Is there a relationship between Berry-Pancharatnam phase and CP violation in quark mixing?
Active 10 years, 11 months ago
Berry-Pancharatnam phase is the phase that quantum systems exhibit when they pass through a sequence of states and return to their original state. It's a complex phase and it is different from the usual complex phases in that it does not depend on the arbitrary complex phases present in quantum states. For an encyclopedia-level introduction to B-P phase for the purpose of this question, see
Péter Lévay, Encyclopedia of Mathematical Physics, Elsevier 2006, "Geometric Phases"
http://arxiv.org/abs/math-ph/0509064v1
When a quark interacts with the weak force it typically changes flavor and emits or absorbs an electron and a neutrino (or anti-electron / anti-neutrino, as appropriate). This process is generally assumed to require a unitary matrix called the CKM matrix. The elements of this matrix can be measured in high dollar physics experiments but only in absolute value. That is, the complex phases are unknown.
So when the data is fit to a unitary CKM matrix, physicists have a choice in how to arrange the complex phases. There are four degrees of freedom in the data and five degrees of freedom in the arbitrary complex phases. The usual method is to use three of the degrees of freedom to (approximately) define the transition probabilities between generations. That is, one has $\{\theta_{12},\; \theta_{13},\; \theta_{23}\}$ for the "mixing angles" between the 1, 2, and 3 generations. (This is only approximate because the probability for going from the 1st to 3rd generation is not equal to the probability for going from the 3rd to the 1st generation.) The fourth degree of freedom is chosen to be $\delta$. If this fourth parameter is zero, then there can be no CP violation. The wikipedia article is a good introduction:
http://en.wikipedia.org/wiki/Cabibbo%E2%80%93Kobayashi%E2%80%93Maskawa_matrix
The $\delta$ angle enters the CKM matrix as a complex phase, that is, it appears as $e^{\pm i\delta}$. The other parameters $\theta_{12},\; \theta_{13},\; \theta_{23}$ are used in cosines and sines and are usually abbreviated for example as $\cos(\theta_{12}) = c_{12}$.
In both these cases the central object, Berry-Pancharatnam phase and the $\delta$ angle, is a complex phase. And neither has anything to do with the arbitrary complex phases of quantum states. So my question is this: Can CP violation be defined in terms of a Berry-Pancharatnam phase?
particle-physics standard-model berry-pancharatnam-phase cp-violation weak-interaction
Carl BrannenCarl Brannen
$\begingroup$ including the expression for the CKM matrix would be help make this question self-contained. Also it would help if you could suggest some explanation for why the $\delta$ can be seen to be a BP phase. Otherwise you're just asking folks to connect two seemingly disparate topics. $\endgroup$
$\begingroup$ @Lubos' answer has clarified the question so disregard my previous comment. $\endgroup$
$\begingroup$ @space_cadet; I'll post more probably tomorrow. $\endgroup$
– Carl Brannen
Dear Carl, first, one has to correct your statement:
The elements of this matrix can be measured in high dollar physics experiments but only in absolute value. That is, the complex phases are unknown.
It's not that the phases - beyond the usual 4 degrees of freedom of the CKM matrix - are "unknown". Instead, these phases are unphysical.
They can be changed to whatever you like by a phase redefinition of the 3+3 = 6 mass eigenstates of the quarks - which doesn't change their being eigenstates. There are only 5 degrees of freedom in the CKM matrix that you may redefine by these 6 phases of the mass eigenstates: that's because the overall change of all 6 eigenstates by the same phase doesn't change the CKM matrix.
Now the main question. Yes, of course, the CP-violating phase may be interpreted as a generalized Berry phase.
However, you must allow the normalization of the states to change during the monodromy. Now, if you get to the positive point, if you make a cyclic round trip around the three generations whose probability amplitude will be proportional to $$V_{12}V_{23}V_{31},$$ the real matrix would mean that the complex phase of the product above would match the phase of a similar product $$V_{13}V_{32}V_{21},$$ up to a sign. However, the phase of the product above becomes something else - a generic complex phase - for the CP-violating CKM matrix. Note that the ratio $$\frac{V_{12}V_{23}V_{31}}{V_{13}V_{32}V_{21}} $$ is invariant under the phase redefinition of all six states, $d,s,b;d',s',b'$ (each of the 6 phases cancels among the numerator and denominator, I think, because each value of the index 1,2,3 appears once as the first index and once as the second index, in both numerator and denominator), so it may be viewed as an invariant definition of the CP-violating phase. I hope that the ratio is complex (unreal) for the actual CKM matrix; correct me if I am wrong.
In this sense, and only in this sense, you may interpret the CP-violating phase as a Berry phase. However, there's no "parallel transport" that would literally keep the character of the initial state physically unchanged. That's because there's no unbroken symmetry between the flavors in different generations.
