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 wiggle room 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 $e^{i\delta_{13}}$ 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.

  • 2
    $\begingroup$ I once asked a related question here. $\endgroup$
    – knzhou
    Commented Jan 30, 2021 at 18:27
  • 1
    $\begingroup$ Of course you are right that in some cases these weird “pseudo Dirac spinors” can’t be redefined to get ordinary Dirac spinors. But in those cases, most physicists would say that this mean trying to group Weyl spinors in pairs to form four-component spinors is more trouble than it’s worth, and they’ll just work directly with the Weyl spinors instead. $\endgroup$
    – knzhou
    Commented Jan 30, 2021 at 18:30
  • $\begingroup$ @knzhou, agreed, if you treat the right-handed and left-handed spinors separately (such as Weyl spinors), you would have more freedom in pseudscalor-like rotations without explicitly invoking the pseudscalor. This is exactly what is going in CKM-related rotations. But there is still not enough freedom in case of three generations to cancel out all the pseudscalors in the family-mixing parameters, therefore you are left with a residual CP violating phase in the electroweak sector. $\endgroup$
    – MadMax
    Commented Jan 30, 2021 at 18:42
  • $\begingroup$ @Oбжорoв the vector triple product is a pseudoscalar, ie a scalar that changes sign under an inversion in coordinates $\endgroup$
    – Triatticus
    Commented Jan 30, 2021 at 21:36
  • $\begingroup$ @MadMax: the derivation is correct. Actually you may want to know that your transformation involving $\exp(i\theta \gamma_5)$ is known as an "axial rotation" or "axial symmetry" in textbooks. It is relevant in the massless case, where it is a symmetry of the massless Dirac equation (but is famously anomalous in QFT...) $\endgroup$
    – user21299
    Commented Jan 31, 2021 at 5:41

2 Answers 2


There seems to be some confusion in the comments so here is a full answer.

First of all set $\eta=(-,+,+,+)$ and $$ \gamma_\mu\gamma_\nu + \gamma_\nu\gamma_\mu=2\eta_{\mu\nu}\,,\quad (\gamma^5)^2=1\,,\qquad \gamma^5 \gamma_\mu + \gamma_\mu \gamma^5=0\,. $$

Consider the standard Dirac equation first: $$ (\partial_\mu \gamma^\mu +m)\psi=0\,. $$ We find $$ 0=(-\partial_\mu \gamma^\mu +m)(\partial_\mu \gamma^\mu +m) \psi=-\eta^{\mu\nu}\partial_\mu\partial_\nu \psi + m^2\psi. $$ So we do not literally have a square root. You need a sign flip to kill the cross terms, similarly to how $x^2+y^2=(x+iy)(x-iy)$. Since we are here using the mostly plus convention, we recognise the Klein-Gordon equation in the last equality here. (See also Peskin & Schroeder ch 3.2, although they use the terrible mostly minus convention.)

What about $$ (\partial_\mu \gamma^\mu +m \exp(i\theta\gamma^5))\psi=0? $$ is this a "square root" in the above sense? Sure; all solutions of the above also solve the KG equation: $$ 0=(-\partial_\mu \gamma^\mu +m \exp(-i\theta\gamma^5))(\partial_\mu \gamma^\mu +m \exp(i\theta\gamma^5))\psi\\ =(-\eta^{\mu\nu}\partial_\mu\partial_\nu +m e^{-i\theta\gamma^5} \partial_\mu\gamma^\mu -m\partial_\mu\gamma^\mu e^{+i\theta\gamma^5}+m^2)\psi\\ =-\eta^{\mu\nu}\partial_\mu\partial_\nu \psi + m^2\psi\,. $$ Since $\gamma^5$ anticommutes with $\gamma^\mu$, the cross-terms cancel.

This is essentially the same as the derivation given in the question (which I agree with).

As I mention in my comment above, the axial rotation that kills the mass term that the question is about is more popular than this modified Dirac equation, if for no other reason than

  1. The axial rotation is a symmetry in the massless case, and
  2. In the massive case you can usually redefine your fermions with the axial rotation to kill the mass term $m\exp(i\theta\gamma^5)$ in favour of $m$.

EDIT: clearly you can classify all similar "Dirac equations" by tabulating matrices that (anti)commute with all $\gamma_{\mu}$. Of course that is textbook material. My recollection is that a Schur's lemma argument shows all such are in the span of $1$ and $\gamma^5$ (in any dimension).


Uniqueness is relatively broader reasearch topic: we need to check what is applicable in which context: take a look at this: The pseudodifferential operator square root of the Klein–Gordon equation - Claus Lammerzahla, 1993, ZARM. Also see: https://www.zarm.uni-bremen.de


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