Magnetic Monopoles, Dirac-Zwanziger Condition

One of the most beautiful facts in quantum mechanics is that the existence of a magnetic monopole would force electric charge to be quantized. If a particle of electric charge (e) moves in the field of a magnetic monopole of magnetic charge (g), then consistency of the quantum theory requires

$\displaystyle eg = \frac{n\hbar c}{2}, \qquad n\in \mathbb Z.$

Equivalently, if we define the dimensionless charge-monopole coupling $\displaystyle q := \frac{eg}{\hbar c},$ then the condition is $\displaystyle 2q\in \mathbb Z$ .

There are two famous derivations of this condition.

The first is the gauge-patch derivation. It says that the magnetic field of a monopole cannot be represented by a single smooth vector potential on the whole sphere surrounding the monopole. We must cover the sphere by two patches, with two vector potentials $A_N$ . and $A_S$ .. On the overlap, these two potentials differ by a gauge transformation. Since a charged quantum wavefunction transforms by a phase under a gauge transformation, the two local wavefunctions must be glued together by a phase-valued transition function. The requirement that this transition function be single-valued around the equator forces $2q\in\mathbb Z$ .

The second is the angular-momentum derivation, often associated with Saha. It says that a charge-monopole pair carries an intrinsic angular momentum. The correct conserved angular momentum is not just $\mathbf r\times \boldsymbol\pi$ , but rather

$\displaystyle \mathbf J = \mathbf r\times \boldsymbol\pi - \frac{eg}{c}\widehat{\mathbf r}.$

The radial projection of $\mathbf J$ is fixed:

$\displaystyle \mathbf J\cdot \widehat{\mathbf r} = -\frac{eg}{c} = -q\hbar.$

But quantum angular momentum projections occur in integer or half-integer multiples of $\hbar$ . Thus $q$ must be an integer or half-integer, again giving $2q\in\mathbb Z$ .

At first glance these look like two unrelated arguments. The first speaks the language of charts, transition functions, gauge potentials, and topology. The second speaks the language of conserved quantities, commutation relations, and angular momentum eigenvalues. The point of this exposition is to show that they are not merely analogous. They are the same computation expressed in two different coordinate systems on the space of ideas.

The gauge-patch derivation computes the integer $2q$ as the winding number of a transition function $S^1\to U(1).$ The angular-momentum derivation computes the same integer as the weight of a $U(1)$ representation sitting inside the rotation group. Both are ultimately describing the same object: a complex line bundle over $S^2$ with connection and first Chern number $2q.$ The charged wavefunction is not an ordinary function on $S^2$ . It is a section of this line bundle. The Dirac quantization condition is precisely the condition that this line bundle exists globally.

The magnetic monopole field

Place a magnetic monopole of magnetic charge $g$ at the origin. Its magnetic field is given by

$\displaystyle \mathbf B = g\frac{\mathbf r}{r^3}=g\frac{\widehat{\mathbf r}}{r^2}.$

In spherical coordinates, the outward area element on a sphere of radius $r$ is $\displaystyle d\mathbf S = r^2\sin\theta ~d\theta ~d\phi ~\widehat{\mathbf r}.$ Therefore the magnetic flux through the sphere is

$\displaystyle \int_{S^2}\mathbf B\cdot d\mathbf S = \int_0^{2\pi}\int_0^\pi \left(g\frac{\widehat{\mathbf r}}{r^2}\right)\cdot \left(r^2\sin\theta ~\widehat{\mathbf r}~d\theta ~d\phi\right) = g\int_0^{2\pi}\int_0^\pi \sin\theta ~d\theta ~d\phi.$

So we get $\displaystyle \int_{S^2}\mathbf B\cdot d\mathbf S = 4\pi g.$ This nonzero flux is the basic source of all the interesting mathematics. In differential-form notation, the magnetic field on the sphere is represented by the curvature two-form $F = g\sin\theta ~d\theta\wedge d\phi.$ Indeed,

$\displaystyle \int_{S^2}F = \int_0^{2\pi}\int_0^\pi g\sin\theta ~d\theta ~d\phi =4\pi g.$

The usual relation between magnetic field and vector potential is $\displaystyle \mathbf B = \nabla \times \mathbf A.$ In differential forms this becomes $F=dA,$ where $A$ is the magnetic vector potential regarded as a one-form. The first question is: can we find one globally defined smooth one-form $A$ on all of $S^2$ such that $F=dA$ ?

