards.
still preening in front of your mirror, Reiku?

16. Originally Posted by Aethelwulf
I'm not trying to cover up anything. I am telling you this is a well known procedure; more fool you for trying to denounce it some how.
Separation of variables is a well known procedure, except you can't do it properly.

This is a problem you regularly have Reiku, you think that if you attempt to use some result from the mainstream then you're not in the wrong but this assumes you can actually use the method properly. Invariably this turns out not to be the case.

Originally Posted by Aethelwulf
I can see why people might say what they say about this site. I heard one not so-polite comment saying it was full retards.
You're not exactly helping.

Originally Posted by Aethelwulf
See, when I tell you this is a well known procedure, you still don't listen.
The problem is you don't know how to do it properly! This is just another piece of evidence you clearly lack the necessary understanding and working knowledge of any of the necessary ground work to work with things like the Dirac equation, which you started talking about in another thread. Part of the reason we're able to call you on this is yes, it is a well known method, we know it. It's sufficiently familiar to people who went to university to do physics, rather than just lie on internet forums like you do, that mistakes in it are easily spotted. Your inability to spot such mistakes illustrates you don't understand the method, you just blindly apply formulae you find and tell yourself you couldn't possibly be wrong in doing so. Wrong. Wrong. Go on, have another go. Tach and I both commented about something you got wrong, why don't you tell us what it is.

17. Originally Posted by Aethelwulf
My name is Aethelwulf,
...Aethelwulf is a sockpuppet for Reiku. Doesn't matter how many times you change your name, the BS you post gives you away.

I don't think you even know what you are talking about half the time.
Reiku, I couldn't care less about what you think.

18. Originally Posted by Aethelwulf
Then you are not contributing, you making a great waste of bandwidth.
Bye, Reiku.
Until you come back as yet another sockpuppet.

19. Originally Posted by Aethelwulf
My name is Aethelwulf, and no, I am actually still mildly amused how you said that the equation you copied saying it wasn't an equation, then not even justifying your comment. You spout off a lot of hot air. I don't think you even know what you are talking about half the time.
Better not to know what one is talking about half the time than all of the time, as seems to be the case with you. Reiku, you've been shown to have made a plethora of mistakes in just about every thread like this you've made. Just like you used to have happen with your previous accounts. On the same subjects. With the same equations. And same mistakes. And same whining. And same IP address range.

When you return from your 3 day holiday (and it wasn't me who reported you this time) try to point out where your mistake is in the bit Tach quoted. You can also try to retort my lengthy explanation of why you were wrong in several ways in regards to your opening posts and how it demonstrates you haven't ever actually learnt quantum mechanics properly, you just attempt to parrot in some way which isn't obvious copy/pasting. Too bad you lack the necessary understanding to correctly alter and reshuffle equations you don't understand.

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