Gauge transformation of $\Psi$

Next, we discuss the gauge transformation of the vector potential $\mathbf{A}$ in the axisymmetric case. It is well-known that magnetic field remains the same under the following gauge transformation:

$$\mathbf{A}^{\ensuremath{\operatorname{new}}} =\mathbf{A}+ \nabla f,$$ (9)

where $f$ is an arbitrary scalar field. Here we require that $\nabla f$ be axisymmetric because, as mentioned above, an axisymmetric vector potential suffices for describing an axisymmetric magnetic field. Note that, in cylindrical coordinates, $\nabla f$ is given by

$$\nabla f = \frac{\partial f}{\partial R} \hat{\mathbf{R}} + \frac... ...c{1}{R} \frac{\partial f}{\partial \phi} \hat{\ensuremath{\boldsymbol{\phi}}} .$$ (10)

Since $\nabla f$ is axisymmetric, it follows that $\partial^2 f / \partial R \partial \phi = 0$, $\partial^2 f / \partial Z \partial \phi = 0$, and $\partial^2 f / \partial \phi^2 = 0$, which implies that $\partial f / \partial \phi$ is independent of $R$, $Z$, and $\phi$, i.e., $\partial f / \partial \phi$ is actually a constant. Using this, the $\phi$ component of the gauge transformation (9) is written

$$A_{\phi}^{\ensuremath{\operatorname{new}}} = A_{\phi} + \frac{1}{R} \frac{\partial f}{\partial \phi}$$

$$= A_{\phi} + \frac{C}{R},$$ (11)

where $C$ is a constant. Note that the requirement of axial symmetry greatly reduces the degree of freedom of the gauge transformation for $A_{\phi}$ (and thus for $R A_{\phi}$, i..e, $\Psi$). Multiplying Eq. (11) with $R$, we obtain the corresponding gauge transformation for $\Psi$,

$$\Psi^{\ensuremath{\operatorname{new}}} = \Psi + C,$$ (12)

which indicates $\Psi$ has the same gauge transformation as the familiar electric potential, i.e., adding a constant. (Note that the definition $\Psi (R, Z) \equiv R A_{\phi}$ does not mean $\Psi (R = 0, Z) = 0$ because $A_{\phi}$ can have a $1 / R$ dependence under the gauge transformation (11)).