The answer is no. Suppose, for contradiction, that a globally smooth one-form $A$ existed on all of $S^2$ with $F=dA.$ Then by Stokes’ theorem,

$\displaystyle \int_{S^2}F = \int_{S^2}dA = \int_{\partial S^2}A.$

But the sphere has no boundary: $\displaystyle \partial S^2=\varnothing.$ Therefore

$\displaystyle \int_{S^2}dA=0.$

This contradicts $\displaystyle \int_{S^2}F=4\pi g,$ unless $g=0$ .

Thus, for a genuine monopole, there is no globally smooth vector potential $A$ on $S^2.$ This is the geometric obstruction. It is not a computational inconvenience. It is a topological fact. The magnetic field is globally defined. The vector potential is only locally defined.

To describe the monopole, we cover the sphere by two open sets: $U_N = S^2\setminus{\text{south pole}},$ $U_S = S^2\setminus{\text{north pole}}.$ On $U_N$ , define $A_N = g(1-\cos\theta) ~d\phi.$ On $U_S$ , define $A_S = -g(1+\cos\theta)~ d\phi.$ These are not globally defined on all of $S^2$ , because $d\phi$ is singular at the poles. But $A_N$ is regular at the north pole, and $A_S$ is regular at the south pole. Let us check this. Near the north pole, $\theta\approx 0$ . Since $1-\cos\theta\sim \frac{\theta^2}{2},$ we have $A_N\sim g\frac{\theta^2}{2} ~ d\phi.$ Although $d\phi$ is singular at the pole, the factor $\theta^2$ suppresses the singularity. Thus $A_N$ is regular near the north pole. Near the south pole, $\theta\approx \pi$ . Since $1+\cos\theta\to 0$ , the potential $A_S=-g(1+\cos\theta)d\phi$ is regular near the south pole.

Now compute their exterior derivatives.

For $A_N$ , $\displaystyle dA_N = d\bigl(g(1-\cos\theta)d\phi\bigr) = g ~d(1-\cos\theta)\wedge d\phi =g\sin\theta~ d\theta\wedge d\phi.$

For $A_S$ , $dA_S = d\bigl(-g(1+\cos\theta)d\phi\bigr) = -g~d(1+\cos\theta)\wedge d\phi=g\sin\theta~ d\theta \wedge d\phi.$

Thus $dA_N=dA_S=F$ and the two local vector potentials give the same magnetic field.

The gauge transformation on the overlap

The two patches overlap on the region away from both poles. On $U_N\cap U_S$ , both $A_N$ and $A_S$ are valid descriptions. They must therefore differ by a gauge transformation.

Compute: $\displaystyle A_N-A_S = g(1-\cos\theta)~d\phi - \bigl[-g(1+\cos\theta)~d\phi\bigr].$

So $A_N-A_S = g(1-\cos\theta)d\phi + g(1+\cos\theta)~ d\phi = 2g~d\phi.$

Thus $A_N=A_S+d\lambda,$ where locally $\lambda=2g\phi.$

This equation is already very close to the quantization condition. The subtlety is that $\phi$ is an angular coordinate. The values $\phi$ and $\phi+2\pi$ represent the same point. Therefore $\lambda=2g\phi$ is not itself a globally single-valued ordinary function on the overlap circle. What must be single-valued is not $\lambda$ , but the quantum-mechanical phase generated by $\lambda$ .

