I often read that the Dirac equation is covariant under Lorentz transformations and that this property makes it the right equation and in some sense beautiful.

The thing is, the equation $$ \left(i\gamma^\mu\partial_\mu-\frac{mc}{\hbar}\right)\psi(x)=0 $$ is not covariant at all unless one assumes that the spinor transforms in a very special, not at all obvious way under Lorentz transformations. In particular $$\psi'(x')\neq\psi(x) \rightarrow \psi'(x')=S(\Lambda)\psi(x)$$

Now all of the references I read go on to use the covariance of the Dirac equation to show the form of this spinor transformation. Clearly this is very circular reasoning. Because in the end they say: Look, with this transformation the Dirac equation is covariant under Lorentz transformations, what a beautiful equation!

Is there a way out? One would have to find an argument for the transformation law of the spinor that does not rely on using the covariance of the Dirac equation. I think one even needs a good argument to conclude that $S(\Lambda)$ is a linear operator.

I'm wondering why one should even transform the spinor at all, why not just transform the $\gamma$ matrices in some funny way. My lecture notes make all of this seem very obvious, but I think it's not at all.

  • $\begingroup$ You probably need to read a bit more, for example Bjorken and Drell's book on relativistic quantum mechanics or some other, you will see that indeed the Dirac equation is what it is due to need for Lorentz invariance. $\endgroup$ Commented Jun 14, 2020 at 12:45
  • $\begingroup$ @ChiralAnomaly Yes this would be great! Thanks for the Reference Nelson. $\endgroup$
    – user224659
    Commented Jun 14, 2020 at 18:10
  • $\begingroup$ For me, before proving that the matrix $S(\Lambda)$ exist and what it is, we could imagine the Dirac equation be non covariant, in spite of its apparent covariance form. The successful search for $S$ is not a circular reasoning. $\endgroup$ Commented Jun 14, 2020 at 23:33

1 Answer 1


For a moment, forget about Lorentz transformations. Let's step back and think more generally.

If a transformation $T$ leaves some thing $\Omega$ invariant, then we may call the transformation $T$ a symmetry of $\Omega$. We may call it a symmetry even if the transformation isn't linear. The thing $\Omega$ could be a solid shape, or it could be something more abstract.

Let's specialize that general idea a little bit. Consider a theory governed by some set of equations of motion. For example, consider Maxwell's equations, which are the equations of motion for the electromagnetic field. The purpose of the equations of motion is to answer this question: of all the behaviors that we can imagine, which ones are physically allowed? A behavior is physically allowed if and only if it satisfies the equations of motion. In this context, we can take $\Omega$ to be the set of all physically-allowed behaviors, and any transformation $T$ that leaves $\Omega$ invariant (map solutions to solutions) can be called a symmetry of the theory.

Now let's specialize that general idea a little more. Consider a "theory" whose equation of motion is the Dirac equation $$ \newcommand{\pl}{\partial} (i\gamma^\mu\pl_\mu-m)\psi(x)=0. \tag{1} $$ Just like in the preceding example, we can think of this equation as telling us which four-component functions $\psi(x)$ are physically allowed. We can take $\Omega$ to be the set of all physically-allowed functions — that is, all solutions of the Dirac equation (1) — and any transformation $T$ that leaves the set $\Omega$ invariant can be called a symmetry of this "theory." (I'm putting "theory" in scare-quotes because it's too simple to describe any interesting real-world phenomena. Maybe "toy theory" would be a better name.)

What symmetries does the theory (1) have? That could be a difficult question to answer completely, so let's make things easier. Instead of asking for all symmetries, let's just ask for symmetries that have some mathematically easy form. Linear is about as easy as we can get, so let's consider transformations of the form $\psi(x)\to S\psi(\Lambda x)$ where $S$ is a matrix and $\Lambda$ is a linear transformation of the coordinates. We don't need to assume that $\Lambda$ is a Lorentz transformation.

Remember what we're asking: we want to know if the transformation $T$ defined by $\psi(x)\to S\psi(\Lambda x)$ maps solutions of (1) to other solutions of (1). That's what we mean by (linear) symmetry. For most choices of the pair $(S,\Lambda)$, it won't be a symmetry, because it will map a solution to a non-solution. To see which choices $(S,\Lambda)$ work, suppose that $\psi(x)$ satisfies (1), and require $$ \newcommand{\pl}{\partial} (i\gamma^\mu\pl_\mu-m)\psi'(x)=0 \hskip1cm \text{with }\ \psi'(x) := S\psi(\Lambda x). \tag{2} $$ If we can find any $(S,\Lambda)$ such that equation (1) implies equation (2), then we've found a symmetry. Notice that we don't change the differential operator $i\gamma^\mu\pl_\mu-m$ at all. We change the function from $\psi(x)$ to $\psi'(x)$, and we ask whether the new function $\psi'(x)$ still satisfies the same equation.

Now, suppose we find a symmetry $(S,\Lambda)$ for which $\Lambda$ happens to be a Lorentz transformation, meaning that the transformation $x\to\Lambda x$ leaves the quantity $-x_0^2+x_1^2+x_2^2+x_3^2$ invariant. Such symmetries of the Dirac equation do exist: for every Lorentz transformation $\Lambda$, there is at least one matrix $S$ such that $(S,\Lambda)$ is a symmetry.

More generally, suppose that the set of allowed behaviors in a field theory includes a symmetry $(A,B,C,...,\Lambda)$ for every Lorentz transformation $\Lambda$, where the matrices $A,B,C,...$ act on the components of the various fields. I don't think we'd be stretching the etiquette of language too far by referring to this property as Lorentz covariance. With this definition, Maxwell's equations in free space are Lorentz covariant. With this same definition, the Dirac equation is also Lorentz covariant.

Is that circular? Well, we can't point to a definition and call it a derivation. That would be circular. But we can adopt a general definition of Lorentz covariance, one that works like we want it to in more familiar cases (like Maxwell's equations), and then derive the fact that the Dirac equation also satisfies that same general definition. That's not circular.

  • $\begingroup$ What you call toy theory is giving me enough headaches already. But this answer is very nice. Now I see that my original issues came from an wrong/incomplete definition of Lorentz covariance. There is no symmetry for coordinate lorentz transformations alone. $\endgroup$
    – user224659
    Commented Jun 15, 2020 at 13:19

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