The canonical commutation relations in gravity are sometimes written $$ [\gamma_{ij}(x),\pi^{kl}(y)]=\frac{i\hbar}{2}(\delta_i^k\delta_j^l+\delta_i^l\delta_j^k)\delta^3(x-y),\tag{0} $$ where $\gamma_{ij}$ is the 3-metric. This means \begin{align} [\gamma_{11}(x),\pi^{11}(y)]&=i\hbar\delta^3(x-y) \tag{1} \\ [\gamma_{12}(x),\pi^{12}(y)]&=\frac{i\hbar}{2}\delta^3(x-y) \tag{2} \end{align} for example.

Eq. (1) looks standard so I would conclude that $\gamma_{11}$ is canonically conjugate (c.c.) to $\pi^{11}$.

Eq. (2), however, has the factor of $\frac{1}{2}$ which doesn't look standard at all. It would seem to suggest that $\gamma_{12}$ and $\pi^{12}$ are not canonically conjugate. Any comments?

EDIT: I think the simple conclusion is that $\gamma_{ij}$ is c.c. to $2\pi^{ij}$ when $i\neq j$ and c.c. to $\pi^{ij}$ when $i=j$.

  • 3
    $\begingroup$ Can you ask a more specific question than "any comments"? This is far from my specialty but I see nothing particularly worrisome about factors of 1/2 appearing in the context of symmetric tensors ($\gamma_{ij} = \gamma_{ji}$, right?) $\endgroup$
    – ACuriousMind
    Jul 14 at 18:00

1 Answer 1

  1. The factor $\frac{1}{2}$ in the symmetrized Poisson bracket relations $$\begin{align} \{\gamma_{ij}(x),\pi^{kl}(y)\} ~=~&\frac{1}{2}(\delta_i^k\delta_j^l+\delta_i^l\delta_j^k)\delta^3(x-y),\cr \{\gamma_{ij}(x),\gamma_{kl}(y)\}~=~&0,\cr \{\pi^{ij}(x),\pi^{kl}(y)\}~=~&0, \end{align} \tag{A} $$ is a natural consequence of the symmetry of the metric tensor $\gamma_{ij}$ (and the symmetry of $\pi^{kl}$), cf. e.g. this related Phys.SE post.

  2. We may write the Schrödinger representation as $$ \pi^{kl}(x)~=~\frac{\hbar}{2i}\left(\frac{\delta}{\delta\gamma_{kl}(x)}+\frac{\delta}{\delta\gamma_{lk}(x)} \right).\tag{B}$$

  3. Alternatively, we may view the tensors $\gamma_{ij}$ and $\pi^{kl}$ as fundamentally non-symmetric tensors, and proceed as follows:

    • Option (i): They satisfy the symmetrized Poisson bracket relation (A). Only later impose the first-class constraints $$\begin{align} \phi_{ij}~:=~&\gamma_{ij}-\gamma_{ji}~\approx~0,\cr \chi^{kl}~:=~&\pi^{kl}-\pi^{lk}~\approx~0,\end{align}\tag{C}$$ where $$\{\phi_{ij}(x),\chi^{kl}(y)\} ~\stackrel{(A)+(C)}{=}~0.\tag{D}$$

    • Option (ii): They satisfy the non-symmetrized Poisson bracket relations $$\begin{align} \{\gamma_{ij}(x),\pi^{kl}(y)\} ~=~&\delta_i^k\delta_j^l\delta^3(x-y),\cr \{\gamma_{ij}(x),\gamma_{kl}(y)\}~=~&0,\cr \{\pi^{ij}(x),\pi^{kl}(y)\}~=~&0. \end{align} \tag{E} $$ Then $$\{\phi_{ij}(x),\chi^{kl}(y)\} ~\stackrel{(C)+(E)}{=}~2(\delta_i^k\delta_j^l-\delta_i^l\delta_j^k)\delta^3(x-y).\tag{F}$$ Only later impose the second-class constraints (C). The Dirac bracket becomes $$\begin{align}\{F,G\}_D ~=~&\{F,G\}\cr ~-~&\frac{1}{4}\int\!d^3z\{F,\chi^{mn}(z)\}\{\phi_{mn}(z),G\}\cr ~+~&\frac{1}{4}\int\!d^3z\{F,\phi_{mn}(z)\}\{\chi^{mn}(z),G\}. \end{align}\tag{G}$$
      One may check that the Dirac bracket (G) reproduces the symmetrized Poisson bracket relations $$\begin{align} \{\gamma_{ij}(x),\pi^{kl}(y)\}_D ~=~&\frac{1}{2}(\delta_i^k\delta_j^l+\delta_i^l\delta_j^k)\delta^3(x-y),\cr \{\gamma_{ij}(x),\gamma_{kl}(y)\}_D~=~&0,\cr \{\pi^{ij}(x),\pi^{kl}(y)\}_D~=~&0. \end{align} \tag{H} $$

    Both options (i) & (ii): The variables $\gamma_{ij}$ and $\pi^{kl}$ are then canonically conjugate variables on the constrained submanifold, cf. OP's question.

  • 1
    $\begingroup$ Yes but then I can't conclude that $\gamma_{12}$ is canonically conjugate (c.c.) to $\pi^{12}$ (because for $q$ and $p$, say, to be c.c. they must satisfy $[q,p]=i$). $\endgroup$
    – dennis
    Jul 14 at 18:04
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jul 14 at 19:45

Your Answer

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

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