Since writing the MNS as a magic unitary matrix, of course I’ve been working on writing the CKM matrix the same way. This involved learning a lot more about 3×3 unitary and 3×3 magic matrices, and writing a Java program to do the heavy lifting.
The first thing one must do to deal with magic unitary matrices is to define a parameterization of these matrices. A full parameterization of all unitary matrices requires 9 real variables. Five of these define the arbitrary complex phases that can be applied to any row or column (yes there are 3 rows and 3 columns, but one of them is redundant). The remaining 4 variables are usually written in the Wolfenstein parameterization. In this parameterization, three variables are mixing angles and the fourth variable defines the CP violation.
If one is given a magic unitary matrix, the effect of multiplying any row or column by a complex phase would be to destroy the magic. Consequently, putting a unitary matrix into magic form (if this can be done) amounts to choosing a set of unique phases. Therefore, the largest number of free parameters we can expect to need to define magic unitary matrices is the 4 used in the Wolfenstein parameterization (which also amounts to choosing a unique set of phases). In fact, I’ve found a parameterization of the unitary magic matrices on 4 real parameters so one supposes that any unitary matrix can be put into a (more or less unique) magic form.
First, let’s write some reduced notation. We will abbreviate the 1-circ and 2-circ matrices as follows:
Note that that “E” and “F” are reversed in the above 2-circ definition. This is to make the Fourier transforms easier to deal with. In our abbreviated notation, we are to solve:
Sums of 1-circulants and 2-circulants
In solving the unitary magic matrix problem, the first thing to note is that the product of a 1-circ and a 2-circ is a 2-circ, and a product of two matrices of the same type gives a 1-circ result. The second thing to note is that the “dagger” operation preserves circularity. The right hand side of the above unitarity equation is a 1-circulant, so we can break the above equation into two equations, based on the circularity of the right and left hand sides:
In the above, I’ve abbreviated the circulant matrices even a little farther.
One can write out the above matrix products in terms of the A, B, C, D, E, and F. From the top equation one finds:
AA* + BB* + CC* + DD* + EE* + FF* = 1,
AB* + BC* + CA* + DE* + EF* + FD* = 0,
AC* + BA* + CB* + DF* + ED* + FE* = 0.
The top equation just says that the sum of the squared magnitudes of the six complex numbers must be 1. The other two equations are more difficult in that they mix terms, but they are identical equations, that is, one is the complex conjugate of the other. So we only need to consider one of them.
Eigenvectors and Eigenvalues
The two equations that mix terms can be written simply if we think of the (ABC) and (DEF) and kets and their complex conjugates as bras. Then the 2nd equation above can be written as:
This is in the form of a matrix element as used in quantum mechanics. In the above, the matrix is a permutation matrix, one that we have called “J” in our previous posts. It is useful to rewrite the (A B C) and (D E F) vectors to a basis of eigenvectors of the J matrix. The matrix is a 1-circulant matrix, such matrices have three eigenvectors:
(1 1 1) with eigenvalue 1
(1 w w*) with eigenvalue w
(1 w* w) with eigenvalue w*,
where w and w* are the complex cube roots of unity.
So to make the lower equation simple, we rewrite into the basis for the permutation matrix (as Kea says, this is a discrete Fourier transform, always a good thing to do):
(A B C) = A_1 (1 1 1) + A_2 (1 w w*) + A_3 (1 w* w),
(D E F) = D_1 (1 1 1) + D_2 (1 w w*) + D_3 (1 w* w).
Note that since our eigenvectors are unnormalized, we will end up with a factor of 3. But since the difficult equations we’re trying to solve are all crapola = 0, we can factor the sqrt(3) out, and we’ll just ignore this detail. And as far as a matrix element goes, the top equation is even simpler, it is the matrix element for the unit matrix. Accordingly, after Fourier transform, it will be unchanged; the squared magnitudes of the Fourier coefficients must still sum to 1.
The permutation matrix leaves the eigenvectors unchanged except for multiplication by the eigenvalues. Accordingly, the 2nd equation, where before the A, B, and C were mixed, become unmixed:
The above equation does not involve the phases of the parameters A_n and D_n. So the only source of complex phase are the w and w*. Accordingly, we can interpret the above as a statement about the magnitudes of the parameters.
On the complex plane, 1, w and w* are unit vectors separated by 120 degrees. For three real multiples of these constants to sum to 0, as in the above equation, amounts to a requirement that the sides of an equilateral triangle all be equal. Therefore we have that:
Since the top 1-circ equation defined the sum of the squared magnitudes of the Fourier coefficients, we can parameterize the magnitudes of the Fourier coefficients with 3 real parameters as follows:
The 2-circulant Part
This completely solves the parameterization problem for the 1-circ part of the problem. Now we move on to the 2-circ part, the part that mixes the (A B C) terms with the (D E F) terms:
As before, we have three independent complex equations, corresponding to the three complex parts of a 2-circulant matrix. After moving the terms around a little, we find:
(AD* + DA*) + (BF* + FB*) + (CE* + EC*) = 0,
(AF* + FA*) + (BE* + EB*) + (CD* + DC*) = 0,
(AE* + EA*) + (BD* + DB*) + (CF* + FC*) = 0.