Charged wavefunctions and gauge transformations

A charged particle couples to the gauge potential through the covariant derivative. In one common convention,

$D = d - \frac{ie}{\hbar c}A.$

The precise sign convention is not important for the quantization condition. What matters is that under a gauge transformation $A\mapsto A+d\lambda,$ the wavefunction transforms by a phase

$\displaystyle \psi\mapsto \exp\left(\frac{ie}{\hbar c}\lambda\right)\psi.$

Therefore, on the overlap $U_N\cap U_S$ , the local wavefunctions $\psi_N$ and $\psi_S$ must be related by

$\displaystyle \psi_N = \exp\left(\frac{ie}{\hbar c}\lambda\right)\psi_S.$

Since $\lambda=2g\phi,$ we get

$\displaystyle \psi_N = \exp\left(\frac{2ieg}{\hbar c}\phi\right)\psi_S.$

Using $q=\frac{eg}{\hbar c},$ this becomes

$\displaystyle \psi_N=e^{2iq\phi}\psi_S.$

This equation is the heart of the gauge-patch derivation. It says: the wavefunction is not a single ordinary function $\psi$ on the whole sphere. Rather, it is a pair of local functions $\psi_N\quad\text{on }U_N, \qquad \psi_S\quad\text{on }U_S,$ related on the overlap by

$\displaystyle \psi_N=e^{2iq\phi}\psi_S.$

This is exactly what it means for $\psi$ to be a section of a complex line bundle over $S^2$ .

Single-valuedness and the Dirac condition

On the equator, $\displaystyle \phi$ is a coordinate on a circle. The points $\displaystyle \phi$ and $\phi+2\pi$ are identical. The transition function $\displaystyle t_{NS}(\phi)=e^{2iq\phi}$ must therefore be a well-defined map $\displaystyle S^1\to U(1).$ That means $t_{NS}(\phi+2\pi)=t_{NS}(\phi).$ Compute $\displaystyle t_{NS}(\phi+2\pi) = e^{2iq(\phi+2\pi)} = e^{2iq\phi}e^{4\pi iq}.$ For this to equal $\displaystyle e^{2iq\phi}$ , we need $e^{4\pi iq}=1.$ This is true exactly when $\displaystyle 4\pi q = 2\pi n, n\in\mathbb Z.$ Therefore $\displaystyle 2q=n.$ Since $q=eg/(\hbar c)$ , this gives

$\displaystyle \frac{2eg}{\hbar c}=n,$

which is the same as

$\displaystyle \boxed{eg=\frac{n\hbar c}{2},\qquad n\in\mathbb Z.}$

This is Dirac quantization.

Notice what has really happened. The function $\displaystyle t_{NS}(\phi)=e^{2iq\phi}$ winds around the unit circle as $\displaystyle \phi$ goes once around the equator. Its winding number is $\displaystyle n=2q.$ Since winding numbers are integers, $\displaystyle 2q$ must be an integer.

Thus the gauge-patch proof says: Dirac quantization is the integrality of the winding number of the transition function.

The same statement can be expressed in terms of the first Chern class. The charged particle sees the connection not simply as $\displaystyle A$ , but as the dimensionless connection $\displaystyle \frac{e}{\hbar c}A.$ Its curvature is $\displaystyle \frac{e}{\hbar c}F.$

The first Chern number is $\displaystyle c_1 = \frac{1}{2\pi}\int_{S^2}\frac{e}{\hbar c}F.$

Using $\displaystyle \int_{S^2}F=4\pi g,$ we get $\displaystyle c_1 = \frac{1}{2\pi}\frac{e}{\hbar c}(4\pi g = \frac{2eg}{\hbar c} =2q.$ Since $c_1$ is an integer for any complex line bundle, $2q\in\mathbb Z.$ So the gauge-patch derivation may be summarized as

$\boxed{ \displaystyle \frac{1}{2\pi}\int_{S^2}\frac{e}{\hbar c}F\in\mathbb Z.}$

This is the integrality of the first Chern class.

The angular-momentum argument

Let us now forget the patch construction temporarily and derive the same condition from angular momentum.

