# Euclidean Methods on Dirac Quantization and Gauge Representation

Dohyun Kim<EMAIL_ADDRESS>Department of Physics, Osaka
University, Osaka 560-0043, Japan

###### Abstract

Geometrical interpretation on U(1) gage theory of Dirac monopole, introduced
here from the line integral[1] form of vector potentials, shows the gauge
representation be multi-valued. In this paper, we construct Euclidean form of
U(1) gauge theory on Dirac monopole based on the geometrical interpretation,
derive the Dirac quantization condition geometrically. Even several
theoretical methods which leads to Dirac quantization $eg=l/2$ has been
developed, these are only possible under the postulate that gauge
representation is single-valued. Using the line integration representation of
vector potential, we derived the gauge equation which satisfies under the
‘phase gauge transformation’ ($\mathbf{A}\to\mathbf{A}^{\prime}=\mathbf{A},\
\psi\to\psi^{\prime}=e^{ie\Lambda}\psi$), also showed that the Dirac
quantization condition can be derived by the geometrical boundary condition on
this gauge equation, without any supposing of canonical quantization
condition. We finally explain geometrical meaning of constant phase of gauge
and how the Dirac quantization holds single valued gauge representation.

##  I introduction

On the Dirac’s paper[2], the vector potential of single magnetic pole of the
$+z$-axis string $\mathbf{n}(0,0,1)$and $-z$-axis string
$\mathbf{n}^{\prime}$(0,0,-1) is given by

$\displaystyle\mathbf{A}_{\mathbf{n}}$
$\displaystyle=-g\frac{1+\cos{\theta}}{r\sin{\theta}}\ \mathbf{e}_{\varphi},$
(1) $\displaystyle\mathbf{A}_{\mathbf{n}^{\prime}}$
$\displaystyle=g\frac{1-\cos{\theta}}{r\sin{\theta}}\ \mathbf{e}_{\varphi}.$

This gives representation of magnetic fields each potential
$\mathbf{A}_{\mathbf{n}}(\mathbf{r})$ and
$\mathbf{A}_{\mathbf{n}^{\prime}}(\mathbf{r})$ makes using the regularized
from of potential[4, 5, 6, 3] for the exclusion of singularity as

$\displaystyle\mathbf{B}_{\mathbf{n}}(\mathbf{r})=g\frac{\mathbf{r}}{r^{3}}-4g\pi\mathbf{n}\theta(z)\delta(x)\delta(y),$
(2)
$\displaystyle\mathbf{B}_{\mathbf{n}^{\prime}}(\mathbf{r})=g\frac{\mathbf{r}}{r^{3}}-4g\pi\mathbf{n}^{\prime}\theta(z)\delta(x)\delta(y)$

where we put $\theta(z)$ as Heavside step function. From the representation of
magnetic field (2), Dirac string($\mathbf{n}$ or $\mathbf{n}^{\prime}$) has
both divergent field $\mathbf{B}_{g}(\mathbf{r})$ and string field
$\mathbf{B}_{\rm str}(\mathbf{r})$. Hereby the string field only holds the
Bianchi identity of vector potential as
$\boldsymbol{\nabla}\cdot\mathbf{B}=4g\pi-4g\pi=0$. This gives the line
integral representation[4, 5, 6] along to the string from the Green function
of poisson equation as

$\displaystyle\mathbf{A}_{\mathbf{n}}(\mathbf{r})$
$\displaystyle=\int_{\tau\to\infty}\boldsymbol{\nabla}\left(\frac{1}{|\mathbf{r}-\mathbf{r}^{\prime}|}\right)\times\mathbf{B}_{\rm
str}\ d^{3}\mathbf{r}^{\prime}$ (3)
$\displaystyle=g\int_{C(\mathbf{n})}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\times
d\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}$

where we put the integral path $C(\mathbf{n})$ as string path which starts
from the single magnetic pole $g$. Therefore the U(1) gauge transformation
gives rotation of string $\mathbf{n}\to\mathbf{n}^{\prime\prime}$ also
represented by line integration on same integral path $C(\mathbf{n})$. Hereby
we put the solid angle
$\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})$ on gauge
transformation
$\mathbf{A}_{\mathbf{n}}(\mathbf{r})\to\mathbf{A}_{\mathbf{n}^{\prime\prime}}(\mathbf{r})$
as:

Figure 1: The solid angle on vector $\mathbf{r}$ of string($\mathbf{n}$ and
$\mathbf{n}^{\prime\prime}$).

