In classical field theory (for a single field $\psi$) the dynamical variables are defined to be functions of the fields $\psi$, $\pi$, $\partial_{x_{i}}\psi$ and maybe $\mathbf{r}$, where $\pi$ is the conjugated field to $\psi$.

For $F=\int\mathcal{F}\,d\mathbf{r}$ and $G=\int\mathcal{G}\,d\mathbf{r}$, where F and G are dynamical variables, the functional Poisson bracket can be defined according to (José and Saletan, “Classical Dynamics: A Contemporary Approach”, cap 9))

$$\left\{ F,G\right\} ^{f}=\intop\left(\frac{\delta F}{\delta\psi}\frac{\delta G}{\delta\pi}-\frac{\delta F}{\delta\pi}\frac{\delta G}{\delta\psi}\right)d\mathbf{x},$$ where the derivatives are functional derivatives. The fields themselves have the canonical property

$$\left\{ \psi(\mathbf{y}),\pi(\mathbf{z})\right\} =\delta(\mathbf{y}-\mathbf{z}),$$ $$\left\{ \pi(\mathbf{y}),\pi(\mathbf{z})\right\} =\left\{ \psi(\mathbf{y}),\psi(\mathbf{z})\right\} =0.$$

So far so good, but I'm not sure how to handle the functional derivatives. I'm interested, for example, in the following bracket

$$\left\{ F(\mathbf{x}),\pi(\mathbf{z})\right\} ^{f}$$ $(F(\mathbf{x})\equiv F(\psi(\mathbf{x}),\pi(\mathbf{x}),\partial_{x_{i}}\psi(\mathbf{x}))$

Using $\frac{\delta\pi(\mathbf{z})}{\delta\psi}=0$ and $\frac{\delta\pi(\mathbf{z})}{\delta\pi}=\delta(\mathbf{y}-\mathbf{x})$, I think the answer is

$$\left\{ F(\mathbf{x}),\pi(\mathbf{z})\right\} ^{f}=\frac{\delta F(\mathbf{z})}{\delta\psi}.$$

Is this result correct?

  • $\begingroup$ I think you have misconfigured your expressions big time... Are you using WP conventions? Do you mean ${\cal F} ({\mathbf x})$ instead? $\endgroup$ Commented Oct 1, 2020 at 19:52
  • $\begingroup$ If so, your final expression is plausible, with an extra factor $\delta ({\mathbf x-z})$ on the rhs. Normally functional derivatives have a functional in the "numerator" and a function in the "denominator". $\endgroup$ Commented Oct 1, 2020 at 20:04
  • 2
    $\begingroup$ Related: Mathematical interpretation of Poisson Brackets and links therein. $\endgroup$
    – Qmechanic
    Commented Oct 1, 2020 at 20:51
  • $\begingroup$ @CosmasZachos that there is an extra Dirac delta involved in the rhs is a possibility I considered. I came here because I could not tell what option is correct (if any). $\endgroup$
    – AndresB
    Commented Oct 1, 2020 at 22:21
  • $\begingroup$ @Qmechanic Thanks for the link. I think what is confusing me is that the Poisson bracket is supposed to pair two functionals but the canonical bracket is given in terms of the functions $\psi(\mathbf{y})$ and $\pi(\mathbf{z})$. $\endgroup$
    – AndresB
    Commented Oct 1, 2020 at 23:22

1 Answer 1


You are on the right track, but some clarification is needed. First as you mentioned, the Poisson bracket is defined over functionals of the local fields, i.e. an object of the form $F=\int d^n x {\cal F}[\psi(x),\pi(x)]$. Note that here we integrate over $x$, so $F$ is not a function of $x$. Accordingly, your last equation should read \begin{align} \{F,\pi(z)\}&=\int d^nx \{{\cal F}[\psi(x),\pi(x)],\pi(z)\}=\int d^nx \frac{\delta {\cal F}}{\delta \psi(x)}\frac{\delta \pi(z)}{\delta \pi(x)}=\int d^nx \frac{\delta {\cal F}}{\delta \psi(x)}\delta^n(z-x)=\frac{\delta {\cal F}[\psi(z),\pi(z)]}{\delta \psi(z)} \end{align} So in your last equation, there is two typos: on the lhs, there is no argument $x$ for $F$, and on the rhs, it is the density $\cal{F}$ instead of the functional $F$.

If you insist on having the left hand side of your last equation as it is, you can start by smearing the density $\cal{F}$ with a delta function, i.e. you take your functional to be $F(x)=\int d^nx' \delta^n(x'-x){\cal F}[\psi(x'),\pi(x')]$. Now repeating the procedure you find \begin{align} \{F(x),\pi(z)\}&=\frac{\delta {\cal F}[\psi(z),\pi(z)]}{\delta \psi(z)}\,\delta^n(x-z). \end{align}

  • $\begingroup$ Hi, thanks for the answer, though there is something I don't understand. The definition of the Poisson bracket given in the book have $\frac{\delta F}{\delta\psi}$ inside the integral and not $\frac{\delta\mathcal{F}}{\delta\psi}$. $\endgroup$
    – AndresB
    Commented Oct 1, 2020 at 21:54
  • $\begingroup$ The point is that $\frac{\delta F}{\delta \psi(x)}=\int dx' \frac{\delta{\cal F}(x')}{\delta \psi(x)}=\int dx' \frac{\delta{\cal F}(x)}{\delta \psi(x)}\delta(x-x')=\frac{\delta {\cal F}(x)}{\delta \psi(x)}$, so they are equal. $\endgroup$
    – Ali Seraj
    Commented Oct 2, 2020 at 8:09
  • $\begingroup$ That's somewhat confusing, I have to ponder on it. Would it then be correct to make the identification $\left\{ \cdot,\pi(\mathbf{z})\right\} =\frac{\delta}{\delta\psi}$ evaluated in z (Equivalent to the relation $\left\{ \cdot,p\right\} =\frac{\partial}{\partial q}$ of hamiltonian mechanics for the particle)? $\endgroup$
    – AndresB
    Commented Oct 2, 2020 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.