A particle of charge $\displaystyle e$ in a vector potential $\displaystyle \mathbf A$ has canonical momentum $\displaystyle \mathbf p=-i\hbar\nabla,$ but the gauge-covariant, or mechanical, momentum is

$\displaystyle \boldsymbol\pi = \mathbf p - \frac{e}{c}\mathbf A.$

The Hamiltonian for a nonrelativistic particle of mass $m$ is $\displaystyle H=\frac{1}{2m}\boldsymbol\pi^2.$ The components of $\displaystyle \boldsymbol\pi$ do not commute. One has

$\displaystyle [\pi_i,\pi_j] = i\hbar\frac{e}{c}\epsilon_{ijk}B_k$

up to sign convention depending on how $\displaystyle \boldsymbol\pi$ is defined. With the monopole field $\displaystyle \mathbf B=g\frac{\mathbf r}{r^3},$ this becomes

$\displaystyle [\pi_i,\pi_j] = i\hbar\frac{eg}{c}\epsilon_{ijk}\frac{r_k}{r^3}.$

The naive orbital angular momentum is $\displaystyle \mathbf L = \mathbf r\times\boldsymbol\pi.$ For an ordinary central force problem, $\displaystyle \mathbf L$ would be conserved. But here $\displaystyle \mathbf L$ is not the correct conserved angular momentum. The correct object is

$\displaystyle \mathbf J=\mathbf r\times\boldsymbol\pi - \frac{eg}{c}\widehat{\mathbf r}.$

In terms of $\displaystyle q=eg/(\hbar c)$ , this is

$\displaystyle \mathbf J=\mathbf r\times\boldsymbol\pi - q\hbar\widehat{\mathbf r}.$

The second term is the characteristic monopole correction. There are several ways to understand the extra term.

Classically, the electromagnetic field of an electric charge and a magnetic monopole carries angular momentum. The field angular momentum is proportional to $\displaystyle \frac{eg}{c}\widehat{\mathbf r}.$ Depending on conventions for the direction of $\displaystyle \widehat{\mathbf r}$ , this gives precisely the additional contribution needed so that the total angular momentum is conserved.

Quantum mechanically, the extra term is required so that the generators of rotations satisfy the usual angular momentum algebra:

$\displaystyle [J_i,J_j]=i\hbar\epsilon_{ijk}J_k.$

The ordinary expression $\displaystyle \mathbf r\times\boldsymbol\pi$ alone does not have the correct algebra in the monopole background. The noncommutativity of the covariant momenta produces extra curvature terms. The correction $\displaystyle -\frac{eg}{c}\widehat{\mathbf r}$ cancels the unwanted curvature contribution and restores the standard rotation algebra. Thus $\displaystyle\mathbf J$ is the true generator of rotations for a charged particle in a monopole background.

Now comes the key elementary computation. Since $\displaystyle \mathbf r\times\boldsymbol\pi$ is perpendicular to $\displaystyle \mathbf r$ , it is also perpendicular to $\displaystyle \widehat{\mathbf r}$ . Therefore $\displaystyle \mathbf r\times\boldsymbol\pi)\cdot\widehat{\mathbf r}=0.$ Hence

$\displaystyle \mathbf J\cdot\widehat{\mathbf r} = \left(\mathbf r\times\boldsymbol\pi - \frac{eg}{c}\widehat{\mathbf r}\right)\cdot\widehat{\mathbf r}.$

The first term vanishes, and $\displaystyle \widehat{\mathbf r}\cdot\widehat{\mathbf r}=1$ . Thus

$\displaystyle \mathbf J\cdot\widehat{\mathbf r}=-\frac{eg}{c}.$

Equivalently,

$\displaystyle \mathbf J\cdot\widehat{\mathbf r}=-q\hbar.$

This says that the angular momentum has a fixed radial component. Geometrically, $\displaystyle \mathbf J$ lies on a cone around $\displaystyle \mathbf r$ , rather than being perpendicular to $\displaystyle \mathbf r$ as ordinary orbital angular momentum would be.

