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i am doing a thesis on the quantization of a real scalar field in a gravitational wave background. I am doing this in lightcone coordinates, so $u$ is $z-t$. I start with an action and define a momentum operator by $\frac{\partial S}{\partial (\partial_u \phi)} = \hat{\pi}$. Then you want to get an Hamiltonian $H = \int (\partial_u \phi) \hat{\pi} - \mathcal{L} d^{D-1}x$. After that you write $\partial_z \hat{\phi} = \frac{\hat{\pi}}{\sqrt{1-h_+^2 cos^2 (\omega u)}}$. Now using the standard canonical commutation relations between $\hat{\phi}$ and $\hat{\pi}$ and the Heisenberg equation of motion, you obtain a different equation for $\hat{\pi}$. How is that possible, i can't see where it went wrong?

\begin{equation} S = \int \mathcal{L}d^D x = S_{UdW} - \int d^{D}x \sqrt{-g} \bigg[ \frac{1}{2} g^{\mu\nu}(\partial_\mu \hat{\phi}) (\partial_\nu \hat{\phi}) +\frac{1}{2}m^2 \phi^2 \bigg] \end{equation} With $m$ the mass of the field $\phi$ and $S_{UdW}$ the action of the Unruh-deWitt particle detector, that we neglect for now. We also multiply with $\sqrt{-g}$ to make $d^D x \sqrt{-g}$ constant under coordinate transformations.

Letting $\mu$ and $\nu$ (non-zero terms are (1,1), (2,2), (3,3), (3,0) and (0,3), respectively) run along every possible combination you can express the action as \begin{equation} \begin{split} S = -\int d^D x \frac{\sqrt{1-h_+^2 cos^2 (\omega u)}}{2}\Bigg[ \frac{(\partial_x \hat{\phi})^2}{1+h_+ cos(\omega u)} + \frac{(\partial_y \hat{\phi})^2}{1-h_+ cos(\omega u) } + (\partial_z \hat{\phi})^2 -2\partial_u \phi \partial_z \hat{\phi} + m^2 \hat{\phi}^2 \Bigg] \end{split} \end{equation}

Defining the momentum operator $\hat{\pi}$ \begin{equation} \hat{\pi} = \frac{\delta S}{\delta (\partial_u \hat{\phi})} = \sqrt{1-h_+^2 cos^2 (\omega u)}\partial_z \hat{\phi} \end{equation} Now looking at the Hamiltonian \begin{equation} H = \int \mathcal{H} d^{D-1} x; \hspace{0,7cm} \mathcal{H} = (\partial_u \phi) \hat{\pi} - \mathcal{L} \end{equation} Plugging the Lagrangian in \begin{equation} H = \int d^{D-1} x \frac{\sqrt{1-h_+^2 cos^2 (\omega u)}}{2}\Bigg[ \frac{(\partial_x \hat{\phi})^2}{1+h_+ cos(\omega u)} + \frac{(\partial_y \hat{\phi})^2}{1-h_+ cos(\omega u) } + (\partial_z \hat{\phi})^2 + (m \hat{\phi})^2 \Bigg] \end{equation} Writing $\partial_z$ as a function of $\hat{\pi}$, and applying partial integration on $(\partial_i \phi)^2$ (for $i$ is $x$ and $y$) and neglecting boundary terms, you can rewrite the Hamiltonian. \begin{equation} \hat{H}(u,\vec{x}) = \int d^{D-1} x \frac{\sqrt{1-h_+^2 cos^2 (\omega u)}}{2}\Bigg[ \frac{\hat{\pi}^2}{1-h_+^2 cos^2 (\omega u)} - \frac{ \hat{\phi} \partial_x^2 \hat{\phi}}{1+h_+ cos(\omega u)} - \frac{ \hat{\phi} \partial_y ^2 \hat{\phi}}{1-h_+ cos(\omega u) } + m^2 \hat{\phi}^2 \Bigg] \end{equation} Using the canonical commutation relations of $\hat{\phi}$ and $\hat{\pi}$ \begin{equation} [\hat{\phi}(u,\vec{x}), \hat{\pi}(u,\vec{x'})] = i\hbar \delta(\vec{x}-\vec{x}'), \hspace{0,7cm} [\hat{\phi}(u,\vec{x}), \hat{\phi}(u,\vec{x}')] = [\hat{\pi}(u,\vec{x}), \hat{\pi}(u,\vec{x}')] = 0 \end{equation} From the Heisenberg equation of motion and we take the $\hat{\phi}$ operator is not a function on $u$, we can get an equation for the 'time'-derivative of $\hat{\phi}$ \begin{equation} \partial_u \hat{\phi}(u,\vec{x}) = -\frac{[\hat{H}(u,\vec{x}'),\hat{\phi}(u,\vec{x})]}{i\hbar} = -\frac{1}{i\hbar} \int d^{D-1} x' \frac{\sqrt{1-h_+^2 cos^2 (\omega u)}}{2} \Bigg[ \frac{\hat{\pi}(u,\vec{x}')(-i\hbar \delta(\vec{x}-\vec{x}'))}{1-h_+^2 cos^2 (\omega u)} \Bigg] \end{equation} Applying the $\delta$-function on $\hat{\pi}(u,\vec{x}')$, one obtains \begin{equation} \partial_u \hat{\phi}(u,\vec{x}) = \frac{ \hat{\pi}(u,\vec{x}) }{2\sqrt{1-h_+^2 cos^2 (\omega u)}} \end{equation}

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