# Quantizing Newtons Laws

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You err when you claim that I didn't read the thread. Iread it and can't find such an example. Instead of my guessing why not simply tell me he post number and make this conversation shorter than it need be?

The only thing I can find is this

First you use $\nabla (\frac{\partial }{\partial t}\Psi) = \frac{\partial }{\partial t}(\nabla\Psi)$ but latter you use $\frac{\partial }{\partial t}(\nabla \Psi) = (\frac{\partial }{\partial t}\nabla) \Psi$ which is incorrect.

One can tell merely by inspection that this expression is correct.

$\nabla \Psi = \frac{\partial \Psi}{\partial x}\hat{e}_x + \frac{\partial \Psi}{\partial y}\hat{e}_y + \frac{\partial \Psi}{\partial z}\hat{e}_z$

$\frac{\partial}{\partial t}(\nabla \Psi) = \frac{\partial \Psi}{\partial t \partial x}\hat{e}_x + \frac{\partial \Psi}{\partial t \partial y}\hat{e}_y + \frac{\partial \Psi}{\partial t \partial z}\hat{e}_z$

$\frac{\partial}{\partial t}(\nabla \Psi) = (\frac{\partial}{\partial t \partial x}\hat{e}_x + \frac{\partial}{\partial t \partial y}\hat{e}_y + \frac{\partial}{\partial t \partial z}\hat{e}_z)\Psi$

$\frac{\partial}{\partial t}(\nabla \Psi) = (\frac{\partial}{\partial t}(\frac{\partial}{\partial x}\hat{e}_x + \frac{\partial}{\partial y}\hat{e}_y + \frac{\partial}{\partial z}\hat{e}_z))\Psi$

$\frac{\partial}{\partial t}(\nabla \Psi) = (\frac{\partial}{\partial t} \nabla)\Psi$

which proves that the expression is correct.

I never said or insinuate that.

Then all you had to say was either yes or no. It was unclear in your previous posts. That's why I asked. Since you didn't directly answer the question with either yes or no then I'll assume the answer is no, you're not saying that Hermitian operators don't follow the associative law. Then that's good since they do.

But you aren't. You're quantizing Hooke's law.

What does it mean to "quantize" an equation/law? Do you mean to cast the classical law/equation into an expressuion containing operators?

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What does it mean to "quantize" an equation/law? Do you mean to cast the classical law/equation into an expressuion containing operators?

I wonder the same. Perhaps they mean make it an eigenvalue equation with a distcrete spectrum for the operator (non-continuous set of eigenvalues)? Don't know, just a guess.

Edited by mississippichem
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I wonder the same. Perhaps they mean make it an eigenvalue equation with a distcrete spectrum for the operator (non-continuous set of eigenvalues)? Don't know, just a guess.

Quantization

http://en.wikipedia.org/wiki/Canonical_quantization

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I wonder the same. Perhaps they mean make it an eigenvalue equation with a distcrete spectrum for the operator (non-continuous set of eigenvalues)? Don't know, just a guess.

There is such a thing a Canonical Quantization, a First Quantization and a Second Quantization method. My approach would have been a type of Canonical Quantization where we introduce the use of h-bar (a purely quantum mechanical) entity and their respective operators for the replacement of classical variables.

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$\frac{d \hat{p}}{d t} = \hat{F}$

If you're attempting to say that given two operators P and F when inserted into Newton's secind law yields $\frac{d \hat{p}}{d t} = \hat{F}$ then you're wrong.

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• 2 weeks later...

I decided to quantize Newtons 2nd Law

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

today. Has anyone done this before? I need some help.

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

Hitting it with a wave function $\Psi = \psi(x)\phi(t)$ gives

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

To solve it we are therefore going to use the separation of variables method. (But I need you guys to make sure I am doing this right)

Divide through by $\Psi$ gives

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

So this is as far as I have got. Should I move the $\frac{\partial}{\partial x}$ term to the right as

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

So that on the left I purely have variables of t and on the right variables of x?

I think this would mean that both sides are independent, meaning that they equal a constant?

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

$\frac{-kx}{(\frac{\partial}{\partial x})} = \lambda$

Then I would like to concentrate on the time-dependant case

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

Will now be

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

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

Then the solution would be $C e^{\frac{-i \mathcal{A}}{\hbar}}$ where I have let $\lambda t = \mathcal{A}$.

Does this seem right, or have I messed up somewhere?

In one of your other posts you wrote that if F = -kx and F = dP/dt, then -kx = dP/dt. This is only true for the special case when the sum of the forces on the system under consideration is equal to -kx. Newton's second law states that if you take the vector sum of all the forces acting on an object of mass m, then that sum is equal to the time derivative of the object's momentum.

I started reading through your post, and noticed that you divided both sides of the equation by the wave function, canceling the spatial part of the wave function on both sides prior to taking the derivative with respect to x of that same function. Once I got to that math error, I quit reading.

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I started reading through your post, and noticed that you divided both sides of the equation by the wave function, canceling the spatial part of the wave function on both sides prior to taking the derivative with respect to x of that same function. Once I got to that math error, I quit reading.

You report the mathematical mistake when he 'obtains' his fourth equation, but his mathematical and physical errors start even before as noticed in this thread.

If you're attempting to say that given two operators P and F when inserted into Newton's secind law yields $\frac{d \hat{p}}{d t} = \hat{F}$ then you're wrong.

You do not know what you are saying, $\frac{d \hat{p}}{d t} = \hat{F}$ is a known quantum mechanical equation.

Edited by juanrga

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