to relating geometrical rotation of string and U(1) gauge transformation. Here
the direction of string-transforming area vector
$\boldsymbol{\Gamma}[r_{p}^{\prime},r_{q}^{\prime}]$ is define as which
follows the Stoke’s theorem on closed path $g\to r_{p}^{\prime}\to
r_{q}^{\prime}$. Then the solid angle
$\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})$ on FIG. 1 is
represented by

$\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})=\int_{\sigma}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}.$ (4)

Now we come to the U(1) gauge representation of string rotation
$\mathbf{n}\to\mathbf{n}^{\prime\prime}$. Let us put the U(1) gauge
transformation of fields $\psi_{\mathbf{n}}(\mathbf{r})$ and
$\mathbf{A}_{\mathbf{n}}(\mathbf{r})$ on string $\mathbf{n}$ for the string
rotation $\mathbf{n}\to\mathbf{n}^{\prime\prime}$ as:

$\displaystyle\psi_{\mathbf{n}}(\mathbf{r})$ $\displaystyle\to\
\psi_{\mathbf{n}^{\prime\prime}}(\mathbf{r})=e^{ie\Lambda(\mathbf{r})}\psi_{\mathbf{n}}(\mathbf{r}),$
(5) $\displaystyle\mathbf{A}_{\mathbf{n}}(\mathbf{r})$ $\displaystyle\to\
\mathbf{A}_{\mathbf{n}^{\prime\prime}}(\mathbf{r})=\mathbf{A}_{\mathbf{n}}(\mathbf{r})+\boldsymbol{\nabla}\Lambda(\mathbf{r})$

where the notation $\psi_{\mathbf{n}^{\prime\prime}}(\mathbf{r})$ and
$\mathbf{A}_{\mathbf{n}^{\prime\prime}}(\mathbf{r})$ show the fields on
rotated string $\mathbf{n}^{\prime\prime}$. If we observe the line integration
representation of vector potential of each string $\mathbf{n},\
\mathbf{n}^{\prime\prime}$ on (3), the gauge transformation of vector
potential now comes to

$\delta\mathbf{A}_{\mathbf{n}}(\mathbf{r})=g\int_{C(\mathbf{n}^{\prime\prime})+C^{-1}(\mathbf{n})}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\times
d\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}=\boldsymbol{\nabla}\Lambda(\mathbf{r})$
(6)

where
$\delta\mathbf{A}_{\mathbf{n}}(\mathbf{r})=\mathbf{A}_{\mathbf{n}^{\prime\prime}}(\mathbf{r})-\mathbf{A}_{\mathbf{n}}(\mathbf{r})$.
As boundary integral on $[r_{p}^{\prime},r_{q}^{\prime}]$ vanished, we shall
use the modified closed path of string
$\boldsymbol{\Gamma}[r_{p}^{\prime},r_{q}^{\prime}]$ on integration (6), the
Stoke’s theorem gives directly:

$\displaystyle\delta\mathbf{A}_{\mathbf{n}}(\mathbf{r})$
$\displaystyle=\lim_{r_{p}^{\prime},r_{q}^{\prime}\to\infty}g\int_{\partial\boldsymbol{\Gamma}[r_{p}^{\prime},r_{q}^{\prime}]}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\times
d\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}$ (7)
$\displaystyle=\lim_{r_{p}^{\prime},r_{q}^{\prime}\to\infty}-g\boldsymbol{\nabla}\times\int_{\boldsymbol{\Gamma}[r_{p}^{\prime},r_{q}^{\prime}]}\boldsymbol{\nabla}\left(\frac{1}{|\mathbf{r}-\mathbf{r}^{\prime}|}\right)\times
d\boldsymbol{\sigma}^{\prime}$
$\displaystyle=g\left[\boldsymbol{\nabla}\left(-\boldsymbol{\nabla}\cdot\int_{\sigma}\frac{d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|}\right)\right]$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+g\int_{\sigma}\nabla^{2}\left(\frac{1}{|\mathbf{r}-\mathbf{r}^{\prime}|}\right)\
d\boldsymbol{\sigma}^{\prime}.$

On second line of calculation (7), we use the Stoke’s theorem in form of

$\int_{\partial\Omega}f\ d\mathbf{r}=\int_{\Omega}\boldsymbol{\nabla}f\times
d\boldsymbol{\sigma},$ (8)

also vector calculus identity
$\boldsymbol{\nabla}\times(\boldsymbol{\nabla}\times\mathbf{A})=\boldsymbol{\nabla}(\boldsymbol{\nabla}\cdot\mathbf{A})-\nabla^{2}\mathbf{A}$
used on third line. By feeding (4) to (6) and (7), we get generalized
representation[4, 5, 6] of gauge $\Lambda(\mathbf{r})$ as

