Finding the expression for probability density (the Klein Gordon equation) 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...

 A: The Schr$\ddot{\rm o}$dinger equation is non-relativistic and for a free particle is derived from the Hamiltonian
\begin{equation}
H\boldsymbol{=} \dfrac{p^2}{2m}
\tag{K-01}\label{eqK-01}    
\end{equation}
by the transcription
\begin{equation}
H\boldsymbol{\longrightarrow} i\hbar\dfrac{\partial}{\partial t}\quad \text{and}\quad \mathbf{p}\boldsymbol{\longrightarrow} \boldsymbol{-}i\hbar\boldsymbol{\nabla}
\tag{K-02}\label{eqK-02}    
\end{equation}
so that
\begin{equation}
i\hbar \dfrac{\partial \psi}{\partial t}\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\psi\boldsymbol{=} 0
\tag{K-03}\label{eqK-03}     
\end{equation}
For a first try to derive a relativistic quantum mechanical equation we make use of the property that according to the theory of special relativity the total energy $\;E\;$ and momenta $\;(p_x,p_y,p_z)\;$ transform as components of a contravariant four-vector
\begin{equation}
p^\mu\boldsymbol{=}\left(p^0,p^1,p^2,p^3\right)\boldsymbol{=}\left(\dfrac{E}{c},p_x,p_y,p_z\right)
\tag{K-04}\label{eqK-04}    
\end{equation}
of invariant length
\begin{equation}
\sum\limits_{\mu\boldsymbol{=}0}^{3}p_{\mu} p^{\mu}\boldsymbol{\equiv}p_{\mu} p^{\mu}\boldsymbol{=}\dfrac{E^2}{c^2}\boldsymbol{-}\mathbf{p}\boldsymbol{\cdot}\mathbf{p}\boldsymbol{\equiv}m^2c^2\tag{K-05}\label{eqK-05}    
\end{equation}
where $\;m\;$ is the rest mass of the particle and $\;c\;$ the velocity of light in vacuum.
Following this it is natural to take as the Hamiltonian of a relativistic free particle
\begin{equation}
H\boldsymbol{=}\sqrt{p^{2}c^2\boldsymbol{+}m^2c^4}
\tag{K-06}\label{eqK-06}    
\end{equation}
and to write for a relativistic quantum analogue of \eqref{eqK-03}
\begin{equation}
 i\hbar \dfrac{\partial \psi}{\partial t}\boldsymbol{=}\sqrt{\boldsymbol{-}\hbar^2c^2 \nabla^{2}\boldsymbol{+}m^2c^4}\,\psi
\tag{K-07}\label{eqK-07}    
\end{equation}
Facing the problem of interpreting the square root operator on the right in eq. \eqref{eqK-07} we simplify 
mathematics by removing this square root operator, so that
\begin{equation}
\left[\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\boldsymbol{-}\nabla^{2}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\right]\psi\boldsymbol{=}0
\tag{K-08}\label{eqK-08}    
\end{equation}
or recognized as the classical wave equation
\begin{equation}
\left[\square\boldsymbol{+}\left(\dfrac{mc}{\hbar}\right)^2\right]\psi\boldsymbol{=}0
\tag{K-09}\label{eqK-09}    
\end{equation}
where(1)
\begin{equation}
\square\boldsymbol{\equiv}\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\boldsymbol{-}\nabla^{2}\boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\dfrac{\partial}{\partial x^\mu} 
\tag{K-10}\label{eqK-10}    
\end{equation}
Equation \eqref{eqK-09} is the Klein-Gordon equation for a free particle. With its complex conjugate we have
\begin{align}
& \dfrac{1}{c^2}\dfrac{\partial^2 \psi\hphantom{^{\boldsymbol{*}}}}{\partial t^2}\boldsymbol{-}\nabla^{2}\psi\hphantom{^{\boldsymbol{*}}}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\psi\hphantom{^{\boldsymbol{*}}}\boldsymbol{=} 0
\tag{K-11.1}\label{eqK-11.1}\\
&\dfrac{1}{c^2}\dfrac{\partial^2 \psi^{\boldsymbol{*}}}{\partial t^2}\boldsymbol{-}\nabla^{2}\psi^{\boldsymbol{*}}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\psi^{\boldsymbol{*}}\boldsymbol{=} 0
\tag{K-11.2}\label{eqK-11.2}
\end{align}
Multiplying them by $\;\psi^{\boldsymbol{*}},\psi\;$ respectively and subtracting side by side we have(2)
\begin{align}
\dfrac{1}{c^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial^2 \psi}{\partial t^2}\boldsymbol{-}\psi\dfrac{\partial^2 \psi^{\boldsymbol{*}}}{\partial t^2}\right)\boldsymbol{-}\left(\psi^{\boldsymbol{*}}\nabla^{2}\psi\boldsymbol{-}\psi\nabla^{2}\psi^{\boldsymbol{*}}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)&\boldsymbol{=} 0\quad \boldsymbol{\Longrightarrow}
\nonumber\\
\dfrac{1}{c^2}\dfrac{\partial}{\partial t}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\boldsymbol{+}\boldsymbol{\nabla \cdot}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)&\boldsymbol{=} 0
\tag{K-12}\label{eqK-12}    
\end{align}
We multiply above equation by $\;i\hbar/2m\;$ in order to have real quantities on one hand and on the other hand to have an identical expression for the probability current density vector as that one from the Schr$\ddot{\rm o}$dinger equation
\begin{equation}
\dfrac{\partial}{\partial t}\left[\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\right]\boldsymbol{+}\boldsymbol{\nabla \cdot}\left[\dfrac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)\right]\boldsymbol{=} 0
\tag{K-13}\label{eqK-13}    
\end{equation}
so
\begin{equation}
\dfrac{\partial \varrho}{\partial t}\boldsymbol{+}\boldsymbol{\nabla \cdot}\boldsymbol{S}\boldsymbol{=} 0
\tag{K-14}\label{eqK-14}     
\end{equation}
where
\begin{equation}
\boxed{\:\:\varrho\boldsymbol{\equiv}\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\:\:}\quad \text{and} \quad \boxed{\:\:\boldsymbol{S}\boldsymbol{\equiv}\dfrac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)\:\:}
\tag{K-15}\label{eqK-15}     
\end{equation}
We would like to interpret $\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)$ as a probability density $\varrho$. However, this is impossible, since it is not a positive definite expression.

