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CKM Matrixby ChrisVer
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#1
Dec513, 08:41 PM

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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 (2N1) less :/ If I write: V_{CKM}= U_{u}^{τ} U_{d} and make a transformation: V_{CKM}= U_{u}^{τ} F_{u}^{τ} F_{d} U_{d} how could it change my free parameters like 2N1? 


#2
Dec513, 09:19 PM

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[tex]\left( \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right) [/tex] See the problem? 


#3
Dec513, 09:53 PM

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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)... 


#4
Dec513, 10:02 PM

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CKM Matrix
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
Dec513, 10:03 PM

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#6
Dec513, 10:30 PM

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#7
Dec513, 10:37 PM

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#8
Dec613, 02:39 AM

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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 nonunitary matrix from the N^2 free parameters, then N^2 is not the right number of free parameters. Do we agree? 


#9
Dec613, 05:27 AM

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[itex]AA^{τ}=A^{τ}A=1[/itex] (τ is the hermitian conjugate) this drops your free parameters by [itex]N^{2}[/itex], 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 offdiagonal elements, thus you fall by [itex]N^{2}[/itex]. So for any unitary matrix [itex]A_{n×n}[/itex] you need [itex]n^{2}[/itex] real parameters to determine it. The other way around is not always true. For example a real NxN matrix has N^{2} elements, but it of course doesn't have to be unitary... 


#10
Dec613, 06:01 AM

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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
Dec613, 09:02 AM

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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 [itex]2N1[/itex] constraints...
The [itex]F_{u} F_{d}[/itex] matrices would bring 2N phases, and i was missing the 1 


#12
Dec613, 09:20 AM

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#13
Dec613, 09:26 AM

P: 1,023

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



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