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Kaon notation

by CookieSalesman
Tags: kaon, notation
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CookieSalesman
#1
Sep3-14, 01:24 AM
P: 20
I was looking at this old chart (before the standard model was completed) and this thing with Kaon notations was confusing. So I understand that the top right number on a particle denotes charge, but what is subscript? Does that denote the number, as in which exact particle it refers to? Because it was regarding Kaon decay stuff, which had to do with more than one Kaon.
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Hepth
#2
Sep3-14, 01:33 PM
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Its possible its the mass, unless you have the chart handy or an example of the numbers I can't help much.

If you look at the pdg :

http://pdg.lbl.gov/2013/tables/rpp2013-sum-mesons.pdf

down some are the Kaons. Here
##K_S## = K - Short (in terms of lifetime, due to mixing)
##K_L## = K - Long (in terms of lifetime, due to mixing)
##K_1(1270)## means that it has J=1, and is an excited state with a mass near 1270.
Meir Achuz
#3
Sep3-14, 02:11 PM
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Quote Quote by CookieSalesman View Post
I was looking at this old chart (before the standard model was completed) and this thing with Kaon notations was confusing. So I understand that the top right number on a particle denotes charge, but what is subscript? Does that denote the number, as in which exact particle it refers to? Because it was regarding Kaon decay stuff, which had to do with more than one Kaon.
Can you show the example from the old chart?

CookieSalesman
#4
Sep3-14, 09:05 PM
P: 20
Kaon notation

Yes, the old chart looked like...
Mesons were listed, and the Kaons were split into three
The top where the negative Kaon equaled the positive Kaon, and following that, MeV rest mass and lifetime, etc, and decay modes.

Then, the neutral K and anti-K were in one category. On the top row was K0 and on the bottom was K0 (With a line over the K, to denote antiparticle

To the right of those two, was K01 (498 rest mass, .7e-10 lifetime) and below that was K02 (498 rest mass, .4e-8 lifetime).

The subscript and superscript are a bit messed up in the post, but on the picture were directly on top of each other.


(It went from left to right)
Sorry if the wording made visuals confusing.


I'm pretty sure Kaon 1 was just the normal neutral Kaon
And I'm pretty sure Kaon 2 was just the anti-neutral, right?
clem
#5
Sep4-14, 11:01 AM
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P: 1,269
"I'm pretty sure Kaon 1 was just the normal neutral Kaon
And I'm pretty sure Kaon 2 was just the anti-neutral, right? "

No. K##^0_1## and K##^0_2## are C=+1 and C=-1 eigenstates.
With C violation, they mix into two mass eigenstates with pure lifetimes (until CP violation)
called K##_S## and K##_L##.
CookieSalesman
#6
Sep4-14, 08:46 PM
P: 20
:O

Woosh.
What is an eigenstate?

And... sorry what is C violation or the rest?
Sorry, I have no idea...
mfb
#7
Sep5-14, 10:41 PM
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P: 12,113
Neutral mesons are a bit tricky.

You can take a strange-quark and an anti-down quark to form a kaon. This has a well-defined quark content (because we used this as definition), and you can also consider its antiparticle. Those particles can change into each other over time - a kaon becoming an antikaon and vice versa. If you write down this time-evolution in quantum mechanics, you can find a way to give two states that do not oscillate in that way - superpositions of "kaon and antikaon". Those are (approximately) called KS (S=short because it has a short lifetime) and KL (L=long).
ChrisVer
#8
Sep6-14, 03:42 AM
P: 1,067
an eigenstate is .... well without knowing what eigenstate is, I'd recommend you start from quantum physics [or even classical mehanics] before going to particle physics. The eigenstates [itex]u[/itex] of an operator [itex]\hat{A}[/itex] are the "solutions" of the eigenvalue-eigenstate equation:
[itex] \hat{A} u = \alpha u [/itex]
Where [itex]\alpha[/itex] is called eigenvalue.
In your case, the two kaons [itex]K_{1,2}[/itex] are C=+1,-1 eigenstates means that when the charge conjugate operator C acts on them, you will get +1 or -1 eigenvalues:
[itex] \hat{C} K_{1} = + K_{1}[/itex]
[itex] \hat{C} K_{2}= - K_{2} [/itex]
Now C violation means that C cannot be conserved [is not a symmetry anymore].. For that means that the Hamiltonian operator [which gives the evolution of your states] doesn't commute with the C operator [itex] [H,C] \ne 0 [/itex], thus there can be a mixing between the [itex]K_{1,2}[/itex] with time.. However you can find some linear expression with them which won't be a eigenstate of C anymore [but a superposition] but will remain fixed with time [they won't oscillate between each other].


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