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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.

my2cts
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2 Answers2

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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.

Chiral Anomaly
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Disclaimer Some of the writing below is taken directly from [1]. I do so under fair use for pedagogic purposes.

Postulate [First Postulate of Special Relativity] The laws of physics take the same form in all inertial frames of reference [0].

Definition [Lorentz covariant quantity, Lorentz covariant equation]

  1. A physical quantity is said to be Lorentz covariant if it transforms under a given representation of the Lorentz group. ...[T]hese quantities are built out of scalars, four-vectors, four-tensors, and spinors...
  2. An equation is said to be Lorentz covariant if it can be written in terms of Lorentz covariant quantities.... The key property of such equations is that if they hold in one inertial frame, then they hold in any inertial frame.... This condition is a requirement according to the principle of relativity; i.e., all non-gravitational laws must make the same predictions for identical experiments taking place at the same spacetime event in two different inertial frames of reference. [2]

As shown in [1] and [3], in a frame of reference the homogeneous wave equation can be written in the form $$ i\,\hbar\,\gamma^\mu\,\frac{\partial\boldsymbol{\psi} }{\partial x^\mu} - m\,c\,\boldsymbol{\psi} = \mathbf{0}.\tag{1} $$ The $\gamma$ matrices satisfy the Clifford algebra $$ \gamma^\mu\,\gamma^\nu + \gamma^\nu\,\gamma^\mu = 2\,g^{\mu\nu}\,I, $$ where $g$ is the metric tensor of special relativity and $I$ is the identity element.

The most general homogeneous Lorentz transformation between two coordinate systems may be written $$ {x'}^\mu = {a^\mu}_\nu\,x^\nu. \tag{2} $$ From Equation (2), we write $$ \frac{\partial}{\partial x^\mu} = \frac{\partial {x'}^\lambda}{\partial x^\mu}\,\frac{\partial}{\partial {x'}^\lambda} = {a^\lambda}_\mu\,\frac{\partial}{\partial {x'}^\lambda}.\tag{3} $$ Using Equation (3), we obtain from Equation (1) that $$ i\,\hbar\,{a^\lambda}_\mu\,\gamma^\mu\,\frac{\partial \boldsymbol{\psi} }{\partial {x'}^\lambda} - m\,c\,\boldsymbol{\psi} = \mathbf{0}.\tag{4} $$ In analogy with Equation (2), we define $$ {\gamma'}^\lambda \equiv {a^\lambda}_\mu\,\gamma^\mu. \tag{5} $$ One can verify with Equation (2) that the $\gamma'$ matrices satisfy the same Clifford algebra as the $\gamma$ matrices. In other words, $$ {\gamma'}^\mu\,{\gamma'}^\nu + {\gamma'}^\nu\,{\gamma'}^\mu = 2\,g^{\mu\nu}\,I. $$ By Pauli's Fundamental Theorem (see [4] for a statement of the theorem and [1] for a proof of the theorem), up to a multiplicative constant there exists only one change-of-basis matrix $S$ such that for each and every $\lambda\in \left\{0,1,2,3\right\}$ we have that $$ {\gamma'}^\lambda = S^{-1}\,\gamma^\lambda\,S. \tag{6} $$ Upon substituting this relation into Equation (4), multiplying on the left by $S$, and changing the dummy index $\lambda$ to $\mu$, we find that $$ i\,\hbar\,\gamma^\mu\,\frac{\partial S\,\boldsymbol{\psi} }{\partial {x'}^\mu} - m\,c\,S\,\boldsymbol{\psi} = \mathbf{0}.\tag{7} $$ Defining the Lorentz transformation law for spinors to be $$ \boldsymbol{\psi}' \equiv S\,\boldsymbol{\psi} , $$ we have from Equation (7) that $$ i\,\hbar\,\gamma^\mu\,\frac{\partial \boldsymbol{\psi}' }{\partial {x'}^\mu} - m\,c\,\boldsymbol{\psi}' = \mathbf{0}.\tag{8} $$

What we have found is that Equation (1) and Equation (8) have the same form upon changing the dependent variables and independent variables from $\boldsymbol{\psi}=\boldsymbol{\psi}(\mathbf{x})$ to $\boldsymbol{\psi}' = \boldsymbol{\psi}'(\mathbf{x}')$ via \begin{align} \boldsymbol{\psi}'(\mathbf{x}') = S\,\boldsymbol{\psi} (\mathbf{x}) \quad\text{and}\quad {x'}^\mu = {a^\mu}_\nu\,x^\nu . \end{align} This finding indicates that the equation of motion given by Dirac's equation conforms to the first postulate of special relativity. Further, all of the quantities in the equation are Lorentz covariant quantities. Hence, the Dirac equation is a Lorentz covariant equation.

As an addendum, it is important to note the following. The physical consequences of the Dirac equation are independent of the frame of reference used to derive them. "Since the trace of a matrix is insensitive to a similarity transformation [see Equation (6)], traces of products of gamma matrices do not depend on the specific representation used. Therefore, physical results determined by the Dirac theory should be expressed in terms of these traces [2]."

Bibliography

[0] https://en.wikipedia.org/wiki/Postulates_of_special_relativity

[1] Bethe and Jackiw, Intermediate Quantum Mechanics, pp. 349-366.

[2] https://en.wikipedia.org/wiki/Lorentz_covariance

[3] Dirac, The Quantum Theory of the Electron

[4] https://math.stackexchange.com/q/5061118

Michael Levy
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