# Quantisation of gauge field in temporal gauge

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.

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