Legend has it that Dirac arrives at the Dirac equation (all equations are in Planck units $c= \hbar = 1$ in this post): $$ i\gamma^\mu\partial_\mu\psi - m\psi =0 $$ by taking the square root of the operator $$ \partial^\mu\partial_\mu $$ of the Klein–Gordon equation which is the relativistic version of the Schrodinger equation: $$ (\partial^\mu\partial_\mu + m^2)\psi =0 $$
However, is Dirac equation the only "square root" of Klein–Gordon equation? Is there any other "square root" of Klein–Gordon equation?
It turns out there is a "square root" of Klein–Gordon equation which is different from Dirac equation: $$ i\gamma^\mu\partial_\mu\psi - me^{\theta i\gamma_5} \psi = 0 $$ where the "complex" fermion mass term is endowed with both CP even scalar part and CP odd pseudoscalar $i\gamma_5$ part: $$ m e^{\theta i\gamma_5} = m\cos\theta + m i\gamma_5 \sin\theta . $$
Can we verify that the above "modified" Dirac equation is indeed the "square root" of Klein-Gordon equation? Let's get cracking on the nitty-gritty: $$ m^2\psi \\ = (me^{-\theta i\gamma_5}) (me^{\theta i\gamma_5})\psi \\ = (me^{-\theta i\gamma_5}) (i\gamma^\mu\partial_\mu)\psi \\ = (i\gamma^\mu\partial_\mu) (me^{\theta i\gamma_5}) \psi \\ =(i\gamma^\mu\partial_\mu)(i\gamma^\nu\partial_\nu) \psi \\ = -\partial^\mu\partial_\mu\psi. $$ Voila! We indeed recover the classic Klein-Gordon equation, without any funny "complex" factor. Note that in the 4th line we leveraged the crucial anti-commuting property between $\gamma_5$ and $\gamma^\mu$.
Can anyone please verify the above derivation is correct or wrong? Or is it actually the same as the original Dirac equation?
Added note
The fun fact is that after an axial rotation of the fermion field $$ \psi \rightarrow e^{-\frac{1}{2}\theta i\gamma_5} \psi. $$ the "complex" mass term in the Lagrangian format can be transformed into a scalar mass term (while leaving the kinetic term intact): $$ m\bar{\psi} e^{\theta i\gamma_5} \psi \rightarrow m\bar{\psi} \psi. $$ In other words, via redefining the fermion field, the CP odd part of the mass $m i\gamma_5 \sin\theta$ can be effectively rotated away. That being said, we know that the fermion mass is generated via the Higgs mechanism. Rotating to the "real" mass is achievable if only there is one Higgs boson. If you fancy a fancy-schmancy beyond standard model involving multiple Higgs bosons, the said rotation can only make one Higgs boson CP even, while leaving the other Higgs boson with a CP breaking pseudoscalar $i\gamma_5$ phase.
Of course these multi-Higgs boson scenarios are not usually considered in entry-level QFT books. Thus the "modified" Dirac equation is usually not mentioned at all. But do you think the usual QFT text books should mention it just for fun?
If you treat the right-handed and left-handed spinors separately (such as Weyl spinors), you would have more freedom in axial rotations without explicitly invoking the pseudoscalar. This is exactly what is going on in the CKM-related rotations. But there is still not enough freedom in case of three generations to cancel out all the pseudoscalars in the family-mixing parameters, therefore you are left with a residual CP violating phase in the electroweak sector.
In a nutshell, the most general "square root" counterpart of the Klein–Gordon equation is the Dirac equation with a "complex" mass term $m e^{\theta i\gamma_5} \psi$. The "real" mass Dirac equation is merely a special case of $\theta = 0$.
More added note:
There is a separate issue of the scalar mass term $$ m\bar\psi \psi $$ being imaginary if you plug in the Grassmann number components from $\psi$ and $\bar\psi$. See details here.