Lets say we are working in a classical scalar field theory and we have two functional $ F[\phi, \pi](x)$ and $G[\phi, \pi](x)$. In most of the references, starting with two functional the Poisson bracket is defined as $$\{F(x),G(y)\} = \int d^3z \left( \frac{\delta F(x)}{\delta \phi(z)}\frac{\delta G(y)}{\delta \pi(z)} - \frac{\delta F(x)}{\delta \pi(z)}\frac{\delta G(y)}{\delta \phi(z)}\right) . $$

But as explained here the functional derivative $\frac{\delta F}{\delta \phi} $ is a distribution rather than a function, so the previous definition does not make much sense. I was wondering then, if the Poisson bracket can be interpreted as the convolution calculated in $(x-y)$ (in the sense of distributions) between the functional derivatives. This works in case of interest such as $\{\phi(x), \pi(y) \}$ but I'm not sure it can be applied for two generic functional (the dependence $(x-y)$ is not explicit). Is there a proof that the Poisson bracket is a convolution? More in general, can field theories be formulated in a formal way in the sense of distributions?


1 Answer 1


I) It is worthwhile mentioning that there exists a basic approach well-suited to physics applications (where we usually assume locality) that avoids multiplying two distributions together. The idea is that the two inputs $F$ and $G$ in the Poisson bracket (PB)

$$\tag{1}\{F,G\} ~=~ \int_M \!dx \left( \frac{\delta F}{\delta \phi(x)}\frac{\delta G}{\delta \pi(x)} - \frac{\delta F}{\delta \pi(x)}\frac{\delta G}{\delta \phi(x)} \right) $$

are assumed to be (differentiable) local functionals.$^1$ When a functional $F$ is differentiable$^2$ the functional derivatives

$$\tag{2}\frac{\delta F}{\delta \phi(x)},\frac{\delta F}{\delta \pi(x)},$$

of $F$ wrt. all fields $\phi(x)$, $\pi(x)$, exist.

If the two inputs $F$ and $G$ are assumed to be differentiable local functionals, the functional derivatives (2) will be local functions$^1$ (as opposed to distributions), and it makes sense to multiply two such functional derivatives together, and finally integrate to get the PB (1). The output $\{F,G\}$ is again a differentiable$^3$ local functional, so that the Poisson bracket $\{\cdot,\cdot\}$ is a product in the set of differentiable local functionals.

II) Some physical quantities are already local functionals $F$, while others are local functions $f(x)$. How do we turn a local function into a local functional? We use a test function $\eta(x)$. If $f(x)$ is a local function, define a corresponding local functional as

$$\tag{3}F[\eta]~:=~ \int_M \! dx f(x)\eta(x). $$

Then it is ready to be inserted in the PB (1).


  1. J.D. Brown and M. Henneaux, On the Poisson brackets of differentiable generators in classical field theory, J. Math. Phys. 27 (1986) 489.


$^1$ For the definition of a local function and a local functional, see e.g. this Phys.SE post and links therein.

$^2$ The existence of a functional derivative (2) of a local functional $F$ depends on appropriate choice of boundary conditions.

$^3$ The differentiability of the PB (1) is guaranteed under appropriate assumptions, cf. Ref. 1, which in turn also discusses the Jacobi identity for the PB (1).

  • $\begingroup$ Thanks for the answer. Probably I'm missing the key point but you are claiming that $\frac{\delta F}{\delta \phi}$ is a local function, so can be expressed in terms of a function of $n+1$ variables (referring to the link you posted). On the other hand it is well known that $\frac{\delta F}{\delta \phi} = \delta(x-y)$ cannot be considered a function (only in certain limits) so the problem persists. $\endgroup$
    – user47224
    Commented Sep 9, 2014 at 15:04
  • $\begingroup$ @user47224 : The functional derivative $\frac{\delta F}{\delta \phi(y)} = \delta(x-y)$ in your example is indeed a distribution. I assume that you take $F=\phi(x)$, which is a local function but not a local functional. $\endgroup$
    – Qmechanic
    Commented Sep 9, 2014 at 15:12
  • $\begingroup$ Sorry if I insist, but: Lets call $\mathcal{D}$ the space of the test functions (smooth and a compact support), $\mathcal{D'} =\{ f:\mathcal{D} \rightarrow \mathbb{R} \}$ its dual space. As can be seen [here][1] in order to introduce the derivative $\frac{d}{d\epsilon}|_{\epsilon = 0} F[\phi + \epsilon g]$ it is necessary that, in my example $F:\mathcal{D}\rightarrow \mathbb{C}$. Moreover the object $<\frac{\delta F}{\delta \phi}; . > :\mathcal{D}\rightarrow \mathbb{C} $ so $\frac{\delta F}{\delta \phi} \in D'$ [1]:en.wikipedia.org/wiki/… $\endgroup$
    – user47224
    Commented Sep 9, 2014 at 16:20
  • $\begingroup$ then the Poisson bracket is defined acting in the following domain $\{ ;\} : \mathcal{D}'\times\mathcal{D}' \rightarrow \mathcal{D}'$. The only operator (that i know) that acts in this way is the convolution $\ast$ $\endgroup$
    – user47224
    Commented Sep 9, 2014 at 16:25
  • $\begingroup$ @user47224 : Note that a local function $f(x)$ is also a distribution (but not the other way around). The set of local functions can be imbedded in the set of distributions. The functional derivative $\frac{\delta F}{\delta \phi(x)}$ is a distribution, but for a local functional $F$, this distribution can be represented by a local function $f(x)$. $\endgroup$
    – Qmechanic
    Commented Sep 9, 2014 at 16:37

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