1) In the continuous limit, you can think of the index i on the coordinates (q_{i},p_{i}) as a multi-index (a , \vec{x}) taking values in the set \{1,2, \cdots , d \} \times \mathbb{R}^{3}. So, you can write (q_{i},p_{i})(t) = (q_{(a , \vec{x})} , p_{(a , \vec{x})}) (t) \equiv (Q^{a} , P_{a})(\vec{x},t) . So, for every a \in \{1 , 2 , \cdots , d \}, the pair (Q^{a},P_{a}) are functions of the space-time point (\vec{x},t) \in \mathbb{R}^{3} \times \mathbb{R} = \mathbb{R}^{(3,1)}, i.e., you have fields. And, of course, the sum \sum_{i = 1}^{\infty} get replaced by \sum_{a = 1}^{d}\int_{\mathbb{R}^{3}} d^{3}\vec{x} .2) Have you heard the saying “Classical Mechanics is a (0+1)-dimensional field theory”? Indeed, one can think of the pairs (q_{i},p_{i})(t) as functions of the “space-time” point t \in \mathbb{R}^{(0,1)}, i.e., you have a (0 +1)-dimensional field theory. Now, in the usual (3 +1)-dimensional field theory, every observable A(\phi , \pi) defines the following (Hamiltonian) vector field X_{A} = \sum_{a = 1}^{d} \int_{\mathbb{R}^{3}} d^{3}x \left( \frac{\delta A}{\delta \phi^{a} (\vec{x},t)}\frac{\delta}{\delta \pi_{a}(\vec{x},t)} - \frac{\delta A}{\delta \pi_{a}(\vec{x},t)}\frac{\delta}{\delta \phi^{a}(\vec{x},t)}\right) , and the Poisson bracket is simply X_{A}(B) = \{A , B \}. So, for field theory on the “space-time” \mathbb{R}^{(0,1)}, the vector field associated with the observable A(q,p) becomes X_{A(q,p)} = \sum_{i = 1}^{d} \int_{\mathbb{R}^{0}} d^{0}x \left( \frac{\delta A}{\delta q^{i}} \frac{\delta}{\delta p_{i}(t)} - \frac{\delta A}{\delta p_{i}(t)}\frac{\delta}{\delta q^{i}(t)}\right) = \sum_{i}^{d} \left( \frac{\partial A}{\partial q^{i}}\frac{\partial}{\partial p_{i}} - \frac{\partial A}{\partial p_{i}}\frac{\partial}{\partial q^{i}} \right) . This vector field gives you the Poisson bracket that you are familiar with X_{A}(B) = \{ A , B\}_{(q,p)}.
3) To apply the canonical formalism to continuous system, it is more convenient to work with a denumerable infinity (of degree of freedom) instead of continuous infinity. To do this, you can divide the continuous (3-dimensional) spatial volume into very small cubical cells. Clearly, one can label these cells by discrete variable. So, one has infinite number of cells of volume \delta V_{i}. In order to apply the canonical formalism, one needs to associate a coordinate pair (q_{i}(t), \dot{q}_{i}(t)) to the ith cell. These can be identified with the average value of \varphi (\vec{x} ,t) and \dot{\varphi}(\vec{x} , t) over the ith cell: \left( q_{i}(t) , \dot{q}_{i}(t) \right) \equiv \frac{1}{\delta V_{i}} \int_{(\delta V_{i})} d^{3}\vec{x} \left( \varphi (\vec{x} , t) , \dot{\varphi}(\vec{x} , t)\right) . This method, described in many textbooks, allows you to translate the usual canonical formalism of discrete systems to field theory. However, there is an alternative method that can help you understand the connection between the continuous and the discrete Poisson brackets quite easily. For simplicity, I will consider real scalar field \varphi (\vec{x},t) described by the Lagrangian L = \int_{\mathbb{R}^{3}} d^{3}x \mathcal{L} \left(\varphi , \dot{\varphi} , \nabla \varphi \right)(\vec{x} ,t) . \ \ \ \ \ \ \ \ \ (3.1) The field conjugate momentum is defined by \pi (\vec{x} , t) = \frac{\partial \mathcal{L}}{\partial \dot{\varphi}}(\vec{x},t) . \ \ \ \ \ \ \ \ \ \ \ (3.2) Let \{ u_{m}(\vec{x}) , \vec{x} \in \mathbb{R}^{3} , \ m = 1,2,3, \cdots \} be a complete set of real orthonormal functions \sum_{m=1}^{\infty} u_{m}(\vec{x}) u_{m}(\vec{y}) = \delta^{3} (\vec{x} - \vec{y}) , \ \ \ \ \ \ \ \ (3.3a) \int_{\mathbb{R}^{3}} d^{3}x \ u_{m}(\vec{x}) u_{n}(\vec{x}) = \delta_{mn} . \ \ \ \ \ \ \ \ (3.3b) Now we can write the following expansions \varphi (\vec{x} , t) = \sum_{m = 1}^{\infty} q_{m}(t) \ u_{m}(\vec{x}) , \ \ \ \ \ \ \ (3.4a) \dot{\varphi} (\vec{x} , t) = \sum_{m = 1}^{\infty} \dot{q}_{m}(t) \ u_{m}(\vec{x}) . \ \ \ \ \ \ \ (3.4b) Using Eq(3.3b), we obtain the inverted expansions
q_{n}(t) = \int_{\mathbb{R}^{3}} d^{3}x \ \varphi (\vec{x} ,t) \ u_{n}(\vec{x}) , \ \ \ \ \ \ (3.5a)
\dot{q}_{n}(t) = \int_{\mathbb{R}^{3}} d^{3}x \ \dot{\varphi} (\vec{x} ,t) \ u_{n}(\vec{x}) . \ \ \ \ \ \ (3.5b) Next we interpret (q_{n} , \dot{q}_{n})(t) as countable set of infinite coordinates and velocities, and apply the usual canonical formalism using the Lagrangian function L(q , \dot{q})(t). Indeed, using (3.1) and the relations (3.4), it is easy to show that
\frac{\partial L}{\partial q_{n}} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_{n}}) = \int_{\mathbb{R}^{3}} d^{3}x \left( \frac{\partial \mathcal{L}}{\partial \varphi} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi)} \right) \right) u_{n}(\vec{x}) . If the left hand side is zero, then the completeness of the set \{u_{n}\} leads to the field equation \frac{\partial \mathcal{L}}{\partial \varphi} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi)} \right) = 0 . \ \ \ \ \ \ (3.6) Conversely, if the field equations (3.6) are satisfied then \frac{\partial L}{\partial q_{n}} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_{n}}) = 0 , \ n = 1,2,3, \cdots . \ \ \ \ (3.7) Thus, at every (\vec{x} , t) the field equation (3.6) is equivalent to the infinite set of equations (3.7) for (q_{n},\dot{q})(t), and vice versa.
