# What's the physical intuition for symplectic structures?

I always thought about symplectic forms as elements of areas in little subspaces because of the Darboux theorem, however I cannot get the physical intuition for it and for the hamiltonian vector field.

To simplify things, let's consider the configuration space $TQ$, we know that $T^{*}Q$ always have a symplectic structure by putting $\omega = d\theta$ where $\theta = \sum p_i dq_i$ is the Liouville one-form, then the hamiltonian vector field is defined by $\omega(X_H, Y) = dH(Y)$ and I can change from the Lagrangian $L : TQ \longrightarrow \mathbb{R}$ to the hamiltonian $H :T^{*}Q \longrightarrow \mathbb{R}$ by the mass (1, 1)-tensor $M=M_i^j$. So what's the physical intuition for $\omega$, $X_H$ and $\theta$? Why do people uses a symplectic structure in mechanics (if it's to define $X_H$, what's the utility of $X_H$?)? Furthermore is the unique utility in changing the lagrangian to the hamiltonian the existence of a symplectic form in $T^{*}Q$?

-
taking the bird's eye view, dynamics are given by a vector field on some manifold (it's just the infinitesimal description of the flow); if we want to derive that vector field from a potential, we need additional structure as the natural differential operator $d$ yields a 1-form –  Christoph Dec 4 '13 at 8:05
–  Qmechanic Feb 17 '14 at 16:21

If you consider the phase space (the space of initial data) ${\cal{M}}$ of a classical system it can be seen as the cotangent bundle $T^{*}Q$ of the configuration space $Q$.

As you say this bundle has a natural symplectic structure $\omega:T{\cal{M}}\times T{\cal{M}}\rightarrow \mathbb{R}$. Now given a Hamiltonian $H:{\cal{M}}\rightarrow \mathbb{R}$ as you say using the inverse of the symplectic structure we can obtain the Hamiltonian vector field $X_{H}=\omega^{-1}(dH,.):T^{*}{\cal{M}}\rightarrow \mathbb{R}$.

Let's consider now coordinates $(q_{1},..,q_{n})$ in $Q$. This set of coordinates give rise to a natural set of coordinates $(q_{1},..,q_{n};p_{1},..,p_{n})$ on ${\cal{M}}$ by taking $(p_{1},..,p_{n})$ to be the components of the cotangent vectors in the coordinate basis associated with $(q_{1},..,q_{n})$.

The symplectic form then takes the form $\omega=\sum_{\mu}dp_{\mu}\wedge dq_{\mu}$ and the inverse takes the form $\omega^{-1}=\sum_{\mu}(\frac{\partial}{\partial q}_{\mu})\otimes (\frac{\partial}{\partial p}_{\mu})-(\frac{\partial}{\partial p}_{\mu})\otimes (\frac{\partial}{\partial q}_{\mu})$.

Then the hamiltonian vector field is denoted by:$X_{h}=\sum_{\mu}(\frac{\partial H}{\partial q}_{\mu})\otimes (\frac{\partial}{\partial p}_{\mu})-(\frac{\partial H}{\partial p}_{\mu})\otimes (\frac{\partial}{\partial q_{\mu}})$.

If you consider now an integral curve of this vector field which means that the curve $\alpha:\mathbb{R}\rightarrow {\cal{M}}$ satisfies $\frac{d \alpha}{dt}=X_{h}$

We obtain

$$\frac{dq_{\mu}}{dt}=\frac{\partial H}{\partial p_{\mu}}\\ \frac{dp_{\mu}}{dt}=-\frac{\partial H}{\partial q_{\mu}}$$

which are Hamilton's equation.

Moreover, we can defined the poisson bracket of two classical observables as $\{f,g\}=\omega^{-1}(df,dg)$ which satisfies for the coordinates $\{q_{\mu},q_{\nu}\}=0,\{p_{\mu},p_{\nu}\}=0,\{q_{\mu},p_{\nu}\}=\delta_{\nu\mu}$. As you can see these relations are similar to the observables in QM. In fact, there are a lot of quantization procedures from classical theories where this is the starting point.

Finally you can define the classical action when the Hamiltonian doesn't depend on time as $S=\int\theta$ with the integral understood to be taken over the manifold defined by holding the energy $E$ constant: $H=E=$const.

Here are two pictures that might help from Roger Penrose's The Road of reality:

The curves that have as tangent vectors the Hamiltonian flow are the solutions to the equations of motion of the system.

-
Thanks for the answer, but it contains almost everything that I already know (except the action you have defined at the end). Maybe I´m being too exigent. Anyway, by your answer it´s not clear if the existence of a symplectic form in $T^{*}Q$ is the unique utility in changing the lagrangian to the hamiltonian. Why Hamiltonian mechanics? What´s the problem with the Lagrangian mechanics (almost everything in quantum field theory is defined by a Lagrangian)? –  user40276 Apr 28 '14 at 12:48
The utility to change to Hamiltonian mechanics is the existence of a Poisson structure which uses the symplectic structure . As I comment, this is the starting point for quantization procedures. There is no problem with Lagrangian mechanics, but also there is no problem with Hamiltonian mechanics. Note also that by doubling the variables to $(q,p)$ the equations are now first order so for example to solve specific problems you might use techniques from semigroup theory. At the same time, the Hamiltonian has a direct physical interpretation as the energy (under certain conditions). –  yess Apr 28 '14 at 13:11
Having said that, I must confess actually there are some problem in the Hamiltonian approach as stated here philsci-archive.pitt.edu/4916 also consider the answer to physics.stackexchange.com/q/89035 –  yess Apr 28 '14 at 13:30