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

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 $$v$$.

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.
• Hi Jacob, thank you very much for your answer. The non-relativistic limit of equations (1) and (2) that you have are exactly what I would have expected. However, I can't help but notice that $R$, $S$, and $v$ could depend on $c$ ($R$ and $S$ most likely do, as a solution to the relativistic Schrödinger equation) so I am not sure how one can argue that this limit is the right one?