$\boldsymbol{\nabla}\Lambda(\mathbf{r})=g\boldsymbol{\nabla}\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})-4g\pi\int_{\sigma}d\boldsymbol{\sigma}^{\prime}\
\delta^{(3)}(\mathbf{r}-\mathbf{r}^{\prime}).$ (9)

Now we can evaluate U(1) gauge $\Lambda(\mathbf{r})$. Since the gradient of
solid angle
$\boldsymbol{\nabla}\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})$
diverges on point $\mathbf{r}_{\sigma}\in\sigma$, we have observation of gauge
equation (9) in form

$\boldsymbol{\nabla}\left(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})-\frac{1}{g}\Lambda(\mathbf{r})\right)=4\pi\int_{\sigma}d\boldsymbol{\sigma}^{\prime}\
\delta^{(3)}(\mathbf{r}-\mathbf{r}^{\prime}).$ (10)

Then, we shall consider two path $\gamma_{1}(t),\gamma_{2}(t)\ (0\leq t\leq
1)$ of line integration on (10) shown by:

Figure 2: The path $\gamma_{1},\gamma_{2}$ of line integral on (10)

which is homotope($\gamma_{1}(0)=\gamma_{2}(0),\gamma_{1}(1)=\gamma_{2}(1)$).
Each path of line integrals on (10) gives

$\displaystyle\delta_{1}\left(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})-\frac{1}{g}\Lambda(\mathbf{r})\right)=4\pi\iint_{\gamma_{1}\times\sigma}d\mathbf{r}\cdot
d\boldsymbol{\sigma}^{\prime}\ \delta^{(3)}(\mathbf{r}-\mathbf{r}^{\prime}),$
(11)
$\displaystyle\delta_{2}\left(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})-\frac{1}{g}\Lambda(\mathbf{r})\right)=4\pi\iint_{\gamma_{2}\times\sigma}d\mathbf{r}\cdot
d\boldsymbol{\sigma}^{\prime}\ \delta^{(3)}(\mathbf{r}-\mathbf{r}^{\prime}).$

These results are supported by the integral term

$\iint_{\gamma\times\sigma}d\mathbf{r}\cdot d\boldsymbol{\sigma}^{\prime}\
\delta^{(3)}(\mathbf{r}-\mathbf{r}^{\prime})$ (12)

converges. Now we get this U(1) gauge $\Lambda(\mathbf{r})$ be discontinuous
and multi-valued function on $\mathbf{r}\not\in\mathbf{r}_{\sigma}$. The
$\gamma_{1}$ like path which passing through the string transforming area
$\sigma$ gives integral term on right side of (11) as $-4\pi$. But, the
$\gamma_{2}$ like path which passing through the outside of string
transforming area $\sigma^{c}$ integral term on right side of (11) gives as 0.
If we put gauge fixing on string rotation
$\mathbf{n}\to\mathbf{n}^{\prime\prime}$ on FIG. 2 as $\psi(\mathbf{r}_{0})\
\to\
\psi^{\prime}(\mathbf{r}_{0})=e^{ieg\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{0})}\psi(\mathbf{r}_{0})$,
the gauge transformation on $\mathbf{r}_{1}$ have following two
representations:

$\displaystyle\psi(\mathbf{r}_{1})\ \to\ \psi^{\prime}(\mathbf{r}_{1})$
$\displaystyle=U_{1}\psi(\mathbf{r}_{1});\ \ \
U_{1}=e^{ieg(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{1})+4\pi)},$
(13) $\displaystyle\psi(\mathbf{r}_{1})\ \to\ \psi^{\prime}(\mathbf{r}_{1})$
$\displaystyle=U_{2}\psi(\mathbf{r}_{1});\ \ \
U_{2}=e^{ieg\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{1})}.$

On the limit $\mathbf{r}_{0}\to\mathbf{r}_{\sigma}-$,
$\mathbf{r}_{1}\to\mathbf{r}_{\sigma}+$, the equation of gauge-variation (11)
becomes

