Say one has a mechanical system with hamiltonian $H$, and two other arbitrary observables $f,g$. $H$ is super useful since $\{H, \cdot\} = \frac{d}{dt}$. But does $\{f,g\}$ give any useful information in and of itself?

I'm currently going through "Lectures on Quantum Mechanics for Mathematics Students" by Faddeev and Yakubovskii (with not terribly much background in classical physics).


Well, $\{f, \cdot \}$, similarly to $\{H,\cdot\}$, computes the derivative of the argument $\cdot$ with respect to the action of the one-parameter group of canonical transformations generated by $f$ (see the note below for the complete definition) $$\phi_a^{(f)} : F \to F\:,\quad a \in \mathbb R\:,$$ satisfying $$\phi_a^{(f)} \circ \phi_b^{(f)}= \phi_{a+b}^{(f)}\:, \quad\phi_{-a}^{(f)} = (\phi_a^{(f)})^{-1} \:, \quad \phi_0^{(f)}= id$$ Here $F$ is the space of phases. Indeed it holds (see below) $$\{f,g\}(x)= \frac{d}{da}|_{a=0} g(\phi_a^{(f)}(x))\:,\tag{1}$$ where $g: F \to \mathbb R$ is sufficiently regular.

Therefore, $\{f,g\}(x)=0$ everywhere in $F$ means that $g$ is invariant under the group of transformations generated by $f$ (the fact that the derivative is computed at $a=0$ is immaterial, as the group structure implies that the derivative vanishes for every value of $a$).

In particular $\{f,H\}=0$ means that the Hamiltonian function is invariant under the action generated by $f$. This fact is remarkable because it gives rise to the Hamiltonian version of Noether theorem.

As a matter of fact, since $\{H,f\}=- \{f,H\}=0$, invariance of $H$ under the action of $f$ is equivalent to the invariance of $f$ under the action of $H$ (i.e. under time evolution). In other words,

$H$ is invariant under the action of the one-parameter group of canonical transformations generated by $f : F \to \mathbb R$ if and only if $f$ is constant along the motion of the physical system.

Finally, let $X_h$ be the vector field over $F$ tangent to the orbits of the curves $\mathbb R \ni a \mapsto \phi_a^{(h)}(x)$ for every $x\in F$ (this vector field is fully defined in the note below). Since $$[X_f,X_g]=X_{\{f,g\}} \tag{1'}\:,$$ $\{H,f\}=0$ implies that, if $t \mapsto x(t)$ solves Hamilton equations, $t \mapsto \phi^{(f)}_a(x(t))$ does for every value of $a$. In other words, $\{H,f\} =0$ also implies that the group of canonical transformations generated by $f$ transforms motions of the physical system to motions of the system as well.

(a) $\{f,g\}=0$

implies, via (1') and using $X_0=0$, that

(b) the action of the group of transformations on the states of the system (points in $F$) and on observables (real valued functions on $F$) generated by $f$ and the one generated by $g$ commute.

Since $X_h=X_l$ if and only if $h=l + const.$, the two statements (a) and (b) are not completely equivalent. This non-equivalence turns out to be fundamental in quantization procedures since it permits to deal with CCR and central extensions of groups.

NOTE regarding used definitions

[1] if $\omega$ is the symplectic form on $F$, the Hamiltonian field associated to $f\in C^\infty(F,\mathbb R)$ is defined as the unique vector field, $X_f$, such that $$\omega_x(X_f,u)= \langle df_x, u\rangle \tag{2}$$ for every vector $u \in T_xF$. $X_f$ is uniquely defined this way since $\omega$ is non-degenerate by definition.

[2] The one-parameter group of canonical diffeomorphisms $\phi^{(f)}$ generated by $f$ is properly defined as follows. $$\mathbb R \ni a \mapsto \phi_a^{(f)}(x) =: y_x(a)\in F \tag{3}\:,\quad \forall x \in F$$ where $y_x$ is the unique (maximal) solution of the Cauchy problem $$\frac{dy}{da} = X_f(y(a))\:, \quad y(0) =x \tag{4}$$ (I am assuming that the solution is complete, as it happens if $f$ is compactly supported of $F$ itself is compact, otherwise some subtleties regrading domains are to be fixed and $\phi_a^{(f)}(x)$ is only locally defined in the variable $a$.)

[3] The Poisson bracket is defined as $$\{f,g\}:= \omega(X_f,X_g) \quad f,g \in C^\infty(F,\mathbb R)\:.\tag{5}$$

With these definitions, (3) and (4) imply, as asserted in the main text, that $X_f$ is tangent to the curves $\mathbb R \ni a \mapsto \phi_a^{(f)}(x)$. Next (4) and (5) easily produce (1). An explicit expression of the action of $\phi^{(f)}$ on a function $g : F \to \mathbb R$, $$\left(\phi^{(f)*}_{a}[g]\right)(x):= g(\phi^{(f)}_{a}(x))$$ is provided by the formula $$\phi^{(f)*}_{a}[g] = \sum_{n=0}^{+\infty} \frac{a^n}{n!}\{f,\:\:\}^n g\:.$$ This identity holds if $f,g$ are real analytic and not only smooth.

It is finally worth stressing that the equations in (4) are nothing but the standard Hamilton equations if $f$ is indicated by $H$.

  • $\begingroup$ This is a nice answer. It would be even easier to understand with a diagram illustrating the vector flows. $\endgroup$ – DanielSank Oct 4 '15 at 19:43
  • $\begingroup$ Thanks. Unfortunately I am not able to draw diagrams in Latex! $\endgroup$ – Valter Moretti Oct 5 '15 at 7:06
  • $\begingroup$ You can attach images to Stack Exchange posts. $\endgroup$ – DanielSank Oct 5 '15 at 7:54
  • $\begingroup$ Well, drawing images is the problem, also I have no time to do it. $\endgroup$ – Valter Moretti Oct 5 '15 at 8:11
  • $\begingroup$ Nice edit - it's definitely a lot more readable and beginner-friendly. I'll leave the bounty up the full week as free advertisement ;). If you have image drafts (using e.g. MS Paint) I might be able to pimp them up to a more presentable form. $\endgroup$ – Emilio Pisanty Oct 5 '15 at 14:20

Yes it does. In fact, it is one of the( if not the) most important conclusions of Quantum mechanics. If {f,g}= 0 it means that the variables have simultaneous eigenvalues i.e. you can measure both of them on same instant of time. but {f,g} can be non-zero which leads to the theoretical conchusion that the eigenstates of f and g are not simutaneously measurable. This is the Hisenberg Uncertainty Principal's most general form.

for the mathematical treatment you may wanna see http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/GenUncertPrinciple.htm

  • 1
    $\begingroup$ But the question is about Poisson brackets in classical mechanics, not commutators in quantum mechanics... $\endgroup$ – Nathaniel Oct 6 '15 at 7:40

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