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I Conservation of strangeness and eigenstates

  1. Jun 24, 2016 #1
    Hi, guys.

    In Povh's book, page 198, he says: "The strong force conserves the strangeness S and so the neutral kaons are in an eigenstate of the strong interaction."

    I do not see why this must be the case. My atempt to understand it:

    $$ŜĤ_s |K_0 \rangle = Ĥ_sŜ |K_0 \rangle$$
    So
    $$Ŝ(Ĥ_s |K_0 \rangle) = -Ĥ_s |K_0 \rangle $$

    Since the ket ##Ĥ_s |K_0 \rangle## has strangeness -1, it belongs to the eigensubspace of ##Ŝ## with eigenvalue -1. I don't know how one conclude, from this, that
    $$ Ĥ_s |K_0 \rangle \, \alpha \, |K_0\rangle $$
    which is what I want to prove.
     
  2. jcsd
  3. Jun 24, 2016 #2

    ChrisVer

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    well, abstractly speaking, can't you write [itex]|K_0>[/itex] state in whichever basis you want? and so the strong interactions' basis... the commutation relation you wrote (and I guess you take as given) will hold in any basis of the state [itex]|K_0>[/itex].
    So you can define that the [itex]K_0[/itex] represents a strong state.
     
  4. Jun 24, 2016 #3

    mfb

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    A more experimental approach: neutral kaons cannot decay via the strong interaction (they are the lightest neutral particles with a strange quark and ##K^0 \to K^+ \pi^-## is not possible either), so no matter how you write them as state, they have to be an eigenstate of the strong interaction.
     
  5. Jun 24, 2016 #4
    "So"? i don't see why...
     
  6. Jun 24, 2016 #5

    ChrisVer

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    whether you define the x axis showing to your left hand, or showing to your right, if there is a left/right symmetry it doesn't really matter.

    If we suppose that K0 in your case is not a strong state, you can "rotate" it into being in the strong state K0'...
    THis happens by diagonalizing the Hamiltonian in the strong basis by [itex]H \rightarrow H_{\text{str-diag}}=U H U^\dagger [/itex] and so will the states: [itex]|K_0> \rightarrow U |K_0>[/itex] which will be your states written in the strong interaction basis.
    The thing is that U is a unitary matrix (since it only changes the basis) and you of course get the same thing..
     
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