$\delta\left(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r})-\frac{1}{g}\Lambda(\mathbf{r})\right)=-4\pi,$
(14)

the representation of U(1) gauge transformation only permitted as

$\displaystyle\lim_{\mathbf{r}_{0}\to\mathbf{r}_{\sigma}-}\psi^{\prime}(\mathbf{r}_{0})$
$\displaystyle=\lim_{\mathbf{r}_{0}\to\mathbf{r}_{\sigma}-}e^{ieg\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{0})}\psi(\mathbf{r}_{0}),$
(15)
$\displaystyle\lim_{\mathbf{r}_{1}\to\mathbf{r}_{\sigma}+}\psi^{\prime}(\mathbf{r}_{1})$
$\displaystyle=\lim_{\mathbf{r}_{1}\to\mathbf{r}_{\sigma}-}e^{ieg(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{1})+4\pi)}\psi(\mathbf{r}_{0})$

where we put the each notation $\mathbf{r}\to\mathbf{r}_{\sigma}\pm$ as
$\mathbf{r}$ approaching to $\mathbf{r}_{\sigma}$ from the below or above side
of area $\sigma$. Now that the representation of gauge $\Lambda(\mathbf{r})$
is single-valued only for $\mathbf{r}\in\mathbf{r}_{\sigma}$, but in generally
it has
$e^{ieg\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{0})}\neq
e^{ieg(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{1})+4\pi)}$.
Limits on (15) only yields trivial identity:

$\lim_{\mathbf{r}_{0}\to\mathbf{r}_{\sigma}-}{\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{0})}=\lim_{\mathbf{r}_{1}\to\mathbf{r}_{\sigma}+}{\Omega_{\mathbf{n}\to\mathbf{n}^{\prime\prime}}(\mathbf{r}_{1})+4\pi}.$
(16)

So the discussion to Dirac quantization on [6] must be modified.

##  II the half-plane gauge representation

Let us construct the representation of half-plane gauge transformation what
makes string rotation:

$\mathbf{n}(0,0,1)\to\mathbf{n}^{\prime}(0,0,-1),$ (17)
$\sigma=\lim_{r_{p}^{\prime},r_{q}^{\prime}\to\infty}\boldsymbol{\Gamma}[r_{p}^{\prime},r_{q}^{\prime}]$
(18)

with string transforming area $\sigma$ as $xz$-half plane(FIG. 3):

(a)

(b)

Figure 3: String transforming area $\sigma$ on string rotation (17).

As here we put the variable conversion as
$\boldsymbol{\beta}\equiv\mathbf{r}-\mathbf{r}^{\prime}$ on (4), the solid
angle $\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r})$ on this half-
plane gauge transformation comes to

$\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r})=\int_{\sigma}\frac{\hat{\boldsymbol{\beta}}\cdot
d\boldsymbol{\sigma}}{\beta^{2}}=\int_{-\pi}^{\varphi-\pi}\int_{0}^{\pi}\sin{\theta^{\prime}}\
d\theta^{\prime}d\varphi^{\prime}=2\varphi.$ (19)

From the representation of gauge (9), space of this system is branch-cut by
string transforming area $\sigma$, the azimuthal angle has restriction of
$\varphi\in(-\pi,\pi)$. If here we put gauge fixing on this string rotation
$\mathbf{n}\to\mathbf{n}^{\prime}$ as

$\Lambda(\mathbf{r})=g\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r}),$
(20)

the gauge transformation of each field on this string rotation comes to

$\displaystyle\psi_{\mathbf{n}}(\mathbf{r})\ \to\
\psi_{\mathbf{n}^{\prime}}(\mathbf{r})$
$\displaystyle=e^{ieg\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r})}\psi_{\mathbf{n}}(\mathbf{r})=e^{2ieg\varphi}\psi_{\mathbf{n}}(\mathbf{r}),$
(21) $\displaystyle\mathbf{A}_{\mathbf{n}}(\mathbf{r})\ \to\
\mathbf{A}_{\mathbf{n}^{\prime}}(\mathbf{r})$
$\displaystyle=\mathbf{A}_{\mathbf{n}}(\mathbf{r})+\boldsymbol{\nabla}\Lambda(\mathbf{r})=\mathbf{A}_{\mathbf{n}}(\mathbf{r})+\boldsymbol{\nabla}\varphi.$

We shall here confirm that the gauge $\Lambda(\mathbf{r})=2g\varphi$ from
calculation of solid angle (19) exactly gives the half-plane string rotation:

$\mathbf{A}_{\mathbf{n}}(\mathbf{r})+\boldsymbol{\nabla}\varphi=g\frac{1-\cos{\theta}}{r\sin{\theta}}\
\mathbf{e}_{\varphi}=\mathbf{A}_{\mathbf{n}^{\prime}}(\mathbf{r}).$ (22)

Let us consider the representation of gauge transformation by generalized
string transforming area. By using same method, we also get the expansion of
gauge transformation (21) for $\varphi_{0}$-rotated string transforming area
$\tilde{\sigma}$ shown on FIG. 4:

(a)

(b)

Figure 4: The rotated string transforming area $\tilde{\sigma}$.

