- #1
FunkHaus
- 10
- 1
Hi,
I've had a question ever since my quantum classes that's pretty simple I guess, but still seems to elude me. So here it is:
One text I used for quantum (Liboff's "Introductory Quantum Mechanics") says that in classical mechanics, there is a "vector of the state" of a system, that contains all the information in a mechanical system. This vector he claims contains the positions and velocities of all particles of a system (this is all just chapter 1). Then he goes on to say that in quantum mechanics, the vector of the state of the system cannot contain all these quantities simultaneously--this is the nature of the Heisenberg uncertainty principle.
But wait a second--why do the positions and velocities of a system yield all the information in classical mechanics? Is Liboff talking about instantaneous positions and velocities, or is he talking about the trajectories (world line functions)
[tex]\vec x(t)[/tex] and [tex]\frac{d \vec x(t)}{dt}[/tex]?
Because, basically, if you have determined the trajectory function
[tex] \vec x(t) [/tex] (a map [tex]\Re^3 \rightarrow \Re[/tex]),
which you can do in mechanics, then you can also find the velocity [tex] \vec v(t) [/tex] at all points, the acceleration [tex] \vec a(t)[/tex]...hell, I could find the 20th derivative of position if I wanted to! So what's with saying the system is "totally specified" by giving positions and velocities? I would argue that all you need is the position function [tex] \vec x(t) [/tex].
So then that's where my quantum mechanics questions come in. What would the Heisenberg uncertainty principle have to say about, for instance, the uncertainty in position and acceleration ? What about jerk? What about the seventh derivative of position? If the seventh derivative of position were the "observable" I wanted to measure in a quantum system, how would I do it? What would the operator associated with this observable be? This is one of the main reasons quantum has always confused me. In classical mechanics, you really can't separate position, velocity, acceleration, etc. They're all just derivatives of the same fundamental world line (trajectory) function [tex] x(t) [/tex].
If anyone has any insight, it would be greatly appreciated. Thanks a lot!
I've had a question ever since my quantum classes that's pretty simple I guess, but still seems to elude me. So here it is:
One text I used for quantum (Liboff's "Introductory Quantum Mechanics") says that in classical mechanics, there is a "vector of the state" of a system, that contains all the information in a mechanical system. This vector he claims contains the positions and velocities of all particles of a system (this is all just chapter 1). Then he goes on to say that in quantum mechanics, the vector of the state of the system cannot contain all these quantities simultaneously--this is the nature of the Heisenberg uncertainty principle.
But wait a second--why do the positions and velocities of a system yield all the information in classical mechanics? Is Liboff talking about instantaneous positions and velocities, or is he talking about the trajectories (world line functions)
[tex]\vec x(t)[/tex] and [tex]\frac{d \vec x(t)}{dt}[/tex]?
Because, basically, if you have determined the trajectory function
[tex] \vec x(t) [/tex] (a map [tex]\Re^3 \rightarrow \Re[/tex]),
which you can do in mechanics, then you can also find the velocity [tex] \vec v(t) [/tex] at all points, the acceleration [tex] \vec a(t)[/tex]...hell, I could find the 20th derivative of position if I wanted to! So what's with saying the system is "totally specified" by giving positions and velocities? I would argue that all you need is the position function [tex] \vec x(t) [/tex].
So then that's where my quantum mechanics questions come in. What would the Heisenberg uncertainty principle have to say about, for instance, the uncertainty in position and acceleration ? What about jerk? What about the seventh derivative of position? If the seventh derivative of position were the "observable" I wanted to measure in a quantum system, how would I do it? What would the operator associated with this observable be? This is one of the main reasons quantum has always confused me. In classical mechanics, you really can't separate position, velocity, acceleration, etc. They're all just derivatives of the same fundamental world line (trajectory) function [tex] x(t) [/tex].
If anyone has any insight, it would be greatly appreciated. Thanks a lot!