Sunday, 14 July 2019

field theory - Mathematical interpretation of Poisson Brackets



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?



Answer



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.


--


$^1$ 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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