This gauge transformation by area $\tilde{\sigma}$ is geometrically equivalent
to FIG. 3. Hence, here we shall take the proper rotation of coordinates
$(x,y)\mapsto(x^{\prime},y^{\prime})$ what makes FIG. 4(a) comes to FIG. 4(b).
The azimuthal angle of vector $\mathbf{r}$ on new coordinates
$(x^{\prime},y^{\prime})$ now comes to

$\varphi^{\prime}=\varphi+\varphi_{0}.$ (23)

On the gauge fixing (20), it leads the representation of this gauge
transformation $\mathbf{n}\to\mathbf{n}^{\prime}$ by rotated string
transforming area $\tilde{\sigma}$ as:

$\displaystyle\psi_{\mathbf{n}}(\mathbf{r})\ \to\
\psi_{\mathbf{n}^{\prime}}(\mathbf{r})$
$\displaystyle=e^{2ieg\varphi^{\prime}}\psi_{\mathbf{n}}(\mathbf{r})=e^{2ieg(\varphi+\varphi_{0})}\psi_{\mathbf{n}}(\mathbf{r}),$
(24) $\displaystyle\mathbf{A}_{\mathbf{n}}(\mathbf{r})\ \to\
\mathbf{A}_{\mathbf{n}^{\prime}}(\mathbf{r})$
$\displaystyle=\mathbf{A}_{\mathbf{n}}(\mathbf{r})+\boldsymbol{\nabla}(\varphi+\varphi_{0})=\mathbf{A}_{\mathbf{n}}(\mathbf{r})+\boldsymbol{\nabla}\varphi$

where it takes the gauge $\Lambda(\mathbf{r})=2g(\varphi+\varphi_{0})$. This
representation is supported by results of (21). Also, the gauge fixing is
still held after the rotation of coordinates
$(x,y)\mapsto(x^{\prime},y^{\prime})$. This expression shows that the constant
term of phase is related to the direction of string transforming area.

##  III the phase gauge transformation

Let us form the gauge representation of phase gauge trnasformation what makes
following 2-step string rotation:

$u:\ \ \ \mathbf{n}(0,0,0)\ \xrightarrow[U_{+}(0)]{}\
\mathbf{n}^{\prime\prime}(0,0,-1)\ \xrightarrow[U_{-}(\varphi_{0})]{}\
\mathbf{n}(0,0,1),$ (25)
$\sigma_{+}=\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}\boldsymbol{\Gamma}_{a}[\xi_{p}^{\prime},\xi_{q}^{\prime}]\
,\ \ \
\sigma_{-}=\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}\boldsymbol{\Gamma}_{b}[\xi_{q}^{\prime},\xi_{p}^{\prime}]$
(26)

with string transforming area $\sigma_{\pm}$ as $xz$-half plane(FIG. 5):

$x$$y$$z$$\varphi$$\xi_{p}^{\prime}$$\xi_{q}^{\prime}$$\sigma_{+}$$\mathbf{r}$$r_{\parallel}$
(a) First step gauge transformation $U_{+}(0)$.

$x$$y$$z$$\varphi$$\varphi_{0}$$\xi_{p}^{\prime}$$\xi_{q}^{\prime}$$\sigma_{-}$$\mathbf{r}$$r_{\parallel}$
(b) Second step gauge transformation $U_{-}(\varphi_{0})$.

Figure 5: The gauge transforming area $\sigma_{\pm}$ on phase gauge.
$x$$\vdots$$\mathbf{r}$$d\boldsymbol{\sigma}_{-}$$(d\boldsymbol{\sigma_{-}}\cdot\hat{\boldsymbol{\beta}})\hat{\boldsymbol{\beta}}$$y$$d\boldsymbol{\sigma}_{+}$$(d\boldsymbol{\sigma}_{+}\cdot\hat{\boldsymbol{\beta}})\hat{\boldsymbol{\beta}}$$\varphi$$\varphi_{0}$
Figure 6: The Diagramme of string transforming area $\sigma_{\pm}$ and vector
$\boldsymbol{\beta}$ on $xy$ plane. System is branch-cut by area
$\sigma_{\pm}$.

