First, C) isn't the Dirac Equation, it's the Klein-Gordon equation

Now, to your main question. A) comes from the classical equation for a free massive particle:

$\dfrac{p^2}{2m} = E$

by making the operator (operating on $\phi$) substitutions:

$p^2 \rightarrow - \hbar^2 \nabla^2$

$E \rightarrow i \hbar \dfrac{\partial}{\partial t}$

C) comes from B) by further recognizing that:

$E^2 \rightarrow -\hbar^2 \dfrac{\partial^2}{\partial t^2}$

Paul,

This particular writing of the problem in the article I have always thought was sloppy as well. The most confusing part of the discussion is the statement "The continuity equation is as before". At first one writes the continuity equation as:

$$\nabla \cdot J + \dfrac{\partial\rho}{\partial t} = 0$$

Although the del operator can be defined to be infinite dimensional, it is frequently reserved for three dimensions and so the construction of the sentence does not provide a clear interpretation. If you look up conserved current you find the 4-vector version of the continuity equation:

$$\partial_\mu j^\mu = 0$$

What is important about the derivation in the wikipedia article is the conversion of the non time dependent density to a time dependent density, or rather:

$$\rho = \phi^*\phi$$

becomes

$$\rho = \dfrac{i\hbar}{2m}(\psi^*\partial_t\psi - \psi\partial_t\psi^*)$$

the intent is clear, the want to make the time component have the same form as the space components. The equation of the current is now:

$$J^\mu = \dfrac{i\hbar}{2m}(\psi^*\partial^\mu\psi - \psi\partial^\mu\psi^*)$$

which now contains the time component. So the continuity equation that should be used is:

$$\partial_\mu J^\mu = 0$$

where the capitalization of $J$ appears to be arbitrary choice in the derivation.

One can verify that this is the intent by referring to the article on probability current.

From the above I can see that the sudden insertion of the statement that one can arbitrarily pick $$\psi$$ and $$\dfrac{\partial \psi}{\partial t}$$ isn't well explained. This part the article was a source of confusion for me as well until one realized that the author was trying to get to a discussion about the Klein Gordon equation

A quick search of web for "probability current and klein gordan equation" finds good links, including a good one from the physics department at UC Davis. If you follow the discussion in the paper you can see it confirms that the argument is really trying to get to a discussion about the Klein Gordon equation and make the connection to probability density.

Now, if one does another quick search for "negative solutions to the klein gordan equation" one can find a nice paper from the physics department of the Ohio University. There we get some good discussion around equation 3.13 in the paper which reiterates that, when we redefined the density we introduced some additional variability. So the equation:

$$\rho = \dfrac{i\hbar}{2mc^2}(\psi^*\partial_t\psi - \psi\partial_t\psi^*)$$

(where in the orginal, c was set at 1)
really is at the root of the problem (confirming the intent in the original article). However, it probably still doesn't satisfy the question,

"can anyone show me why the expression for density not positive
definite?",

but if one goes on a little shopping spree you can find the book Quantum Field Theory Demystified by David McMahon (and there are some free downloads out there, but I won't link to them out of respect for the author), and if you go to pg 116 you will find the discussion:

Remembering the free particle solution $$\varphi(\vec{x},t) = e^{-ip\cdot x} = e^{-i(Et- px)}$$ the time derivatives are $$\dfrac{\partial\varphi}{\partial t} = -iEe^{-i(Et- px)}$$ $$\dfrac{\partial\varphi^*}{\partial t} = iEe^{i(Et- px)}$$ We have $$\varphi^*\dfrac{\partial\varphi}{\partial t} = e^{i(Et- px)}[-iEe^{-i(Et- px)}] = -iE$$ $$\varphi\dfrac{\partial\varphi^*}{\partial t} = e^{-i(Et- px)}[iEe^{i(Et- px)}] = iE$$ So the probability density is $$\rho = i(\varphi^*\dfrac{\partial\varphi}{\partial t} - \varphi\dfrac{\partial\varphi^*}{\partial t}) = i(-iE-iE) = 2E$$ Looks good so far-except for those pesky negative energy solutions. Remember that $$E = \pm\sqrt{p^2+m^2}$$ In the case of the negative energy solution $$\rho = 2E =-2\sqrt{p^2+m^2}<0$$ which is a negative probability density, something which simply does not make sense.

Hopefully that helps, the notion of a negative probability does not make sense because we define probability on the interval [0,1], so by definition negative probabilities have no meaning. This point is sometimes lost on people when they try to make sense of things, but logically any discussion of negative probabilities is non-sense. This is why QFT ended up reinterpreting the Klein Gordan equation and re purposing it for an equation that governs creation and annihilation operators.

## Best Answer

The probability current is just that - the rate and direction that probability flows past a point. It is analogous to electric current or to a fluid current, and the continuity equation is the same as for those concepts.

For example, if the the probability current is high on the left-hand side of a region and low on the right hand side, more probability is flowing in from the left than out from the right, and the total probability for the particle to be found in that region is increasing.

To calculate this, the probability that a particle is found in a region is

$$\int_\text{region} \phi^* \phi \,\, \mathrm{d}x$$

The time derivative of this is the rate that the probability for the particle to be in that region changes.

$$\int_\text{region} \left(\frac{\partial \phi^*}{\partial t} \phi + \phi^*\frac{\partial \phi}{\partial t}\right) \,\,\mathrm{d}x$$

We know what the time derivative of $\phi$ is, though, from the Schrodinger equation. If you plug that in and assume the potential is real, this simplifies to

$$\frac{i \hbar}{2m} \int_\text{region} \left((\nabla^2\phi^*) \phi - \phi^* (\nabla^2 \phi) \, \,\right)\mathrm{d}x$$

If you integrate this by parts, you see it's the same as the integral of the flux of the probability current over the surface. Thus the probability current is a flow of probability the same way the electric current is a flow of charge.

The continuity equation is just the differential form of this same relation. Since we had to use the Schrodinger equation to find $\dfrac{\partial \phi}{\partial t}$, we've shown that the continuity equation follows from Schrodinger's equation.