Limit of quantum mechanics as h -> 0

In summary, the Heisenberg equation of motion can be approximated by the classical Poisson bracket. The equation you attempted to derive is actually$$\frac{d\hat{p}}{dt}=F(\hat{x})$$where $$F(x)=-\frac{\partial V(x)}{\partial x}$$Note that ##\hat{p}## and ##\hat{x}## are quantum operators, so the first equation is not yet classical. To get something classical-like you have to take the quantum average of it$$\frac{d\langle\psi|\hat{p}|\psi\rangle}{dt}=
  • #1
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TL;DR Summary
I recently saw an explanation for how quantum mechanics approaches classical mechanics at the limit of Planck's constant becoming 0 using the Heisenberg equation of motion but am confused about what it is about this limit that reduces the equation of motion to its classical limit.
Starting from the Heisenberg equation of motion, we have

$$ih \frac{\partial p}{\partial t} = [p, H]$$
which simplifies to $$ih \frac{\partial p}{\partial t} = -ih\frac{\partial V}{\partial x}$$
but this just results in ## \frac{\partial p}{\partial t} = -ih\frac{\partial V}{\partial x}## and I'm not sure where the limit of the Planck's constant was even used. Can anyone point out my mistake or help me understand?
 
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  • #2
offscene said:
I recently saw an explanation f

where?
 
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  • #3
offscene said:
but this just results in ## \frac{\partial p}{\partial t} = -ih\frac{\partial V}{\partial x}##
No, it results in ## \frac{\partial p}{\partial t} = -\frac{\partial V}{\partial x}##.
 
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  • #4
Classical Poisson bracket { } , https://en.wikipedia.org/wiki/Poisson_bracket, corresponds with quantum commutator [ ] with
[tex] \frac{[\ \ ]}{i\hbar} \rightarrow \{\ \ \}[/tex]
in classical limit. ##\frac{\partial V}{\partial x}## comes from classical Poisson bracket.
 
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  • #5
The equation you attempted to derive is actually
$$\frac{d\hat{p}}{dt}=F(\hat{x})$$
where
$$F(x)=-\frac{\partial V(x)}{\partial x}$$
Note that ##\hat{p}## and ##\hat{x}## are quantum operators, so the first equation is not yet classical. To get something classical-like you have to take the quantum average of it
$$\frac{d\langle\psi|\hat{p}|\psi\rangle}{dt}=\langle\psi|F(\hat{x})|\psi\rangle$$
which is called the Ehrenfest theorem. But this is still not the classical equation. The classical equation is obtained if the right-hand side can be approximated as
$$\langle\psi|F(\hat{x})|\psi\rangle \approx F(\langle\psi|\hat{x}|\psi\rangle)$$
It is this last approximation that requires the small ##\hbar## limit, which I leave as an exercise for you. (Hint: Assume that ##\langle x|\psi(t)\rangle=\psi(x,t)## is a narrow wave packet, thus resembling a classical particle with well defined position.)
 
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  • #6
offscene said:
TL;DR Summary: I recently saw an explanation for how quantum mechanics approaches classical mechanics at the limit of Planck's constant becoming 0 using the Heisenberg equation of motion but am confused about what it is about this limit that reduces the equation of motion to its classical limit.

Starting from the Heisenberg equation of motion, we have

$$ih \frac{\partial p}{\partial t} = [p, H]$$
which simplifies to $$ih \frac{\partial p}{\partial t} = -ih\frac{\partial V}{\partial x}$$
but this just results in ## \frac{\partial p}{\partial t} = -ih\frac{\partial V}{\partial x}## and I'm not sure where the limit of the Planck's constant was even used. Can anyone point out my mistake or help me understand?
I wrote an insight about it. I believe I called it: The classical limit of commutators, or something similar
 
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