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Whenever we use temporal gauge and quantise gauge field we implement Gauss law. I have seen some papers but the point is not cleared to me that why we implement Gauss law there. Please explain this if possible.

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The reason for this is that Gauß' law is not an equation of motion, but a constraint. The Lagrangian $$ L = -\frac{1}{4}\int F^{\mu\nu}F_{\mu\nu}\mathrm{d}^4 x $$ gives us the canonical momenta $\pi^\mu = F^{\mu0}$ and hence a primary constraint $\pi^0\approx 0$ ($\approx$ means an equality that hold on the constraint surface/upon use of the equations of motion). The Hamiltonian reads $$ H = \int \left(\frac{1}{2}\pi^i \pi_i + \frac{1}{4}F^{ij}F_{ij} - A_0 \partial_i\pi^i\right)\mathrm{d}^3x$$ where the Roman indices run only over spatial indices and so we incur the secondary constraint $$ \{\pi^0, H\} = \{\pi^0, -A_0\partial_i\pi^i\} = \partial_i\pi^i \approx 0,$$ which, since $\pi^i = F^{0i} = E^i$, is just Gauß' law. As $\{\pi^0, \partial_i\pi^i\} = 0$, both constraints are first-class and both generate gauge transformations.

The constraint $\pi^0$ generates gauge transformations $A_0 \mapsto A_0 + c$ for arbitrary functions $c$, and so the temporal gauge condition $A_0 = 0$ fixes the gauge for this constraint completely. What remains is a theory with only the spatial variables $(A_i,\pi^i)$ and the first-class constraint $\partial_i\pi^i$ generating residual gauge transformations $A_i \mapsto A_i + \partial_i\alpha$ for arbitrary functions $\alpha$, which you have to still deal with in quantization.

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  • $\begingroup$ If you further impose $\partial_i A^i \approx 0$, then you can use the Dirac bracket on the reduced phase space. $\endgroup$
    – DanielC
    Commented Apr 4, 2021 at 16:41

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