# What is Hamiltonian action?

## Main Question or Discussion Point

$$W = \int_{t_0}^t \mathcal{L}\,dt$$

Apparently Schrodinger used it along with the Hamilton-Jacobi equation to derive the Schrodinger equation so it's a pretty important part of quantum physics history.

According to wikipedia it's a function which takes the trajectory of a system and returns a real number. But this seems quite vague to me, what is the significance of the scalar it outputs?

I've also looked in several books, but the mathematics of analytical mechanics is mind-boggling, I don't really have the time right now to fully digest the whole of the subject. The only definitions I've gotten from various book seem to be very vague and mathematical.

Thanks :)

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it is energy weighted with time

Hamilton's principle of least action states that a mechanical system will behave in such a way as to minimize this action integral. The Lagrangian L that is integrated is T-V, the kinetic energy of the system minus the potential energy of the system.
Starting from this, all of classical mechanics can be derived very elegantly (as an alternative to starting with Newton's laws), so it's not just important in quantum mechanics but also in classical mechanics.

In practice, you don't really need to work with the action but using the Euler-Lagrange equations $$\frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = 0$$ (with the dot representing the derivative with respect to time) you can derive the equations of motion for the system. The Euler-Lagrange equations are a consequence of the least action principle, which you will learn when you study the calculus of variations.

You don't really need an intuitive idea for what this action is, all that is important is that it automagically assumes a minimum, and that once you know the Lagrangian (kinetic and potential energy of your system), you can derive the equations of motion without messing around with forces how you are used to from newtonian mechanics.

For example a particle moving vertically near the Earth's surface has the kinetic energy $$T = \frac{m}{2} \dot{z}^2$$, a potential energy $$V = mgz$$ and therefore a Lagrangian $$L = \frac{m}{2} \dot{z}^2 - mgz$$. In this case, $$\frac{\partial L}{\partial z} = -mg, \frac{\partial L}{\partial \dot{z}} = m \dot{z}$$, therefore $$mg + m\ddot{z} = 0$$ or $$\ddot{z} = -g$$, which of course you knew already, but the point is that this works for much more complicated systems too.

Hope this helps, if not I guess you'll have to wait until you take a class on classical mechanics again.

Thanks niklaus, that helped a lot

it still not says anything what the action "is"...

also niklaus post can be read at wikipedia basically

Well, if the hamiltonian action only really has an abstract meaning then I guess I'll have to settle with that

dx
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A deeper understanding of the action is provided by quantum mechanics: the action for a process is the phase of the quantum mechanical amplitude of that process.

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