From Quantum Mechanics to Classical Mechanics [duplicate]

Question

Is it possible, and has it been attempted, to use quantum mechanics to deduce Newtonian, macroscopic level mechanics laws as was the case of statistical mechanics deriving thermodynamic relations?

Answer 1

Sure there was ! Just like we deduce the laws of Newton from relativity.

There is a famous theorem in quantum mechanics named Ehrenfest's theorem, which states that quantum mechanical expectation-values follow classical laws. So after averaging out the quantum-behaviour you just get classical mechanics.

For the correspondence with the classical mechanics one usually takes the limit $\hbar\rightarrow0$ (on which I will elaborate more further).

To make the link with the classical world you are free to choose the formalism in which you work. In Feynman's formalism of quantum mechanics (the path-integral way) this correspondence becomes fairly easy. Feynman states that the amplitude for a (quantum mechanical) particle to go from a state $|x_at_a\rangle$ to a state $|x_bt_b\rangle$ is given by summing over all possible paths between the begin- and endstate:$$\langle x_bt_b|x_at_a\rangle=\int\mathcal{D}x\exp\left(\frac{i}{\hbar}S\right),$$ where $S$ is the action that you know from classical mechanics. In the limit $\hbar\rightarrow0$ (or if you take really big actions compared to $\hbar$ which is always the case in the classical world), then the value of $$\exp\left(\frac{i}{\hbar}S\right)$$ will be stationary when $\delta S=0$ (the action is minimal), and will oscillate quite hard (nearby paths will give destructive interference) when you deviate from this extremum. So to sum it up, in the limit of large actions (or $\hbar\rightarrow 0$) you get back your Lagrangian formalism.

You could also work back from the operator-quantum mechanics, there you see that in the limit $\hbar\rightarrow0$ the commutation $$[\hat{x},\hat{p}]=i\hbar\rightarrow 0$$ which should classically hold (in a classical world all your values commute since they are numbers).

Now to give some information on the second link given bij @Qmechanic: The logic in quantum mechanics is given by Hilbert-space logic: your states are given by wavefunctions which are part of a complex vectorspace (you can have superpositions of states) named $L^2$, while the logic of classical mechanics is given by set-logic: your state represents a single point in phase-space, so a state corresponds to a SINGLE point. So when you take the limit $\hbar\rightarrow 0$ you need to be aware of this fact because simply taking the limit $\hbar\rightarrow 0$ is maybe a bit crude to say

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Extra method: Wignerfunctions The fact that you have an uncertainty principle also gives you the Heisenberg-theorem which states that $$\Delta x\Delta p = \frac{\hbar}{2}.$$This gives that you can't do quantum mechanics in the phase-space (since you can't pinpoint a quantummechanical state to one point). One way that makes this possible is to represent the quantummechanical state with a quasi-probatility, the most used one is the Wigner-function. These Wignerfunctions have a finite width in the phase-space due to the Heisenberg uncertainty. Now it seems that the minimal height goes as $\sim 1/\hbar$ and the minimal width goes as $\sim \hbar$. So in the limiet $\hbar\rightarrow 0$ you get delta-functions and hence exact localized states in phase-space (which is of course what you classically have).

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Derivation of Newton's second law The easiest way to get Newton's second law out of the above is by using Feynman's formalism, taking the limit $\hbar\rightarrow 0$ we argued that the only path's that contributed were given by the paths for which $\delta S=0$ (the action-functional has an extremum). The action-functional is given by: $$S=\int\limits_{t_a}^{t_b}L[x,\dot{x},t]\text{d}t,$$ saying that this action is extremal gives the Euler-Lagrange equations of motion:$$\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}.$$ The Lagrangian itself is defined by (you could derive it's form from the Newton-symmetries and the fact that you should retrieve Newton): $$L[x,\dot{x},t]=T-V,$$ where $T$ is the kinetic energy and $V$ is the potential energy. If we now take a non-relativistic point-particle this will give us that: $$L=\frac{m\dot{x}^2}{2}-V(x),$$ with $V(x)$ a portential of your choice. Applying the Euler-Lagrange equations yields: $$m\ddot{x}=-\frac{\partial V}{\partial x},$$ which is exactly Newton's second law since the force is given by ''minus the gradient of the potential''. Filling in for example $V=mgx$ (nog $x$ is the height) we get:$$m\ddot{x}=-mg,$$ which gives a free falling body.

Answer 2

Sure. In the limit $\hbar \to 0$ the Feynman path integral reduces to classical mechanics. You can also take the WKB approximation, which when $\hbar \to 0$ gives classical mechanics in the form of the Hamilton-Jacobi equation.