In standard QED, we couple the electron to electromagnetism by replacing $$\partial_\mu \to \partial_\mu + i e A_\mu.$$ Upon taking the classical limit, we find that this gives electrons an electric charge of $e$ and zero magnetic charge.

Accommodating magnetic monopoles looks difficult, since their presence makes the vector potential behave weirdly. From a little searching, the actual theory of magnetic monopoles looks very complicated, but I just want to know what term one would write down.

  1. What term can be added to the Lagrangian to give something a magnetic charge?

  2. With magnetic monopoles, does the gauge group of E&M remain $U(1)$?


1 Answer 1


You cannot just add a term to the Lagrangian to give the usual electromagnetic gauge theory magnetic charge. The reason is rather simple: The equation of motion for a magnetic four-current $j_m$ is $\mathrm{d}F = j_m$. But $\mathrm{d} F = \mathrm{d}\mathrm{d} A = 0$ independently of the equations of motion. So simply adding a term doesn't work.

The first way out is viewing monopoles as topological objects, whose location is removed from the spacetime on which the gauge theory is considered (or "where the gauge field is singular"). How this leads to e.g. the Dirac string and charge quantization I describe in this answer of mine. No term is added to the Lagrangian, the appearance of the monopole is of a purely topological nature, and $\mathrm{d}F = 0$ everywhere where it is defined.

Another way out is to introduce a manifestly electro-magnetic dual Lagrangian with an electric four-potential $A$ and a magnetic four-potential $B$ unfortunately with a non-canonical choice of spacelike four-vector $n$ where the Lagrangian is now \begin{align} L = \frac{1}{2n^2} & \left( - \mathrm{d}A(n) \cdot {\star}\mathrm{d}B(n) + \mathrm{d} B(n) \cdot \mathrm{d}{\star}A(n) - \mathrm{d}A(n)\cdot{\star}\mathrm{d}A(n) - \mathrm{d}B(n)\cdot {\star}\mathrm{d}B(n)\right) \\ & - A\cdot j_e - B\cdot j_m \end{align} where $j_e,j_m$ are the electic/magnetic currents, $X(n)$ denotes the contraction of the form $X$ with the vector $n$ in its first slot/index, and all the $\cdots$ are the usual Minkowski inner products of (co)vectors. The field strength is now given by several equivalent formulae, one of them is $F = \mathrm{d}A - (n\cdot\partial)^{-1} ({\star} n\wedge j_m)$, where the expression $(n\cdot\partial)^{-1}$ the integral operator whose kernel is the Green's function of $(n\cdot \partial) f = 0$ (at least, for consistency, this is the original field strength $\mathrm{d}A$ for $j_m = 0$!). This solution (which I may have butchered transmitting it into my notation) is due to Zwanziger in Local-Lagrangian Quantum Field Theory of Electric and Magnetic Charges.

A third way is to postulate that the electromagnetic $\mathrm{U}(1)$ comes from breaking a $\mathrm{SU}(2)$ through a Higgs-like mechanism, then there are 't Hooft-Polyakov monopoles, whose far field looks like a magnetic monopole for the unbroken $\mathrm{U}(1)$ but which does not require the removal of the location of the monopole (because it is not actually located at a point, the field is non-singular everywhere). This however introduces further modifications into the theory because now you have additionally the massive bosons of the broken symmetry.

  • $\begingroup$ If you consider a GUT with magnetic monopoles then what emerges at low energy is ordinary QED with magnetic charges and the usual $D_\mu=\partial_\mu+ieA_\mu$ coupling. The gauge potential is multi-valued, but the Dirac string is unobservable by the Dirac condition. $\endgroup$
    – Thomas
    Jun 7, 2016 at 18:00

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