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Possible decay states strong interaction, parity conservation

  1. Mar 14, 2015 #1
    The question is for which of the ##1P## meson states - ##1^{1}P_{1}, 1^{3}P_{0},1^{3}P_{1}, 1^{3}P_{2} ## ##D_{s}## states decaying to a ##1S## state is the decay: ##D_{s}**^{+} -> D_{s}^{+}\pi^{0} ## possible?

    Solution

    So the strong interaction conserves parity. Parity of meson is given by ## (-1)^{l+1} ##, for the ##1s## states, ##l=0## and so ##p=1##.

    The solution than states as both the decay products have zero spin the total angular momentum of the decaying particle must be equal to the orbitial angular momentum of ##D_{s}^{+}\pi^{0}## system.

    So I agree with this last comment, but I have no idea why both decay products have ##0## spin????
    I know that a meson is a quark and its antiquark, and so the spin adds to either ##1## or ##0##. But how do we know which it is?

    The solution then uses parity of the ##D_{s}^{+}\pi^{0}## system is ##P = (−1)^{l} × −1 × −1 ## *2.
    Im confused where this comes from - so I know that the parity of a meson in it's lowest states - i.e- ##l=0## is ##-1##. And I know that for a system of particles ##P=(-1)^{l}##, but, I've never seen how to consider the parity of two particles decayed. Is this how you 'add' the parities ?

    And so how exactly should you think of the system of decay products. So a particle has an intrinsic parity. Is this a particle or a quark? I.e is ##P=(-1)^{l}## coming from thinking of a system of quarks or a system of the two particles ##D_{s}^{+}\pi^{0}##?

    (From which the solution follows from the fact that we require ##P=1##, again I'm okay with this once I understand expression *2.)

    Thanks, your help is really appreciated !!
     
  2. jcsd
  3. Mar 14, 2015 #2

    mfb

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    You can look that up (and one can argue that "spin zero" is part of the definition of those particles).

    Parity is multiplicative - you multiply all parity values to get the total quantum number.
    Both descriptions work, but as you deal with hadrons here it is easier to consider those.
     
  4. Mar 15, 2015 #3
    As in from a table of mesons?
    And theres no other way of knowing?
     
  5. Mar 15, 2015 #4
    Okay, I've had a look a the next part of the question which is to do the same (find the possible states) for the decay ##D_{s}^{+}** -> D_{s}*^{+} \pi ^{0} ##.

    And I follow the solution , if it is the case that the ##D_{s}*^{+} ## has ##s=1## , (it doesnt actually state this but I'm pretty sure it's being used).

    Is this the case in general, the unexcited version of the particle has ##s=0## and the excited ##s=1##?

    Thanks .
     
  6. Mar 15, 2015 #5

    mfb

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    Right.
    You have to find out what the symbol "##\pi^0##" means - you either know it or you have to look it up.

    What is the s=0, s=1 now?
     
  7. Mar 16, 2015 #6
    Sorry? I'm unsure what you are asking?
     
  8. Mar 16, 2015 #7

    mfb

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    Where does it come from and what does it mean?
     
  9. Mar 16, 2015 #8
    Sorry in post 4 ##s## stands for spin.
     
  10. Mar 16, 2015 #9

    mfb

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    Sometimes spin 1 states are called excited states, sometimes they have their own particle name. That is not completely consistent.
     
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