- #1
snoopies622
- 840
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I'm not perfectly clear on the connection between Poisson brackets in classical mechanics and commutators in quantum mechanics.
For any classical mechanical system, if I can find the Poisson bracket between two physical observables, is that always the value of the corresponding commutator in the quantum mechanical formulation of the same system (divided by [itex] i \hbar [/itex] )?
If so, there must be some mathematical reason for this. Some kind of homomorphism between the two systems, perhaps?
I know that Poisson brackets and commutators share some algebraic properties and that
Hamilton's [tex]
\frac{df}{dt}= \{ f,H \} + \frac { \partial {f} }{\partial {t}}
[/tex]
looks similar to Heisenberg's [tex]
\frac{d \hat{A} }{dt}= \frac {i}{\hbar} [ \hat{A},\hat{ H} ] + \frac { \partial { \hat{A} } }{\partial {t}} [/tex]
but for that method to work every time, it feels to me like there must be something more.
For any classical mechanical system, if I can find the Poisson bracket between two physical observables, is that always the value of the corresponding commutator in the quantum mechanical formulation of the same system (divided by [itex] i \hbar [/itex] )?
If so, there must be some mathematical reason for this. Some kind of homomorphism between the two systems, perhaps?
I know that Poisson brackets and commutators share some algebraic properties and that
Hamilton's [tex]
\frac{df}{dt}= \{ f,H \} + \frac { \partial {f} }{\partial {t}}
[/tex]
looks similar to Heisenberg's [tex]
\frac{d \hat{A} }{dt}= \frac {i}{\hbar} [ \hat{A},\hat{ H} ] + \frac { \partial { \hat{A} } }{\partial {t}} [/tex]
but for that method to work every time, it feels to me like there must be something more.