Explanation of a failed approach to relativize Schrodinger equation

[nacadaryo]

I'm reading the Wikipedia page for the Dirac equation

[itex]\rho=\phi^*\phi\,[/itex]

...

[itex]J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*)[/itex]

with the conservation of probability current and density following from the Schrödinger equation:

[itex]\nabla\cdot J + \frac{\partial\rho}{\partial t} = 0.[/itex]

The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression

[itex]\rho = \frac{i\hbar}{2m}(\psi^*\partial_t\psi - \psi\partial_t\psi^*).[/itex]

which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression

[itex]J^\mu = \frac{i\hbar}{2m}(\psi^*\partial^\mu\psi - \psi\partial^\mu\psi^*)[/itex]

The continuity equation is as before. Everything is compatible with relativity now, but we see immediately that the expression for the density is no longer positive definite - the initial values of both ψ and [itex]∂_t ψ[/itex] may be freely chosen, and the density may thus become negative, something that is impossible for a legitimate probability density. Thus we cannot get a simple generalization of the Schrödinger equation under the naive assumption that the wave function is a relativistic scalar, and the equation it satisfies, second order in time.

I am not sure how one gets a new [itex]\rho[/itex] and [itex]J^\mu[/itex]. How does one do to derive these two? And can anyone show me why the expression for density not positive definite?

 

[mfb]

I think those formulas are guessed, and it can be shown that they satisfy the relevant equations afterwards. As you can see from the nonrelativistic probability flow, it is nothing completely new...

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

Consider a function ##\psi## where ##\rho>0## and look at ##\psi^*##.
##\rho(\psi)=-\rho(\psi^*)##

 

[Bill_K]

Take ψ to be a plane wave, ψ(x,t) = e^{i(k·x - ωt)}, which will be a solution provided ω² = k² + m². For this solution, ρ = ħω/m, obviously positive/negative whenever ω is positve/negative. For the general solution which is a superposition of plane waves, ρ is an integral over the positive frequency solutions minus an integral over the negative frequency ones.

According to the continuity equation, Q = ∫ρ d³x is a conserved quantity. Although the 'derivation' of this usually consists of simply writing it down, its existence is no accident. For a complex scalar field, Q represents the total charge. 'Charge' can mean either the ordinary electric charge or some other charge such as strangeness.