What does it mean for two variables to be canonically conjugate? The word "canonical" has been used in many of my classes (canonical
ensemble, canonical transformations, canonical conjugate variables) and I am not really sure what it means physically.
More specifically, in the context of the Hamiltonian formulation of mechanics , what does canonically conjugate variables mean physically?
why is it that because $\{x,p_x\}= 1$ ,they are canonically conjugate variables? what does the Poisson bracket value really mean?
and why is it that canonically conjugate variables, when we go to quantum mechanics, have operators that do not commute , $[\hat{x},\hat{p_x}] = i \hbar$, which leads to uncertainty relations.
There seems to be a deeper connection here that I do not want to skip over, please recommend me readings as my search efforts have not led me anywhere.
 A: One can see canonical conjugate variables as a pair of variables which generate the displacement of each other. For instance, consider the quantum mechanical case
$$
[\hat{x},\hat{p}]=i\hbar
$$
If we take momentum as the generator, $\exp(i \epsilon\hat{p}/\hbar)$, the transformed of $\hat{x}$ under this operation
$$
\hat{x} \to \hat{x}(\epsilon)=\exp(i \epsilon\hat{p}/\hbar)\hat{x}\exp(-i \epsilon\hat{p}/\hbar)
$$
satisfies the differential equation
$$
\frac{\partial \hat{x}(\epsilon)}{\partial \epsilon}=-\frac{i}{\hbar}[\hat{x}(\epsilon),\hat{p}]=-\frac{i}{\hbar}\exp(i \epsilon\hat{p}/\hbar)[\hat{x},\hat{p}]\exp(-i \epsilon\hat{p}/\hbar)=1
$$
which has the solution (note that $\hat{x}(0)=\hat{x}$)
$$
\hat{x}(\epsilon)=\hat{x}+\epsilon.
$$
Similarly, if we consider the transformation generated by $\hat{x}$, $\exp(i \epsilon\hat{x}/\hbar)$, we find
$$
\hat{p}(\epsilon)=\exp(i \epsilon\hat{x}/\hbar)\hat{p}\exp(-i \epsilon\hat{x}/\hbar)=\hat{p}-\epsilon
$$
In the classical case, the situation is analogous with the canonical transformations generated by each of the variables in terms of the Poisson brackets:
$$
\frac{\partial x(\epsilon)}{\partial \epsilon}=\{x(\epsilon),p\}=1\Rightarrow x(\epsilon)=x+\epsilon\\
\frac{\partial p(\epsilon)}{\partial \epsilon}=\{p(\epsilon),x\}=-1\Rightarrow p(\epsilon)=p-\epsilon
$$
This implies that both variables play a reciprocal role, they can be exchanged by a suitable transformation, and their evolution is liked in a precise way. Namely, the time variation of one of them is given by the energy (Hamiltonian) variation of with respect to the other:
$$
\frac{d x}{dt}=\{x,H\}=\frac{\partial H}{\partial p}\\
\frac{d p}{dt}=\{p,H\}=-\frac{\partial H}{\partial x}
$$
These are known as Hamilton's equations. Of course, here $H$ must be written explicitly in terms of the conjugated pair, in this case $x$ and $p$.
A: I suspect you're probably looking in the wrong places to find the answer to this question. I believe the term canonical comes from mathematics rather than physics, I once heard a mathematician(/mathematical physicist) say something along the lines of: "canonical means that once you understand this, this is the most natural form of whatever you are looking at."
However, I found a more robust answer than one vague memory of mine with a quick search on a maths wiki here: https://ncatlab.org/nlab/show/canonical+form
Another place you may search is in category theory (which nlab is a wiki for), this is a branch of mathematics that may have originally been developed to rigorously define what mathematicians meant whenever they said something was 'natural' (and has been broadly successful with that from what I heard). I suspect the canonical form is a basis/presentation of your operator algebra that happens to be something called a "universal property" for a sufficiently defined problem.
Unfortunately, I have never given the meaning of canonical a deep enough thought to give a rigorous definition here or to give anything better than a handwavy description of how it appears in this framework. But hopefully this might be enlightening to you (and I'll update this answer if I ever come back to this).
A: In Hamiltonian dynamics, a change of variables $(q,p) \longrightarrow (Q,P)$ that leaves the form of the equations of motion unchanged (the symplectic structure is conserved) is called a canonical transform.
$$\begin{cases} \frac{dq}{dt}= \frac{\partial H(p,q)}{\partial p}  \\ \frac{dp}{dt}=- \frac{\partial H(p,q)}{\partial q}\end{cases}  \longrightarrow \begin{cases} \frac{dQ}{dt}= \frac{\partial H(P,Q)}{\partial P}  \\ \frac{dP}{dt}=- \frac{\partial H(P,Q)}{\partial Q}\end{cases}$$
Variables (q,p) or (Q,P) used in the Halmitonian are called conjugate variables. If they are formed by a canonical transform, they are said to be canonically conjugate.
Conjugate variables have the interesting property that their Poisson brackets is one.
Poisson brackets encapsulate all the information about the dynamics of the system. Consider a dynamical functions f(q,p,t) that can be expanded in series of time:
$$ f(q,p,t)= \sum_{n=0}^ \infty     b(q,p)^{(n)}   \frac{t^{n}}{n!} $$
Where for simplicity we did not write the argument t=0. Let's define a new operator:[H] by:$b(q,p)^{(1)}=\{b,H\}=[H]b$
Then using Poisson brackets, we can evaluate the time derivatives in the above series.
$$  b(q,p)^{(1)}=\{b,H\}=[H]b$$
$$b(q,p)^{(2)}=\{b^{(1)},H\}=\{\{b,H\},H\}=[H]^{2}b$$
$$ b(q,p)^{(n)}=\{b^{(n-1)},H\}=[H]^{(n)}b$$
The time evolution of f(q,p,t) is given by:
$$ f(q,p,t)= \sum_{n=0}^ \infty     [H]^{n} b(q,p)  \frac{t^{n}}{n!}= e^{t[H]} b(q,p)$$
Of course, it is very formal, but Mathematicians use the symplectic structure of Hamiltonian dynamics to derive theorems like the Liouville theorem.
