In non-relativistic quantum mechanics we have the continuity equation

$$ \frac{\partial \rho ( \vec r, t)}{\partial t} + \vec \nabla \cdot \vec j ( \vec r, t) =0 $$

where $\rho (\vec r,t ) = \psi ^* (\vec r,t ) \psi ( \vec r,t ) = |\psi (\vec r,t )|^2$ which is positive and provides a probability density

From the time-dependent Schrodinger equation we write

$$ \vec j (\vec r, t) = -\frac{i\hbar}{2m} \left [ \psi ^* (\vec r,t ) (\vec \nabla \psi ( \vec r,t )) - ( \vec \nabla \psi ^* (\vec r,t)) \psi (\vec r,t ) \right ] $$

Now looking at relativity we have

$$ j^\mu = ( c\rho,\vec j ) $$

with probability density $\rho$ and probability current $\vec j$ so then here the continuity equation is

$$ \partial _\mu j^\mu =0 $$

💼 Case: from this lets try to derive the expression of $j^\mu$ in terms of the wave function as above

• Previously we needed the time-dependent Schrodinger equation now we need the Klein-Gordon

  $$ \frac{1}{c^2} \frac{\partial ^2 }{\partial t^2 } \psi = \vec \nabla ^2 \psi - \left ( \frac{mc}{\hbar} \right ) ^2 \psi $$

• Taking the complex conjugate we get

  $$ \frac{1}{c^2 }\frac{\partial ^2 }{\partial t^2 }\psi ^=\nabla ^2\psi ^ -\left ( \frac{mc}{\hbar} \right ) ^2 \psi ^* $$

• Now multiplying both by their conjugate counterpart we get

  $$ \begin{aligned} \frac{1}{c^2} \psi ^\frac{\partial ^2 }{\partial t^2 } \psi &= \psi ^ \vec \nabla ^2 \psi - \left ( \frac{mc}{\hbar} \right ) ^2 \psi ^\psi \\ \frac{1}{c^2}\psi \frac{\partial ^2 }{\partial t^2 } \psi^ &= \psi \vec \nabla ^2 \psi^* - \left ( \frac{mc}{\hbar} \right ) ^2 \psi \psi ^* \end{aligned} $$

• Subtracting then naturally leads to the continuity equation

  $$ \begin{aligned} \frac{1}{c^2} \left ( \psi ^* \frac{\partial^2 }{\partial t^2 } \psi - \psi \frac{\partial ^2}{\partial t^2} \psi ^* \right )&=\psi ^* \nabla ^2 \psi - \psi \nabla ^2 \psi ^* \\ \frac{\partial}{\partial t}\underbrace{\left [ \frac{i\hbar}{c} \left ( \psi ^* \frac{\partial }{\partial t } \psi - \psi \frac{\partial }{\partial t} \psi ^* \right ) \right ]}{\rho}+ \vec \nabla \cdot\underbrace{\left [ -i\hbar c ( \psi ^* \vec \nabla \psi - \psi \vec \nabla \psi ^*)\right ]}{\vec j} &=0 \\ \frac{\partial \rho}{\partial t} + \vec \nabla \cdot \vec j &=0 \\ \partial _\mu j^\mu &=0 \end{aligned} $$

• We thus by comparison can define our four vector as

  $$ \boxed{j^\mu = i\hbar c \left ( \psi ^* \partial ^\mu \psi -\psi \partial ^\mu \psi ^* \right )} $$

💎 Conclusion: $\rho=\frac{i\hbar}{c} \left ( \psi ^* \frac{\partial }{\partial t } \psi - \psi \frac{\partial }{\partial t} \psi ^* \right )$ is no longer positively defined and thus is not as straightforward as a probability density anymore

💃 Example: consider the free particle plane wave solution

$$ \psi(\vec r,t )=Ne^{i(\vec p \cdot \vec r - E_{\vec p}t)/\hbar} $$