Conditions when using Magnetic Scalar potentials Under what circumstances may the magnetic field H be written as H = −∇φ,
where φ is a scalar potential which satisfies Laplace’s equation?
Note that I'm perfectly aware that this implies the curl of B must equate to 0, meaning the sum of displacement current and (both polarisation and free) current equate to 0.  However, does this automatically imply that the displacement current and polarisation/free current equate to 0?
 A: The magnetic scalar potential $\phi$ is for instance used in the design of magnets with different number of poles, i.e. dipoles, quadrupoles and sextupoles etc. These are used in order to generate static magnetic fields (so displacement current zero) for the guidance and focalisation of charged particle beams, for instance in accelerators. So inside of the empty part of such a magnet $\nabla \times H =0$, so $H=-\nabla \phi$, whereas in the (non-empty part) magnetic poles/material $H=0$ is assumed, because $\mu_r\gg1$ inside the magnetic material.  For the design of magnets finally the equation $\Delta \phi =0$ is solved with boundary conditions for $\phi$ at the inside border of magnetic poles.  Or more formally:
$$-\Delta \phi = \nabla \cdot H = \frac{1}{\mu_0}\nabla \cdot B - \nabla \cdot M  =  - \nabla \cdot  M|_{\mathrm{@border} \neq 0\,\,/\,\, \mathrm{inside} =0 }  $$
where $M$ is the magnetic polarisation. However, $\nabla \cdot M$ is only non-zero at the border between the (typically vacuum) interior of the magnet and the magnet poles, i.e. it defines the boundary conditions at the border of the area the scalar potential is computed. Everywhere else  $\nabla \cdot M$ is zero. 
The problem of the computation of the magnetic field is relegated to the solution of the Laplace-equation with boundary conditions which is a standard engineering problem, i.e. made easy. That's the purpose of the magnetic scalar potential $\phi$.
