- #1
hyksos
- 37
- 12
- TL;DR Summary
- Path Integral relationship to Principle of Least Action
Several weeks ago I had considered the question as to how one can start from the Schroedinger Equation, and after several transformations, derive F=ma as a limiting case. At some point in my investigations of this derivation, it occurred to me that this is simply too much work. While in principle, this derivation is entirely possible, it is very long and has to be to performed in stages. I concluded this is too much to tackle in one go.
(in rough terms, you would go from from Schroedinger Eq --> Path Integral --> Least Action --> Euler-Lagrange Eq --> Lagrangian --> ##\left(\frac{\delta \mathcal{L}}{\delta q} \ =\frac{d}{dt} mv\right)## --> ##F=ma## )
Instead of tackling this monumental task in one swoop, I have decided to make a thread here on just one stage. I want to see how to derive the classical form of the principle of Least Action by starting from the Path Integral. (Path Integral --> Least Action) At first we notice that PI does not imply POLA directly. Instead, POLA is what falls out when you take a limit of PI. Thus POLA acts as a "limiting case" of the more general PI.
I found the following form of PI, and it contains h-bar in several places. Although I disagree with the use of L(q,qdot) since that could cause problems.
$$\bra{q_{F} \ }\ket{e^{-i\hat{H} t/\hbar } |q_{I}} =\int _{\begin{array}{ c } q( t) =q_{F}\\
q( 0) =q_{I}\end{array}} Dq\ \ exp\left[ \ \ \ \frac{i}{\hbar }\int ^{t}_{0} dt^{\prime } L( q,\dot{q} \ )\right]$$
From there we want to take the limit as ##\hbar \ \rightarrow 0## . The justification is that for large objects like trees and basketballs, ##\hbar## can be approximated as if it were ##\hbar## = zero when considering their aggregate number of constituent particles. The goal is to reach this equality
$$\delta \int ^{t_{2}}_{t_{1}} L( q,\dot{q} ,t) dt\ =\ 0$$
The best response is to link to this derivation already completed by someone else , either somewhere on the PhysicsForums or offsite. Thanks.
(in rough terms, you would go from from Schroedinger Eq --> Path Integral --> Least Action --> Euler-Lagrange Eq --> Lagrangian --> ##\left(\frac{\delta \mathcal{L}}{\delta q} \ =\frac{d}{dt} mv\right)## --> ##F=ma## )
Instead of tackling this monumental task in one swoop, I have decided to make a thread here on just one stage. I want to see how to derive the classical form of the principle of Least Action by starting from the Path Integral. (Path Integral --> Least Action) At first we notice that PI does not imply POLA directly. Instead, POLA is what falls out when you take a limit of PI. Thus POLA acts as a "limiting case" of the more general PI.
I found the following form of PI, and it contains h-bar in several places. Although I disagree with the use of L(q,qdot) since that could cause problems.
$$\bra{q_{F} \ }\ket{e^{-i\hat{H} t/\hbar } |q_{I}} =\int _{\begin{array}{ c } q( t) =q_{F}\\
q( 0) =q_{I}\end{array}} Dq\ \ exp\left[ \ \ \ \frac{i}{\hbar }\int ^{t}_{0} dt^{\prime } L( q,\dot{q} \ )\right]$$
From there we want to take the limit as ##\hbar \ \rightarrow 0## . The justification is that for large objects like trees and basketballs, ##\hbar## can be approximated as if it were ##\hbar## = zero when considering their aggregate number of constituent particles. The goal is to reach this equality
$$\delta \int ^{t_{2}}_{t_{1}} L( q,\dot{q} ,t) dt\ =\ 0$$
The best response is to link to this derivation already completed by someone else , either somewhere on the PhysicsForums or offsite. Thanks.