# CKM matrix parameters

1. Mar 11, 2012

### LAHLH

Hi,

I understand that a 3x3 unitary matrix needs 9 real parameters to be specified (18 real parameters to start with then 9 equations of constraint arising from unitarity), but what I'm struggling to understand is how we can make phase changes of the form:

$$\mathrm{e}^{-i\beta_I} V_{IJ} \mathrm{e}^{\alpha_J}$$

to make the first row and column of $V_{IJ}$ real leaving us with only 9-5=4 independent parameters $\theta_1,\theta_2,\theta_3, \delta$, is there an easy way to see this can be done?

thanks

2. Mar 12, 2012

### blechman

This comes from the freedom to make phase redefinitions of the quark fields themselves:

$$u_I\rightarrow e^{i\alpha_I}u_I$$
$$d_I\rightarrow e^{i\beta_I}d_I$$

That will reproduce your formula. Doing this looks like you can eliminate six phases, except that one of them is an overall phase that does not do anything and thus cannot be used to remove a phase in the CKM matrix (remember from quantum mechanics: overall phases are not physical). So that's why you can only eliminate 5 phases and not 6 (naively, a unitary matrix has 6 phases and 3 angles).

The way I always thought of it was not in terms of CKM, but in terms of Yukawa matrices (describing the coupling of the Higgs boson to the quarks). There are two 3x3 Yukawa matrices, each complex, giving a total of 36 (real) parameters. But there is also a U(3)^3 flavor symmetry that allows you to set many of these parameters to zero. Each U(3) sets 9 parameters to zero, so that allows you to kill 27 parameters; but you cannot use BARYON number to set a parameter to zero, so actually you can only set 26 parameters to zero. That leaves 10 nonzero parameters in the Yukawa sector (6 quark masses, 3 CKM angles and 1 CKM phase).

I like that argument better since it makes it very clear that the CKM matrix is coming from the Higgs sector. But if you're not familiar with Yukawa matrices and the Standard Model Lagrangian, perhaps that is too much...

3. Mar 15, 2012

### LAHLH

I think my question might actually be simpler than you are suspecting. I'm happy with why we're allowed to make these global changes by the freedom to make redefinitions in the fields. My question really could have been stated as just a mathematical one (without even mentioning CKM or the SM), about how we can take some arbitrary 3x3 unitrary matrix, apply some phase factors to it to obtain

$$\mathrm{e}^{-i\beta_i}U_{ij} \mathrm{e}^{+i\alpha_j}$$

to make the first row and column real, and eliminate 5 of the 9 real parameters needed to specify 3x3 unitary matrix. Is there some conventional parametrisation that one writes an arbitrary 3x3 unitary matrix in terms of the 9 needed parameters, for which this then becomes obvious?

4. Mar 15, 2012

### blechman

A U(3) matrix has 6 phases and 3 angles. Just write a 3x3 matrix and parametrize each element in terms of a magnitude and a phase and use $UU^\dagger=1$ - you will find 3 linearly independent relationships between the 9 phases you wrote down in order to make this work. You can always choose an overall phase convention to make the phases of the first row and column nonzero, as the problem is underconstrained (3 equations, 9 unknowns).

Then let:

$$\beta_i = \arg U_{i1},\quad\alpha_j=-\arg U_{1j}$$

$\alpha_1$ and $\beta_1$ are redundant, so you only need one of them, hence you can only remove 5 phases.

5. Mar 18, 2012

### LAHLH

$$\left[ \begin{array}{ccc} a_1 e^{ib_1} & a_2 e^{ib_2} & a_3 e^{ib_3} \\ a_4 e^{ib_4} & a_5 e^{ib_5} & a_6 e^{ib_6} \\ a_7 e^{ib_7} & a_8 e^{ib_8} & a_9 e^{ib_9} \end{array} \right]$$

Obviously this starts with 18 real parameters, then we use unitarity $UU^\dagger=1$ and we get a total of 9 linearly independent equations of constraint, meaning only 18-9=9 of our real parameters are linearly indep:

