Does there exist a square root of Euler-Lagrange equations $\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}-\frac{\partial \mathcal{L}}{\partial \phi} = 0$ in the sense that $(x+iy)(x-iy) = x^2+y^2$? By which I mean, does there exist a partial differential equation that squares to $\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}-\frac{\partial \mathcal{L}}{\partial \phi} = 0$? Just like how Dirac operator $(i\gamma^\mu\partial_\mu-m)\psi = 0$ squares to Klein-Gordon operator $(-\eta^{\mu\nu}\partial_\mu\partial_\nu + m)\psi = 0$. Can it be used to derive Dirac equation from Klein-Gordon Lagrangian?


2 Answers 2


The "square root" in the sense of $-(a^2 +b^2)=(ia-b)(ia+b)\,\,$ of the Klein-Gordon's differential operator is the Dirac differential operator. That is the main message:

$$(i\gamma^\mu\partial_\mu-m)(i\gamma^\nu\partial_\nu +m) = -\gamma^\mu\gamma^\nu\partial_\mu\partial_\nu -m^2 = -(\gamma^{\left[\mu\right.}\gamma^{\left.\nu\right]} + \gamma^{\left(\mu\right.}\gamma^{\left.\nu\right)})\partial_\mu\partial_\nu -m^2 = -\gamma^{\left(\mu\right.}\gamma^{\left.\nu\right)} \partial_\mu\partial_\nu -m^2 = -\eta^{\mu\nu}\partial_\mu\partial_\nu -m^2 = -(\Box +m^2)$$

In the algebra we threw away $\gamma^{\left[\mu\right.}\gamma^{\left.\nu\right]}\partial_\mu\partial_\nu$ because $\gamma^{\left[\mu\right.}\gamma^{\left.\nu\right]}$ is antisymmetric whereas the sucession of two partial derivatives is symmetric. It is also essential for this decomposition that the "mass operator" commutes with $i\gamma^\mu\partial_\mu$.

The analogy of a square root can be still extended a little bit further. Instead of writing the first term of the KG-equation (short Klein-Gordon: KG) $\partial^\mu \phi \partial_\mu\phi $ we can apply partial integration and write:

$$\partial^\mu \phi \partial_\mu\phi = -\phi \partial^\mu\partial_\mu \phi + \partial^\mu (\phi \partial_\mu \phi)$$

and just neglect the total partial derivative $\partial^\mu (\phi \partial_\mu \phi)$. Such a Lagrangian also leads to the KG-equation. In that case the KG-Lagrangian looks like this:

$${\cal{L}} = -\phi (\Box + m^2)\phi$$

In order to reach as close as possible to the Dirac Lagrangian we actually better start off from the Lagrangian of the complex KG-field:

$${\cal{L}} = -\phi^\dagger (\Box + m^2)\phi$$

We replace $\phi \rightarrow \psi$ and $\phi^\dagger \rightarrow \bar{\psi}$ as a first step and then choose of the 2 factors $i\gamma^\mu\partial_\mu\pm m$ the one with the minus sign :

$${\cal{L}} = \bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$

Therefore, it is not the Lagrangian that is square-rooted, but the differential operator inside.

Needless to say that we cannot just keep the complex scalar field sandwiching the differential operator as the $\gamma^\mu$ matrices have not only vector character but are also "bispinor tensors"(the indices are usually suppressed in order to avoid clutter) which requires spinors as sandwich in order to make the Lagrangian Lorentz-covariant.

It was the genius of P.Dirac to recognize this thereby giving up the scalar character of the involved fields.

  • $\begingroup$ Essentially for setting up a field theory is about above all which field to choose, a scalar field theory will be always scalar, even if you take the square root of it. And would $\sqrt{L }$ supposed to be? It would not have the right dimensions. Moreover, $L$ has to be Lorentz invariant, which is a very strong constraint. The Lagrangian has to be at least quadratic in its degrees of freedom, otherwise the EQM give absurd results. So square-rooting $\phi D\phi$ would make $L$ linear in $\phi$ with a absurd EQM. $\endgroup$ Mar 4 at 16:22
  • $\begingroup$ Actually formally one can write the KG-equation as $i\partial_t \phi= \sqrt{-\nabla^2 +m^2}\phi$ and develop the rhs (apart from $\phi$) as Taylor series. However, the resulting equation is non-local and not manifestly Lorentz-covariant. However, locality is an important feature of modern field theory, a principle that cannot easily abandoned. Therefore it is not the way to go. $\endgroup$ Mar 4 at 21:40
  • $\begingroup$ Thank you very much for your so valuable thoughts. However, your answer, due to imprecision in the formulation of my question, unintentionally tells me things I am already aware of. The true question is, does there exist a square root of Euler-Lagrange equations in the sense it forms a factorization, just like how you proved that Dirac operator squares to Klein-Gordon? $\endgroup$
    – user356874
    Mar 4 at 22:08
  • $\begingroup$ There is no much choice for finding another decomposition. Imagine the search of the roots of a quadratic form. Decomposing the the KG-differential operator in 2 linear factors is quite the same stuff. The pair of roots of a quadratic form is unique, it is weary to search for another pair. Of course one might wonder if the representation (up to equivalence) of the Clifford algebra is really unique. In 4D it is indeed the case which is well documented in the literature. If there were another decomposition, it would be already lively discussed in the physics literature which is not the case. $\endgroup$ Mar 5 at 0:41

The Euler Lagrange equations are the equations used to minimise the action. So, they do not exist in versions (i.e. square root of EL eqs etc). What you should be after if you would want (for some reason) a Dirac equation for a scalar field is a Lagrangian whose EL EoM is the Dirac equation.

This is however troublesome, as in the Dirac equation you have matrices and the solutions are spinor fields, which is not in agreement with having scalar fields which is your case now


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