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People say that the probability density function of the continuity equation for the Dirac equation is definite positive. I wanted to see it myself and followed the same path as Bjorken & Drell's did in their RQM book. They say "We left-multiply the Dirac equation with the Hermitian conjugate wave function, then we take the Hermitian conjugate of the Dirac equation and we right-multiply it with the wavefunction by subtracting the second equation from the first one we get the continuity equation as follows".

I'll add a screenshot from the book below.

What is that I don't understand is why they take the derivative operators as if they are Hermitian operators. Those operators are anti-hermitian thus left side of the equation after subtraction is, \begin{align} i\hbar(\Psi^\dagger\frac{d\Psi}{dt}-\frac{d\Psi^\dagger}{dt}\Psi) \end{align}

at this point, due to minus sign, we can't use product rule in derivative to simplify this side as, \begin{align} i\hbar\frac{d}{dt}(\Psi^\dagger\Psi) \end{align}

I looked for other books too all seem to follow the same process like this one, I am not sure what I am missing there.

Book Screenshot,

enter image description here

In the form with a minus sign, the probability density might not be positive definite for an arbitrary wavefunction.

Where do you think the problem is?

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  • $\begingroup$ @ZeroTheHero Probability current depends on spatial derivatives sir, this is the left side of the equation, the right side is probability current but I didn't write that part here because it is unnecessary for my question. Yes, everybody claims that probability density is \Psi^\dagger \Psi that is what I aim to reach to but they use product rule for derivatives to reach that identity. I can't see how we can use product rule since there is a minus between them, not a positive sign. $\endgroup$ – Omer Faruk May 11 at 9:41
  • $\begingroup$ Yes your are right. sorry for my bad. I will remove my comment. $\endgroup$ – ZeroTheHero May 11 at 13:03

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