Question

Can a wavefunction be written as an active transformation of the metric tensor?

Answer 1

Let's examine the Dirac Lagrangian of quantum mechanics:

ψ¯(iγμ∂→μ−mc2)ψ=0
Where the arrow implies direction of action for the derivative. The barred wavefunction can be written as:

ψ¯=ψ†γ0
Where the dagger denotes the conjugate transpose. Then we have:

ψ†γ0(iγμ∂→μ−mc2)ψ=0
It is very common in discussions of conservecd current to also examine the Adjoint Dirac equation:

ψ†γ0(iγμ∂←μ+mc2)ψ=0
Where the derivative is now acting to the left. Adding both of these together we obtain a conservation of current equation:

∂μ(ψ†γ0γμψ)=0
This is all standard, my question here is can we define another, equivalent Dirac Lagrangian of the form:

ψ†(iγμ∂→μ−mc2)ψ,=0
Where:

ψ,=γ0ψ
Which together with it's adjoint equation:

ψ†(iγμ∂←μ+mc2)γ0ψ=0
ψ†(iγμ∂→μ−mc2)γ0ψ=0
also defines a conserved current

∂μ(ψ†γμγ0ψ)=0
Now if we combine all four of our expressions we obtain:

∂μ(ψ†γ0γμψ)+∂μ(ψ†γμγ0ψ)=0
or, perhaps more familiar:

∂μ(ψ†g0μψ)=0
Which is just the condition that:

∂μ(ψ†gνμψ)e0=0
From this standpoint, ψ seems to simply be playing the role of a coordinate transformation, or more properly an active transformation, changing the physical properties of the metric.

Note that the current is an integrand, and one can apply Gauss's law here to obtain a second order equation. So here's my question again:

Are we changing anything physically by making the switch:

ψ⟹γ0ψ
instead of:

ψ†⟹ψ†γ0
In the Dirac Lagrangian?

Answer 2

This is because the phase of a wavefunction can only be determined relative to the phase of another particle’s wavefunction, when they occupy the same point in space.