Since this phase gauge transformation only holds the direction of Dirac
string, it leads the U(1) gauge representation on this string rotation by

$\displaystyle\psi(\mathbf{r})$ $\displaystyle\xrightarrow[u]{}\
\psi^{\prime}(\mathbf{r})=e^{ie\Lambda_{0}}\psi(\mathbf{r}),\ \ \ \ \ \ \
\mathbf{A}(\mathbf{r})\ $ $\displaystyle\xrightarrow[u]{}\
\mathbf{A}^{\prime}(\mathbf{r})=\mathbf{A}(\mathbf{r})$ (27)

for a constant phase $\Lambda_{0}$. With the half-plane gauge representation
derived on II now we come to the representation of phase $\Lambda_{0}$ on U(1)
gauge (27). By taking the representation of half-plane gauge representation
for each $U_{+}(0)=e^{2ieg\varphi}$,
$U_{-}(\varphi_{0})=e^{-2ieg(\varphi+\varphi_{0})}$ driven on (21) and (24) on
gauge fixing (20), we thus get the representation of phase gauge
transformation $u$ as:

$u:\ \ \psi(\mathbf{r})\ \xrightarrow[u]{}\
U_{+}(0)U_{-}(\varphi_{0})\psi(\mathbf{r})=e^{-2ieg\varphi_{0}}\psi(\mathbf{r}).$
(28)

We may also take different approach to this phase gauge transformation. If we
remain the term of boundary path
$G_{\xi_{q}^{\prime}\to\xi_{p}^{\prime}}^{a}$,
$G_{\xi_{p}^{\prime}\to\xi_{q}^{\prime}}^{b}$ (each dashed path shown on FIG.
5), the integral representation of gauge transformation (7) comes to

$\displaystyle\mathbf{0}$
$\displaystyle=\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}\left(g\sum_{i}\int_{\partial\boldsymbol{\Gamma}_{i}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\times
d\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}-g\int_{G_{\xi_{q}^{\prime}\to\xi_{p}^{\prime}}^{a}+G_{\xi_{p}^{\prime}\to\xi_{q}^{\prime}}^{b}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\times
d\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}\right)$ (29)
$\displaystyle=g\boldsymbol{\nabla}\left(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r})+\Omega_{\mathbf{n}^{\prime}\to\mathbf{n}}(\mathbf{r})-\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}\int_{\mathcal{G}_{\xi^{\prime}}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}\right)$

with $i=a,b$ on $\mathbf{r}\not\in\sigma_{\pm}.$ So the relation (29) also
leads

$\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r})+\Omega_{\mathbf{n}^{\prime}\to\mathbf{n}}(\mathbf{r})-\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}\int_{\mathcal{G}_{\xi^{\prime}}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}=\alpha_{0}$
(30)

for constant $\alpha_{0}$. Here we denote the area
$\mathcal{G}_{\xi^{\prime}}$ as a surface111The boundary term on (29) exactly
vanished:
$g\int_{G_{\xi_{q}^{\prime}\to\xi_{p}^{\prime}}^{a}+G_{\xi_{p}^{\prime}\to\xi_{q}^{\prime}}^{b}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\times
d\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}\
\xrightarrow[\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty]{}\ 0.$ with boundary
$\partial\mathcal{G}_{\xi^{\prime}}=G_{\xi_{q}^{\prime}\to\xi_{p}^{\prime}}^{a}+G_{\xi_{p}^{\prime}\to\xi_{q}^{\prime}}^{b}$
shown on FIG. 7:

$x$$y$$z$$\varphi$$\varphi_{0}$$\boldsymbol{\Gamma}_{a}$$\xi_{p}^{\prime}$$\xi_{q}^{\prime}$$\boldsymbol{\Gamma}_{b}$$\mathbf{r}$$r_{\parallel}$$\mathcal{G}_{\xi^{\prime}}$
Figure 7: Path on integral (29) and (30).

