Classical Mechanics from Quantum Mechanics

In this article we will try to ‘produce’ classical mechanics from quantum mechanics.

Quantum Mechanics

We present a simplified framework for quantum mechanics.

A quantum system is simply a projective Hilbert space $P\mathcal{H}$. The (quantum) states are its elements. The symmetries of the system are (projectivised) unitary operators which are representations of a symmetry group $G$.

We will focus only on Lie group symmetries. Since it is cumbersome to take $U(1)$-central extensions into account we will assume also that the symmetry groups are finite dimensional, connected, simply connected, with trivial second group cohomology $H^2(G,U(1))$.


While traditionally observables are defined as self-adjoint operators on the Hilbert space $\mathcal{H}$ associated to $P\mathcal{H}$, we will take a philosophical stance that all observables are built from the generators of the symmetries of the quantum system. In our case, given a symmetry group $G$, observables are built from its Lie algebra $\text{Lie}(G)=\mathfrak{g}$.

Lie’s third theorem guarentees that the representation theory of $G$ and that of $\mathfrak{g}$ are the same. Given a representation $U(g)$ of $g\in G$ which is a unitary operator, we can write it as a product of $e^{tX}$’s, where $X\in\mathfrak{g}$. We only need to consider those generators $X\in\mathfrak{g}$’s of the Lie algebra and the Lie group elements that are in the one-parameter groups generated by those generators, namely $g=e^{tX}$.

Given $g=e^{tX}$, its representation can be written as $U(g) = e^{it H^X}$, where $H^X$ is the self-adjoint operator $H^X = \hat{U}(X)$ where $\hat{U}$ is the representation of the Lie algebra corresponding to $U$. These $H^X$ are the observables that we will be focusing on, since all other observables (in our definition) are built from them. We will call them primary observables, and the corresponding Lie algebra elements proto primary observables.

Evolution of Observables

Given a proto primary observable $X$, we have the one-parameter group generated by it, whose elements are $e^{sX}$. Correspondingly given a primary observable $H^X$, we have a one-parameter group of unitary operators $e^{is H^X}$.

The evolution of an observable $H^Y$ in the direction of $H^X$ is given by

$$ H^Y(k+s) = e^{-is H^X} H^Y(k) e^{is H^X} $$

from which we can derive the Heisenberg equation of evolution

$$ \frac{d}{ds} H^Y(s) = -i[H^X, H^Y]. $$

A special case is the Heisenberg equation of motion, where $H^X = H$ is the energy operator, i.e. the generator of time translation $H$. More concretely, if we take special relativity into account, it should be the representation of the Poincare algebra generator that corresponds to the energy component in a 4-momentum.

Classical Mechanics

We will start just with a smooth manifold $M$, the phase space, or in analogy to the quantum situation (classical) state space. We worked in the Heisenberg picture in quantum mechanics, hence here we will also work with a ‘Heisenberg picture’.

A classical symmetry is just a diffeomorphism $\rho: M\rightarrow M$. Symmetries form a group $G$, then we have symmetry group actions $\rho:G\times M\rightarrow M$.

A proto primary observable $X$ is a vector (or we can regard it as a vector field) in the tangent space $T_e G$. In the quantum situation we produced a primary observable $H^X$ from its unitary representation. Now we want to produce something from $X$, on the state space $M$, without losing much informations. A good choice would be the fundamental vector field $X^\sharp \in \Gamma(TM)$, defined as $X^{\sharp}_p = \frac{d}{ds} \rho(e^{sX},p)$, of $X$. We will call $X^\sharp$’s pre-observables.

Hamilton’s equation, version 1

The infinitsimal evolution of a pre-observable $Y^\sharp$ in the direction of $X^\sharp$ should be given by the Lie derivative,

$$ \mathcal{L}_{X^\sharp} Y^\sharp = [X^\sharp,Y^\sharp]. $$

Notice that this is formally similar to the Heisenberg equation of evolution

$$ \frac{d}{ds} H^Y(s) = i[H^Y,H^X]. $$

Also since taking fundamental vector fields $X\mapsto X^\sharp$ is a Lie algebra (anti-)homomorphism, we have

$$ \mathcal{L}_{X^\sharp}Y^\sharp = -[X,Y]^\sharp $$

However pre-observables are not (classical) observables, which should be smooth functions.

