Let B be a space of physics we have and T be the duration. Let L be a lagrangian density of the field such that the action is a functional of ϕ:R4→R: S=∫B×Td4xL(ϕ(t,→x),∂μϕ(t,→x)).
We can then derive the equations of motion: δSδϕ=0.
Otherwise we can define the hamiltonian density H=π˙ϕ−L=H(π,ϕ,∂iϕ) whereas π=∂L∂˙ϕ and i=1.2.3. Then the hamiltonian is a functional of (R3→R):
H(t)=∫Bd3xH. Let ϕ(t) and π(t) be 2 functions R3→R. We define the Poisson bracket for 2 functionals A[ϕ,π] and B[ϕ,π]: {A,B}=δAδπδBδϕ−δAδϕδBδπ and we have the canonical relation (for t,→x,→y fixed): {π(t,→x),ϕ(t,→y)}=δ(3)(→x−→y).
How can we show that the equation of motions is now
˙π={H,π};˙ϕ={H,ϕ} ?
Answer
First, note that the EOM are the Euler-Lagrange equations:
δSδϕ=∂μ(∂L∂ϕ,μ)−∂L∂ϕ=˙π+∂i(∂L∂ϕ,i)−∂L∂ϕ where I isolated the μ=0 term, and used the definition of π.
Next, use L=π˙ϕ−H: δSδϕ=˙π+∂i(∂L∂ϕ,i)−∂L∂ϕ=˙π−∂i(∂H∂ϕ,i)+∂H∂ϕ=˙π+δHδϕ because π and ˙ϕ are not functions of ϕ.
Thus, we get ˙π=−δHδϕ={H,π}.
The other relation is easier: {H,ϕ}=δHδπ Next, use H=π˙ϕ−L: {H,ϕ}=δHδπ=˙ϕ because L is not a function of π.
ADDENDUM
Let f(ϕ,π) and g(ϕ,π) be two function on "Phase space". Then, by definition, \{f(\boldsymbol x),g(\boldsymbol y)\}\equiv\int\mathrm d\boldsymbol z\ \frac{\delta f(\boldsymbol x)}{\delta \phi(\boldsymbol z)}\frac{\delta g(\boldsymbol y)}{\delta \pi(\boldsymbol z)}-\frac{\delta g(\boldsymbol y)}{\delta \phi(\boldsymbol z)}\frac{\delta f(\boldsymbol x)}{\delta \pi(\boldsymbol z)} where everything is evaluated at the same time t (not writen for clarity). This means that \{H(t),\pi(t,\boldsymbol x)\}=\int\mathrm d\boldsymbol z\ \frac{\delta H(t)}{\delta \phi(\boldsymbol z)}\frac{\delta \pi(\boldsymbol x)}{\delta \pi(\boldsymbol z)}-\frac{\delta \pi(\boldsymbol x)}{\delta \phi(\boldsymbol z)}\frac{\delta H(t)}{\delta \pi(\boldsymbol z)}=\int\mathrm d\boldsymbol z\ \frac{\delta H(t)}{\delta \phi(\boldsymbol z)}\delta(\boldsymbol x-\boldsymbol z) which equals \frac{\delta H}{\delta \phi}, as expected.
ADDENDUM II
Proof of \frac{\delta H}{\delta \phi}=\frac{\partial \mathscr H}{\partial \phi}-\partial_i\left(\frac{\partial\mathscr H}{\partial\phi_{,i}}\right):
The dynamical variables of the Lagrangian are \phi and \partial_\mu\phi. In the Hamiltonian formulation, we change \partial_0\phi\leftrightarrow\pi, so that the dynamical variables of the Hamiltonian are \phi,\,\partial_i\phi and \pi.
With this in mind, the Hamiltonian is, by definition, H=\int\mathrm d\boldsymbol x\ \mathscr H(\phi(\boldsymbol x),\phi_{,i}(\boldsymbol x),\pi(\boldsymbol x))
Therefore, \delta H=\int\mathrm d\boldsymbol x\ \delta\mathscr H(\phi(\boldsymbol x),\phi_{,i}(\boldsymbol x),\pi(\boldsymbol x)) where \delta\mathscr H(\phi,\phi_{,i},\pi)=\frac{\partial \mathscr H}{\partial \phi}\delta\phi+\frac{\partial \mathscr H}{\partial \phi_{,i}}\delta\phi_{,i}+\frac{\partial \mathscr H}{\partial\pi}\delta\pi
Next, as we want \frac{\delta H}{\delta \phi}, we want to leave \pi unchanged, so \delta\pi=0 (this is analogous to ordinary partial derivatives: when you calculate \frac{\partial f(x,y)}{\partial x} you want to make a small displacement of x, while leaving y unchanged)\phantom{}^1.
Anyway, in the integral over \mathrm d\boldsymbol x, we can integrate by parts the \delta \phi_{,i}=\partial_i\delta \phi to make the derivative act on \frac{\partial\mathscr H}{\partial\phi_{,i}}:
\delta\mathscr H(\phi,\phi_{,i},\pi)=\frac{\partial \mathscr H}{\partial \phi}\delta\phi-\partial_i\left(\frac{\partial \mathscr H}{\partial \phi_{,i}}\right)\delta\phi+\text{surface terms}
Finally, back to H: \delta H=\int\mathrm d\boldsymbol x\ \left(\frac{\partial \mathscr H}{\partial \phi}-\partial_i\left(\frac{\partial \mathscr H}{\partial \phi_{,i}}\right)\right)\delta\phi where I assumed that surface terms dont contribute. The expression in the parentheses is, by definition, the functional derivative of H w.r.t. \phi.
\phantom{}^1: If we took \delta\pi\neq 0 and \delta\phi=0, we would get \frac{\delta H}{\delta \pi} instead. All this is possible because the system is unconstrained, which is not true in some theories (such as the Dirac Lagrangian); in these cases, you can't use Poisson brackets, but Dirac brackets instead.
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