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We know the Lorentz group is $O(3,1)$ in 4 dimensional spacetime.

We know that there are 4 disconnected components in Lorentz group $O(3,1)$, and https://math.stackexchange.com/q/2204349/

$$\pi_0(\mathrm{O}(1,3)) \cong \mathbb{Z}_2\times\mathbb{Z}_2.$$

My question is that in QFT we have a discrete charge conjugation symmetry $C$, parity $P$ and time reversal $T$.

We know the $P$ and $T$ flips the 4 disconnected components of $O(3,1)$ into each other.

How about the charge conjugation $C$, does it sit in the Lorentz group? Or does it only act on the matter field? How do we understand $C$ in the Lorentz group $O(3,1)$?

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    $\begingroup$ Charge lives in spacetime but is not part of it. $\endgroup$ – G. Smith Sep 23 at 3:57
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In QED, for example, charge conjugation commutes with the Lorentz group. It's an "internal" symmetry, not part of Lorentz (or Poincaré) symmetry.

However, a different kind of connection exists between charge conjugation and the Lorentz group, via the CPT theorem. The CPT theorem says that every relativistic QFT (satisfying certain axioms) has a symmetry that, among other things, reflects a timelike direction and an odd number of spatial directions. We call it the "CPT" theorem, but that's a little misleading, because it should be regarded as more fundamental (that is, more broadly applicable) than C, P, or T individually. In other words, we really ought to define charge conjugation in terms of CPT, not the other way around.

Here's the point: The general proof of the CPT theorem makes use of the complex Lorentz group (through the fact that correlation functions can be analytically continued to complex values of their spacetime arguments), and in this sense there is a kind of connection between C and Lorentz symmetry. This is reviewed in

and in the classic text by Streater and Wightman, PCT, Spin and Statistics, and All That.

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  • $\begingroup$ thanks very much vote up $\endgroup$ – annie heart Sep 24 at 1:46

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