$a_1^2+a_2^2+a_3^2=1$
$a_4^2+a_5^2+a_6^2=1$
$a_7^2+a_8^2+a_9^2=1$
$a_1a_4\cos{(b_1-b_4)}+a_2a_5\cos{(b_2-b_5)}+a_3a_6\cos{(b_3-b_6)}=0$
$a_1a_4\sin{(b_1-b_4)}+a_2a_5\sin{(b_2-b_5)}+a_3a_6\sin{(b_3-b_6)}=0$
$a_1a_7\cos{(b_1-b_7)}+a_2a_8\cos{(b_2-b_8)}+a_3a_9\cos{(b_3-b_9)}=0$
$a_1a_7\sin{(b_1-b_7)}+a_2a_8\sin{(b_2-b_8)}+a_3a_9\sin{(b_3-b_9)}=0$
$a_4a_7\cos{(b_1-b_7)}+a_5a_8\cos{(b_2-b_8)}+a_6a_9\cos{(b_6-b_9)}=0$
$a_4a_7\sin{(b_1-b_7)}+a_5a_8\sin{(b_2-b_8)}+a_6a_9\sin{(b_6-b_9)}=0$

3 of these constrain only magnitudes, $a_i$; the other 6 constraint both phases and magnitudes.

If you look at (#4)^2+(#5)^2, you get an equation constraining just the magnitudes, a1,a2,a3,a4,a5,a6; similarly (#6)^2+(#7)^2 constraints just a1,a2,a3,a7,a8,a9, and (#8)^2+(#9)^2 constraints just a4,a5,a6,a7,a8,a9. How do I get the three equations for just the phases $b_i$ out of these that you mention?

6. Mar 19, 2012

### blechman

Divide equations 4,6,8 by equations 5,7,9 (you have some typos in those equations, btw):

$$\frac{\cos{(b_1-b_4)}+r\cos{(b_2-b_5)}}{\sin{(b_1-b_4)}+r\sin{(b_2-b_5)}}=\cot(b_3-b_6),\qquad r=\frac{a_2a_5}{a_1a_4}$$

$$\frac{\cos{(b_1-b_7)}+s\cos{(b_2-b_8)}}{\sin{(b_1-b_7)}+s\sin{(b_2-b_8)}}=\cot(b_3-b_9),\qquad s=\frac{a_2a_8}{a_1a_7}$$

$$\frac{\cos{(b_4-b_7)}+t\cos{(b_5-b_8)}}{\sin{(b_4-b_7)}+t\sin{(b_5-b_8)}}=\cot(b_6-b_9),\qquad t=\frac{a_5a_8}{a_4a_7}$$

Notice that since I can fix 6 of the $a_i$, I can choose to fix r,s,t, so these are 3 equations for 6 phase differences, which is what I wanted.

Last edited: Mar 19, 2012
7. Mar 21, 2012

### LAHLH

Thanks, I can't see the typo, my equations seem consistent with the ones you wrote?

Anyhow, I see now how you can fix 6 of the $a_i$ (through equations 1-3 and by 4^2+5^2, 6^2+7^2, 8^2+9^2) and therefore making r,s,t constants we have 3 equations of constraint on the 9 phases. This means only 9-3=6 of the phases are independent.

So I think ultimately we have 3 independent magnitudes and 6 independent phases..making up our 9 independent parameters.

Now coming back to your point:

OK so we leave the phase of the 5 elements constituting the first row/column non-zero (plus one of the remaing four elements, and the other 3 will be determined by these via 3 constraints on phase)

We would surely need, say: $\beta_1=\arg U_{11}$; $\beta_2=\arg U_{21}$; $\beta_3=\arg U_{31}$; and $\alpha_1=0$; $\alpha_2=-\arg U_{12}+\arg U_{11}$; $\alpha_3=-\arg U_{13}+\arg U_{11}$;

Otherwise with your choice $U_{12}\to \mathrm{e}^{-i\beta_1}U_{12}\mathrm{e}^{+i\alpha_2} \to \mathrm{e}^{-iarg(U_{11})}U_{12}\mathrm{e}^{-iarg(U_{12})}$ which is not real.

so anyway, I agree that all 5 of the phases in the first row can be set to zero making the first row and column real (just like we could have done for any complex matrix) But we had 6 phases independent for a 3x3 unitary, so only one of the other four elements of the matrix has a non-zero phase too,say for arguments sake this is $U_{23}$, we have already picked our values of $\beta_2, \alpha_3$ to kill the phases of the first row and column so we can't kill the phase of this element...Thus finally I see why only 5 of the 6 phases can be killed, leaving just the one phase $\delta$.

