Aethelwulf Posted June 23, 2012 Share Posted June 23, 2012 (edited) I decided to quantize Newtons 2nd Law [math]-kx = \frac{\partial P}{\partial t}[/math] today. Has anyone done this before? I need some help. I start with quantizing the equation, naturally; [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial}{\partial t}) = -kx[/math] Hitting it with a wave function [math]\Psi = \psi(x)\phi(t)[/math] gives [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial \phi(t)}{\partial t}\psi(x)) = -kx \psi(x)\phi(t)[/math] 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 [math]\Psi[/math] gives [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = -kx[/math] So this is as far as I have got. Should I move the [math]\frac{\partial}{\partial x}[/math] term to the right as [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = \frac{-kx}{(\frac{\partial}{\partial x})}[/math] 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? [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = \lambda[/math] [math]\frac{-kx}{(\frac{\partial}{\partial x})} = \lambda[/math] Then I would like to concentrate on the time-dependant case [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}) = \lambda \phi(t)[/math] Will now be [math](\frac{\partial \phi(t)}{\phi(t)}) = \frac{-i\lambda}{\hbar}\partial t[/math] [math]\int (\frac{\partial \phi(t)}{\phi(t)}) = \int \frac{-i\lambda}{\hbar}\partial t \rightarrow \frac{-i\lambda}{\hbar} t +C[/math] Then the solution would be [math]C e^{\frac{-i \mathcal{A}}{\hbar}}[/math] where I have let [math]\lambda t = \mathcal{A}[/math]. Does this seem right, or have I messed up somewhere? Edited June 23, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

imatfaal Posted June 23, 2012 Share Posted June 23, 2012 I decided to quantize Newtons 2nd Law [math]-kx = \frac{\partial P}{\partial t}[/math] today. Has anyone done this before? I need some help. I start with quantizing the equation, naturally; [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial}{\partial t}) = -kx[/math] Hitting it with a wave function [math]\Psi = \psi(x)\phi(t)[/math] gives [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial \phi(t)}{\partial t}\psi(x)) = -kx \psi(x)\phi(t)[/math] 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 [math]\Psi[/math] gives [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = -kx[/math] So this is as far as I have got. Should I move the [math]\frac{\partial}{\partial x}[/math] term to the right as [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = \frac{-kx}{(\frac{\partial}{\partial x})}[/math] 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? [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = \lambda[/math] [math]\frac{-kx}{(\frac{\partial}{\partial x})} = \lambda[/math] Then I would like to concentrate on the time-dependant case [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}) = \lambda \phi(t)[/math] Will now be [math](\frac{\partial \phi(t)}{\phi(t)}) = \frac{-i\lambda}{\hbar}\partial t[/math] [math]\int (\frac{\partial \phi(t)}{\phi(t)}) = \int \frac{-i\lambda}{\hbar}\partial t \rightarrow \frac{-i\lambda}{\hbar} t +C[/math] Then the solution would be [math]C e^{\frac{-i \mathcal{A}}{\hbar}}[/math] where I have let [math]\lambda t = \mathcal{A}[/math]. Does this seem right, or have I messed up somewhere? Youve lost me at the first line - I still think of N 2 Law as f=ma! Could you explain your symbols and assumptions so the members can keep up. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 23, 2012 Author Share Posted June 23, 2012 Youve lost me at the first line - I still think of N 2 Law as f=ma! Could you explain your symbols and assumptions so the members can keep up. [math]F=Ma[/math] is not true in relativity. The force equation which remains unchanged in relativity is [math]F=d(m_{rel}v)/dt[/math] The reason why is because in [math]F=Ma[/math], mass is generally a constant which is not true in relativity. I had a little look on line and there is a wiki article which mentions it: http://en.wikipedia.org/wiki/Mass_in_special_relativity Oh, and -kx is just hookes law for force. 1 Link to comment Share on other sites More sharing options...

timo Posted June 23, 2012 Share Posted June 23, 2012 From F=ma and a²+b²=c² we unambigiously conclude that [math] F = m \sqrt{b^2 - c^2} [/math]. Since we are in a relativistic framework, we know that E=mc². Therefore, [math] F^2 = m^2 b^2 - mE [/math]. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 23, 2012 Author Share Posted June 23, 2012 From F=ma and a²+b²=c² we unambigiously conclude that [math] F = m \sqrt{b^2 - c^2} [/math]. Since we are in a relativistic framework, we know that E=mc². Therefore, [math] F^2 = m^2 b^2 - mE [/math]. That doesn't make any sense to me. I take it that was an attempt of a joke. Link to comment Share on other sites More sharing options...

