Condtion on transformation to solve the Dirac equation

[Markus Kahn]

TL;DR Summary

Given a spinor $\psi$ that solved the Dirac equation and a Lorentz transformation $x\mapsto x^\prime :=\Lambda^{-1}x$ such that $\psi^\prime (x^\prime)=S\psi(x)$ I'd like to show that $S$ has to satisfy
$$\Lambda^\mu{}_\nu S^{-1}\gamma^\nu S = \gamma^\mu$$
for $\psi^\prime (x^\prime)$ to be a solution of the Dirac equation.

The problem is given in the summary.

My attempt: Assume that $\psi^\prime (x^\prime)$ is a solution of the Dirac equation in the primed frame, given the transformation $x\mapsto x^\prime :=\Lambda^{-1}x$ and $\psi^\prime (x^\prime)=S\psi(x)$, we have
$$
\begin{align*}
0&=(\gamma^\mu \partial_\mu^\prime - m)\psi^\prime (x^\prime) =(\gamma^\mu \partial_\mu^\prime - m)S\psi(x) \\
&=S(S^{-1}\gamma^\mu S \partial_\mu^\prime -m)\psi(x)\\
&=S({S^{-1}\gamma^\mu S \Lambda^{\nu}{}_\mu}\partial_\nu -m)\psi(x).
\end{align*}
$$
For the last equation to reduce to the Dirac equation, and therefore make $\psi(x)$ a solution of the Dirac equation, we need
$${S^{-1}\gamma^\mu S \Lambda^{\nu}{}_\mu}=\gamma^\nu.$$

My Professor did something similar, but not quite the same and I have a hard time judging if what we do is truly equivalent or if I'm just running in circles...
My Professors explanation: We first note that $\psi^\prime(x^\prime)=S\psi(x)$ is equivalent to $\psi^\prime(x) = S\psi(\Lambda x)$. With this we find
$$
\begin{aligned}
0 &=\left(\gamma^{\mu} \partial_{\mu}-m\right) \psi^{\prime}(x)=\left(\gamma^{\mu} \partial_{\mu}-m\right) S \psi(\Lambda x) \\
&=\left(\gamma^{\nu} S \Lambda_{\nu}^{\mu} \partial_{\mu} \psi-S m \psi\right)(\Lambda x) \\
&=S\left(S^{-1} \gamma^{\nu} S \Lambda_{\nu}^{\mu} \partial_{\mu} \psi-\gamma^{\mu} \partial_{\mu} \psi\right)(\Lambda x) \\
&=S\left(\Lambda^{\mu}{}_{\nu} S^{-1} \gamma^{\nu} S-\gamma^{\mu}\right)\left(\partial_{\mu} \psi\right)(\Lambda x)
\end{aligned}
$$
So the term in the bracket must vanish for invariance of the Dirac equation.

The issue I have with his approach is that I don't really understand the first step, i.e. I don't see why
$$0 =\left(\gamma^{\mu} \partial_{\mu}-m\right) \psi^{\prime}(x)$$
should hold.

 

[PeroK]

The issue I have with his approach is that I don't really understand the first step, i.e. I don't see why
$$0 =\left(\gamma^{\mu} \partial_{\mu}-m\right) \psi^{\prime}(x)$$
should hold.

If you look at the working after this, what must be meant is:
$$0 =\left(\gamma^{\mu} \partial'_{\mu}-m\right) \psi^{\prime}(x')$$

 

[vanhees71]

The transformation law is
$$\psi'(x')=S \psi(x)=S \psi(\Lambda x'),$$
if $x'=\Lambda^{-1} x$ with $\Lambda$ a proper orthochronous Lorentz transformation. So I'd prefer the first demonstration, which is much clearer in distinguishing the primed and unprimed quantities referring to the two different inertial reference frames.

 

[Markus Kahn]

@PeroK Could you elaborate? I'm asking because I don't see how what you write adds up with the very next line in my professors reasoning, i.e.
$$\begin{aligned}

\left(\gamma^{\mu} \partial_{\mu}^\prime-m\right) \psi^{\prime}(x^\prime)\neq\left(\gamma^{\mu} \partial_{\mu}-m\right) S \psi(\Lambda x) \\

\end{aligned}
$$

@vanhees71 Just to make sure, the part I labeled as "My attempt" has no circular reasoning and is sound? I'm asking because Peskin (and other places) explain it just like my Professor did, so I automatically assumed I'm making some kind of hidden assumption...

 

[vanhees71]

I think, it's sound and solid. The problem is with the primes in the argument of the field.

If you want to describe finally the transformation behavior of the quantized field, i.e., the field operators, it reads
$$\hat{U}(\Lambda) \psi(x) \hat{U}^{\dagger}(\lambda) = S \psi(\Lambda x).$$
Here $\hat{U}(\Lambda)$ is the unitary operator representing Lorentz transformations, and the transformation law describes a local realization of the Poincare group on the field operators, i.e., the field operators transform as the classical fields. That may trigger some textbook authors to use this notation also for the unquantized theory, but as I said, I prefer your notation. When I learned QFT this point confused me a lot until I understood the reason behind this somewhat less clear notation.

 

[PeroK]

@PeroK Could you elaborate? I'm asking because I don't see how what you write adds up with the very next line in my professors reasoning, i.e.
$$\begin{aligned}

\left(\gamma^{\mu} \partial_{\mu}^\prime-m\right) \psi^{\prime}(x^\prime)\neq\left(\gamma^{\mu} \partial_{\mu}-m\right) S \psi(\Lambda x) \\

\end{aligned}
$$

He has used $x$ to denote the variable in the transformed frame and $\Lambda x$ to denote the variable in the original frame. He's not using a prime on the spacetime coordinate or derivatives. He doesn't, as far as I can see, use any notation to distinguish between $\partial$ and $\partial'$. He just uses the context.

He only uses a single prime on $\psi$ to indicate that he's starting in the transformed frame. That's the way some people think about these things! If it was up to me, I'd put primes everywhere.

PS

Your notation is $x \ \rightarrow \ x' = \Lambda^{-1}x$.

His notation is $\Lambda x \ \rightarrow \ x = \Lambda^{-1} \Lambda x$