How Do You Derive the Relativistic Probability Current Formula?

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SUMMARY

The discussion focuses on deriving the relativistic probability current formula, specifically transitioning from the equation j(𝑥,𝑡) = Re[Ψ*(𝑥,𝑡)(-i/m)∇Ψ(𝑥,𝑡)] to j = (-i/2m)[Ψ*(∇Ψ) - (∇Ψ*)Ψ]. The user successfully identifies the relationship needed to bridge the gap as Re(z) = (z + z*)/2, which is crucial for simplifying the expression. This derivation is essential for understanding quantum mechanics and the behavior of wave functions.

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[SOLVED] Relativistc probability current

Homework Statement



I've got two formulae here and I'm not sure how you get from the first to the second:

\textbf{j}(\textbf{x},t) = Re[\Psi^{*}(x,t)(-i/m)\nabla\Psi(x,t)]

= \frac{-i}{2m}[\Psi^{*}(\nabla\Psi) - (\nabla\Psi^{*})\Psi]

Can anyone suggest a relationship I need to bridge the gap? Thanks.

Homework Equations





The Attempt at a Solution

 
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Try Re(z)=(z+z*)/2.
 
Got it! Thanks Dick :)
 

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