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?


3 Answers 3


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:

\begin{align}(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) \end{align}

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. $\endgroup$ Commented Mar 4, 2023 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. $\endgroup$ Commented Mar 4, 2023 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
    Commented Mar 4, 2023 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. $\endgroup$ Commented Mar 5, 2023 at 0:41
  • $\begingroup$ If we lift our attention to bigger matrices or generally speaking other algebras the coefficients forming a non trivial square root of the KG operator live on some kind of curve (or algebraic variety if you'll indulge). I wonder from a purely mathematical standpoint if a family of differential operators whose coefficients live on a particular variety itself carries interesting mathematical structure. $\endgroup$ Commented Jun 9 at 18:51

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


Finding Square roots:

These square roots do exist however the solutions can get very complex quickly.

First we recall the general form of the LHS of the Euler-Lagrange equations with standard boundary conditions of an operator depending on variables $t,f,f',...f^{(n)}$

$$ \frac{\partial}{\partial t} - \frac{d}{dt}\left( \frac{\partial }{\partial f} \right) + \frac{d^2}{dt^2}\left( \frac{\partial }{\partial f'} \right) - ... $$

If our operator depends ONLY on $t$ then this is just:

$$ \frac{\partial }{\partial t} $$

And therefore this has an operator square root via the fractional derivative:

Now we wade a bit deeper into the water and consider an operator that depends exclusively on $t,f$ and no higher order terms. The Euler lagrange equations in this case are of the form

$$ \frac{\partial}{\partial t} - \frac{d}{dt}\left( \frac{\partial }{\partial f} \right) $$

Which can be expanded out as:

$$ \Omega = \frac{\partial}{\partial t} - \frac{\partial^2 }{\partial f \partial t} - \frac{\partial^2}{\partial f^2}f' $$

So this is a linear superoperator. It takes a non linear operator (where an operator itself a function that takes in a function and outputs a function) and produces another non linear operator. That also obeys given two operators $O_1, O_2$ we have that $\Omega[O_1 + O_2] = \Omega[O_1] + \Omega[O_2]$.

So one can try to find a matrix of operators $A,B,C$ s.t. given

$$ L = A*I + B \frac{\partial}{\partial t} + C \frac{\partial }{\partial f} $$

Then $L^2 = \Omega$. In particular this produces a system of equations $A^2 = 0_n$, $AB + BA = I_n$, $AC+CA=0_n$, $BC+CB=-I_n$ and $C^2 = -f'*I_n$. Where $0_n$ and $I_n$ refer to the 0 and identity matrix over $n \times n$ matrices.

I'm not sure when but there definitely exists some $n$ for which this has a solution. And if you wanted to, you could go looking for it, but even with $2\times 2$ matrices this becomes a system of 10 quadratic algebraic equations with 12 unknowns (so very likely solvable but very tedious to solve)

The interaction of $\frac{\partial}{\partial t}$ with $f(t)$ makes this a lot more complicated but still probably solvable. Instead of $10$ quadratic equations in $12$ numerical unknowns. You instead have $10$ second order PDEs in $12$ functional variable unknowns.

Considering higher order euler lagrange dependencies similarly follows. With multiple variables the problem gets even more complex and in general an interesting mathematical question could be given an operator depending on $q$ functional variables $f_1 ... f_q$ of differential order $c_1 ... c_q$ what is the minimum $n$ so that its square root can be taken of the Euler lagrange equations over the $n \times n$ real matrices.

Your particular instance is $q=4$ all of order $1$.

Connection to Diracs Equation:

It's not obvious to me what the relationship between factoring the euler lagrange equations and dirac's equation is. Let me explain why: 1. Dirac's Equations is a square root of a PDE operator (That accepts a function and outputs a function).

The euler lagrange equations are superoperators (that accept an operator and produce an operator).

Your question is in some sense asking what's the relationship between a particular function and a number. Philosophically one might ask what is the relationship between $e^x$ and $e$, but it would be very hard to find coherent and useful ideas explaining the relationship between $e^x$ and say the number $3$.

Similarly one might ask does knowing about the relationships between $\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}, \frac{d}{dx}$ allow you to know anything about $x^{\sqrt{2}}, x^2 $ the latter being a function $g$ s.t. $g^2 = g(g(x)) = x^2$.

You're question in particular lives on the next rung of this tower. You are not comparing a number and a function. You are not comparing a function and an operator. You are comparing a linear operator (dirac's equation) to a superlinear operator (square root of Euler lagrange).


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