Is the phase space in classical mechanics a linear vector space (LVS)? If yes,, can we define operators, inner products in this space?

Edit: I have seen Liouville operator $\mathcal{L}$ in Classical mechanics defined as $$i\mathcal{L}\equiv \dot{q}\frac{\partial}{\partial q}+\dot{p}\frac{\partial}{\partial p}\tag{1}$$ which act on functions $f(q,p)$ of phase space variables $(q,p)$. One also defines inner product on phase space as $$(f,g)=\int dq\int dp f^*(q,p)g(q,p).\tag{2}$$ Using (2) one also proves the Hermiticity of $\mathcal{L}$ in Classical mechanics. For a link see this.

Since this is analogous to the mathematical operators in a Linear vector space, I wonder whether a Phase space is a LVS.

  1. A phase space is not necessarily a linear vector space or an affine space. More generally, it is a Poisson manifold or a symplectic manifold. It is often not possible to globally assign a linear or affine structure to a manifold.

  2. In quantization, one often constructs a Hilbert space ${\cal H}=L^2(X,\Sigma,\mu)$ as a $L^2$-space over a measure space $(X,\Sigma,\mu)$. E.g. a symplectic manifold $(M,\omega)$ comes equipped with a canonical volume form $\Omega=\omega^{\wedge n}$ (cf. e.g. this Phys.SE post), and is hence a measure space. The Hilbert space ${\cal H}=L^2(X,\Sigma,\mu)$ is indeed a linear vector space. And it is possible to consider linear operators thereon.

  • $\begingroup$ 1. Can operators be defined on a Poisson manifold? 2. Is a LVS a special case of a manifold? For example, the link in the EDIT part defines Liouville operator in Classical mechanics. $\endgroup$ – SRS Jan 2 '17 at 11:30
  • $\begingroup$ 1. Well, for starters, operators do not need to be linear, cf. my Phys.SE answer here. 2. A finite-dimensional linear vector space is a manifold. $\endgroup$ – Qmechanic Jan 2 '17 at 12:55
  • $\begingroup$ @Qmechanic $L^2(M)$ needs a positive measure to be defined. On a generic manifold there is not. In the considered case however, if the Poisson structure over the $2n$-dimensional manifold $M$ is induced by a symplectic form $\omega$, there is the canonical measure $\mu = \omega \wedge \cdots (n \: times ) \wedge\cdots \omega$. With this measure, the scalar product of $L^2(M)$ is invariant under canonical transformations $\endgroup$ – Valter Moretti Jan 2 '17 at 16:11
  • $\begingroup$ @Valter Moretti: Good points. I updated the answer. $\endgroup$ – Qmechanic Jan 2 '17 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.