From my understanding of spinors, they arise as projective representations of $SO_0(1,3)$ that do not correspond to representations of $SO_0(1,3)$. But still one says here - and virtually everywhere - that the Dirac equation is Lorentz invariant and even gives an explicit "representation" for the Lorentz group acting on it.
In my head, I am telling myself, that I should read the action as being one of $Spin(1,3)$ instead. Then, fixing a covering map $\rho : Spin(1,3) \to SO_0(1,3)$, I can push any $s \in Spin(1,3)$ down to a $\Lambda \simeq \rho(s) \in SO_0(1,3)$ telling me how to transform the derivatives appearing in the Dirac equation. But then, in the parlance of the linked question, $U(\Lambda)$ would have to be multi-valued - or one would have to choose one of many possible values for $U(\Lambda)$.
Of course different choices of $U(\Lambda)$ can only differ by a complex phase $\exp(i \phi)$ for some $\phi \in \mathbb{R}$, which is why I have the feeling that this is intimately connected with the $U(1)$ gauge invariance of QED:
Because we can at least locally around a point $x$ choose a $\phi(y)$ for every $y$ in some neighbourhood of $x$, the Dirac equation can only ever become Lorentz i.e $SO_0(1,3)$-invariant if we promote it to an equation that is also locally $U(1)$ invariant which leads to QED in the familiar way.

From these thoughts, I conclude that the Dirac equation is not Lorentz invariant but rather it is $Spin(1,3)$-invariant. It becomes Lorentz invariant only when it is extended to something making it locally $U(1)$ invariant e.g QED.

Am I off track or on the spot here? I have failed to find anything written on this topic :/


2 Answers 2

  1. Just because the Dirac equation is invariant under projective representations of the Lorentz group does not mean that it is not Lorentz invariant. In general, in any quantum mechanical systems that enjoy some symmetry given by a group $G$, the states transform under projective (anti-)unitary representations of $G$.
  2. You don't need local $\text{U}(1)$ gauge symmetry to fix the ambiguity of the unitary operator $\text{U}(\Lambda)$; the Dirac equation already comes equipped with a global $\text{U}(1)$ symmetry $\psi\to e^{i\alpha}\psi$ that can be used to pick out a unique representative for $\text{U}(\Lambda)$.
  • $\begingroup$ 1. True, but I argued above, that the Dirac equation cannot be Lorentz ($SO_0(1,3)$) invariant. 2. Not globally, i.e for all of $SO_0(1,3)$ $\endgroup$
    – iolo
    Jul 10, 2019 at 13:09
  • $\begingroup$ Isnt 1 only true for compact groups? There are ni finit3 dimensial (anti-) unitary representations of non compact groups if i remember correctly. $\endgroup$
    – lalala
    Jan 3 at 6:21

Dirac equation is invariant under global Lorentz symmetry and $U(1)$ gauge symmetry is not a symmetry of Dirac equation. Rather it is invariant under global $U(1)$ symmetry.

And when one constructs the relation between the $SL(2,C)$ (the double covering group of $SO_{+}(1,3)$) one gets $$M(\Lambda)=\pm \displaystyle\frac{1}{(\det[\Lambda^{\mu}_{\nu}\sigma_{\mu}\bar{\sigma^{\nu}}])^{1/2}}\Lambda^{\mu}_{\nu}\sigma_{\mu}\bar{\sigma^{\nu}}\quad,$$ where $M(\Lambda) ~ \in SL(2,C) $ and $ \Lambda \in SO_{+}(1,3) $.

And the above relation is a homomorphism as it's clear by the $\pm$ sign.

  • 1
    $\begingroup$ Your $M$ is not a homomorphism becaus it is not single-valued. It corresponds to the inverse of my $\rho$ and if it were a homomorphism, it would be a global section of $\rho$ making $\rho: SL(2,\mathbb{C}) \to SO_0(1,3)$ a trivial bundle - which it is not! Instead, $M$ (with a choice of sign) is only a local homomorphism. Thus, we would only have an action of some neighbourhood of the identity of $SO_0(1,3)$ on spinors, but not all of $SO_0(1,3)$ $\endgroup$
    – iolo
    Jul 10, 2019 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.