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Why are there not up and down quantum numbers?

by center o bass
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center o bass
#1
Mar21-12, 09:49 AM
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Hi. I am a bit confused by the fact that there are quantum numbers related to every flavour of quark except the up and down. A consequence of this is that the neutral pion is it's own anti-particle while the neutral kaon is not. Why is this so?
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Nabeshin
#2
Mar21-12, 11:32 AM
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A neutral pion is either [itex] u \bar{u} [/itex] or [itex] d \bar{d} [/itex]. In either case, the antiparticle gives the identical thing (remember, order isn't important!).

A neutral kaon, on the other hand, is [itex]d \bar{s} [/itex], whose antiparticle is [itex] \bar{d} s[/itex], clearly not the same thing.

A caveat is that you can have the kaon as a superposition, like [itex] \frac{1}{\sqrt{2}} \left( d \bar{s} \pm \bar{d} s \right) [/itex], in which case it IS its own antiparticle.
fzero
#3
Mar21-12, 12:23 PM
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You can assign either upness or downness as a quantum number. These flavor quantum numbers can be viewed as the remnants of an [itex]SU(N_f)[/itex] flavor symmetry that would be present in the absence of quark masses. Maximally breaking this symmetry via quark masses leaves the [itex]U(1)^{N_f-1}[/itex] symmetry associated to the flavors. Typically we choose the up quark to be neutral, with each other quark having charge one under a separate factor.

Edit: I should add that downness isn't such an interesting quantum number since the electric charge already distinguishes between the up and down quark (so no analogy of the neutral kaon as already pointed out.)

AdrianTheRock
#4
Mar21-12, 04:30 PM
P: 136
Why are there not up and down quantum numbers?

Quote Quote by Nabeshin View Post
A neutral pion is either [itex] u \bar{u} [/itex] or [itex] b \bar{b} [/itex]...
I assume the latter is a typo, and what you meant was [itex] d \bar{d} [/itex].
Nabeshin
#5
Mar21-12, 06:00 PM
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Quote Quote by AdrianTheRock View Post
I assume the latter is a typo, and what you meant was [itex] d \bar{d} [/itex].
ty :)
tom.stoer
#6
Mar21-12, 06:05 PM
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the reason is simple; you don't call it up-ness or down-ness but isospin
samalkhaiat
#7
Apr1-12, 05:43 PM
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Quote Quote by Nabeshin View Post
A caveat is that you can have the kaon as a superposition, like [itex] \frac{1}{\sqrt{2}} \left( d \bar{s} \pm \bar{d} s \right) [/itex], in which case it IS its own antiparticle.

You cannot have such superposition and the neutral Kaon can never be its own antiparticle! Because strong interaction conserves strangeness while the weak interaction does not, the neutral Kaon eigenstates with respect to these interactions are very different from each other; the strong-interaction eigenstates [itex]K_{0} \sim d \bar{s}[/itex] and [itex]\bar{K}_{0} \sim s \bar{d}[/itex] can MIX through weak transitions such as the one with two pions as intermediate state. In this [itex]K_{0}- \bar{K}_{0}[/itex] mixing, the [itex]\bar{K}_{0}[/itex] is the [itex]CP[/itex] conjugate of [itex]K_{0}[/itex],
[tex]|\bar{K}_{0}\rangle = CP|K_{0}\rangle .[/tex]
The mixing can be described by “effective Hamiltonian” having two eigenstates with very different lifetimes,
[tex]K_{S} = \frac{1}{\sqrt{2(1 + |\epsilon |^{2})}}\{ (K_{0} + \bar{K}_{0}) + \epsilon (K_{0} - \bar{K}_{0}) \}[/tex]
[tex]K_{L} = \frac{1}{\sqrt{2(1 + |\epsilon |^{2})}}\{ (K_{0} - \bar{K}_{0}) + \epsilon (K_{0} + \bar{K}_{0}) \}.[/tex]

Sam


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