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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 $i=j$.

Qmechanic
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dennis
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    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?) – ACuriousMind Jul 14 '23 at 18:00

1 Answers1

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  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.

Qmechanic
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