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We know (say from Griffiths E&M Problem 8.12) that the electric $q_e$ and magnetic charge $q_m$ (with a distance $\vec{z}$ apart) can store the angular momentum in the space: $$ \vec{L}=\int d^3 V (\vec{r} \times (\epsilon_0 \vec{E} \times \vec{B}))=\frac{\mu_0}{4 \pi} q_e q_m \vec{z} = n \frac{\hbar}{2}, \quad n \in \text{integers}. $$ Here the $(\epsilon_0 \vec{E} \times \vec{B})$ is the EM momentum obtained from the Poynting vector. This value in the quantum theory should be quantized $n \frac{\hbar}{2}$ because the angular momentum is quantized in Planck units. This result is general true regardless which $ q_e$ or $ q_m $ charges we take; thus we should better have both $ q_e$ or $ q_m $ charges quantized also in integers. (Any loop holes?)

The above approach to derive is based on

a classical EM derivation + a quantum theory constraint on angular momentum $\vec{L}=n \frac{\hbar}{2}, \quad n \in \text{integers}$.

This is also known as Dirac quantization.

This result implicitly assumes the underlying gauge theory is a $U(1)$ gauge theory instead of the $\mathbb{R}$ gauge theory. This means that the underlying $U(1)$ gauge transformation demands an identification $\theta \in U(1)$: $$ \theta = \theta + 2 \pi $$ where $\theta$ is the parameter in front of the 2-dimensional operator $\exp(i \theta \int * dA)$.

My questions here are that:

  1. Based on the above derivation $\frac{\mu_0}{4 \pi} q_e q_m \vec{z} = n \frac{\hbar}{2}$, how do we see the quantization of $q_e$ or $ q_m $ charges go away to un-quantized, if we change the compact $U(1)$ gauge theory to a $\mathbb{R}$ gauge theory? So $ \theta \in \mathbb{R}$ is no longer identified with other values of $ \theta $.

  2. how do the quantization of $q_e$ or $ q_m $ charges go, if we change the compact radius of the $U(1)$ gauge theory by $k$ times? So $$ \theta = \theta + 2 k \pi. $$

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  • $\begingroup$ There are no magnetic charges in $\mathbb{R}$ gauge theory because there is no possible non-trivial winding of the gauge group around non-trivial (in the deRham sense) cocycles. $\endgroup$ – Richard Myers Mar 1 at 19:58
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First of all, let me just point out that the radius of the $U(1)$ is related to the gauge coupling, so rescaling the radius is essentially the same as rescaling the gauge coupling of the theory. The only place I've seen this pointed out is in references dealing with $S$-duality in electrodynamics since part of the transformation is on the gauge coupling, and hence it matters there because the radius of the $U(1)$ also gets changed as a result under the $S$-transformation. Basically the idea is if you choose the normalization of the gauge field $A$ such that the magnetic charges are integers (this does require a normalization choice and is not implied, as you conjecture, by Dirac's condition), the normalization factor then also multiplies any gauge transformation $d\Lambda$. From this the radius can be deduced, though I don't remember the specifics off the top of my head, just that the result is a factor of $1/e^2$ where $e$ is the gauge coupling.

The more interesting point to make in this answer is that there are no magnetic charges in $\mathbb{R}$ gauge theory. As a result, the Dirac quantization condition is trivial in that case. The difference is a matter of the gauge group's topology, so there isn't a smooth deformation between the two as the original question seems like one might hope. As I mentioned in a comment, the difference is essentially down to the fact that in an $\mathbb{R}$ gauge theory, there can be no non-trivial winding around non-trivial 2-cycles of the gauge parameter.

This is simplest to see in the canonical example of a monopole sitting stationary at the origin. Topologically, the space is $\mathbb{R}^3/\{\boldsymbol 0\}\times\mathbb{R}$. That is, $\mathbb{R}^4$ with the origin in a spatial slice excised. This is the basic example of how a monopole can exist without breaking gauge invariance (Gauss' laws). Because we have deleted a point, we can no longer cover all of spacetime in a single coordinate chart. Usually the way we handle this is to look at just one spatial slice (this is a static situation so the temporal factor can largely be ignored) and in particular a 2-sphere enclosing the deleted point. The space can be covered by two coordinate charts, one valid over the north pole of the sphere (and for all radii) and one valid over the south pole which overlap at the equator.

Since the magnetic charge is defined to be (I have made some normalization choices here for simplicity, these won't matter to the point I'm looking to make) $$ q_m=\int_\Sigma F^{(2)} $$ where $\Sigma$ is any closed 2-dimensional surface and $F^{(2)}$ is the 2-form field strength, we can choose $\Sigma=S^2$ to be the sphere enclosing the deleted point. But we can write this sphere as being the north hemisphere plus the south hemisphere, $S^2=N\cup S$ and write $$ q_m=\int_{N\cup S}d A^{(1)}=\int_N dA^{(1)}+\int_S dA^{(1)}. $$ Applying Stokes' theorem here and noting that $\partial N=E$ and $\partial S=-E$ where $E$ is the equator of the 2-sphere (the important part here really is that the orientations are opposite, the exact signs depend on how you orient $E$), the above may be written $$ q_m=\int_E(A^{(1)}_N-A^{(1)}_S) $$ where $A^{(1)}_N$ is the vector potential in the northern patch and $A^{(1)}_S$ is the potential in the southern patch. From one coordinate patch to another, the vector potential can only ever differ up to a gauge transformation, so there exists some gauge parameter $\Lambda$ such that $A^{(1)}_N-A^{(1)}=d\Lambda$. In terms of this, the above integral may be written (again, Stokes theorem) as $$ q_m=\Lambda(0)-\Lambda(2\pi) $$ where I have written $\Lambda$ as a function of the angular coordinate around the equator.

Now, this is all a lot of words to get to a final expression, $\Lambda(0)-\Lambda(2\pi)$, which at face value should just be zero. The reason it may not be zero, however, is because $\Lambda$ may not be single-valued. This is where the gauge group we are working with finally enters the picture. You see, what we really demand is that the gauge parameter be a smooth function on spacetime valued in the gauge group. If that gauge group is $\mathbb{R}$, then that means that since $0$ and $2\pi$ represent the same point in spacetime, they must also represent the same point in $\mathbb{R}$. For $\mathbb{R}$, that means they are literally the same, so $\Lambda(0)=\Lambda(2\pi)$ and hence $q_m=0$. This is precisely the statement that there are no magnetic monopoles in $\mathbb{R}$ gauge theory.

Just to drive this point home, let's see what were to happen if we took the group to be $U(1)$ instead. Here, because $0$, and $2\pi$ represent the same spacetime point, $\Lambda(0)$ and $\Lambda(2\pi)$ must again still represent the same point, but this time in $U(1)$. Within $U(1)$, we have the identification $\Lambda=\Lambda+2\pi$, and hence for all $n\in\mathbb{Z}$, $\Lambda+2\pi n$ represent precisely the same point. Hence, we then must have that $\Lambda(0)=\Lambda(2\pi)+2\pi n$ for any integer $n$, this integer being precisely the winding around the $U(1)$.

Putting this into our expression for the magnetic charge, we immediately obtain $q_m=2\pi n$ (again, remember, I made some normalization choices from the beginning for simplicity). So this is how in $U(1)$ we are able to obtain magnetic monopoles. The key is the difference in topology between $U(1)$ and the $\mathbb{R}$.

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