Determining correction terms for an equivalent form to the relativistic Schrödinger equation

Question

I am reading the book Stochastic Processes in Quantum Physics by Masao Nagasawa and I am having a bit of trouble getting my head around the following result. I would much appreciate any help. The thing is that Theorem 7.2.1 says that if $\psi(x,t)=\exp(R(x,t)+iS(x,t))$, where $R$ and $S$ are real valued, satisfies the relativistic Schrödinger equation $$i\partial_t \psi=\left(\sqrt{-c^2\Delta+m^2c^4 }-mc^2\right)\psi$$ then $\phi_{\pm}(x,t)$ defined as $\phi_{\pm}(x,t)=\exp(R(x,t)\pm S(x,t))$ satisfy $$\pm\partial_t \phi_{\pm}-v(x,t)\phi_{\pm}=\left(\sqrt{-c^2\Delta+m^2c^4 +c\{\nabla, \textbf{B}(x,t)\}+v(x)^2}-mc^2\right)\phi_{\pm}$$ where ${v}(t, x)$ and $\textbf{B}(t, x)$ are determined by the equations $$ c\textbf{B} \cdot \nabla R-v \frac{\partial R}{\partial t}+\frac{\partial^{2} S}{\partial t^{2}}+2 \frac{\partial R}{\partial t} \frac{\partial S}{\partial t}=0 \tag{1} $$ $$ c\textbf{B}\cdot \nabla S-v\left(\frac{\partial S}{\partial t}-mc^2\right)+c^{2} \Delta R+c^{2} (\nabla R)^{2}+\left(\frac{\partial S}{\partial t}\right)^{2}=0 \tag{2} $$ under the gauge condition $$ c\nabla \cdot \textbf{B}=\frac{\partial v}{\partial t} $$

What I don't understand is the following claim: on page 239 it says that one can determine the correction terms $\textbf{B} (t, x)$ and $v(t, x)$ of vector and scalar potentials using equations (1) and (2). That is what don't comprehend, I can't see how one could compute $\textbf{B}$ and $v$ from these equations. Maybe I could add that I am ultimately interested in taking the non-relativistic limit $c\to+\infty $, so what would be ideal would be to have a closed expression for the correction terms $\textbf{B}$ and $v$.

Answer 1

This was a bit too long of a response for a comment but I hope it can help:

In the limit $c \to \infty$, your equations reduce to

$$ \textbf{B} \cdot \nabla R=0 \tag{1} $$ $$ mv+ \Delta R+ (\nabla R)^{2}=0 \tag{2} $$ with the non-relativistic gauge condition $ \nabla \cdot {\bf B} = 0$. So we can generally take ${\bf B}=\nabla \times A$ for some vector potential $A$. However, I notice that this is a strictly one dimensional problem, so we have ${\bf B}=k + f(t)$, a constant plus some function in time. Additionally, $R$ is one dimensional, so we can re-write these equations as $$ (k+f(t)) \cdot R'(x) = 0 \tag{1*} $$ $$ mv+R''(x)+(R'(x))^2=0 \tag{2*} $$ it looks like equation 1 can simplify to $R(x)=n$, another constant. All told, this constant solution seems quite mundane, so hopefully someone can shed some further light on this problem.