# Tag Info

0

Let us consider what would happen if CP was actually a symmetry. Then, arguing non-perturbatively, the overlap between a $K^0$ state and a $\bar K^0$ after time $t$ would be $$\langle \bar K^0| \exp(-i H t) K^0 \rangle.$$ Let $U$ denote the operator implementing CP symmetry, such that $U^{-1}=U$. Then, since $[U,H]$ by assumption,...

7

The popular press's description of this experiment is wildly wrong. It's hard to tell whether they just got it completely wrong on their own, or Scheck got it wrong and they're accurately describing what he said, or if it's some combination of the two. Scheck is a co-author but not the first author, and none of the ridiculous things they represent him as ...

55

To be honest, much of this feels like very irresponsible journalism, partly on the part of the BBC and very much so on the part of Science alert. If you're looking for an accessible resource to what the paper does, the cover piece on APS Physics and the phys.org piece are much more sedate and, I think, much more commensurate with what's actually reported ...

29

The articles are a little on the hysterical side, but I think they are just saying that violation of CP-symmetry means there must be violation of T-symmetry. T-symmetry means that physical laws are unchanged if we reverse the direction time flows. Classical theories obey T-symmetry, and it seems intuitively obvious that quantum mechanics would as well. But ...

0

Generating CP asymmetry from scattering process is indeed possible. Actually, already in leptogenesis people have considered such processes because they dominate at high energy ($T>M_N$) but it turns out that they have negligible effects on the final asymmetry (e.g., hep-ph/0309342). A field $\psi$, which has CP violating interactions with fermions $f$, ...

4

The phase of a single constant $y$ is unphysical and may be completely eliminated by the redefinition $$y \to t\cdot \exp(i\alpha), \quad \phi\to \phi\cdot \exp(-i\alpha)$$ which doesn't change the Lagrangian and for a given $\alpha$, $y$ will be real positive. One could also get rid of the phase by transforming the fermions. So there can't be a CP-...

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