# Is the phase space a linear vector space?

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 Answer

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.

• 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. – SRS Jan 2 '17 at 11:30
• 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. – Qmechanic Jan 2 '17 at 12:55
• @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 – Valter Moretti Jan 2 '17 at 16:11
• @Valter Moretti: Good points. I updated the answer. – Qmechanic Jan 2 '17 at 18:34