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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?

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user47224
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1 Answers1

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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).

References:

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

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$^1$ 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).

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  • 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. – user47224 Sep 09 '14 at 15:04
  • @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. – Qmechanic Sep 09 '14 at 15:12
  • 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]:http://en.wikipedia.org/wiki/Functional_derivative#Functional_derivative – user47224 Sep 09 '14 at 16:20
  • 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 $\ast$ – user47224 Sep 09 '14 at 16:25
  • @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 $f(x)$. – Qmechanic Sep 09 '14 at 16:37
  • Ok,now I understand your point. If I'm not wrong the locality prescription assures that the Poisson bracket is well defined as a convolution. – user47224 Sep 09 '14 at 17:38