Now we shall evaluate the constant $\alpha_{0}$ using Gauss’ law on integral
(30) by

$\displaystyle\alpha_{0}$
$\displaystyle=\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}\oint_{\boldsymbol{\Gamma}_{a}[\xi_{p}^{\prime},\xi_{q}^{\prime}]+\boldsymbol{\Gamma}_{b}[\xi_{q}^{\prime},\xi_{p}^{\prime}]-\mathcal{G}_{\xi}^{\prime}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}$ (31)
$\displaystyle=\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}-\int_{\text{internal
space}}\nabla^{\prime
2}\left(\frac{1}{|\mathbf{r}-\mathbf{r}^{\prime}|}\right)\ d\Sigma^{\prime}$
$\displaystyle=4\pi\theta_{\pi}(\varphi_{0}-\varphi)$

with azimuthal step function $\theta_{\pi}(\varphi_{0}-\varphi)$:

$\theta_{\pi}(\varphi_{0}-\varphi)=\begin{cases}1\ \ \ \ \ \ \
&(\varphi_{0}<\pi-\varphi)\\\\[8.61108pt] 0\ \ \ \ \ \ \
&(\varphi_{0}>\pi-\varphi)\end{cases}.$ (32)

Combining the (32) and (30), we may also get the integral form of solid angle
$\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r})+\Omega_{\mathbf{n}^{\prime}\to\mathbf{n}}(\mathbf{r})$.
Since the Dirac monopole is abelian, it leads the integral form of phase gauge
transformation $u$ as

$u:\ \ \ U_{+}(0)U_{-}(\varphi_{0})=\exp\left(4\pi
ieg\theta_{\pi}(\varphi_{0}-\varphi)+\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}ieg\int_{\mathcal{G}_{\xi^{\prime}}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}\right)$
(33)

on gauge fixing (20). Comparing these two representation (28) and (33) of
phase gauge transformation $u$ , we get

$e^{-2ieg\varphi_{0}}=\exp\left(4\pi
ieg\theta_{\pi}(\varphi_{0}-\varphi)+\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}ieg\int_{\mathcal{G}_{\xi^{\prime}}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}\right).$
(34)

##  IV The Dirac Quantization

In this section, we investigate geometrical approach to Dirac quantization
under phase gauge transformation. The representation222Here we consider the
phase gauge transformation with path on FIG. 5. of phase gauge representation
$u$ is given by

$\displaystyle\psi(\mathbf{r})$ $\displaystyle\xrightarrow[u]{}\
\psi^{\prime}(\mathbf{r})=e^{-2ieg\varphi_{0}}\psi(\mathbf{r}),$ (35)
$\displaystyle\mathbf{A}(\mathbf{r})$ $\displaystyle\xrightarrow[u]{}\
\mathbf{A}^{\prime}(\mathbf{r})=\mathbf{A}(\mathbf{r})$

on gauge fixing (20). So the phase gauge transformation only gives the
constant phase difference to $\psi(\mathbf{r})$. The key idea to Dirac
quantization consists in finding a proper path for phase gauge transformation
$\not{u}$ what transforms

$\displaystyle\psi(\mathbf{r})$ $\displaystyle\xrightarrow[\not{u}]{}\
\psi(\mathbf{r}),\ \ \mathbf{A}(\mathbf{r})\ $
$\displaystyle\xrightarrow[\not{u}]{}\ \mathbf{A}(\mathbf{r})$ (36)

Now we shall consider following two paths shown on FIG. 8 and FIG. 9:

$x$$y$$z$$\varphi$$\varphi_{0}$$\boldsymbol{\Gamma}_{a}$$\xi_{p}^{\prime}$$\xi_{q}^{\prime}$$\boldsymbol{\Gamma}_{b}(\varphi_{0})$$\mathbf{r}$$r_{\parallel}$
Figure 8: Integral path of phase gauge transformation on path
$\displaystyle\partial\mathbf{D}_{-}=\lim_{\varphi_{0}\to
2\pi-}\partial\boldsymbol{\Gamma}_{1}+\partial\boldsymbol{\Gamma}_{2}(\varphi_{0})$.

We may investigate the representation of phase gauge transformation on path
$\partial\mathbf{D}_{-}$ shown FIG. 8 using (33). Since each azimuthal step
function $\theta_{\pi}(\varphi_{0}-\varphi)$ and boundary integral term on
path $\partial\mathbf{D}_{-}$ goes to

$\theta_{\pi}(\varphi_{0}-\varphi)=1,\ \
\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}\int_{\mathcal{G}_{\xi^{\prime}}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}=-4\pi,$
(37)

we thus have the representation of phase gauge transformation on path
$\partial\mathbf{D}_{-}$ by

$\displaystyle{u}_{D-}:$ $\displaystyle\exp\Bigg{(}4\pi
ieg\theta_{\pi}(\varphi_{0}-\varphi)$ (38) $\displaystyle\left.\ \ \ \ \
+\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}ieg\int_{\mathcal{G}_{\xi^{\prime}}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}\right)_{D_{-}}=\mathbf{1}.$

