I know that phase space is the Hamiltonian description of a system, where we deal with position and momentum in equal footing. My question is in this phase space are those position and momentum are basis for that system?

As far as I know they are independent, in Hamiltonian dynamics, but how can I say that they are orthogonal basis functions? As alwayas we draw position and momentum line perpendicularly!

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    $\begingroup$ What do you mean by basis for the system? What quantities are you writing out as linear superpositions of the generalized positions and momenta? $\endgroup$ – Aaron Stevens Jan 3 '19 at 4:45
  • $\begingroup$ Related: physics.stackexchange.com/q/450479/2451 $\endgroup$ – Qmechanic Jan 3 '19 at 8:11

If the question is if the Hamiltonian phase space has the structure of a vector space equipped with a scalar product, the answer is negative, in general.

It is true that generalized coordinates and momenta of a system with $n$ spatial degrees of freedom are locally represented by $2 n$ real numbers and $\mathbb R^{2 n}$ can be seen as a $2 n$ dimensional vector space. But the impossibility, in general, of a global mapping of the phase space on $\mathbb R^{2 n}$ prevents the possibility of identifying coordinates and momenta as a vector.

As a simple example, which shows why there is such limitation, is the phase space of a rigid body in 3D. Coordinates represent three independent angles. However the set of rotation is not a vector space because in general $3D$ finite rotations around different axes do not commute.

It turns out that the most natural structure of the Hamiltonian phase space is that of a symplectic differentiable manifold. Which implies that, besides the failure of a general identification with a vector space, even at local level, the most important property of Hamiltonian coordinates is not related to the concept of angle and scalar product, but to the concept of local volume.

  • $\begingroup$ It might be worth pointing out that its even easier to see that the global phase space is not a vector space simply because the base topological manifold in this case is not homeomorphic to $R^n$. Its fairly easy to see that if one of the coordinates is an angle it must "wrap around" in a way that $R^n$ does not. $\endgroup$ – jacob1729 Jan 3 '19 at 12:57

I'm afraid you're making a soup of concepts which have nothing to do with one another. My suspicion derives e.g. from your use of words like "orthogonal" and "basis", which I would see better in context about QM.

As to drawing "position and momentum line perpendicularly" you're attributing weight to an innocent practice: if I have a 2D space, it's usual to draw a map of its coordinates in a Cartesian plane. But you shouldn't give the drawing more properties than it's meant to have, e.g. a euclidean structure (orthogonality).

What has instead a meaning in that plane is the area enclosed by a closed curve (the integral $\oint\!p\,dq$). Maybe you don't think of an area without a euclidean structure, but this is actually possible. And that area is interesting because it's invariant under canonical transformations. I assume you'll see that going on with Hamiltonian mechanics.


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