Now that we have Lagrangian formalism, we can pass to the Hamiltonian formalism and address the issue of Poisson brackets. To do this we define momenta p_{n}(t) canonically conjugate to the coordinates q_{n}(t) by the “usual” relation p_{n}(t) \equiv \frac{\partial L}{\partial \dot{q}_{n}(t)} = \int_{\mathbb{R}^{3}} d^{3}x \frac{\partial \mathcal{L}}{\partial \dot{\varphi}(\vec{x},t)} \frac{\partial \dot{\varphi}(\vec{x},t)}{\partial \dot{q}_{n}(t)} . Then, from the definition (3.2) of \pi (\vec{x},t) and the expansion (3.4b), we obtain p_{n}(t) = \int_{\mathbb{R}^{3}} d^{3}x \ \pi (\vec{x},t) \ u_{n}(\vec{x}) . \ \ \ \ \ \ \ (3.8a) This can be inverted to give \pi (\vec{x},t) = \sum_{n}^{\infty} \ p_{n}(t) \ u_{n}(\vec{x}) . \ \ \ \ \ \ \ \ (3.8b) We now have enough relations to describe any observable A as a function of the pair (q_{n} , p_{n})(t) or as a functional \mathcal{A} of (\varphi , \pi)(\vec{x},t). Indeed, A(q , p)(t) = \mathcal{A}(\varphi , \pi )(\vec{x},t) . So, we can calculate \frac{\partial A}{\partial q_{n}(t)} = \int_{\mathbb{R}^{3}} d^{3}x \ \frac{\delta \mathcal{A}}{\delta \varphi (\vec{x},t)} \frac{\partial \varphi (\vec{x},t)}{\partial q_{n}(t)} , which is \frac{\partial A}{\partial q_{n}(t)} = \int_{\mathbb{R}^{3}} d^{3}x \ \frac{\delta \mathcal{A}}{\delta \varphi (\vec{x},t)} \ u_{n}(\vec{x}) . And, similarly for the observable B, we find \frac{\partial B}{\partial p_{n}(t)} = \int_{\mathbb{R}^{3}} d^{3}y \ \frac{\delta \mathcal{B}}{\delta \pi (\vec{y},t)} \ u_{n}(\vec{y}) . Thus \sum_{n}^{\infty} \frac{\partial A}{\partial q_{n}} \frac{\partial B}{\partial p_{n}} = \int_{\mathbb{R}^{6}} d^{3}x d^{3}y \ \frac{\delta \mathcal{A}}{\delta \varphi (\vec{x},t)} \frac{\delta \mathcal{B}}{\delta \pi (\vec{y},t)} \sum_{n} u_{n}(\vec{x}) u_{n}(\vec{y}) . Then, the completeness relation (3.3a) leads to \sum_{n}^{\infty} \ \frac{\partial A}{\partial q_{n}} \frac{\partial B}{\partial p_{n}} = \int_{\mathbb{R}^{3}} d^{3}x \ \frac{\delta \mathcal{A}}{\delta \varphi (\vec{x},t)} \frac{\delta \mathcal{B}}{\delta \pi (\vec{x},t)} . This gives us the following Poisson bracket on the infinite-dimensional phase-space of our field theory
\sum_{n}^{\infty} \left( \frac{\partial A}{\partial q_{n}} \frac{\partial B}{\partial p_{n}} - \frac{\partial A}{\partial p_{n}} \frac{\partial B}{\partial q_{n}}\right) = \int_{\mathbb{R}^{3}} d^{3}x \left( \frac{\delta \mathcal{A}}{\delta \varphi} \frac{\delta \mathcal{B}}{\delta \pi} - \frac{\delta \mathcal{A}}{\delta \pi} \frac{\delta \mathcal{B}}{\delta \varphi} \right) . The same method can be applied for field theory with finite-component field \varphi^{a}(\vec{x},t). In the Poisson bracket, this requires you to sum over the fields index a.