Quantization from angular momentum eigenvalues

In quantum mechanics, if $\displaystyle \mathbf J$ satisfies

$\displaystyle [J_i,J_j]=i\hbar\epsilon_{ijk}J_k,$

then it generates representations of the rotation algebra. The possible angular momentum quantum numbers are

$\displaystyle j=0,\frac12,1,\frac32,2,\dots$

and for a given $\displaystyle j$ , the projection of angular momentum along any chosen axis has eigenvalues

$\displaystyle m\hbar, \quad m=-j,-j+1,\dots,j.$

Thus angular momentum projections come in integer or half-integer multiples of $\displaystyle \hbar$ . But the radial projection of $\displaystyle \mathbf J$ is fixed at $\displaystyle -q\hbar.$ Therefore $\displaystyle q$ must be integer or half-integer:

$\displaystyle q\in\frac12\mathbb Z.$

Equivalently,

$\displaystyle 2q\in\mathbb Z.$

Using $\displaystyle q=eg/(\hbar c)$ , we again find

$\displaystyle \boxed{eg=\frac{n\hbar c}{2}.}$

This is the angular-momentum derivation of Dirac quantization.

The comparison

The gauge-patch proof gave $\displaystyle t_{NS}(\phi)=e^{2iq\phi},$ and single-valuedness around $\displaystyle \phi\mapsto\phi+2\pi$ required $\displaystyle e^{4\pi iq}=1.$ The angular-momentum proof gave $\displaystyle \mathbf J\cdot\widehat{\mathbf r}=-q\hbar,$ and angular momentum quantization required $\displaystyle q\in\frac12\mathbb Z.$

These are the same condition, because

$\displaystyle q\in\frac12\mathbb Z \quad\Longleftrightarrow\quad 2q\in\mathbb Z \quad\Longleftrightarrow\quad e^{4\pi iq}=1.$

But this still does not explain why the same number $\displaystyle q$ appears in both computations. To understand that, we must identify what angular momentum is doing geometrically.

In ordinary quantum mechanics on a sphere, a wavefunction is a complex-valued function $\displaystyle \psi: S^2 \to \mathbb C.$ A rotation $\displaystyle R\in SO(3)$ acts by $\displaystyle (U(R)\psi)(x)=\psi(R^{-1}x).$ The infinitesimal generators of this action are the usual orbital angular momentum operators. But in the monopole problem, $\displaystyle\psi$ is not a globally defined function. Instead, it is a section of a complex line bundle $\displaystyle L\to S^2$ . In local charts, we write it as a pair $\displaystyle (\psi_N,\psi_S),$ with the gluing law $\displaystyle \psi_N=e^{2iq\phi}\psi_S.$ This changes the meaning of rotations.

A rotation moves a base point $\displaystyle x\in S^2$ to another base point $\displaystyle Rx$ . But the wavefunction also has a phase living in the fiber above each point. To rotate a section of a line bundle, one must specify how the fiber over $\displaystyle x$ is identified with the fiber over $\displaystyle Rx$ . That identification is controlled by the connection $\displaystyle A$ . Therefore, the true rotation generators are not ordinary derivatives. They are covariant derivatives plus correction terms. This is precisely what the monopole angular momentum operator $\displaystyle \mathbf J$ encodes. In short: $\displaystyle \mathbf J$ is the infinitesimal rotation operator acting on sections of the monopole line bundle. The extra term $\displaystyle -q\hbar\widehat{\mathbf r}$ is the infinitesimal trace of the same twisting that appears in the transition function $\displaystyle e^{2iq\phi}.$

The deepest connection can be seen by considering rotations about the radial direction $\displaystyle \widehat{\mathbf r}$ . At a point $\displaystyle x\in S^2$ , a rotation about the axis through $\displaystyle x$ leaves the point $\displaystyle x$ fixed. It does not move the base point on the sphere. So what can such a rotation do to the wavefunction? Since it leaves the base point unchanged, it can only act on the fiber above that point. That is, it acts as a phase. In other words, rotation around the radial axis is, from the viewpoint of the line bundle, a gauge transformation. The generator of this rotation is $\displaystyle \mathbf J\cdot\widehat{\mathbf r}=-q\hbar.$ . Therefore a rotation by angle $\displaystyle \alpha$ about $\displaystyle \widehat{\mathbf r}$ acts by the phase $\displaystyle \exp\left(-\frac{i}{\hbar}\alpha \mathbf J\cdot\widehat{\mathbf r}\right)$ . Substituting $\displaystyle \mathbf J\cdot\widehat{\mathbf r}=-q\hbar,$ we get a phase of the form $\displaystyle e^{\pm iq\alpha}.$

