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CP-Violation /Strong CP Problem

  1. Dec 12, 2007 #1
    Okay, so I am making a website for my professor as work for him. He works on the ADMX and I need to learn about CP-violation to be able to write a description of it on the website.

    I've already scoured wikipedia, read part of Intro to Elementary Particles by Griffiths, talked to my prof's grad student, and him. I am starting to understand it more and more, but a little bit more help on the subject would be very appreciated.

    My understanding so far is that:

    The Parity operator reverses the sign of whatever it is acting on, so for example, it acting on some Psi(x) wave-function would reverse the sign of all of the "x"s in the equation. It was thought to be conserved until it was shown that the weak force does not conserve it. The example they (wikipedia) use is that the neutron decays into a proton only when it can change the flavor of one of its quarks and in the process emit a W boson. I get that. Protons are made of different building blocks than neutrons. But then they say that weak interactions only work on left-handed particles. From there I get lost, since I've never met a neutron with hands.

    As for charge, I think I get it. Charge conservation means that if you take the charge conjugate of whatever, nothing changes and everything works out. If you want to have the opposite operation, you need the opposite ingredients.

    Multiplied together you get CP, which was thought that by multiplying C by P you get a new symmetry (the REAL symmetry) which would be conserved. Hah. So then they talk about QCD and I get confused. My professor explained stuff about that to me, but it was right before finals and now I've forgotten most of it.

    The gist is that even though weak interactions maximally break P symmetry by only acting on left-handed particles, CP violation is very small, and that's the big question. My understanding of this breaks down when they hit QCD, since I don't know what it is and all of the buzz words make me want to cry.

    Any help would be greatly appreciated. :D
  2. jcsd
  3. Dec 20, 2007 #2


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    How many neutrons have you met?! :smile:

    "Handedness" refers (*roughly*) to "helicity", that is: the direction of the spin of the particle. A "left-handed" particle means that the spin vector is ANTI-aligned with the momentum and "right-handed" means that the spin is aligned with the momentum. There's a little more to it, but that's the general idea (it's true for massless fermions, but mass makes things a little more touchy). It should be enough for the moment.

    People used to think that it does not make a difference what the helicity is, but it turns out that this is not true! The weak interactions ONLY talk to left-handed particles (as you say).

    If you replace every particle by an anti-particle, you should get the same thing happening (if C is conserved). It's not in the weak interactions: there are no "left-handed antineutrinos" in the standard model!

    Put QCD aside for the moment. It is true that CP is very nearly conserved (left handed particles become right handed antiparticles and vice versa). But there is a small amount of CP violation in the weak interaction. A RULE OF THUMB for CP violation is the following: anytime you have an COMLEX coupling constant, you would expect CP violation to occur somewhere - and that CP violation is related to the PHASE of that coupling. But remember that in QM, you always take the magnitude-squared of an amplitude, so even if there's a phase, you don't always see it. So you must construct an observable that is actually sensitive to the phase (interference experiments). In the SM, these are rare and small. The famous example is [itex]K^0 \bar{K}^0[/itex] mixing (look it up for more details). The thing is that these effects are small even for sizable phases! And this what you were aluding to earlier, and it is more-or-less well understood.

    That was the electroweak CP violation. Now where does QCD come in? Well, it turns out that QCD also has a phase, called the "theta parameter", denoted [itex]\theta_{\rm QCD}[/itex]. This parameter contributes to, among other things, the electric dipole moment of the neutron. We know that this is a VERY small quantity (in fact, it's consistent with zero!) but in order for this contribution to be smaller than the upper bound for the neutron EDM, [itex]\theta_{\rm QCD}[/itex] must be less than something like [itex]10^{-10}[/itex]. From a theoretical point of view, we expect it to be around 1. What do you think. Are we close?!

    This is often called the "Strong CP problem". It's separate from the electroweak CP violation, which seems to account for all observed CP violation in the universe as we know it! So the big question is: why is [itex]\theta_{\rm QCD}[/itex] so incredibly tiny?! There are many proposed solutions (PQ symmetry and axions being the most famous that I know of, but there are others!) But I'll leave that for another day.

