in $R^4$ and satisfying the condition (\[current\_cont\_sc\]). For all $x_1\not=0$ the operator $\tilde D^2$ takes the form $$\begin{aligned}
\label{covder2}
\tilde D^2&=& (\partial_1 \pm i\breve n B x_2)^2 
\nonumber\\
&& + \partial_2^2 +\partial_3^2 + (\partial_4+i\breve n Bx_3)^2 
\nonumber\end{aligned}$$ where plus corresponds to the anti-self-dual configuration and minus is for the self-dual one. By inspection one can see that the eigenfunctions satisfy the relation $$\begin{aligned}
 \label{conditions}
\Phi^{(+)}(x_1,x_\perp)=\Phi^{(-)}(-x_1,x_\perp),\end{aligned}$$ where $(\pm)$ denotes the duality of the background field for a given $x_1$.

Respectively, the square integrable solutions are $$\begin{aligned}
\label{eigenf_sc}
\Phi^{(\pm)}_{kl}(x)&=&\phi^{(\pm)}_k(x_1,x_2)\chi_l(x_3,x_4)
\\
\phi^{(\pm)}_k(x_1,x_2)&=&\int dp_1 f(p_1)e^{\pm ip_1x_1-\frac{1}{2}|\breve n|B(x_2+p_1/|\breve n|B)^2}
\nonumber\\
&\times& H_k\left(\sqrt{|\breve n|B}\left[x_2+\frac{p_1}{|\breve n|B}\right]\right)
\nonumber\\
\chi_k(x_3,x_4)&=&\int dp_4 g(p_4)e^{ip_4x_4-\frac{1}{2}|\breve n|B(x_3+p_4/|\breve n|B)^2 }
\nonumber\\
&\times& H_l\left(\sqrt{|\breve n|B}\left[x_3+\frac{p_4}{|\breve n|B}\right]\right),
\nonumber\end{aligned}$$ where $H_m$ are the Hermite polynomials. The eigenvalues are $$\begin{aligned}
\label{eigenv_sc}
\lambda_{kl}=2|\breve n|B(k+l+1), \, \, k,l=0,1,\dots.\end{aligned}$$ The amplitudes $f(p_1)$ and $g(p_4)$ have to provide square integrability of the eigenfunctions in $x_1$ and $x_4$. In order to satisfy condition (\[current\_cont\_sc\]) one has to restrict the amplitude $f(p_1)$ additionally. The integral current through the domain wall is continuous if both $f$ and $H_k$ are odd or even functions simultaneously under the combined change $p_1\to-p_1$ and $x_2\to-x_2$ $$\begin{aligned}
 \label{cont_cond_sc}
f(-p_1)H_k(-z)=f(p_1)H_k(z).\end{aligned}$$ This property also guarantees the absence of the gauge specific contribution 