imatfaal Posted June 23, 2012 Share Posted June 23, 2012 So you are using capital P as a symbol for momentum - didn't know that one, I always think of that as pressure. And why are you introducing springs into Newtons second law? And in your second equation you still seem to have change of position and time on the LHS and the unusual spring resorative force on the RHS - but where has the mass (relativistic or not) gone? Link to comment Share on other sites More sharing options...

elfmotat Posted June 23, 2012 Share Posted June 23, 2012 (edited) I don't get it, why don't you just use [math]F=-\frac{\partial H}{\partial x}[/math]? You could take the gradient of the Schrodinger equation to get: [math]-i\hbar \frac{\partial }{\partial t}\nabla\Psi=\hat{F}\Psi [/math] , where F-hat is defined as a force operator -∇H. Then, using the fact that the momentum operator is [math]\hat{p}=-i\hbar \nabla[/math], everything simplifies to: [math]\frac{\partial \hat{p}}{\partial t}\Psi=\hat{F}\Psi[/math] This is all very messy and unnecessary by the way. Edited June 23, 2012 by elfmotat 1 Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 23, 2012 Author Share Posted June 23, 2012 So you are using capital P as a symbol for momentum - didn't know that one, I always think of that as pressure. And why are you introducing springs into Newtons second law? And in your second equation you still seem to have change of position and time on the LHS and the unusual spring resorative force on the RHS - but where has the mass (relativistic or not) gone? There's nothing mystical behind the use of Hookes law here. Just looking at my notes, it's just the way I defined the force. All your relativistic dynamics are still wrapped up in the [math]dp/dt[/math] (used a little p now to halt any confusion). Yes, Equation two should be written like that. The first term [math]-i\hbar \frac{\partial}{\partial x}[/math], do you recognize it? It's the momentum operator, we get this when we quantize the equation. That means what we really have [math]\frac{\partial \hat{p}}{\partial t}[/math]. This is why there is a mixture of the change in position and time on the LHS like you spotted. I don't get it, why don't you just use [math]F=-\frac{\partial H}{\partial x}[/math]? You could take the gradient of the Schrodinger equation to get: [math]-i\hbar \frac{\partial }{\partial t}\nabla\Psi=\hat{F}\Psi [/math]. This is all very messy and unnecessary by the way. What's H? Anyway, assuming your approach is correct, that's fine and all dandy. But I wanted to specifically quantize the second law. Ah H is meant to be the Hamiltonian yes? Should it be defined as a Hamiltonian, I thought W=Fx? Link to comment Share on other sites More sharing options...

elfmotat Posted June 23, 2012 Share Posted June 23, 2012 What's H? Anyway, assuming your approach is correct, that's fine and all dandy. But I wanted to specifically quantize the second law. Ah H is meant to be the Hamiltonian yes? Should it be defined as a Hamiltonian, I thought W=Fx? I added to my post to get that equation into a form you might find more familiar. And yes, H is the Hamiltonian. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 23, 2012 Author Share Posted June 23, 2012 I added to my post to get that equation into a form you might find more familiar. And yes, H is the Hamiltonian. Following my own example, I end up with [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}) = \lambda \phi(t)[/math] The left hand side is the energy operator [math]\hat{E} \phi(t) = -i \hbar \frac{\partial}{\partial t} \phi(t) = \lambda \phi(t)[/math] Does the fact you end up (in your simple method) using a force operator make my method wrong? Link to comment Share on other sites More sharing options...

swansont Posted June 24, 2012 Share Posted June 24, 2012 I wanted to specifically quantize the second law. But you aren't. You're quantizing Hooke's law. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Author Share Posted June 24, 2012 But you aren't. You're quantizing Hooke's law. If F=-kx and F=dp/dt then I am actually quantizing both. But, I concentrated on the time dependant part as well, sothe -kx is completely lost at the end anyway. Oh that's nice, I knew this could be done linking the left relativistic force of Newton for the Hookes law ''The pair comes from taking Hooke's law to be the force appearing on the right of the relativistic expressions: dp/dt or dp/dtau'' http://arxiv.org/abs/1205.5299 Link to comment Share on other sites More sharing options...

