Download Wissink P640 – Subatomic Physics I Fall 2007 Problem Set # 1

Wissink
P640 – Subatomic Physics I
Fall 2007
Problem Set # 1
SOLUTIONS
1. Iso-Confused!
In lecture we discussed the “family” of π-mesons, which have spin J = 0 and isospin
I = 1, i.e., they form the isospin triplet (π + , π 0 , π − ). There are also two low-mass vector
mesons, meaning mesons that have a spin of 1: the isospin-0 ω 0 , and the isospin-1 ρ,
which comes in three charge states just like the pion. Using angular momentum algebra,
isospin conservation, and the Generalized Pauli Exclusion Principle, explain the following
observations regarding the strong decay modes of the neutral vector mesons:
(a) The ρ0 decays predominantly to π + π − , but the decay ρ0 → π 0 π 0 is not observed.
(b) The decay ω 0 → π + π − is strongly suppressed compared to ω 0 → π + π − π 0 , even
though the latter has much less kinematic phase space available.
SOLUTION
For both of these cases, we are dealing with mesons (π, ρ, ω), which are bosonic and
therefore must be described by wave functions that are fully symmetric. If we treat isospin
as a ‘good’ quantum number, then our wave functions will be a product of a spatial state,
a spin state, and an isospin state – so to be symmetric, either all three of these pieces must
be symmetric, or exactly one must be symmetric and the remaining two anti-symmetric.
(a) We first consider the decay ρ → π π. Knowing the spins of these particles, conservation
of total angular momentum (spin + orbital) tells us
Jρ = Sπ ⊕ Sπ ⊕ Lππ
Because pions are spin-0, we see immediately that the relative orbital angular momentum
Lππ = Jρ = 1; and because L is odd, the final 2π state is anti-symmetric in space.
Moreover, both pions having spin-0 means the final state will be symmetric in spin, and
so we conclude that the 2π final state must always be anti-symmetric in isospin. The
whole problem thus reduces to determining whether or not there are possible 2π states
that are anti-symmetric in isospin under particle interchange.
Isospin follows the same algebra as spin, so we know that two isospin-1 particles can couple
to a total isospin Itot = 0, 1, or 2; but only the I = 1 states are anti-symmetric (we showed
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this in class for two spin-1 particles using Young Tableaux). Thus, the decay ρ0 → π + π −
conserves angular momentum, spin, and isospin, and is completely allowed, as long as
the two final-state pions couple to Itot = 1. For ρ0 → π 0 π 0 , however, the two outgoing
pions are in the same isospin substate, i.e., both are described by |I I3 i = |10i, so any
state formed will be symmetric in isospin under particle interchange ⇒ this decay mode
should be highly suppressed.
You could also approach this problem (though less rigorously) by not invoking isospin
at all. It still must be true that the 2π final state will be space-A (L = 1) and spin-S.
In this approach, the π 0 π 0 state is symmetric ⇒ can’t proceed, while the π + π − state
involves non-identical particles ⇒ no Pauli Principle at work here. However, we have lost
the insight that the decay is possible only when the charged pions couple to Itot = 1.
(b) We next consider the decay of the ω 0 meson to 2 pions. Because the ω, like the ρ,
is a spin-1 particle, all of the angular momentum arguments presented in part a apply
here as well for the 2π decay, so we must require the two outgoing pions couple to isospin
Itot = 1. But the ω, unlike the ρ, is an isospin-0 particle, so the decay ω 0 → π π (for either
charged or neutral pions) has I = 0 in the initial state and must have I = 1 in the final
state ⇒ can’t proceed. In reality, we know isospin conservation is violated by weak and
electromagnetic interactions, and so we might expect this decay to occur at 1% level, i.e.,
on the order of αe.m. . In fact, this decay has about a 1.7% branching ratio (PDG).
Finally, we consider the more “exotic” decay ω 0 → π + π − π 0 – exotic only in that we now
have three particles in the final state. Starting with spin, we first note that three spinless
particles are always fully symmetric in spin under interchange. (Using Young Tableaux,
you can convince yourself easily that any number of spinless particles can only combine
to form a single, fully symmetric singlet state.) Also, because the initial state has I = 0,
we must couple up three isospin-1 particles to form Itot = 0 in order to conserve isospin.
This is possible, and yields a state fully anti-symmetric in isospin; this is just the isospin
analog to the anti-symmetric, colorless singlet obtained by color-coupling three quarks,
each of which has three possible color substates (same Young Tableaux → 3 boxes and 3
substates). Thus, with a 3π state that is spin-S and isospin-A, we need to find states that
are space-A as well, to keep the ω wave function symmetric. To accomplish this, we finally
invoke angular momentum conservation, which requires that Jtot in the final state equals
that in the initial state, which is Jω = Sω = 1. But once again, the fact that the pions
are spin-0 means that Jtot in the final state is just the total orbital angular momentum
of the 3π state. So the relevant question is: can three particles have an Ltot = 1 and be
anti-symmetric? The answer is yes – it can be shown rigorously with angular momentum
coupling, or you can just think about it this way: Let two of the pions have an Lππ = 1.
Now let the third pion have an L = 1 relative to the center of mass of the first two. To find
Ltot , we couple these two L = 1 configurations, which gives us Ltot = 0, 1, 2 as possibilities,
but only the Ltot = 1 piece is anti-symmetric, as is easily shown using Young Tableaux (in
fact, we did this case in class, coupling two spin-1 objects).
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2. Determining the Deuteron
(a) The spin-1/2 nucleon N has a J = 3/2 excited state – often viewed as a strong πN
resonance – labeled the ∆ state. Show that a 2-baryon combination of N ’s and/or ∆’s
can not have quantum numbers J = 0 and I = 0. Note that because of this, the deuteron
ground state (the only 2-baryon bound state!) is I = 0 but has a spin of 1. (b) Experiment
suggests that the deuteron ground state wave function has small components in which both
nucleons are excited to ∆’s. Using spectroscopic notation (2S+1 LJ ), specify all allowed ∆∆
states of positive parity that could contribute to this part of the wave function.
