Why are magnetic monopoles "incompatible" with quantum mechanics? I'm reading this Physics Today article on magnetic monopoles, and I'm a bit confused by a discussion of the necessity of Dirac strings for compatibility with quantum mechanics. I'll reproduce the relevant discussion here:

At first sight, magnetic monopoles seem to be incompatible with
quantum mechanics. This is because, in quantum mechanics,
electromagnetic fields have to be described in terms of a scalar
potential $\phi$ and vector potential, $\vec{B} = \nabla \times \vec{A}$, and it
follows [that] the field must then be sourceless, $\nabla \cdot \vec{B} = 0$

It's clear, of course, that if $\vec{B} = \nabla \times \vec{A}$, then $\nabla \cdot \vec{B} = 0$. I don't understand, however, what quantum mechanics has to do with any of this.
Under classical electrodynamics, the magnetic vector potential is defined to be that whose curl gives the magnetic field. Together with electric potential $\phi$, we may specify the electric field.
It is stated earlier in the article that magnetic monopoles are compatible with classical electrodynamics. It goes on to suggest that they are (seemingly) incompatible with quantum mechanics, but then argues this using what seems to be classical electrodynamics.
How is this related to QM?
 A: You are correct. This really has nothing to do with quantum mechanics. The existence of magnetic monopoles would require a change in Maxwell's equations (i.e., the divergence of the magnetic field would be nonzero).
Schrödinger's equation is more general than Maxwell's equations and can accommodate the changes you make to them. If you do make this monopole modification, then you would need to alter both the classical and quantum electromagnetic Hamiltonian. However, once you do, then Schrödinger's equation will work as usual.
A: The argument here is supposed to go something like this:

*

*Classical electrodynamics, as long as you don't insist on a Lagrangian formulation, is fully described by Maxwell's equations in terms of the field strength tensor/the electric and magnetic fields. The 4-potential of electrodynamics is a convenient description of electromagnetism, but adding a magnetic charge current to Maxwell's equations and thereby destroying the sourcelessness of the magnetic field that is necessary for the 4-potential to exist is straightforward and poses no obvious problems on the level of Maxwell's equations (see also this answer of mine).


*Quantum mechanics doesn't work with "Maxwell's equations", it's based on quantizing an action formulation (e.g. Lagrangian formulation via path integral quantization or Hamiltonian formulation via canonical quantization) of electromagnetism where the dynamical variable is the 4-potential. Introducing a magnetic charge makes it impossible to define a 4-potential, and therefore there is no straightforward modification of the "quantizable version" (i.e. the action formulation) of electromagnetism that is compatible with the existence of magnetic charges. (The observability of the Aharonov-Bohm effect is often interpreted as the experimental sign of this reliance on the 4-potential, but this is a different can of worms)
I discuss various ways to treat magnetic monopoles in an action formulation in more detail e.g. in this answer of mine
A: Magnetic monopoles are incompatible with $\nabla \cdot B=0$ hence with $B = \nabla \times A$. They are incompatible with the Maxwell equations and with the electromagnetic potential. The electromagnetic potential is required for Lagrangian and Hamiltonian mechanics on which quantum mechanics as we know it is based.
A: For your first question, on the link between quantum mechanics and potentials is that currently, there is no formulation of QM without these potentials and only in terms of the gauge invariant EM field. For example, the Schrödinger equation of a free spinless particle in an EM field is:
$$
i\frac{\partial \psi}{\partial t} = (-i\vec \nabla-q\vec A)^2\psi+qV\psi
$$
and in general, QFT's (including the famously accurate QED) are always formulated using these potentials.
The first difference between classical electrodynamics and is that in the former necessity of fields is not imposed (even if they are mathematically convenient), making monopoles no big issue. This is what the author hints at in the modified version of Maxwell's equations.
Even if we restore gauge invariance in the classical case, there is also another more subtle difference in the nature of allowed gauge transformations which are more restrictive in QM. Classically, $\vec A$ is defined up to an additive rotationless vector field which can be locally expressed as the gradient of a function. In QM, $\vec A$ is usually interpreted as the gauge field of $U(1)$, so the additive field must be globally the gradient of a function, with the caveat that it is defined up to a $2\pi$ multiple since it is a phase. For example, using cylindrical coordinates, $\vec A \to \vec A+ \vec \nabla (r\phi)$ would be a valid gauge transformation in either case but $\vec A \to \vec A+ \vec\nabla (r\phi/2)$ would not be one in QM. This is what leads to the famous quantization conditions that Dirac later found.
Finally, for your final remark, I don't quite understand it as the author mentions Dirac solution, reconciling monopoles with QM, so apart for the lacking experimental evidence, he actually argues for their compatibility.
Hope this helps and tell me if you find some mistakes.
