Why use potential functions to solve Maxwell's equations?

I'm reading "Antennas and Wave Propagation" by Harish and Sachidananda. They introduce the phasor form of Maxwell's equations:

$$\nabla\times E=-j\omega\mu H$$ $$\nabla\times H = j\omega\epsilon E + J$$ $$\nabla\cdot D = \rho$$ $$\nabla\cdot B = 0$$

First of all they observe that by taking the curl of the first equation and rewriting the right hand side with the expression for $\nabla\times H$, we can show that $E$ satisfies a wave equation: $\nabla^2E+aE=bJ$, for some constants $a$ and $b$. They also note that in the same way we can show that $H$ satisfies the wave equation.

Then they introduce the notion of vector and scalar potential, and propose to find $V$, the scalar potential of $E$, and $A$, the vector potential of $H$. The motivation for doing this seems to be that finding $A$ and $V$ is easier because they both satisfy... the wave equation.

Why solve the wave equation to get the potentials, and then work out $E$ and $H$ from there? Why not just solve the wave equations for $E$ and $H$ directly?

• Related: physics.stackexchange.com/q/22747/2451 , physics.stackexchange.com/q/397633/2451 and links therein. – Qmechanic Jul 25 '18 at 15:00
• @Qmechanic The math is a little above my head there. In particular I've never heard the terms "degrees of freedom" and "gauge transformation". – Jack M Jul 25 '18 at 15:02
• Obviously having only (3+1) wave equations for the $\textbf{A},V$potentials is simpler than using the otherwise equally complicated (3+3) equations for the vectors $\textbf{E},\textbf{H}$. This simplification happens because the two vector Helmholtz equations are not really independent and the equivalent potential formulation proves that. – hyportnex Jul 25 '18 at 15:27
• @hyportnex Oh, so it's a matter of revealing the true nature of the problem? We're expressing the equations in terms of a minimal set of independent variables, rather than a set of bound variables? – Jack M Jul 25 '18 at 15:59
• I do not know what "true nature" might mean in this context but the vector Helmholtz equations are not independent because the solutions must also satisfy the Maxwell equations $curl\textbf{E}=-i\omega\mu_0 \textbf{H}$ and $curl\textbf{H}= i\omega\epsilon_0\textbf{E}$. The vector and scalar potentials are almost independent subject to a so-called gauge invariance, see en.wikipedia.org/wiki/Gauge_fixing. – hyportnex Jul 25 '18 at 16:10