# Is the Dirac monopole quantization condition out by 1/2?

Consider an electron with electric charge $$e$$ that is moved around a closed horizontal circular path $$C$$ centered around a magnetic monopole with magnetic charge $$g$$. Assume that the monopole produces a radial magnetic field $$\vec{B}$$ perpendicular to the surface of a sphere $$S$$ inscribed in the path $$C$$.

The electron wavefunction acquires a phase shift $$\phi$$ given by $$\phi=\frac{e}{\hbar}\int_C\vec{A}\cdot d\vec{x},$$ where $$\vec{A}$$ is the electromagnetic vector potential defined by $$\vec{B}=\nabla\times\vec{A}$$.

Using Stokes' theorem we find that the phase change $$\phi$$ is given by the sum of the magnetic flux through the upper hemisphere $$U$$ and the lower hemisphere $$L$$: $$\phi=\frac{e}{\hbar}\int_U \vec{B}\cdot d\vec{S}+\frac{e}{\hbar}\int_L \vec{B}\cdot d\vec{S}.$$ The magnetic flux on the entire sphere $$S$$ due to the magnetic charge $$g$$ at its center is given by $$\int_S \vec{B}.d\vec{S} = \mu_0 g.$$ Therefore the phase shift of the electron wavefunction $$\phi$$ is given by $$\phi=\frac{\mu_0 e g}{\hbar}.$$ Now the standard definition of the Dirac quantization condition assumes that the electron wavefunction is unchanged after each orbit around the magnetic monopole. However the magnetic field of the monopole can interact with the electron causing its magnetic moment and therefore its spin to align radially. This will cause the electron spin to rotate as the electron orbits the monopole. As the electron is a fermion one spin rotation ($$2\pi$$) flips the sign of its wavefunction. It takes two spin rotations ($$4\pi$$) for the electron wavefunction to get back to its original state. Therefore the Dirac quantization condition should be given by $$\frac{\mu_0 e g}{\hbar}=4\pi n$$ or $$\frac{e g}{4\pi\epsilon_0\hbar c^2}=n,$$ where $$\epsilon_0 \mu_0 = 1/c^2$$ and $$n$$ is an integer.

Is this correct?

Correction

I agree with Meng Cheng that Stokes’ Theorem should only be applied to one hemisphere. I agree with ACuriousMind that spin shouldn’t enter into the argument. Putting those two conditions together I still come to the conclusion that the Dirac monopole condition is out by a factor $$1/2$$.

• Why would the fermion nature of the electron matter at all? What half-integer spin means is that under application of the rotation operator for a full rotation, the quantum state picks up a negative sign. Even though your electron travels in a circle, travelling along a circle is not at all the same operation as applying a full rotation operator. As an analogy: Spinning a ball on your fingertips is not the same operation as carrying it with you as you run in circles. Commented Sep 15, 2022 at 18:38
• Would the radial magnetic field from the monopole induce the electron to align its magnetic moment and thus its spin radially? As the electron orbits the monopole its spin would rotate. Commented Sep 15, 2022 at 20:11
• I agree with Meng Cheng that Stokes Theorem should only be applied to one hemisphere. I agree with ACuriousMind that spin shouldn’t enter into the argument. Putting those two conditions together I still come to the conclusion that the Dirac monopole condition is out by a factor $1/2$. Commented Jan 22, 2023 at 12:04

Note that an error in your reasoning is the wrong use of Stokes theorem. The expression for $$\phi$$ should only contain the northern or southern hemisphere, but not both.