(1)
We define 
\begin{align}
\blacktriangleright x^\mu\boldsymbol{=}\left(ct,\mathbf{x}\right)&\blacktriangleright \nabla^\mu\boldsymbol{=}\partial^\mu\boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\boldsymbol{=}\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\boldsymbol{-}\boldsymbol{\nabla}\right)
\nonumber\\
&\blacktriangleright \nabla_\mu\boldsymbol{=}\partial_\mu\boldsymbol{=}\dfrac{\partial}{\partial x^\mu}\boldsymbol{=}\left(\dfrac{1}{c}\dfrac{\partial}{\partial  t},\boldsymbol{+}\boldsymbol{\nabla}\right)\blacktriangleright\square \boldsymbol{=}\nabla^\mu\nabla_\mu \boldsymbol{=}\partial^\mu\partial_\mu \boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\dfrac{\partial}{\partial x^\mu}
\nonumber  
\end{align}

(2)
If $\;\psi\;$ and $\;\mathbf{a}\;$ are scalar and vector functions in $\;\mathbb{R}^{3}$ then
\begin{equation}
\boldsymbol{\nabla \cdot}\left(\psi\mathbf{a}\right)\boldsymbol{=}\mathbf{a}\boldsymbol{\cdot}\boldsymbol{\nabla}\psi\boldsymbol{+}\psi\boldsymbol{\nabla \cdot}\mathbf{a}
\nonumber     
\end{equation}

A: You start off as described in the footnote 7 (We assume the validity of the Klein-Gordon equation for $\phi$ and $\phi^\ast$): 
$$0 = -i\phi^{\ast} (\Box +m^2)\phi +i \phi(\Box +m^2)\phi^{\ast} = i \left[\phi^{\ast}\partial_\mu\partial^\mu \phi - \phi\partial_\mu\partial^\mu\phi^{\ast}\right] =i \left[ \partial_\mu\phi^{\ast} \partial^\mu \phi +\phi^{\ast}\partial_\mu\partial^\mu\phi - \partial_\mu\phi\partial^\mu\phi^{\ast} - \phi\partial_\mu \partial^\mu \phi^{\ast}\right] =  i\left[\partial_\mu(\phi^{\ast}\partial^{\mu}\phi - \phi\partial^{\mu}\phi^{\ast})\right] = \partial_\mu j^{\mu}$$
where we used the definition $$j^\mu = i[\phi^{\ast}\partial^\mu\phi - \phi\partial^\mu\phi^{\ast}]$$   and   $$\Box =-\partial_\mu\partial^\mu$$ Then with $\mu=(0,i)$ and  $(i=1,2,3)$ $$\partial^i =-\nabla$$ you get the relation you wanted to proof (using  $j^\mu =(\rho, \bf{j})$ as  $j^\mu$ is a 4-vector ): 
$$\bf{j} = -i[\phi^{\ast}\nabla\phi - \phi\nabla\phi^{\ast}]$$  respectively $$j^0\equiv \rho = i\left[ \phi^{\ast}\frac{\partial\phi}{\partial t} - \phi\frac{\partial\phi^{\ast}}{\partial t}  \right]$$
As the found $j^{\mu}$ fulfills the continuity equation $0=\partial_\mu j^{\mu}$ it is the current density for the Klein-Gordon field $\phi$. It can of course also be found by using the Noether theorem. 
