# Which quantum system evolves with maximum action instead of with minimum action?

+ 1 like - 0 dislike
159 views

The action is stationary in every evolution; in particular, it can be extremal (maximal or minimal). And indeed, in classical physics, various systems exist that evolve with maximum action.

In quantum physics, is there any system that evolves using maximal action? The examples of quantum systems in textbooks - atoms, photons, spins, etc. - all seem to be examples of minimal action. Is there a quantum system that evolves along maximal action?

edited Nov 15, 2017

A problem in your formulation of the question is that in the quantum case, the action is no longer stationary since quantum deviations arise; so strictly speaking your question cannot be answered. Are you asking for a quantum system whose classical equivalent maximizes the action at its solutions?

Hm, the principle of least action is valid also in quantum mechanics - isn't it? Of course the initial condition is a wave function, but why can the question not be answered?

The real problem of this question is that the least action principle is not about calculating the action itself, but about deriving equations of motion. Thus, once the equation solutions are found, nobody calculates the action, be it classical or quantum mechanical. Nobody calculates it with the equation solutions and nobody compares it with the action with "virtual solutions".

Often those equations of motion were first obtained ("guessed" in Feynman's terminology) from some experimental data and new ideas and only after that from the "least action principle".

An artificial character of the latter is seen in specifying the known initial and final "positions" $x(t_1)$ and $x(t_2)$ although in physics the equations are only supplied with the initial data $x(t_0)$ and $\dot{x}(t_0)$ and the future is calculated from them.

+ 1 like - 0 dislike

In fact, the principle of least action in practice means a stationary action rather than the least one. So it may well happen (in Classical Mechanics) that the action is not minimal, but stationary or maximal, whatever. Anyway, the classical action $S (t_1,t_2)$ is a functional of unknown variables $\tilde{x}_i$, not a functional of equation solutions $x_i (t)$.

Now, we can calculate the action as a function of time $t$ with substituting the known solutions in the integral: $S (t_1,t)$. Its numerical value is of no use in CM.

Now let us turn to QM in the Feynman formulation via path integral over all paths between two fixed points $X_1$ and $X_2$. It is known that the maximal contribution to the transition amplitude is brought with the paths close to the classical trajectory (if any). If the QM system is in a coherent state (see applets with wave packet motions à la Schroedinger), the paths far from the classical ones contribute too little to the amplitude (the action is too big and the integrand oscillations cancel each other). But if the system itself is far from a coherent state, say, it is in the ground or a stationary state (a standing wave), then all paths contribute nearly equally, I would say. The actions are big for them. In this sense the system is more quantum than classical.

answered Nov 14, 2017 by (42 points)

The action is neither a functional of solutions nor of the variables but of paths.

Your contribution does not answer the original question - to give a quantum system whose classical equivalent maximizes the action at its solutions.

@ArnoldNeumaier: Paths of what? I denoted the particle coordinate on paths as $\tilde{x}_i$; they also called virtual trajectories (or virtual paths, if you like). Anyway, they are subject to variations in the least action principle in order to obtain some restrictions on them - equations of motion.

I answered that in QM there are states "very far from classical" or from quasi-classical trajectories described with coherent states. A standing wave does not describe any motion with a certain value of momentum itself and with a zero value of the momentum mean value. All paths contribute nearly equally to the transition amplitude; the corresponding actions are nearly equally "big". This is what came to my mind. If you have something else to add, please, do it.

A path is a mathematical concept, namely a continuous function of a single variable. Paths don't need to be paths of something.

If the action is that of a single particle in 3D then the action is a functional of a path in 3D, namely of the whole virtual trajectory $x(t)$ of the particle in question, where $t$ ranges over the time interval $[t_1,t_2]$ over which the action is integrated.

@ArnoldNeumaier: I just denoted the virtual trajectories as $\tilde{x}$ in order to distinguish them from the unique solution of equations $x(t)$. My notation implies the whole path, as you insist. I do not invent anything new here.

But your text says: ''the classical action is a functional of unknown variables, not a functional of equation solutions'', and I complained that in place of variables it should say paths or virtuual paths or virtual trajectories.

By the way, everyone writes the virtual paths arising as arguments in the action as $x(t)$ or $q(t)$ and not with wiggles. The solutions are just the stationary paths, hence need no different notation.

@ArnoldNeumaier: My notation $\tilde{x}$ implies the whole virtual trajectory as a "variable". Such a trajectory may be very far from the real trajectory.

Normally $x(t)$ denotes the real trajectory and the variations are made around it: $\tilde{x}=x(t)+\delta x(t)$.

In QM many virtual paths are involved on the equal footing.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.