EM CP-Violation, why not?

[ChrisVer]

Why can't there be a term in the SM lagrangian for the U(1)_Y of the form:

$F_{\mu \nu} \tilde{F}^{\mu \nu}$ ?

As there is for the strong interactions?

(Although I've seen such terms appearing in the axion models, such as the KSVZ where by introducing an additional very heavy quark Q with charge $e_Q$, you can have the coupling of the axion field $\alpha$ with light quarks via the EM anomalies: $L_{EM-anom} = \frac{a}{f_a} 3 e_Q^2 \frac{\alpha_{fine-str}}{4 \pi} F_{\mu \nu} \tilde{F}^{\mu \nu}$ )

 

[Orodruin]

For an abelian theory, $F^{\mu\nu}\tilde F_{\mu\nu}$ is a total derivative:
$$
\partial_\mu \epsilon^{\mu\nu\rho\sigma} A_\nu \partial_\rho A_\sigma =
\epsilon^{\mu\nu\rho\sigma} [(\partial_\mu A_\nu)(\partial_\rho A_\sigma) + A_\nu \partial_\mu \partial_\rho A_\sigma].
$$
The last term disappears due to the derivatives commuting and $\epsilon$ being asymmetric. The first term is proportional to $F^{\mu\nu}\tilde F_{\mu\nu}$.

 

[ChrisVer]

The problem is not about total derivatives, because even in strong CP-problem, the term of $\bar{\theta}$ : $G \tilde{G}$ is a total derivative/can be expressed as such. t'Hoft however showed that this total derivative integral doesn't vanish for every gauge...so I guess, It has to do with the gauge transformations in some way...

 

[Haelfix]

It's perfectly legitimate to worry about a CP violating theta term for say the electroweak theory, with the different group structure, although it turns out that such a term is unobservable, and can be rotated away by a suitable chiral transformation. However for the Abelian theory, post 2 is essentially all there is too it.

 

[ChrisVer]

I will try to write it down in maths?

In the QCD, a resolution to the $U_A(1)$ problem, is provided by the chiral anomaly for axial currents.
The axial current assosiated with the $U_A(1)$ gets quantum corrections from the triangle graph which connects it to two gluon fields with quarks going around the loop. This anomaly gives a non-zero divergence of the axial current:

$\partial_\mu J^\mu_5 = \frac{g_s^2 N}{32 \pi^2} G^{\mu \nu}_a \tilde{G}_{a \mu \nu} \ne 0$

This chiral anomaly affects the action:

$\delta Z \propto \int d^4 x \partial_\mu J^\mu_5 = \frac{g_s^2 N}{32 \pi^2} \int d^4 x G^{\mu \nu}_a \tilde{G}_{a \mu \nu}$

And it can be further shown that the $G \tilde{G}$ can be expressed in terms of a total divergence (just like the QED field strength tensors), $G^{\mu \nu}_a \tilde{G}_{a \mu \nu}= \partial_\mu K^\mu$

with $K^\mu = \epsilon^{\mu \rho \sigma \omega} A_{a \rho} [ G_{a \sigma \omega} - \frac{g_s}{3} f_{abc} A_{\sigma b} A_{\omega c} ]$

the problem then comes when you insert this in the action integral above and you reach:

$\delta Z \propto \frac{g_s^2 N}{32 \pi^2} \int \sigma_\mu K^\mu \ne 0$

The last was shown by t'Hoft, because the right boundary condition to use is that $A$ is a pure gauge field at spatial infinity, either then A=0 or a gauge transformation of 0...

Now what's the difference with the same thing you can obtain for the action in QED?
@Orodruin in his post, showed exactly that $F \tilde{F} = c \partial_\mu T^\mu$

So in the action, you will have contributions of the form:

$\delta Z' \propto \int d^4 x \partial_\mu T^\mu = \int d \sigma_\mu T^\mu$

Why in this case the infinity is taken to be T=0 and not a general gauge transformation of T: $T' = T + \partial a$ so a gauge transformation of 0?

I hope I made clear my problem?
Thanks...

 

[samalkhaiat]

If you do the integral $\int d \sigma^{ \mu } K_{ \mu }$ for $SU(2)$, the calculation will tell you why it vanishes for $U(1)$.

 

[Haelfix]

I will try to write it down in maths?
I hope I made clear my problem?
Thanks...

Good! So my claim is that the surface term in the Abelian theory vanishes. To see this, you can try direct computation, you can show it by asymptotic analysis, or you can be really clever and argue it away by topological arguments.

I will give you a hint on how to do it the second way. Note that to keep the action finite, we require that the (F Fbar) term decreases faster than O(1/r^2) where we set our boundary conditions to be the (euclidean) hypersphere as the radius r goes off to infinity. Show that this means that the total derivative goes as O(1/r^5) and that therefore the surface term vanishes.