    Hope that helps!
  4. Jan 6, 2008 #3
    Ack, after nobody replied for a few days I thought this thread had just died. Thanks for replying! :D


    How do you define the spin vector? I'm starting 2nd quarter QM this Monday, where we get in to spins, but the little that I know says that the spin of an electron can either be up or down. I get that the directions are arbitrary, but my point is that there are only 2 spins, so how can you make any sense of it?
  5. Jan 6, 2008 #4


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    the spin vector is the direction of the spin. If the spin is in same direction as the particles momentum, then the particle have positive helicity (right handed), but if the spin vector is in the opposite direction as the momentum, then the particle has negative helicity (left handed).

    Spin is always relative something else, some other symmetry axis or similar. Here it is the momentum vector that is the thing that you compare the spin vector with.
  6. Jan 7, 2008 #5


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    pushing forward with this comment from Malawi_glenn: yes, spin is defined with respect to some vector called the "spin vector". In principle, it is arbitrary - it could be any vector at all! Of course, some choices are better than others. Two very useful choices in particle physics include:

    (1) "Helicity basis" - this is where the spin vector is proportional to the momentum vector of the particle (this is when you can talk of spin being "aligned" or "anti-aligned" with the momentum, and is the one you'll probably see in your QM class).

    (2) "Beamline basis" - when two particles collide and other particles come out (like in a particle accelerator), it is sometimes useful to define the outgoing particle spin with respect to the initial beam axis. This is a very common choice for spin measurements in particle physics detectors.

    There are many others, of course. Which you chose is a matter of convenience, depending on the exact process you are looking at. Part of the challenge in designing an experiment to measure spin is to be able to cleverly chose your "spin axis".

    Anyway, this might be a bit more advanced than what you want; if it confuses you, don't worry about it.
  7. Jan 9, 2008 #6
    Okay okay, I guess my confusion with the spin vector is I thought that spin was a scalar. I found out today that there is a spin for each axis, but you just generally talk about the z-axis spin, right?

    Okay, so if a particle has spin 1/2 in each axis, then what happens if it goes in positive x and y direction, but negative z direction? Or did I completely misunderstand?
  8. Jan 10, 2008 #7


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  9. Jan 11, 2008 #8
    I guess my question is, if you find an electron's spin in the Z direction to be 1/2, and the electron is going in the negative x-direction and negative y-direction, but positive z-direction, then what?

    Does Helicity only depend on the spin you actually know? As far as I understand, if you measure x-Spin, z-Spin is meaningless until you measure it again, right?

    It gets me even more confused when you say you can decide what each axis is (which is true). If you have some Cartesian co-ordinate system, measure Spin-z and get 1/2, then rotate the co-ordinate system by 90º in the x, y, and z directions, then Spin-z has to be re-measured, right? It's just weird that something that doesn't commute with the other axises isn't really discrete... or something.

    I should probably ask my professor this.
  10. Jan 11, 2008 #9


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    I still dont understand what you mean by "then what" ?
  11. Jan 11, 2008 #10


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    If you measure the spin in the z-direction, then THAT is your spin vector, irregardless of the momentum!! The fact that the particle is moving NOT in the z direction is irrelevant. This is NOT the helicity basis I mentioned above. This is some other basis.

    In particular, you spin vector was chosen by your measurement. In this case, the z-direction.

    That's correct.

    I don't get this. If you rotate your coordinate system, you haven't done anything!! You don't need to remeasure a quantity because of a mathematical trick. Perhaps I am misunderstanding you...
  12. Jan 11, 2008 #11


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    The word "Helicity" is very specific: it is the spin ALONG the direction of momentum. That is - the spin vector is the (unit) momentum vector: [itex]\vec{s}=\vec{p}/|\vec{p}|[/itex]. This does **NOT** have to correspond to a "spin measurement". For example, the problem you give above: it is *wrong* to say that "the helicity of the particle is ..." when you didn't perform your spin measurement in the helicity basis!
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