The sign depends on conventions; the quantization condition does not. Thus the monopole coupling $\displaystyle q$ is the weight with which the stabilizer rotation acts on the internal $\displaystyle U(1)$ fiber. For consistency under closed rotations, this weight must be quantized. This is the angular-momentum version of the transition-function argument.

There is a very clean mathematical way to see the unity of the two arguments. The sphere can be represented as a homogeneous space: $\displaystyle S^2\cong SU(2)/U(1)$ . Here $\displaystyle SU(2)$ acts by rotations, and the subgroup $\displaystyle U(1)$ is the stabilizer of a chosen point, say the north pole. This stabilizer consists of rotations around the vertical axis. To build a line bundle over $\displaystyle S^2$ , one may specify how the stabilizer $\displaystyle U(1)$ acts on the fiber over the north pole. A one-dimensional representation of $\displaystyle U(1)$ has the form

$\displaystyle e^{i\alpha}\mapsto e^{in\alpha}, \quad n\in\mathbb Z.$

The integer $\displaystyle n$ labels the line bundle. For the monopole bundle, $\displaystyle n=2q=\frac{2eg}{\hbar c}.$ Thus the angular-momentum derivation is really saying that the fiber over a point must transform under a genuine representation of the stabilizer $\displaystyle U(1)$ . Such representations are labeled by integers. Therefore $\displaystyle 2q$ must be an integer.

The gauge-patch derivation says the same thing in another way. It says that the transition function on the overlap is $\displaystyle e^{2iq\phi},$ which is a well-defined map $\displaystyle S^1\to U(1)$ only if $\displaystyle 2q\in\mathbb Z$ .

In the gauge-patch proof, the phase is the transition function around the equator. In the angular-momentum proof, the phase is the action of the stabilizer rotation on the fiber: The reason the same quantization appears is that the equator transition function and the stabilizer representation are two descriptions of the same line bundle. The patch proof constructs this line bundle using local charts and transition functions. The angular-momentum proof constructs the same object using the representation theory of rotations.

If the bundle were trivial, a wavefunction would just be a global complex-valued function on $\displaystyle S^2$ . But for a monopole, the bundle is twisted. Locally it still looks like a product: $U_N\times\mathbb C, \quad U_S\times\mathbb C$ . The twisting is encoded in how these local products are glued on the overlap: $\psi_N=e^{2iq\phi}\psi_S.$ If $2q=0$ , there is no twisting. If $2q=1$ , the fiber phase winds once around as we go around the equator. If $2q=2$ , it winds twice. And so on. A non-integer number of windings is impossible for a continuous single-valued map from a circle to a circle. This is the topological reason for quantization. The angular-momentum proof sees the same twisting not by walking around the equator, but by asking how rotations act on the fibers. The stabilizer of a point is a circle group. A representation of a circle group must have integer weight. That integer is the same winding number.

Beyond one electric charge and one monopole: dyons

Having understood Dirac quantization for one electric charge and one monopole, the general case is conceptually simple. A dyon is a particle with charge vector $\displaystyle \Gamma=(e,g).$ For two dyons, $\displaystyle \Gamma_1=(e_1,g_1),\qquad \Gamma_2=(e_2,g_2)$ the product $\displaystyle eg$ is replaced by the antisymmetric pairing

$\displaystyle \langle \Gamma_1,\Gamma_2\rangle=e_1g_2-e_2g_1$

The Dirac-Schwinger-Zwanziger condition is

$\displaystyle e_1g_2-e_2g_1=\frac{n\hbar c}{2},\qquad n\in\mathbb Z$

This reduces to Dirac’s original condition when $\displaystyle \Gamma_1=(e,0),\qquad \Gamma_2=(0,g)$ because then