From pre-observables to observables

Maybe the simplest way to find a smooth function associated to a vector field over $M$ is by introducing a 2-form $\omega$. We expect that there is a function $H_X\in C^\infty(M)$, s.t.

$$ \iota_{X^\sharp}\omega = d H_X. $$

We see that $dH_X$ is exact, hence we expect $\iota_{X^\sharp}\omega$ to be exact. This in general cannot be true, hence we make some assumption on $X^\sharp$, and hence on the group action $\rho$ that this is that case. Or we may assume that $\iota_{X^\sharp}$ is closed, while $H^1(M,\mathbb{R})=0$, hence it is also exact.

$H_X$ are called observables.

The 2-form $\omega$ is a structure that we introduced on the state space $M$. The fundamental vector field generates symmetries on $M$, hence we should have for any $X\in\mathfrak{g}$, the associated $X^\sharp$ should satisfy

$$ \mathcal{L}_{X^\sharp} \omega =0 $$

Since by Cartan’s formula we have

$$ \mathcal{L}_{X^\sharp} \omega = d(\iota_{X^\sharp}\omega) + \iota_{X^\sharp} d\omega = \iota_{X^\sharp}d\omega $$

we should expect that $\omega$ is closed. This means we expect that $\omega$ is a presymplectic form. In this case $\rho$ is a Hamiltonian group action and the observable is simply a momentum (the component of the moment map $\mu$ on $X$) associated to the Hamiltonian group action.

If we further want there to be a 1-to-1 correspondence between $X^\sharp$ and $d H_{X}$, we assume that $\omega$ is non-degenerate, i.e. it is symplectic.

Two interesting Lie algebra homomorphisms

In the symplectic case, we have a symplectic manifold $(M,\omega)$ as the state space, and we for a Hamiltonian group action $\rho$, for any $X\in \mathfrak{g}$ we have a observable (momentum) $\mu^X$.

We make a change in our notation. Corresponding to an observable $H$, the fundamental vector field (or Hamiltonian vector field of the Hamiltonian function $H$) we write as $X^\sharp_H$. The sharp $\sharp$ is there to stress that $X^\sharp_H$ is a fundamental vector field. The corresponding Lie algebra element will be written as $X_H$.

Recall that the Hamilton’s equation, version 1, is simply the expression for the Lie derivative of a vector field $\mathcal{L}_{X^\sharp_f} X^\sharp_g = [X^\sharp_f,X^\sharp_g]$. Now for symplectic state space and Hamiltonian group actions, we have

$$ \mathcal{L}_{X^\sharp_f} X^\sharp_g = - [X^\sharp_f, X^\sharp_g] = X_{\omega(X^\sharp_f,X^\sharp_g)} = X^\sharp_{\{f,g\}} = -[X_f,X_g]^\sharp $$

where $\{f,g\}$ is the Poisson bracket associated to the symplectic form. We now have two Lie algebra homomorphisms $X^\sharp_f \mapsto f + \mathbb{R}$ and $X_f \mapsto X^\sharp_f$.

Hamilton’s equation, version 2

The evolution of the observable $f$ in the direction of $g$ would be given by

$$ \mathcal{L}_{X^\sharp_g} f = \{g,f\}. $$

which follows from an easy computation.


  1. Given a presymplectic manifold, if the $G$-action is Hamiltonian, i.e. admits a moment map, we have a ‘pairing’ between $X\in\mathfrak{g}$ and $f_X \in C^\infty(M)$ via $\iota_{X^\sharp} \omega = d f_X$. Note that $d\iota_{X^\sharp}\omega = \mathcal{L}_{X^\sharp}\omega - \iota_{X^\sharp}d\omega=0$, so $\iota_{X^\sharp} \omega$ is trvially closed.
  2. If the manifold is symplectic, the ‘pairing’ between $d f_X$ and $\iota_{X^\sharp} \omega$ is indeed a genuine pairing, i.e. it is 1-to-1.

Poisson framework

With Poisson structure introduced the ‘pairing’ of the Lie algebra $\mathfrak{g}$ of a Lie group $G$ acting on a manifold $M$ and the funcions on $M$ becomes explicit.

