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Parity formulae, orbital angular momentum, mesons

  1. Mar 14, 2015 #1
    So a particle has intrinsic parity ##\pm 1 ## .
    The parity of a system of particles is given by product of intrinsic parities and the result is: ##(-1)^l ## (1).


    1) How does this result follow?
    and what exactly is ##l## here? so it's the orbital angular momentum, so say a particle is made up of 3 quarks, then it's described to be in a certain state, ##1p## , ##1s## states etc, so ##l## is the orbital angular momentum of this state?

    2) For a meson ##p=(-1)^{l+1}##
    The reasoning in my book being that the quark and antiquark have opposite intrinsic parities,

    So I assume this followss from (1), although I am not seeing how??

    But, in the result (1) , this is a general result for any system of particles right? So quarks and antiquarks of the same type could be included in any system yet ## (-1)^l ## still holds?

    Thanks in advance.
  2. jcsd
  3. Mar 18, 2015 #2
  4. Mar 18, 2015 #3


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    [itex]l[/itex] is the angular momentum number. The particles in general will be described by spherical harmonic functions [itex]Y_{lm}(\theta, \phi)[/itex] which have the given parity:
    [itex]\hat{P} Y_{lm} (\theta, \phi) \equiv Y_{lm} (\pi-\theta, \phi+ \pi)= (-1)^l Y_{lm} (\theta, \phi)[/itex]

    Take for example a meson.
    The meson will have a quark and antiquark [itex]q, \bar{q}[/itex].
    So the intrinsic parity of the one will be +1 and the parity of the other will be -1.
    On the other hand, the meson can have some spatial angular momentum number [itex]l[/itex].
    So the total parity will be: [itex] (+1) \times (-1) \times (-1)^l = (-1)^{l+1}[/itex]
  5. Mar 18, 2015 #4


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    So I guess you make a wrong assumption before your eq.1.
    "The parity of a system of particles is given by product of intrinsic parities "
    This is wrong (probably a misconception).

    Again take a meson... it is a [itex]q \bar{q}[/itex] state with some angular momentum [itex]l[/itex] due to the quarks. Since the meson has an orbital angular momentum [itex]l[/itex], its angular distributions are described by the spherical harmonics [itex]Y_{lm}[/itex]. It's from this relation that you get the eq.1 [itex](-1)^l[/itex], and not by getting the intrinsic parities.
    Sec. 14.2
  6. May 2, 2015 #5

    Is this always the case with a particle and antiparticle, that they have opposite parities?
  7. May 2, 2015 #6
    For fermions, yes. For bosons, no.
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