# CKM Matrix

1. Dec 5, 2013

### ChrisVer

I am having some difficulties in understanding something I found. Speaking about CKM matrix for N quark generations...
I know that the Cabibbo Kobayashi Maskawa matrix is unitary, so it has N^2 free real parameters.
Although I cannot understand how by an overall face they get (2N-1) less :/

If I write:
VCKM= Uuτ Ud
and make a transformation:
VCKM= Uuτ Fuτ Fd Ud
how could it change my free parameters like 2N-1?

2. Dec 5, 2013

Staff Emeritus
OK, so by that argument, a 2x2 should have 4. Let me just pick one.

$$\left( \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right)$$

See the problem?

3. Dec 5, 2013

### ChrisVer

what u wrote is not a unitary matrix -obviously-
Pauli matrices are unitary for example... so they can be described by N^2 real parameters (N=2, so N^2=4)
while at the begining you have 2N^2 (2x2 complex entry matrices, so each entry has 2 parameters) because of unitarity it drops to N^2...
[a+ib, c+id ; f+ik , h+ig]
you know that b=g=0 due to unitarity.
and that (f+ik)*= c+id. or c=f and k=-d
so the parameters fell from 8 (a,b,c,d,f,k,h,g) to 4 (a,c,k,h)...

Last edited: Dec 5, 2013
4. Dec 5, 2013

### dauto

The CKM matrix is only meaningful as a factor in a term in the lagrangean.
That term will also have fields (in fact 2N quark fields).
You can remove degrees of freedom from the matrix by transferring phase factors from the matrix to the fields and absorbing the phases into the fields - that is, you redefine the fields to include the phase in their definition.
You would think naively that that process would allow you to remove 2N phases from the matrix but if you chose all the phases transferred to the quarks to be identical to each other, the CKM matrix won't be affected at all. That means that one overall arbitrary common phase for the all the quarks cannot be fixed which means you really end up removing only 2N -1 degrees of freedom from the CKM matrix.

5. Dec 5, 2013

### dauto

That matrix does indeed have four parameters and as pointed out above, it isn't unitary. Unitary does not mean diagonal.

6. Dec 5, 2013

### ChrisVer

Not really dauto... in fact it has 4 parameters, but you have no way to say they are independent :p it can equivalently be 2 (since it's symmetric)

7. Dec 5, 2013

### dauto

point taken

8. Dec 6, 2013

Staff Emeritus
I know the matrix that I posted was not unitary. That was the point. That's why I wrote "see the problem?"

If I can construct a non-unitary matrix from the N^2 free parameters, then N^2 is not the right number of free parameters. Do we agree?

9. Dec 6, 2013

### ChrisVer

A $n×n$ matrix with complex entries has $2n^{2}$ parameters from which you can fully determine it. Now if a matrix $A$ is unitary:
$AA^{τ}=A^{τ}A=1$ (τ is the hermitian conjugate)
this drops your free parameters by $N^{2}$, because while before you had all complex entries now you can get rid of the "complex" part of the diagonal, and relate the up to the down off-diagonal elements, thus you fall by $N^{2}$.
So for any unitary matrix $A_{n×n}$ you need $n^{2}$ real parameters to determine it.

The other way around is not always true. For example a real NxN matrix has N2 elements, but it of course doesn't have to be unitary...

10. Dec 6, 2013

Staff Emeritus
Let me say it again. I know the matrix that I posted was not unitary. That was the point. That's why I wrote "see the problem?"

The question is why N^2 is wrong. That is why I showed a counterexample. And then wrote "see the problem".

11. Dec 6, 2013

### ChrisVer

My question was that I didn't understand how, out of the transformations stated above, the overall phase is being absorbed in the quark fields IN SUCH a way that you get $2N-1$ constraints...
The $F_{u} F_{d}$ matrices would bring 2N phases, and i was missing the 1

12. Dec 6, 2013

### dauto

Did you see my post? I tried to explain the missing 1.

13. Dec 6, 2013

### ChrisVer

Yeah I did it already...I just reexplained the question coz there was a confusion...