Poisson moment map and Lie algebra homomorphism

Given a manifold $M$ and a Lie group action $\rho: G\times M \rightarrow M$, we have obtained a Lie algebra homomorphism $\rho_\ast:\mathfrak{g}\rightarrow \Gamma(TM): X\mapsto X^\sharp$. This mapping is called the infinitesimal action of $\mathfrak{g}$ on $M$. Now there’s a theorem of Palais saying that from a homomorphism of this kind we have a unique group action.

Theorem(Palais) . Let $G$ be a simply connected Lie group with the Lie algebra $\mathfrak{g}$. Let $M$ be a smooth compact manifold. If there is a homomorphism of Lie algebras $\mu:\mathfrak{g}\rightarrow \Gamma(TM)$, then there is a unique action $\rho:G\times M\rightarrow M$ s.t. $\rho_\ast=\mu$.

Given a Poisson manifold $(M,\pi)$ where $\pi$ is the bivector field. We say that a symmetry group action is Poisson if we have a Poisson morphism, called a moment map $\mu:M\rightarrow\mathfrak{g}^\ast$. This means $$ \mu^\ast (\{f,g\}_{\mathfrak{g}^\ast}) = \{\mu^\ast(f),\mu^\ast(g)\}_{M}. $$

Associated to $\hat{\lambda}:\mathfrak{g}\rightarrow C^\infty(M)$ which is linear, there is a unique map $\lambda:M\rightarrow\mathfrak{g}^\ast$ s.t. $\langle\lambda(p),X\rangle = \hat{\lambda}(X)(p)$. It can be proved that the smooth map $\mu:M\rightarrow \mathfrak{g}^\ast$ is a moment map iff the associated map $\hat{\mu}:\mathfrak{g}\rightarrow C^{\infty}(M)$ is a Lie algebra homomorphism.

Now if we put $\chi:C^\infty(M)\rightarrow \Gamma(TM): f\mapsto X_f = \{f,\cdot\}$ (recall that $X_f$, if it can be seen as a fundamental vector field $X^\sharp_f$ that is Hamiltonian, when acting on a function $g$ is nothing but $\mathcal{L}_{X_f}$, up to a sign), this is a Lie algebra morphism. Taking composition we get $\chi\circ\hat{\mu}:\mathfrak{g} \rightarrow \Gamma(TM)$. The theorem of Palais ensures that we have an action $\rho: G\times M\rightarrow M$ with $\rho_\ast = -\chi\circ\hat{\mu}$ for $M$ compact and $G$ simply connected. Furthermore, $\rho$ is $G$-equivariant:

Created by potrace 1.16, written by Peter Selinger 2001-2019

This means all Poisson moment maps comes with a group action. Therefore we can define the observables to be the components of a moment map $\mu$ on Lie algebra elements $X\in\mathfrak{g}$,

$$ \mu^X = \langle \mu, X\rangle. $$

We see that observables are the evaluation of moment maps on proto pre-observables.

Summarizingly, if the Lie group $G$ is simply connected and the Poisson manifold $M$ is compact,

  1. Associated to a linear association $\hat{\mu}:\mathfrak{g}\rightarrow C^\infty(M)$ of proto pre-observables with functions, there is always a unique map $\mu: M\rightarrow \mathfrak{g}^\ast$.
  2. The map $\mu$ is a moment map iff $\hat{\mu}$ is a Lie algebra homomorphism.
  3. Every moment map comes with an equivariant group action.
  4. Observables are components of moment maps on proto pre-observables.
  5. Given a moment map, observables and pre-observables are related through $f\mapsto X_f = \{f,\cdot\}$. Given a moment map $\mu$, an $X_f$ can be seen as a fundamental vector field if $f$ is in the image of the associated homomorphism $\hat{\mu}:\mathfrak{g} \rightarrow C^\infty(M)$.

Co-moment map and Gelfand transform

The Lie algebra morphism $\hat{\mu}:\mathfrak{g}\rightarrow C^\infty(M)$ is sometimes called co-mment map. The existence of the moment map $\mu:M\rightarrow\mathfrak{g}^\ast$ allows the characterization of $M$ as a symplectic realization of the Poisson manifold $\mathfrak{g}^\ast$ endowed with the Lie-Poisson structure.

The co-moment map $\hat{\mu}$ is a sort of Gelfand transform. The algebra $\mathfrak{g}$ is relized through $\hat{\mu}(\mathfrak{g})$ as an algebra of observables $C^\infty_\mathfrak{g}(M)$.

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