juanrga Posted June 24, 2012 Share Posted June 24, 2012 (edited) I decided to quantize Newtons 2nd Law [math]-kx = \frac{\partial P}{\partial t}[/math] Newton law does not use partial derivatives. So you are using capital P as a symbol for momentum - didn't know that one, I always think of that as pressure. Curious, because pressure is almost always denoted by p as in "pdV". You could take the gradient of the Schrodinger equation to get: [math]-i\hbar \frac{\partial }{\partial t}\nabla\Psi=\hat{F}\Psi [/math] , where F-hat is defined as a force operator -∇H. Then, using the fact that the momentum operator is [math]\hat{p}=-i\hbar \nabla[/math], everything simplifies to: [math]\frac{\partial \hat{p}}{\partial t}\Psi=\hat{F}\Psi[/math] [math]-\nabla H\Psi \neq \hat{F}\Psi[/math] unless [math]\nabla\Psi=0[/math] but then the left hand side of your Schrödinger equation is zero. First you use [math]\nabla (\frac{\partial }{\partial t}\Psi) = \frac{\partial }{\partial t}(\nabla\Psi)[/math] but latter you use [math]\frac{\partial }{\partial t}(\nabla \Psi) = (\frac{\partial }{\partial t}\nabla) \Psi[/math] which is incorrect. Indeed [math]\frac{\partial \hat{p}}{\partial t}=0[/math] [math]F=Ma[/math] is not true in relativity. The force equation which remains unchanged in relativity is [math]F=d(m_{rel}v)/dt[/math] The reason why is because in [math]F=Ma[/math], mass is generally a constant which is not true in relativity. I had a little look on line and there is a wiki article which mentions it: http://en.wikipedia....cial_relativity The same link that you give states clearly that the relativistic generalization of the Newtonian law [math]\mathbf{f} = m \mathbf{a}[/math] is [math]F^\mu = m A^\mu[/math] The 3-projection is [math]\mathbf{F} = m \mathbf{A}[/math] Edited June 24, 2012 by juanrga Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 24, 2012 Author Share Posted June 24, 2012 (edited) I might from time to time use partial derivatives instead of full derivatives when I shouldn't. It's a nasty habit I need to snap out of. ''The same link that you give states clearly that the relativistic generalization of the Newtonian law [math]\mathbf{f} = m \mathbf{a}[/math] is'' To be honest I wasn't working from any link. I found that link on the spur of the moment because someone asked what dP/dt was. I actually found a different link where it is customary to link hookes law with the definition of the second law http://arxiv.org/abs/1205.5299 Edited June 24, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

elfmotat Posted June 24, 2012 Share Posted June 24, 2012 [math]-\nabla H\Psi \neq \hat{F}\Psi[/math] unless [math]\nabla\Psi=0[/math] but then the left hand side of your Schrödinger equation is zero. What are you talking about? I defined the operator F to be [math]\hat{F}\Psi \equiv -\nabla \hat{H}\Psi[/math]. As far as I can see that's a perfectly valid definition. First you use [math]\nabla (\frac{\partial }{\partial t}\Psi) = \frac{\partial }{\partial t}(\nabla\Psi)[/math] Yes, partial derivatives commute: [math]\frac{\partial^2 f}{\partial x \partial t}=\frac{\partial^2 f}{\partial t \partial x}[/math] but latter you use [math]\frac{\partial }{\partial t}(\nabla \Psi) = (\frac{\partial }{\partial t}\nabla) \Psi[/math] which is incorrect. How so? Operators are associative, i.e. [math]\hat{A} (\hat{B}\Psi) = (\hat{A} \hat{B})\Psi[/math]. I see nothing invalid about what I did. Indeed [math]\frac{\partial \hat{p}}{\partial t}=0[/math] How do you figure? [math]\frac{\partial \hat{p}}{\partial t}=-i\hbar(\frac{\partial^2 }{\partial t\partial x},\frac{\partial^2 }{\partial t\partial y},\frac{\partial^2 }{\partial t\partial z})[/math] Link to comment Share on other sites More sharing options...

juanrga Posted June 25, 2012 Share Posted June 25, 2012 What are you talking about? I defined the operator F to be [math]\hat{F}\Psi \equiv -\nabla \hat{H}\Psi[/math]. As far as I can see that's a perfectly valid definition. My complaint was not about the definition. Read what I wrote. Yes, partial derivatives commute: [math]\frac{\partial^2 f}{\partial x \partial t}=\frac{\partial^2 f}{\partial t \partial x}[/math] No. I did not wrote commutation of derivatives but something different. Read what I wrote. How so? Operators are associative, i.e. [math]\hat{A} (\hat{B}\Psi) = (\hat{A} \hat{B})\Psi[/math]. I see nothing invalid about what I did. No your relation is not generally true. Associativity means [math]\hat{A} (\hat{B}\hat{C}) = (\hat{A} \hat{B})\hat{C}[/math] How do you figure? [math]\frac{\partial \hat{p}}{\partial t}=-i\hbar(\frac{\partial^2 }{\partial t\partial x},\frac{\partial^2 }{\partial t\partial y},\frac{\partial^2 }{\partial t\partial z})[/math] I merely computed the partial time derivative of the momentum. It is zero. Link to comment Share on other sites More sharing options...

