probabilityquantum mechanicsunitarity
In quantum mechanics, the continuity equation $-{d\rho}/{dt}=\nabla\cdot{J}$ holds for a probability density $\rho$ and probability current $J$. But what does it mean, from a physical point of view?
I imagine it means that a particle can not appear or disappear in a given volume $V$, there must be a "particle flux" in the walls of $V$ for particles entering or leaving $V$.
If I´m wrong, please tell me.
I'll work in the $+---$ convention so $\mathbf{k}\cdot\mathbf{x}-k_0t=-k\cdot x$. You should find that $\psi=\exp{\mp ikx}$ obtains $j^\mu=\pm m^{-1}k^\mu$. Complex conjugation of $\psi$ still gives a solution for the KGE (the TDSE doesn't work like this) and changes the signs of energy and probability. The problem with trying to magic away half the solutions is it breaks invariance under time reversal. This is related to the fact that $E^2=p^2+m^2$ is invariant under $E\to -E$.
The general solution can be written as a Fourier transform. If you generalise a linear combination of planar-wave solutions with constant coefficients so that, after quantization, these coefficients are operator-valued, you encounter another difficulty: $\hat{\psi}$ won't be Hermitian. By contrast, if you keep all solutions the integrand has two terms: one with annihilation operators, the other with creation operators. If either term is deleted, $\psi$ will annihilate the vacuum bra but not the vacuum ket or vice versa, depending on the term retained.
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}
Best Answer
The simplest version refers to a single particle. It may be more obvious in integral form: apply the Stokes theorem. The the equation states that the time derivative of the probability of the particle being measured in V is equal to the rate at which probability flows into V.
So your version is close.
See https://en.wikipedia.org/wiki/Probability_current#Continuity_equation_for_quantum_mechanics