Source: Quantum Field Theory for the Gifted Amateur by Tom Lancaster, Stephen J. Blundell.
I am struggling to understand the logical step from the outline of the 'proof' in the footnote, to the fact that the probabilty density must look like eq. 6.12. Can anyone supply a supplemental text that walks through this more plainly? Moreover, I find my secondary source's derivation at a level above me as well.
6.2 Probability currents and densities
One of the reasons that Schrödinger wasn't happy with the Klein-Gordon equation after he'd derived it was that something rather nasty happens when you think about the flow of probability density. The probability of a particle being located somewhere depends on $\phi^{*}(x)\phi(x)$ and so if this quantity is time-dependent then particles must be sloshing around. The probability density $\rho$ and probability current density5 $\boldsymbol{j}$ obey a continuity equation \begin{equation} \dfrac{\mathrm d\rho}{\mathrm dt}+\boldsymbol{\nabla} \cdot \boldsymbol{j}=0, \tag{6.9}\label{6.9} \end{equation} which is more easily written in four-vector notation as \begin{equation} \partial_{\mu}j^{\mu}=0. \tag{6.10}\label{6.10} \end{equation} If, as is usual in non-relativistic quantum mechanics,6 we take the spatial part to be \begin{equation} \boldsymbol{j}(x)=-\mathrm i\left[\phi^{*}(x)\boldsymbol{\nabla}\phi(x)-\phi(x)\boldsymbol{\nabla}\phi^{*}(x)\right], \tag{6.11}\label{6.11} \end{equation} then, for eqn 6.10 to work,7 we require the probability density to look like8 \begin{equation} \rho(x)\boldsymbol{=}\mathrm i\left[\phi^{*}(x)\dfrac{\partial\phi(x)}{\partial t}\boldsymbol{-}\dfrac{\partial\phi^{*}(x)}{\partial t}\phi(x)\right]. \tag{6.12}\label{6.12} \end{equation} The resulting covariant probability current for the Klein–Gordon equation is then given by \begin{equation} j^{\mu}(x)=\mathrm i\{\phi^{*}(x)\partial^{\mu}\phi(x)-\left[\partial^{\mu}\phi^{*}(x)\right]\phi(x)\}, \tag{6.13}\label{6.13} \end{equation} which, as the notation suggests, is a four-vector. Substituting in our [...]
$^7$ It will work, and you can prove it as follows. Take the Klein-Gordon equation (eqn 6.5) and premultiply it by $\phi^{*}(x)$. Then take the complex conjugate of eqn 6.5 and premultiply by $\phi(x)$. Subtracting these two results will give an equation of the form of eqn 6.9 with $\boldsymbol{j}$ and $\rho$ as given.
Secondary Source: Quantum Field Theory by Lewis H. Ryder.
...where the Schr$\ddot{\rm o}$dinger equation and its complex conjugate have been used. What are the corresponding expressions for the Klein-Gordon equation? To be properly relativistic, $\rho$ should not, as in (2.18), transform as a scalar, but as the time component of a 4-vector, whose space component is $\mathbf{j}$, given by (2.19). Then $\rho$ is given by \begin{equation} \rho(x)=\dfrac{\mathrm i \hbar}{2m}\left(\phi^{*}\dfrac{\partial\phi}{\partial t}-\phi\dfrac{\partial\phi^{*}}{\partial t}\right) \tag{2.20}\label{2.20} \end{equation} and with \begin{equation} j^{\mu}=\left(\rho,\mathbf{j}\right)=\dfrac{\mathrm i \hbar}{m}\phi^{*}\left(\overset{\leftrightarrow}{\partial_{0}},\overset{\leftrightarrow}{\boldsymbol{\nabla}}\right)\phi =\dfrac{\mathrm i \hbar}{m}\phi^{*}\overset{\leftrightarrow}{\partial^{\mu}}\,\phi \tag{2.21}\label{2.21} \end{equation} where \begin{equation} A\overset{\leftrightarrow}{\partial^{\mu}}B\stackrel{\text{def}}{=}\tfrac12\left[A\partial^{\mu}B\boldsymbol{-}(\partial^{\mu}A)B\right], \tag{2.22}\label{2.221} \end{equation} and we have used (2.9), we have the continuity equation \begin{equation} \partial_{\mu}j^{\mu}=\dfrac{\mathrm i \hbar}{2m}\left(\phi^{*}\square \,\phi-\phi\,\square\,\phi^{*}\right)=0, \tag{2.23}\label{2.3} \end{equation} since $\phi^{*}$ also obeys the Klein-Gordon equation. Then $\rho$ and $\mathbf{j}$ are the probability density and current we want. But this immediately presents a problem, because $\rho$, given by equation (2.20), unlike expression (2.18) for the...