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

Let's examine the Dirac Lagrangian of quantum mechanics:

$$\bar{\psi}\left(i\gamma^{\mu}\overrightarrow{\partial}_{\mu}-mc^{2}\right)\psi=0$$

Where the arrow implies direction of action for the derivative. The barred wavefunction can be written as:

$$\bar{\psi}=\psi^{\dagger}\gamma^{0}$$

Where the dagger denotes the conjugate transpose. Then we have:

$$\psi^{\dagger}\gamma^{0}\left(i\gamma^{\mu}\overrightarrow{\partial}_{\mu}-mc^{2}\right)\psi=0$$

It is very common in discussions of conservecd current to also examine the Adjoint Dirac equation:

$$\psi^{\dagger}\gamma^{0}\left(i\gamma^{\mu}\overleftarrow{\partial}_{\mu}+mc^{2}\right)\psi=0$$

Where the derivative is now acting to the left. Adding both of these together we obtain a conservation of current equation:

$$\partial_{\mu}\left(\psi^{\dagger}\gamma^{0}\gamma^{\mu}\psi\right)=0$$

This is all standard, my question here is can we define another, equivalent Dirac Lagrangian of the form:

$$\psi^{\dagger}\left(i\gamma^{\mu}\overrightarrow{\partial}_{\mu}-mc^{2}\right)\psi^{,}=0$$

Where:

$$\psi^{,}=\gamma^{0}\psi$$

Which together with it's adjoint equation:

$$\psi^{\dagger}\left(i\gamma^{\mu}\overleftarrow{\partial}_{\mu}+mc^{2}\right)\gamma^{0}\psi=0$$

$$\psi^{\dagger}\left(i\gamma^{\mu}\overrightarrow{\partial}_{\mu}-mc^{2}\right)\gamma^{0}\psi=0$$ also defines a conserved current

$$\partial_{\mu}\left(\psi^{\dagger}\gamma^{\mu}\gamma^{0}\psi\right)=0$$ Now if we combine all four of our expressions we obtain:

$$\partial_{\mu}\left(\psi^{\dagger}\gamma^{0}\gamma^{\mu}\psi\right)+\partial_{\mu}\left(\psi^{\dagger}\gamma^{\mu}\gamma^{0}\psi\right)=0$$ or, perhaps more familiar:

$$\partial_{\mu}\left(\psi^{\dagger}g^{0\mu}\psi\right)=0$$

Which is just the condition that:

$$\partial_{\mu}\left(\psi^{\dagger}Ig^{\nu\mu}\psi\right)e_{0}=0$$

Where I is the identity $$4x4$$ matrix. From this standpoint, $$\psi$$ 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:

$$\psi\Longrightarrow\gamma^{0}\psi$$

$$\psi^{\dagger}\Longrightarrow\psi^{\dagger}\gamma^{0}$$

In the Dirac Lagrangian?

• Just a note, a lagrangian is not equal to zero, so the right hand side is a nonzero object – Triatticus Nov 28 '18 at 15:18
• @Triatticus I suppose I should have thrown it all under an integral over space and said it's on-shell. – R. Rankin Nov 29 '18 at 1:22

Secondly, your equation $$\psi^{\dagger}\left(i\gamma^{\mu}\overrightarrow{\partial}_{\mu}-mc^{2}\right)\psi^{,}=0$$ is wrong, since $$\gamma^0$$ and $$\gamma^\mu$$ do NOT commute.
Thirdly, the conservation rule as you derived $$\partial_{\mu}\left(\psi^{\dagger}\gamma^{0}\gamma^{\mu}\psi\right)=0$$ is based purely on classical equation of motion (Dirac equation). Such conservation laws may NOT survive in a full quantum theory. The axial current in ABJ anomaly is a perfect example.
• +1 I'm just playing with the Dirac equation here, That's why i didn't tag it as quantum field theory, I tagged it as the Dirac equation. I'm aware they don't commute, however the current is generally derived by multiplying the equation on the left by $\gamma^0$ why can't we right multiply it instead? – R. Rankin Nov 29 '18 at 1:28