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In Hamiltonian mechanics, one can utilize Poisson brackets $\{ ·,·\}$ to express some manner of connection between two quantities. For example, for generalized coordinates $q_i$ and their corresponding canonical momentum $p_i$ one can write: \begin{equation} \{ q_i, p_j \}_{q,p} = \sum_k^{N} \left(\frac{\partial q_i}{\partial q_k} \frac{\partial p_j}{\partial p_k} - \frac{\partial p_j}{\partial q_k} \frac{\partial q_i}{\partial p_k} \right) = \delta_{ij} \end{equation} Similarily, $\{q_i, q_j\}=0$ as well as $\{p_i, p_j\}=0$. So far it makes sense, but what confuses me is the connection between the Poisson bracket between two quantities and the associated commutator of the same quantities. As can be shown, the commutator can be written;

\begin{equation} \left[F, G\right] = \lambda \{ F, G\} \end{equation} For some constant $\lambda$, so \begin{equation} \left[q_i, p_j \right] = \lambda \delta_{ij} \end{equation} To me, this seems to imply that hamiltonian mechanics allows for a generalized coordinate $q_i$ and its corresponding momentum to not commute, i.e. $q_i p_i \neq p_i q_i$, and whereas this makes sense in the context of quantum mechanics where $\lambda = i \hbar$ with operators $\hat{x}$ and $\hat{p}$, I struggle to make sense of this in classical physics. Thus, I have several questions:

  1. Is it so that $\lambda = 0$ in all of classical physics?
  2. If not, what is even meant by $q_i p_i$ vs $p_i q_i$? Are the generalized coordinates/canonical momenta to be interpreted as position/momentum operators as in QM, and if so then how would one express them (matrices, derivatives, etc)?
  3. If they do not commute, then can we follow the same derivation as in QM and derive a general uncertainty principle between these two quantities? What would such an uncertainty principle mean in the context of classical physics?

Apologies if these are stupid questions, I just find it quite confusing as it seems like one should be able to draw very clear parallels to quantum mechanics if $\lambda \neq 0$, but allowing such implies absurdities in classical mechanics.

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    $\begingroup$ It is not true in general that the quantum-mechanical commutator $[F,G]$ is always just the classical Poisson bracket $\{F, G\}$ times $\lambda=i\hbar$. There may also be further corrections of $\mathcal{O}(\hbar^{2})$, because of operator ordering ambiguities. Note, for example, that $\{x,xp\}=\{x,px\}$, but $[x,xp]\neq[x,px]$. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ In quantum physics the commutator of two operators is defined simply in terms of their product as $[\hat A,\hat B]:=\hat A\hat B-\hat B\hat A$. Thus the statement $[\hat A, \hat B] \neq 0$ is equivalent to $\hat A\hat B \neq \hat B\hat A$. In classical physics the Poisson bracket of two functions is defined by the expression given in the question, but this is not directly related to "multiplication"; even if $\{F,G\}\neq0$, where $F=F(q_i,p_i)$, $G=G(q_i,p_i)$, it still obviously holds $FG=GF$, as both are c-number valued functions. $\endgroup$ Commented 5 hours ago

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The derivation in your link really doesn't make any sense for the $F_i$ and $G_i$ being ordinary real-valued functions on classical phase space. Obviously $\lambda = 0$ in classical mechanics, since all real-valued functions on phase space commute as their multiplication is the pointwise multiplication of real numbers.

The expression $FG - GF$ is simply identically zero for all real-valued functions $F,G$, saying it is "proportional to $\{F,G\}$" has literally no mathematical content.

It is also not the case that you would get quantum mechanics by assuming that $F$ and $G$ are some other (not real-valued) functions on phase space, as the operators of quantum mechanics that do not commute are not functions on phase space in any straightforward sense - although there is the phase space formulation of quantum mechanics: It has to deform the Poisson bracket into the Moyal bracket(s) (or respectively the pointwise multiplication of functions into the star product(s)) to model the non-commutativity of the quantum-mechanical operators, while the operators remain represented by ordinary real-valued functions.

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There is already a good answer from ACuriousMind.

  1. Concerning OP's title question: Yes, for functions $f,g\in C^{\infty}(M)$ of the classical phase space $M$ (which is a Poisson manifold), it is possible to replace the commutative, associative pointwise product $$\cdot~:~ C^{\infty}(M)\times C^{\infty}(M)~\longrightarrow ~C^{\infty}(M)$$ with a non-commutative, associative star product $$\star~:~ C^{\infty}(M)\times C^{\infty}(M)~\longrightarrow~ C^{\infty}(M),$$ e.g. the Groenewold-Moyal star product star product.

  2. The idea is to turn the function-to-operator quantization map $$\Phi~:~ (C^{\infty}(M),+,\star)~\longrightarrow~ ({\cal A},+,\circ)$$ into an algebra isomorphism, cf. the topic of deformation quantization.

  3. By the way, in contrast the Koopman-von Neumann theory is an operator formulation of classical Hamiltonian mechanics where position and momentum operators commute.

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If classical physics also includes classical field theory (as it should), then there is a way to have the same non-commuting operators. However, one does not have the same relationship between momentum and wave vectors via the de Broglie relationship, which only applies in quantum physics. So in the context of classical field theory, modeled in terms of operators, one would have a similar commutation relation between position and the wave vector.

The context is the free propagation of classical optical fields, as represented by scalar diffraction theory. Say we want to compute the average position of a beam of light in a given plane. Then we can define a position operator $\hat{\mathbf{x}}$ that multiplies the modulus square of the function of the beam by the coordinates in that plane and integrates over it.

Alternatively we want to compute the direction of propagation of the beam when it passes through the plane. Then we define an operator $\hat{\mathbf{k}}$ in the Fourier domain that multiplies the modulus square of the angular spectrum of the beam by the wave vector and integrates over the wave vectors.

These two operators don't commute. They produce a commutation relation that is analogous to the Heisenberg uncertainty relation: $$ [\hat{\mathbf{x}},\hat{\mathbf{k}}] = i c \mathbb{I} , $$ where $c$ is a numerical constant that depends on the detail of the definitions of the operators.

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