What is the relativistic Hamiltonian for a charged particle in an EM field, using the magnetic scalar potential? The Hamiltonian for a relativistic charged particle moving in a static electromagnetic field is the well known: $$H=c\sqrt{\left(\mathbf{P}-q\mathbf{A}\right)^{2}+m^{2}c^{2}}+q\phi$$
where,\begin{align*}
\mathbf{B} & =\nabla\times\mathbf{A},\\
\mathbf{E} & =-\nabla\phi.
\end{align*}
Now let's suppose that one wants to write the magnetic field in terms of the magnetic scalar potential, $\phi_M$, rather than of $\mathbf{A}$, that is for a magnetic field written as:
$$\mathbf{B}=-\nabla\phi_{M}.$$
What would the Hamiltonian look like, in terms of $\phi_M$ ?
 A: One way would be to express the vector potential as function[al] of the scalar potential and substitute in the Hamiltonian.
Magnetic field (induction) can be expressed as gradient of a potential function only in a limited region of space where the integral
$$
\int_1^2 \mathbf B\cdot d\mathbf s 
$$
between two points does not depend on the path; it may depend only on the endpoints.
Suppose we have such a region and such a potential function so magnetic field is a gradient.
Then, we can ( in principle ) solve the equation
$$
- \nabla \phi = \nabla \times \mathbf A
$$
for unknown function $\mathbf A(\mathbf x)$. Suppose we found a solution which is expressed as function[al] of $\phi$ and space coordinates. Then we can substitute for $\mathbf A$ in the standard Hamiltonian you mentioned and thus obtain Hamiltonian that refers to $\phi$ and space coordinates only.
For example, if the magnetic field is uniform, so we can express it as $\mathbf B=[0,0,B_0]$, the potentials that would give the magnetic field correctly could be defined as
$$
\phi = -B_0z
$$
$$
\mathbf A = \bigg[-\frac{1}{2}B_0 y,\frac{1}{2}B_0 x,0\bigg]
$$
However, it is easy to see that vector potential at some point is not simply a function of scalar potential at the same point; if we are to relate the two, we need to involve also the spatial coordinates explicitly:
$$
\mathbf A = \bigg[-\frac{\phi}{2} \frac{y}{z},\frac{\phi}{2}\frac{x}{z},0\bigg ]
$$
Now we can substitute for $\mathbf A$ in the Hamiltonian.
Finding $\mathbf A$ was simple here but for general magnetic field, the relation would be more complicated, probably integral: $\mathbf A$ would be a functional of $\phi$. Such complicated expression would be hard to work with in the Hamiltonian. Better use directly the vector potential.
A: As far as I know, each vector field can be unambiguously split into a gradient and a rotational part:
$$\mathbf{F} = \nabla \phi + \nabla \times \mathbf A$$
This is called the Helmholtz decomposition: https://en.wikipedia.org/wiki/Helmholtz_decomposition
Due to your assumption $\mathbf{B} = \nabla \times \mathbf A$ it follows that the gradient part is identically zero.