Cheers LM
Luboš MotlLuboš Motl
$\begingroup$ The product V12 V23 V31 does not move a fermion back to its original state. Instead, this product is the product of three amplitudes, d to c, s to t, and b to u. To get a BP factor requires that you move back an forth between flavors. So it requires 4 transitions, half of which are complex conjugated (because they go in the reverse flavor direction). See my answer below. $\endgroup$
$\begingroup$ On further thought, I see that V12 V23 V31 is being thought of as a "cyclic round trip through the generations". My complaint is that it isn't keeping track correctly of quark type, that is, it's not physical because the output (left side) of V21 is not a suitable input (right side) for V32. The output of V21 is an s quark, while the right hand side of V32 is a c quark. These are both of the 2nd generation, but are not the same quark. I'll add more... $\endgroup$
$\begingroup$ In terms of complex phases, (i.e. ignoring magnitudes as is appropriate to Dr. Motl's comment) the product V12 V23 V31 / V13 V32 V21 can be rewritten as V12 V23 V31 V13* V32* V21*. This can be rearranged to form a cycle. (The stars amount to defining the reverse transition, i.e. <a|b>* = <b|a>, but we also should take into account the charge +2/3 or -1/3. Another way of saying this is to note that for us, <a|a> is not 1, and <a|a>* is not <a|a>. Our notation is lacking.) The cycle obtained is V12 V32* V31 V21* V23 V13* and indeed this will produce a Berry-Pancharatnam phase. (more to add) $\endgroup$
$\begingroup$ It's probably best to rewrite this in my notation... $\endgroup$
$\begingroup$ 123=dsb or =uct, depending on the side of the CKM matrix or quark charge -1/3 or 2/3. Then V12 V32* V31 V21* V23 V13* = dc cb bu us st td. Note that the adjacent quarks match. Products like "ud du" do not change the phase and so can always be inserted where convenient. So get a phase equal to that of (dc cb bu ud) (du us st td). This is the product of two J_CPs, so it will either give twice the phase of J_CP or be real. $\endgroup$
Dr. Motl's answer is fairly complete. I'll add a few details, and introduce the Jarlksog invariant $J_{CP}$, show that it gives a measure of CP violation, etc.
Any product of pure density matrices that begins and ends with the same pure density matrix $\hat{x}$ is some complex number $k$ times that pure state. We write:
$\hat{x}\hat{y}\hat{z}...\hat{x} = k_{xyz...x}\;\hat{x}$
where $\hat{x},\hat{y},\hat{z}...$ are pure density matrices and $k_{xyz...x}$ is a number. If the left side happens to be zero, we define $k$ to be zero as well. The $k$ are observables. For example, the transition probability between $\hat{x}$ and $\hat{y}$
Leaving off the neutrinos, an up, charm, or top quark $\{u,c,t\}$ can emit a $W^+$ and become a down, strange, or bottom quark $\{d,s,b\}$:
$\{u,c,t\} \to W^+\;\;+\;\;\{d,s,b\}.$
Similarly the $\{d,s,b\}$ can emit a $W^-$:
$\{d,s,b\} \to W^-\;\;+\;\;\{u,c,t\}.$
The two types of quarks $\{d,s,b\}$ and $\{u,c,t\}$ define two bases for the 3-dimensional Hilbert space. The transition amplitudes define a unitary matrix known as the CKM matrix:
$V_{CKM} = \left(\begin{array}{ccc} \langle u|d\rangle &\langle u|s\rangle &\langle u|b\rangle \\ \langle c|d\rangle &\langle c|s\rangle &\langle c|b\rangle \\ \langle t|d\rangle &\langle t|s\rangle &\langle t|b\rangle \end{array}\right)$
In elementary particles literature the CKM matrix is defined with the weak force boson interaction included so $(u,c,t)^t = V_{CKM}\gamma^0(1-\gamma^5)/2(d,s,b)^t$. See, for example, Byron P. Roe. Particle Physics at the New Millenium. Springer-Verlag, 1996. Our abbreviation is the usual quantum information theory liberty of ignoring force bosons; either way one obtains the same $V_{CKM}$.
To find Berry-Pancharatnam phases in $V_{CKM}$ we must consider transitions between pairs of states such as $\{d,s\}$ and $\{u,c\}$. To work this out, let's define the projection operators using hats, that is, define $\hat{s} = |s\rangle \langle s|$, etc. Then the observable for the transition sequence $d\to c \to s \to u \to d$ is a complex number $k_{duscd}$ defined by:
$k_{duscd}\hat{d} = \hat{d}\hat{u}\hat{s}\hat{c}\hat{d}\;\;$ or
$k_{duscd} = \langle d|u\rangle \langle u|s\rangle \langle s|c\rangle \langle c|d\rangle = V_{du}\;V_{cu}^*\;V_{sc}\;V_{dc}^*$
where $V_{jk}$ are the entries in the CKM mixin matrix. The $k_{duscd}$ is a Jarlskog invariant. See Cecilia Jarlskog, "Commutator of the quark mass matrices in the standard electroweak model and a measure of maximal CP nonconservation", Phys. Rev. Lett., 55:1039–1042, 1985. Note that $k_{duscd}^* = k_{dcsud}$; complex conjugation reverses the ordering. For a measure of CP or T violation, this is exactly what we want, that is, CP violation will be the difference between the forward going process and the backwards going one.
Since $\{d,s,b\}$ form a complete basis we have:
$\hat{s} = 1 - \hat{d}-\hat{b}.$
Substituting the above we find:
$k_{duscd}-k_{dcsud} = (k_{du1cd}-k_{dudcd}-k_{dubcd}) - (k_{dc1ud}-k_{dcdud}-k_{dcbud}),$
$=0-k_{dudcd}-k_{dubcd}-0+k_{dcdud}+k_{dcbud},$
$=k_{dcbud}-k_{dubcd}.$
Thus $J_{CP}$ for transitions between $\{d,b\}$ and $\{u,c\}$ is equal to the $J_{CP}$ for transitions between $\{d,s\}$ and $\{u,c\}$. More generally, $J_{CP}$ is an invariant of the $3\times 3$ CKM matrix, that is, it does not depend (except for sign) on the choice of pairs of states considered. And since we've written it in terms of pure density matrices, there is no dependence on the arbitrary complex phases of the rows and columns of the $V_{CKM}$ matrix. All CP violations in the quarks are proportional to $J_{CP}$.
$\begingroup$ Just in case you're interested, you can use \langle and \rangle to put Dirac notation, for example: $\langle a|b \rangle$ $\endgroup$
– Robert Smith
$\begingroup$ I am interested, but thought I'd tested them and they didn't work. I'll fold them in to the next edit. $\endgroup$
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