(Chandrasekhar and Kendall, 1957) I am considering the Helmholtz vector equation (in $\mathbb R^3$): $$\nabla^2 \bf H+\alpha^2 H=0.$$ Well known solutions include, \begin{align} \bf L&=\nabla \psi \\ \bf T&=\nabla\times (\bf a\psi)\\ \bf S&=\nabla\times\bf T/\alpha \end{align} where $\psi$ is a scalar function such that $\nabla^2 \psi+\alpha^2 \psi=0$.
Why are the above three solutions are the solution for Helmholtz vector equation?
PS: $\alpha =$constant and $\nabla.{\bf H}=0,\ \bf a$ is a unit vector.
Given $\nabla^2\phi+\alpha^2\phi=0$, we can multiply by constant unit vector $\bf \hat{a}$ we will get $$\nabla^2\phi\bf\hat{a}+\alpha^2\phi\bf\hat{a}=0$$ using vector identity $\bf\nabla^2H=\nabla(\nabla.H)-\nabla\times\nabla\times H$, we have $$\nabla(\nabla.\phi\bf\hat{a})-\nabla\times\nabla\times\phi\bf\hat{a}+\alpha^2\phi\bf\hat{a}=0$$ Now, on taking curl, we get $$\nabla\times(\nabla(\nabla.\phi\bf\hat{a}))-\nabla\times(\nabla\times\nabla\times\phi\bf\hat{a})+\nabla\times(\alpha^2\phi\bf\hat{a})=0$$ and for some scalar function $f$, we have vector identity, $\nabla\times\nabla f=0$, here $f=\nabla.\phi\bf\hat{a}$ $$-\nabla\times\nabla\times(\nabla\times\phi\bf\hat{a})+\nabla\times\alpha^2\phi\bf\hat{a}=0$$ using vector identity again, we get $$\nabla^2(\nabla\times\phi\bf\hat{a})+\alpha^2(\nabla\times\phi\bf\hat{a})=0$$ Now if we write $\nabla\times\phi\bf\hat{a}$ as $\bf H$, we have $$\bf\nabla^2H+\alpha^2H=0$$ Same goes for $\bf S$.
NOTE: Solution of scalar Helmholtz equation is a subset of the solution of Vector Helmholtz equation.