elfmotat Posted June 25, 2012 Share Posted June 25, 2012 My complaint was not about the definition. Read what I wrote. All you said was . I said it is by definition. If your problem is not with my definition, then what exactly is it? No. I did not wrote commutation of derivatives but something different. Read what I wrote. I know you didn't, but I was making explicit the fact that what I did was completely valid. No your relation is not generally true. Associativity means [math]\hat{A} (\hat{B}\hat{C}) = (\hat{A} \hat{B})\hat{C}[/math] Even if it weren't true in general (which it is), what I did would still be valid because of the commutativity of partial derivatives. I merely computed the partial time derivative of the momentum. It is zero. I don't see how you could possibly think that's true. Classically it certainly wouldn't be true in general, and only would be when F=0. Regardless, the partial of the momentum operator is simply not zero in general as I showed above. I can't figure out why you think it should be. Link to comment Share on other sites More sharing options...

juanrga Posted June 25, 2012 Share Posted June 25, 2012 (edited) All you said was . I said it is by definition. If your problem is not with my definition, then what exactly is it? I wrote more than that inequality. I wrote an important condition for the wavefunction and you quoted it in your first reply #15. Now you ignore that part of what I said. I don't see how you could possibly think that's true. Classically it certainly wouldn't be true in general, and only would be when F=0. Regardless, the partial of the momentum operator is simply not zero in general as I showed above. I can't figure out why you think it should be. I already remarked in a previous post that the Newton law uses total derivatives. At least the OP already apologized by using partial derivatives in the notation: "It's a nasty habit". Evidently the partial derivative of momentum operator is zero with independence of F, because the expression is [math]\frac{d \hat{p}}{d t} = \hat{F}[/math] This is all well-known: http://en.wikipedia....renfest_theorem Edited June 25, 2012 by juanrga Link to comment Share on other sites More sharing options...

elfmotat Posted June 25, 2012 Share Posted June 25, 2012 I wrote more than that inequality. I wrote an important condition for the wavefunction and you quoted it in your first reply #15. Now you ignore that part of what I said. "unless " But that simply isn't true. by definition. Unless you in some way object to this definition, your "condition" makes absolutely no sense. I already remarked in a previous post that the Newton law uses total derivatives. At least the OP already apologized by using partial derivatives in the notation: "It's a nasty habit". Right, but we aren't exactly talking about Newton's law here, are we? We're talking about the partial time derivative of an operator, not a variable or a function. Don't you understand the distinction? Even if that weren't the case, you can't tacitly claim that ∂f/∂t=0. I.e. consider the momentum function p(t)=p_{0}+kt (representing constant force). Do you think ∂p/∂t=0? Evidently the partial derivative of momentum operator is zero with independence of F, because the expression is [math]\frac{d \hat{p}}{d t} = \hat{F}[/math] This is all well-known: http://en.wikipedia....renfest_theorem First of all, the Ehrenfest theorem has to do with expectation values which isn't what we're talking about. Second, do you or do you not recognize that ? Link to comment Share on other sites More sharing options...