So we have our 3 independent magnitudes and one phase.

8. Mar 25, 2012

### Doofy

I've been trying to work through this stuff to convince myself of it but I'm struggling to get these 3 equations.

When I divide (4) by (5) the closest I have been able to get is a rather ugly expression:

$$- cot(b_{3} - b_{6}) = [cos(b_{1} - b_{4}) + rcos(b_{2} - b_{5})][\frac{1}{sin(b_{1} - b_{4}) + rsin(b_{2} - b_{5}) +qsin(b_{3} - b_{6})}][\frac{sin(b_{1} - b_{4}) + rsin(b_{2} - b_{5})}{qsin(b_{3} - b_{6})} + 1]$$

where $q = \frac{a_{3}a_{6}}{a_{1}a_{4}}$. I can't seem to turn the product of the 2nd and 3rd square brackets into the required $\frac{-1}{sin(b_{1} - b_{4}) - rsin(b_{2} - b_{5})}$

How did you get your equations?

9. Mar 25, 2012

### blechman

I have no idea what you did!

Move the last term of each equation to the RHS and simply put one equation over the other! The (a_3*a_6) cancel out. Then divide numerator and denominator by (a_1*a_4) and you're done.

Similar for the other equations.

10. Mar 26, 2012

### Doofy

I have no idea either, lol. thanks.

11. Mar 31, 2012

### Doofy

How do you get rid of the $b_i$ when you do (4)^2 + (5)^2 ?

for (4) and (4)^2 I get:
$\alpha_1 \alpha_4 cos(\beta_1 - \beta_4) + \alpha_2 \alpha_5 cos(\beta_2 - \beta_5) = -\alpha_3 \alpha_6 cos(\beta_3 - \beta_6)$

$(\alpha_1 \alpha_4 )^2 cos^2(\beta_1 - \beta_4) + (\alpha_2 \alpha_5)^2 cos^2(\beta_2 - \beta_5) + 2(\alpha_1 \alpha_4 \alpha_2 \alpha_5)cos(\beta_1 - \beta_4) cos(\beta_2 - \beta_5) = (\alpha_3 \alpha_6)^2 cos^2(\beta_3 - \beta_6)$

for (5) and (5)^2 I get:
$\alpha_1 \alpha_4 sin(\beta_1 - \beta_4) + \alpha_2 \alpha_5 sin(\beta_2 - \beta_5) = -\alpha_3 \alpha_6 sin(\beta_3 - \beta_6)$

$(\alpha_1 \alpha_4 )^2 sin^2(\beta_1 - \beta_4) + (\alpha_2 \alpha_5)^2 sin^2(\beta_2 - \beta_5) + 2(\alpha_1 \alpha_4 \alpha_2 \alpha_5)sin(\beta_1 - \beta_4) sin(\beta_2 - \beta_5) = (\alpha_3 \alpha_6)^2 sin^2(\beta_3 - \beta_6)$

so for (4)^2 + (5)^2 I get:
$(\alpha_1 \alpha_4 )^2 + (\alpha_2 \alpha_5)^2 + 2\alpha_1 \alpha_4 \alpha_2 \alpha_5 (cos(\beta_1 - \beta_4) cos(\beta_2 - \beta_5) + sin(\beta_1 - \beta_4) sin(\beta_2 - \beta_5) )$

$= (\alpha_1 \alpha_4 )^2 + (\alpha_2 \alpha_5)^2 + 2\alpha_1 \alpha_4 \alpha_2 \alpha_5 (cos(\beta_1 - \beta_4 - \beta_2 + \beta_5)) = (\alpha_3 \alpha_6)^2$

What am I doing wrong here?

12. Apr 1, 2012

### LAHLH

Doofy you are correct, I had made a mistake here in my own notes. I'm not sure how to resolve this now to fix the a_i's...

One thing that could be done with equatons 1)-3) is notice each is an equation for a unit 2-sphere and reparametrise as $a_1=\sin{\theta_1}\cos{\phi_1}$, $a_2=\sin{\theta_1}\sin{\phi_1}$, $a_3=\cos{\theta_1}$, and similary for equations #2,#3. So now we've used these three equations to reduce our 9 a_i's to six. One could then sub these into the remaining equations #4-#9. Not sure if this will be of use, but I haven't got the time to start thinking about this in detail again right now..