Generalizing Heisenberg Uncertainty Priniciple Writing the relationship between canonical momenta $\pi _i$ and canonical coordinates $x_i$
$$\pi _i =\text{  }\frac{\partial \mathcal{L}}{\partial \left(\frac{\partial x_i}{\partial t}\right)}$$
Then using the Lagrangian density for classical electrodynamics
$$\mathcal{L} = \frac{1}{2}\left(\epsilon _0E^2- \frac{1}{\mu _0}B^2\right)- \phi \rho _{\text{free}} + A\cdot J_{\text{free}} + \mathbb{E}\cdot \mathbb{P} + \mathbb{B}\cdot \mathbb{M}$$
Q1: Does it make sense to substitute the canonical scalar field $\phi$ (or electric scalar potential) for $\pi _i$ ; and $\phi \rho _{\text{free}}$ for canonical coordinates $x_i$. Such that these satisfy Heisenberg's Uncertainty relation?
Q2: Is there a generalized method for taking a Lagrangian and deriving conjugate variables, which then satisfy Heisenberg's Uncertainty relation?
 A: The fields in the electromagnetic Lagrangian satisfy the uncertainty principle with their conjugate momenta, but the result is a distributional uncertainty principle as appropriate for the distributional quantum fields.
From the Lagrangian, the canonical conjugate momentum to the field $A$ is the electric field $E$. If you take a gauge choice where A is purely spatial ($\phi=0$ gauge), you get the Bohr-Rosenfeld uncertainty relation between a measurement of A and a measurement of E in a region.
The commutation relation for the quantum field is
$$ [A_i(x),E_j(y)]  = \delta(x-y)\delta_{ij}$$
And the resulting uncertainty principle comes from smearing both fields using test functions. Define
$$ A^f_i = \int f(x) A_i(x) dx $$
$$ E^g_j = \int g(y) E_j(y) dy$$
Where f,g are positive $C^\infty$ bump functions, then
$$ [A^f, E^g] = \delta_{ij} \int dx f(x)g(x)$$
This leads to a field theoretic uncertainty principle
$$ \delta A_i^f \delta E_i^g \ge (\int dx f(x)g(x))$$
This uncertainty principle demands that the uncertainty products diverge in a particular way as the region becomes small, and Bohr and Roesenfeld verify that the physical uncertainty must be present when you use small  charged quantum pith-balls to measure the value of E and A as best you can. This is automatic from a Lagrangian formulation of quantum mechanics, since you derive the commutation relation from the Lagrangian in a bosonic path integral. So this is one of those things that became trivial after Feynman's Lagrangian formulation became standard. It was controversial in Bohr and Rosenfeld's time, because field quantization made photons.
For a general bosonic Lagrangian, the canonical commutators
$$ [\phi(x),\pi(y)] = \delta(x-y) $$
Where $\pi(x) = {\partial L\over \partial \dot{\phi} }$ leads to an uncertainty relation between $\phi$ and $\pi$ in the exact same way. These field uncertainty relations are in every way analogous to the usual momentum-position uncertainties in ordinary quantum mechanics.
A: It doesn't really make sense to do this, for a few reasons: first, classical field theory has no concept of an uncertainty. It's not even clear what $\sigma_\phi$ and $\sigma_\pi$ would mean in that case. Besides, these quantities are necessarily going to commute. The uncertainty principle relates the product of the variances to the commutator, so the only result you'll get is that $\sigma_\phi\sigma_\pi \ge 0$. It's completely trivial. (If you're dealing with matrix-valued classical fields, then I guess you could get a nonzero commutator, but you still have the issue of why you're computing the $\sigma$s.)
It might make more sense in quantum field theory, where you could try to apply the generalized uncertainty relation,
$$\sigma_{\phi_i}\sigma_{\pi^{j\mu}} \ge \left|\frac{1}{2i}\langle[\phi_i,\pi^{j\mu}]\rangle\right|$$
There's a problem, though: the thing on the right (the commutator) is a field, which has a potentially different value at every point in spacetime, whereas the thing on the left is just a number. So this doesn't really work. It's only in basic quantum mechanics (or for only special operators), where the operator fields $\phi$ and $\pi$ are constant over all of space, that you can ignore the position dependence on the right.
A: It seems that OP's generalized Heisenberg uncertainty principle as formulated in the second subquestion(v5) is nothing but canonical quantization of an arbitrary classical Lagrangian field theory.
Formally, at the classical level, this involves performing a Dirac-Bergmann analysis to obtain a Hamiltonian formulation via a (not necessarily regular) Legendre transformation. This may$^{\dagger}$ introduce first class and second class constraints. 
In case of first class constraints, this indicates that the system possesses gauge symmetry. In case of second class constraints, the canonical Poisson bracket should be replace with the Dirac bracket. 
Quantum mechanically, at the leading order in $\hbar$, the bracket should be replaced with $(i\hbar)^{-1}$ times the corresponding commutator, in accordance with the quantum mechanical correspondence principle, thereby leading to (generalized) Heisenberg uncertainty relations. At subleading orders in $\hbar$, the issue of operator ordering ambiguities arises.
Related topics are geometric and deformation quantization.
Reference:


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*M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.


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$^{\dagger}$ E.g., in the Hamiltonian formulation of classical electrodynamics, Gauss' law becomes a first class constraint.