juanrga Posted June 26, 2012 Share Posted June 26, 2012 (edited) "unless " But that simply isn't true. by definition. Unless you in some way object to this definition, your "condition" makes absolutely no sense. The definition was [math]\hat{F}\equiv - \nabla \hat{H}[/math] (check your own #7). This is the same than [math]\hat{F}\Psi \equiv - (\nabla \hat{H}) \Psi[/math]. At the other hand, it is trivial to check that [math]\nabla \hat{H}\Psi = (\nabla \hat{H})\Psi + \hat{H} (\nabla \Psi)[/math] equals [math](\nabla \hat{H}) \Psi[/math] only when [math]\nabla \Psi = 0[/math]. Right, but we aren't exactly talking about Newton's law here, are we? We're talking about the partial time derivative of an operator, not a variable or a function. Don't you understand the distinction? Are you aware of the title of this thread? "Quantizing Newton laws". The quantum version of the classical Newtonian law is [math]d\hat{p}/dt = \hat{F}[/math] I wrote this expression two or three times before. Evidently [math]\hat{p}[/math] is an operator. First of all, the Ehrenfest theorem has to do with expectation values which isn't what we're talking about. Second, do you or do you not recognize that ? I was not citing the Ehrenfest theorem, what I cited was the equation of motion [math]d\hat{p}/dt = \hat{F}[/math] and the well-known result [math]\partial\hat{p}/ \partial t = 0[/math]. Check my previous post. You can find both in the link given in my previous post. In fact the wikipedia page even devotes a footnote to explain why [math]\partial\hat{p}/ \partial t = 0[/math]. In general [math](\hat{A}\hat{B})f \neq \hat{A}(\hat{B}f)[/math]. In particular [math]\frac{\partial\hat{p}}{\partial t} = -i\hbar\frac{\partial}{\partial t} \left( \nabla \right) \neq -i\hbar \left( \frac{\partial^2}{\partial t\partial x},\frac{\partial^2}{\partial t\partial y},\frac{\partial^2}{\partial t\partial z} \right)[/math] The left hand side [math]\frac{\partial\hat{p}}{\partial t} = -i\hbar\frac{\partial}{\partial t} \nabla = 0[/math]. The right hand side [math]-i\hbar \left( \frac{\partial^2}{\partial t\partial x},\frac{\partial^2}{\partial t\partial y},\frac{\partial^2}{\partial t\partial z} \right) \neq 0[/math] Apart from the mistakes regarding operator associativity, algebra, derivatives, etc., you started from a wrong classical equation ([math]{\partial p}/{\partial t} = F[/math]) and wrote a wrong quantum equation ([math]{\partial \hat{p}}/{\partial t} = \hat{F}[/math]). In fact if you were to take averages on your wrong quantum equation you would obtain [math]\langle \frac{\partial \hat{p}}{\partial t} \rangle = \langle \hat{F} \rangle[/math] but as is well-known (check my posts or check the wikipedia [the equation just after "Suppose we wanted to know the instantaneous change in momentum [math]\hat{p}[/math]"]) [math]\langle \frac{\partial \hat{p}}{\partial t} \rangle = 0[/math] which means that you obtain the wrong result [math]0 = \langle \hat{F} \rangle = F[/math]. In short. The quantum Newton law is [math]\frac{d\hat{p}}{dt} = \hat{F}[/math] taking the classical limit (Eherenfest theorem) gives the well-known Newton law [math]\frac{dp}{dt} = F[/math] Edited June 26, 2012 by juanrga -1 Link to comment Share on other sites More sharing options...

CaptainPanic Posted June 26, 2012 Share Posted June 26, 2012 ! Moderator Note I notice that there is a disagreement. That's fine. Also, I notice that at least two people seem to get a little agitated by this disagreement. That's not fine.So, here's a moderator top tip: keep it polite, please. I don't want to have to intervene in a more serious way in this thread. 2 Link to comment Share on other sites More sharing options...

pmb Posted June 26, 2012 Share Posted June 26, 2012 No your relation is not generally true. Please provide a counter example. Associativity means [math]\hat{A} (\hat{B}\hat{C}) = (\hat{A} \hat{B})\hat{C}[/math] What kind of operators are you talking about? Hermitian? If so and given this statement and the comment above, are you saying that Hermitian operators don't follow the associative rule? Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 26, 2012 Author Share Posted June 26, 2012 Juangra Enough. You have highjacked this thread enough to disturb enough of even the mods. Please retire to a new discussion. Link to comment Share on other sites More sharing options...

juanrga Posted June 27, 2012 Share Posted June 27, 2012 (edited) Please provide a counter example. I gave a specific example. Read the thread. are you saying that Hermitian operators don't follow the associative rule? I never said or insinuate that. Read the thread. Juangra Enough. You have highjacked this thread enough to disturb enough of even the mods. Please retire to a new discussion. Nobody here "highjacked this thread" and no moderator said so. I wrote the correct quantum version of Newton second law and gave a link showing how the quantum version provides the correct classical limit via the Ehrenfest theorem. In fact, the equations for the quantum harmonic oscillator are well-known. You only need to substitute the value of the force F in the equations that I wrote in #18 (and repeated in #20) and you obtain [math]\frac{d \hat{p}}{d t} = -kx[/math] with harmonic constant [math]k = mw^2[/math] You can again check this equation online (although virtually any QM textbook gives this equation) http://en.wikipedia....tator_relations You and another poster claimed [math]{\partial \hat{p}}/{\partial t} = -kx[/math] but your equation is wrong. EDIT: I attach the image with the quantum equations from the Wiki link Edited June 27, 2012 by juanrga Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted June 27, 2012 Share Posted June 27, 2012 ! Moderator Note juanrga, Please go back and have another read of Captain's mod tip. Link to comment Share on other sites More sharing options...

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