What I tried to point out is that the complex coupling constant from the CP violating phase in CKM implies non-hermitian Hamiltonian between individual quark interactions and therefore the observable such as charge is not constant in time evolution so that it violates the conservation of charge.
Okay, the not completely obvious answer to this question came to me.
If a force is conservative with respect to an observable, then any closed path causes no change in the observable. For example, energy is conserved in Newton's gravity, so when a planet returns to the same position in an orbit it has the same energy.
For the case of a table of coupling constants, one finds out whether or not it is conservative by tracing different paths through the coupling constants. In the case of the CKM or MNS matrices, one does this by tracking a state through the matrix, through multiple transitions.
When you trace through a series of transitions in a table of coupling constants, you multiply the consecutive coupling constants. Each coupling constant will contribute a probability and a phase. Now the probabilities have nothing to do with conservation per se. They simply indicate the probability that this particular phase is applicable.
But the product of the phases may or may not be conserved. As an example, suppose that you had a table of coupling constants between a red, green, and blue state and they were of the form:
\left(\begin{array}{ccc}1&e^{+i\delta}&e^{-i\delta}\\<br />
e^{-i\delta}&1&e^{+i\delta}\\<br />
e^{+i\delta}&e^{+i\delta}&1\end{array}\right)
The above is to be interpreted as a table of (R,G,B) by (R,G,B) coupling constants.
If you made the RGR sequence of coupling transitions, then you would pick up a + delta in the first step from R to G (i.e. position 1,2 in the matrix), and then a - delta in the step from G to R. The net change in phase is zero, the path is conservative.
But if you made the RGBR sequence, you would return to R with a phase of 3 delta, and this is the negative of what the RBGR phase would give. Therefore, phase is not conserved by the above coupling constants unless delta is of form 2n \pi/3.
Now the above case was simplified in that I assumed that the same matrix could be used for repeated application. That is, I assumed that the outgoing state was of the same type as the incoming state. This is the case explicitly with my model of the lepton masses, but the case with the CKM and MNS matrices is not so clear.
The CKM and MNS matrices tell you how to move from the flavor eigenstates to the mass eigenstates. In order to presume that this can be repeated, and therefore that one could fail to conserve phase, one must imagine that one is performing a sequence of measurements that alternate between weak force and mass measurement.
Now the presumed "tribimaximal" form of the MNS matrix is real, so there are no phases to not conserve. It is the CKM matrix that includes phases, so the question is this: Is there a non conserved phase in the CKM matrix?
Now getting back to JC Yoon's original comment, the fact that phase is not conserved through the mixing matrices does not imply that charge is not conserved. Phase is still a global symmetry of the system and so is conserved.
As an example of a much simpler system that changes phase (but clearly conserves charge), consider the sequence of Stern-Gerlach filters that keep particles with spin-1/2 oriented in the z, y, x and z directions. The overall picked up phase is pi/4. Ignoring amplitudes (i.e. real factors) and keeping only complex phases:
(1+\sigma_z)(1+\sigma_x)(1+\sigma_y)(1+\sigma_z) = e^{i\pi/4}(1+\sigma_z)
We can see this explicitly by building up a table of coupling constants. Here I'll keep the amplitudes, and use matrix positions 1, 2 and 3 to stand for x, y and z:
\frac{1}{2}\left(\begin{array}{ccc}2&1+i&\sqrt{2}\\<br />
1-i&2&\sqrt{2}\\<br />
\sqrt{2}&\sqrt{2}&2\end{array}\right)
The individual entries in the above are of the form <a|b>, where a and b take x, y and z.
And certainly a Stern-Gerlach experiment conserves charge.
Carl
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