Let $\phi$ be a scalar field, and $\Pi$ be the conjugate momentum of $\phi$. Let $\cal L=\cal L(\phi, \partial_\mu \phi)$ be the Lagrangian density.

Define the stress-energy tensor as $$ T^{\mu\nu}=\Pi^\mu \partial^\nu \phi - g^{\mu\nu}\cal L$$ and define $$ H=\int d^4\mathbf x (\Pi^0\partial^0 \phi - \cal L), \quad P^r=\int d^3\mathbf x \Pi^0\partial^r\phi.$$

For functionals of the field $A,B$, we define the Poisson bracket as $$ \{A,B\}= \int d^3\mathbf x \left( \frac{\delta A[\phi_t]}{\delta\phi(x)} \frac{\delta B[\Pi_t]}{\delta \Pi(x)}- \frac{\delta B[\phi_t]}{\delta\phi(x)} \frac{\delta A[\Pi_t]}{\delta \Pi(x)} \right),$$ where $\phi_t(\mathbf x):=\phi(t,\mathbf x)$ is the section of $\phi$, and analogous notation is used for $\Pi_t$.

The question in the book is to prove that $\{\phi, P^r\}=\Pi^r$, and $\{\Pi, P^r\}=\partial^r\Pi$. Here $\Pi=\Pi^0$ is the canonical momentum and $\Pi^r (r=1,2,3)$ is defined as $$ \Pi^r=\frac{\partial \cal L}{\partial(\partial_r\phi)}.$$

My question is that, how can these expressions make sense? The Poisson bracket is defined for a functional of field, and neither $\phi$ nor $\Pi$ is a functional of field (they are just fields itself)!

  • $\begingroup$ $\phi$ and $\Pi$ are indeed fields, but they may also be viewed as functionals. $\endgroup$ – Qmechanic Jun 18 '19 at 12:59
  • $\begingroup$ @Qmechanic Could you elaborate more on this please? If $\phi$ is viewed as a functional $J_\phi$, say, then what is the value of $J_\phi(\psi)$ which should be a c-number for a field $\psi$? $\endgroup$ – eigenvalue Jun 18 '19 at 13:02
  • $\begingroup$ @Qmechanic I edited the question. Isn't $\Pi^r$ the standard notation? $\endgroup$ – eigenvalue Jun 18 '19 at 13:19
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    $\begingroup$ This is analogous to the statement $\{x,p\}=1$ in classical mechanics. The Poisson bracket is defined for functions of $x,p$, but the bracket is nevertheless well-defined, at least formally. For, after all, $(x,p)\mapsto x$ and $(x,p)\mapsto p$ are meaningful functions (at least when the phase space is flat). $\endgroup$ – AccidentalFourierTransform Jun 18 '19 at 14:22
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    $\begingroup$ The correct statement is $\{\phi(x),P\}=\pi(x)$, where $\phi(x)$ is the functional $\phi\mapsto \phi(x)\in\mathbb C$, which depends parametrically on $x$ (more precisely, $F_x[\phi]:=\phi(x)$; this is basically the $\delta_x$ functional). $\endgroup$ – AccidentalFourierTransform Jun 18 '19 at 16:14

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