The terms in parenthese are formulas for twice the real part of the various products beginning with AD*. Therefore we can rewrite these more tersely as:
Re( AD* + BF* + CE*) = 0,
Re( AF* + BE* + CD*) = 0,
Re( AE* + BD* + CF*) = 0.
The above three equations can also be written out in matrix form. As before, we find a set of 3 equations that look like matrix calculations for a quantum mechanics problem. But this time, the operator turns out to be the odd permutation matrices, the matrices we’ve previously referred to as “R”, “G”, and “B”. For example, the top equation becomes:
The middle equation becomes the matrix element of G, and the bottom equation becomes the matrix element of B.
R, G, and B are odd permutation operators; they swap two elements of the vectors. So the (1 1 1) Fourier basis vector, which has all its elements equal, is left unchanged by these operators. The other basis vectors, (1 w w*) and (1 w* w) are swapped by R, G, and B, and, in addition, the vectors are multiplied by 1, w or w*:
R (1 w w*) = 1 (1 w* w),
R (1 w* w) = 1 (1 w w*),
G (1 w w*) = w*(1 w* w),
G (1 w* w) = w (1 w* w),
B (1 w w*) = w (1 w* w),
B (1 w* w) = w*(1 w w*).
This is not quite as convenient and simple as if they were eigenvectors of R, G, and B, but it is good enough. We end up with three Fourier transformed equations. As before, these all must be imaginary:
D_1* A_1 + 1 D_2* A_3 + 1 D_3* A_2,
D_1* A_1 + w D_2* A_3 + w* D_3* A_2,
D_1* A_1 + w* D_2* A_3 + w D_3*A_2.
Adding all three together, and noting 1+w+w* = 0, we find that D_1* A_1 is imaginary.
To get information on the other Fourier coefficients, take the sum and differences between the second and third equations. Since w+w* = -1, we have that
D_2*A_3 + D_3*A_2
is imaginary. And since w-w* is purely imaginary, we find that
D_2*A_3 – D_3*A_2
must be real.
As before, we treat D_2*A_3 and D_3*A_2 as vectors in the complex plane. Their sum is imaginary and their difference is real. For their sum to be imaginary, their real parts must be equal in magnitude and opposite in sign. And for their difference to be real, their imaginary parts must be equal. Therefore, we can write:
D_2* A_3 = R exp( i k)
D_3* A_2 = R exp( pi – i k).
Taking the magnitude of the above two equations, and remembering our parameterization of the magnitudes, we find that
and therefore, if we allow for the magnitudes to be positive or negative (to account for the two solutions in x to tan(y) = x).
When one writes a magic matrix as the sum of a 1-circulant and a 2-circulant, one always has a symmetry (or redundancy) in that the sum is left unchanged by the transformation:
A_n = A_n + alpha,
D_n = D_n – alpha
for alpha any complex constant. To eliminate this redundancy, we can take alpha = D_1 and therefore assume D_1 = 0. This means we must have , and we have eliminated another one of the theta parameters; we have fully parameterized the magnitudes of A_n and D_n.
It remains to parameterize the phases. Since the magnitudes for D_2 and D_3 are the same, as are the magnitudes of A_2 and A_3, the cross terms given become pure phase equations. One ends up with three phase angle parameters. A complete parameterization is:
where are the parameters. Note, in taking the above back through the inverse Fourier transform, one will have to deal with the factor of 3 that was left off earlier.
In the physics literature, the usual parameterization of a 3×3 unitary matrix is the Wolfenstein. The objective of the Wolfenstein parameterization is to make the matrix as real as possible. There are three Wolfenstein parameters that do not cause the matrix to become complex, while the 4th parameter gives the unavoidable complexity of a (CP violating) matrix. It is interesting that the parameterization I’ve found also splits its four parameters into three () and a 1 (), but instead the three parameters are complex phases and the one parameter defines magnitudes which are real.
The CKM Matrix
Once one has a parameterization of the magic unitary 3×3 matrices, one is naturally drawn to see what the CKM matrix looks like in magic form. To do the calculation, I naturally wrote a java computer program. The program also checked my mathematics and corrected various errors. So in honor of the Java programming language, I’ll reproduce the actual screen shot (click to size it larger):
The magic unitary version of the CKM matrix is highlighted with the red bars.
While working all this out, I also found a few other things that I’m not sure has been mentioned before:
a) Every magic matrix can be written as the sum of a 1-circulant and a 2-circulant, and of course every sum of a 1-circulant and a 2-circulant is magic.
b) Every magic unitary 3×3 matrix has a complex phase as the sum of its rows and columns. To see this, note that the sum over a row or column is always A+B+C+D+E+F. When one converts to the Fourier transform, only the A_1 and D_1 contribute to this sum as the other terms appear multiplied by 1, w and w*, which sum to zero. Of course one can choose that phase to be 0, as the above paramaterization does.
c) And as noted, it appears that every unitary 3×3 matrix can be put into magic form, though so far I do not have a proof of this, only the fact that the number of parameters matches that of the Wolfenstein parameterization.