Similarly, we also consider the phase gauge transformation on path
$\partial\mathbf{D}_{+}$:

$x$$y$$z$$\varphi$$\varphi_{0}$$\boldsymbol{\Gamma}_{a}$$\xi_{p}^{\prime}$$\xi_{q}^{\prime}$$\boldsymbol{\Gamma}_{b}(\varphi_{0})$$\mathbf{r}$$r_{\parallel}$
(a) $\varphi_{0}\to 0+$ Figure 9: Integral path of phase gauge transformation
on path $\displaystyle\partial\mathbf{D}_{+}=\lim_{\varphi_{0}\to
0+}\partial\boldsymbol{\Gamma}_{1}+\partial\boldsymbol{\Gamma}_{2}(\varphi_{0})$.

On the phase gauge transformation with path $\partial\mathbf{D}_{+}$, (37)
comes to

$\theta_{\pi}(\varphi_{0}-\varphi)=0,\ \
\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}\int_{\mathcal{G}_{\xi^{\prime}}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}=0,$ (39)

then we also have the representation of phase gauge transformation on path
$\partial\mathbf{D}_{-}$:

$\displaystyle{u}_{D+}:$ $\displaystyle\exp\Bigg{(}4\pi
ieg\theta_{\pi}(\varphi_{0}-\varphi)$ (40) $\displaystyle\left.\ \ \ \ \
+\lim_{\xi_{p}^{\prime},\xi_{q}^{\prime}\to\infty}ieg\int_{\mathcal{G}_{\xi^{\prime}}}\frac{(\mathbf{r}-\mathbf{r}^{\prime})\cdot
d\boldsymbol{\sigma}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}\right)_{D+}=\mathbf{1}.$

Thus, here we confirm that both path
$\partial\mathbf{D}_{-},\partial\mathbf{D}_{+}$ of phase transformation gives
gauge transformation (36). So the proper path for phase gauge transformation
$\not{u}$ what transforms like (36) is given by
$\partial\mathbf{D}_{-},\partial\mathbf{D}_{+}$. Also, the path
$\partial\mathbf{D}_{-}$ leads to Dirac quantization from the gauge equation
(34) and gauge representation (38):

$u_{D-}:\ \ \ \lim_{\varphi_{0}\to 2\pi-}e^{-2ieg\varphi_{0}}=e^{-4\pi
ieg}=1,$ (41)

which leads

$eg=\frac{l}{2},\ \ \ \ (l=\pm 1,\pm 2,\cdots).$ (42)

Here we notice that Dirac quantization (42) is derived geometrically under
U(1) gauge invariance without a supposing of canonical quantization
$[\hat{q}_{i},\hat{p}_{j}]=i\delta_{ij}$.

##  V Discussion

In the previous section, we have seen that only the U(1) gauge invariance
leads to Dirac quantization condition $eg=l/2$ through the geometric tools.
This geometrical approach to Dirac quantization (42) gives some meaningful
interpretation about a physical meaning of gauge representation followed.

The representation of U(1) gauge (13) now becomes single-valued quantity:

$U_{1}=e^{ieg(\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r})+4\pi)}=e^{ieg\Omega_{\mathbf{n}\to\mathbf{n}^{\prime}}(\mathbf{r})}=U_{2}$
(43)

from the result of Dirac quantization (42). Thus the gauge of Dirac monopole
have single-valued representation everywhere. It is only supported by Dirac
quantization (42). The single-valued gauge representation is not a trivial
postulate, only the results derived by Dirac quantization condition $eg=l/2.$

On a half-plane gauge transformation(section II), even we shall add the
constant gauge $\Lambda(\mathbf{r})\to\Lambda(\mathbf{r})+\Lambda_{0}$, the
gauge transformation of field $\mathbf{A}(\mathbf{r})$ does not changed. But
the phase of field $\psi(\mathbf{r})$ is changed, this phase difference is
related to the azimuthal angle $\varphi_{0}$ of string transforming area(see
FIG. 4):

$\varphi_{0}=\frac{\Lambda_{0}}{2g}-\frac{\pi n}{eg}\in(-\pi,\pi)$ (44)

for a proper integer $n$. Here we get the relation (44) by comparing
$e^{ie\Lambda(\mathbf{r})}\to e^{ie(\Lambda(\mathbf{r})+\Lambda_{0})}$ and the
representation of gauge transformation (24).

## References

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