$\displaystyle \langle \Gamma_1,\Gamma_2\rangle=eg$

The pairing is not symmetric because the two-body problem has an orientation. If $\displaystyle r=r_1-r_2$ then the interaction of $\displaystyle e_1$ with $\displaystyle g_2$ and that of $\displaystyle e_2$ with $\displaystyle g_1$ enter with opposite orientations. Geometrically, this is the oriented area form on the electric-magnetic charge plane. It is the analogue of $\displaystyle x_1y_2-x_2y_1$ for two vectors in the plane. So DSZ quantization says: the oriented electric-magnetic area between any two allowed charge vectors is quantized.

For the ordinary charge-monopole system, the conserved angular momentum contains the radial term $\displaystyle -\frac{eg}{c}\hat r$ . For two dyons, the same formula holds after the replacement $\displaystyle eg\longmapsto e_1g_2-e_2g_1$ . Thus

$\displaystyle J=r\times\pi-\frac{e_1g_2-e_2g_1}{c}\hat r$

Taking the radial projection gives

$\displaystyle J\cdot\hat r=-\frac{e_1g_2-e_2g_1}{c}$

Since quantum angular momentum projections are quantized in half-integer units of $\displaystyle \hbar$ , this forces

$\displaystyle e_1g_2-e_2g_1=\frac{n\hbar c}{2},\qquad n\in\mathbb Z$

The underlying bundle is the relative monopole line bundle for the two-dyon system. Remove the coincidence point where the two particles meet. The relative coordinate is $\displaystyle r=r_1-r_2\in \mathbb R^3\setminus{0}$ Radially, this deformation-retracts to the sphere of directions $\displaystyle S^2$ So the relevant topology is not in the radius, but in the angular variable $\displaystyle \hat r\in S^2$ Thus the bundle lives over $\displaystyle S^2_{\mathrm{rel}}$ the sphere of relative directions between the two dyons.

The two-dyon system reduces, in relative coordinates, to the same geometry as a charge moving in a monopole background. But the effective monopole strength is not $\displaystyle eg$ . It is the mutual pairing

$\displaystyle \langle \Gamma_1,\Gamma_2\rangle=e_1g_2-e_2g_1$

The wavefunction is not a scalar function on the relative configuration space. It is a section of a complex line bundle whose first Chern number is the DSZ integer. The DSZ condition says that the relative quantum state of two dyons is a section of a monopole line bundle over the sphere of relative directions, and the Chern number of that bundle is the integer-valued electric-magnetic pairing.

The two-dyon case therefore introduces no new conceptual mechanism. It only replaces the elementary product $\displaystyle eg$ by the natural antisymmetric pairing $\displaystyle \langle \Gamma_1,\Gamma_2\rangle=e_1g_2-e_2g_1.$ The relative coordinate $\displaystyle r=r_1-r_2$ takes values in $\displaystyle \mathbb R^3\setminus{0}$ , whose angular part is the sphere $\displaystyle S^2_{\mathrm{rel}}.$ Over this sphere sits a complex line bundle $\displaystyle L_{12}\to S^2_{\mathrm{rel}}$ and the relative wavefunction is not a global function, but a section $\displaystyle \psi\in \Gamma(S^2_{\mathrm{rel}},L_{12}).$ Its first Chern number is

$\displaystyle c_1(L_{12})=\frac{2\langle \Gamma_1,\Gamma_2\rangle}{\hbar c}.$

Thus the DSZ condition is simply $\displaystyle c_1(L_{12})\in\mathbb Z.$

This is the common content of the two derivations. The gauge-patch argument sees this integer as the winding number of the transition function. The angular-momentum argument sees the same integer as the weight with which the stabilizer $\displaystyle U(1)$ acts on the fiber. They are two coordinate descriptions of the same line bundle.

So the final picture is: quantized charge is equal integral Chern class of the quantum line bundle. For one charge and one monopole, this gives Dirac’s condition. For two dyons, it gives the Dirac-Schwinger-Zwanziger condition. In both cases, quantization is the price of asking the quantum wavefunction to exist globally.