Why does a term proportional to $\left(F,\,\tilde{F}\right)\propto Tr\left[ F_{\mu\nu}\tilde{F}^{\mu\nu}\right]$ in the Lagrangian of the pure Yang-Mills theory violate CP?
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2 Answers
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If we write the field strength in terms of "electric" and "magnetic" fields $\vec{E}$ and $\vec{B}$, the relevant expression can be written as
$$\text{Tr}F_{\mu\nu}\tilde{F}^{\mu\nu}=4\,\text{Tr}\vec{E}\cdot\vec{B}.$$.
Under parity transformations, $\vec{E}\rightarrow-\vec{E}$ and $\vec{B}\rightarrow\vec{B}$, while under charge conjugation, $\vec{E}\rightarrow-\vec{E}$ and $\vec{B}\rightarrow-\vec{B}$. Together, this leads to a transformation behaviour of the term in the Lagrangian under $CP$-transformations given by
$$\text{Tr}F_{\mu\nu}\tilde{F}\rightarrow-\text{Tr}F_{\mu\nu}\tilde{F}.$$
This clearly shows that the term is not invariant.
answered May 4, 2014 at 12:32
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1$\begingroup$ It should be noted that you use the "active" transformation - in the passive one $\vec{E} \to \vec{E}, \vec{B} \to -\vec {B}$. (It hit me in the eyes because classically we usually say that the "axial vector" $\vec{B}$ changes sign under parity.) $\endgroup$– VoidCommented Jul 23, 2014 at 18:00
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3$\begingroup$ I have one question about the terminology. Since the $\vec{E} \cdot \vec{B}$ term only breaks the parity transformation (and time reverse transformation of course, I mean it doesn't break charge conjugation), why the literature often stress that "this term breaks $C P$ transformations" (It's correct but it take innocent $C$ into consideration) not "break $P$ and $T$ transformations"? $\endgroup$– luyuwuliCommented Aug 4, 2014 at 8:04
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1$\begingroup$ This is a nice explanation for QED. However, how can we explain for the non-abelian case? For example for SU(2) or SU(3) where the first equation in function of E and B is no longer true because has an extra non-abelian term. $\endgroup$ Commented Apr 8, 2021 at 17:57
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The usual action for Yang-Mills theory is, using differential forms $$S = \int \operatorname{tr} (F \wedge \star F)$$ where $\star$ is the Hodge dual. Now note that the integral of a differential form is always defined with respect to an orientation, and the Hodge dual is also defined with respect to an orientation. Parity is reversing the orientation, which changes the sign of both the integral and the Hodge dual, so the signs cancel.
However the term $$S_\theta = \int \theta \operatorname{tr} (F \wedge F)$$ only has one sign change when you reverse the orientation, so it is not $P$ invariant.
answered May 4, 2014 at 12:36
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2$\begingroup$ This is a nice explanation with respect to parity. Can one somehow deduce the charge conjugation properties in the language of differential forms? $\endgroup$ Commented May 4, 2014 at 12:51
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4$\begingroup$ Parity and time-reversal are geometric, charge conjugation is not, so it's not really a property of forms. With a choice of gauge, $F$ is Lie algebra-valued, taking values in $\mathfrak g$. You can choose the generators of $\mathfrak g$ to be Hermitian matrices, so that on quantization the field is its own anti-particle. $\endgroup$ Commented May 4, 2014 at 13:55