Quark mixing factor in CKM matrix

[Amith2006]

I find that the quark mixing factor say for example $V_{ub}$ is the same for:
u $\Leftrightarrow$ b
$u\Leftrightarrow\bar{b}$
$\bar{u}\Leftrightarrow$ b
$\bar{u}\Leftrightarrow\bar{b}$
Does this have something to do with weak interaction being unable to distinguish these from one another?
Thanks in advance.

 

[Orodruin]

There is no (known) interaction which turns a quark into an anti-quark as that would violate baryon number.

For the coupling between $u$ and $b$ being equal (in magnitude) to the coupling between $\bar u$ and $\bar b$, this is a simple consequence of the Lagrangian density being Hermitian.

 

[Amith2006]

There is no (known) interaction which turns a quark into an anti-quark as that would violate baryon number.

For the coupling between $u$ and $b$ being equal (in magnitude) to the coupling between $\bar u$ and $\bar b$, this is a simple consequence of the Lagrangian density being Hermitian.

I can understand that the coupling between u and b is equal (in magnitude) to the coupling between $\bar{u}$ and $\bar{b}$ due to the Hermitian property but in the decay of charged Kaon,
$K^+ \rightarrow \mu^+ + \nu_\mu$
the hadronic part of the current has $V_{c\bar{s}}$ which is the same coupling as $V_{cs}$.

 

[mfb]

It is not a transition between them, that is the point.
You can rotate the diagram to get u -> s+W+, for example (no c involved in a kaon decay). The vertex stays the same, so you need V_us.

 

[Amith2006]

Sorry $V_{us}$. Ah, now I get it. Thanks.