SOLUTION
(a) Our first task is to prove that no combination of two N ’s and/or ∆’s can have the
quantum numbers I = 0 and J = 0. We can rule out the N ∆ combination based on isospin
alone: we know
IN ∆ = IN ⊕ I∆ = 3/2 ⊕ 1/2 = 1 or 2
This is just a specific application of a very general rule: two angular momenta (or two
isospins – same algebra) can only couple to zero if they are equal to each other. Thus,
both N N and ∆∆ states can couple up to I = 0, as well as higher isospins. The I = 0
state thus formed is anti-symmetric under interchange, as we showed in lecture (or you
can easily show using Young Tableaux). And because these are systems of two identical
fermions, the total wave function must also be anti-symmetric ⇒ the product of spatial
and spin states must be symmetric. But note: if the state is space-S (even L), this means
it must also be spin-S (odd S for fermions), while if the state is space-A (odd L) it must
be spin-A (even S). Thus, in either case, L can not equal S (one is odd, the other is even),
and so they can not couple to J = 0 ⇒ no 2-baryon combination of N ’s and ∆’s can have
both I = 0 and J = 0.
(b) We are given that the deuteron ground state is isospin-0, but has a spin of 1. Thus,
based on our results for part a, we see there are ∆∆ configurations that could contribute
to this state, with the constraint that because I = 0, either L is even and S is odd, or L
is odd and S is even. The requirement of positive parity limits us to the even-L solutions,
and so S is odd ⇒ S can be 1 or 3. Finally, we know that J = L ⊕ S = 1. Thus, the usual
rules of angular momentum coupling show that the complete set of allowed configurations
for each L are given by (in clunky spectroscopic notation):
L=0
L=2
S=1
S = 1, 3
3
L=4
S=3
7
3
3
S1
D1 , 7 D1
G1
3. The Wonderful World of Color – plus Strangeness!
(a) No light baryons (that is, those containing only u or d valence quarks) have been
observed with quantum numbers spin J = 3/2 and isospin I = 1/2, or with J = 1/2 and
I = 3/2. Argue why these two combinations are not allowed for a color singlet system
of three light quarks arranged in a relative s-state (L = 0). (b) Now consider baryons
composed of u, d, or s quarks coupled to total spin J = 3/2, though still in an L = 0 state.
List the valence quark content of all allowed baryons with these quantum numbers, and
identify the actual (observed) particle corresponding to each. For this last step, list the
‘name,’ mass, and charge state, using the PDG compilation as needed.
SOLUTION
(a) For a system of three u or d quarks, we can schematically describe the total wave
function as
|Ψiqqq = |spacei|spini|isospini|colori
In an isospin formulation, we view u and d quarks as identical particles, though occupying
different isospin (or flavor) substates. Therefore the above wave function, describing a set
of three identical fermions, must be fully anti-symmetric under particle interchange. Given
that we are interested in a color singlet (so Ψ is color-A) of quarks in a relative s-state (so
L = 0 and Ψ is space-S), we conclude that the product of spin and isospin states
must be fully symmetric for all allowed configurations.
In this case, the implications are clear, because the spin and isospin couplings follow
the exact same algebra: for either quantum number, we are coupling three S (or I) =
1/2 particles, giving us an Stot (or Itot ) of 1/2 or 3/2. For each quantum number, the
3/2 states are symmetric, while the 1/2 states are of mixed symmetry, as we showed in
lecture by developing the Young Tableaux for this case. Thus, to keep the product of spin
and isospin symmetric, we must have Stot = Itot , and so (dropping the tot subscripts,
and setting J = S when L = 0) we see that the combinations (J = 3/2, I = 1/2) and
(J = 1/2, I = 3/2) are not allowed. The remaining two combinations, (J = 1/2, I = 1/2)
and (J = 3/2, I = 3/2), where J = I, ensure that wave functions |Ψiqqq are fully antisymmetric. These two combinations, in fact, correspond to known families of particles:
the nucleons and Deltas encountered in the previous problem.
(b) We now complicate things a bit, and add a new isospin substate, corresponding to
the possibility of having a strange, or s quark. (Note that the letter s is now standing
for strange, symmetric, spin, and “L = 0” – wonderful!) In modern usage, the concept of
“isospin,” when used to describe quarks, is replaced by the term “flavor,” but the algebra
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remains the same, and so the relevant Young Tableaux, for example, are unchanged. And,
other than the flavor state, the properties of Ψ worked out in part a still apply here: for
a triplet of quarks to form a color singlet (color-A) in a relative s-state (space-S) with
J = 3/2 (spin-S), Ψ can be fully anti-symmetric only if the state is also flavor-S. Thus,
the whole problem reduces to determining all possible arrangements of three particles (the
quarks), each of which has three possible flavor substates (1 = u, 2 = d, or 3 = s), that are
fully symmetric under particle interchange. Using Young Tableaux, and considering only
the symmetric configurations ⇒ 3 horizontal boxes, we can easily work out the well-known
decuplet of spin-3/2 baryons, whose experimental discovery (especially that of the triplystrange Ω− ) really turned the quark model from being cute numerology into a quantitative
tool. Working out the tableaux, and searching the PDG compilation for the lowest mass
J = 3/2 state for each arrangement, we find:
[1 1
1]
|uuui
q = +2
∆++ (1232)
[1 1
2]
|uudi
q = +1
∆+ (1232)
[1 2
2]
|uddi
q=0
∆0 (1232)
[2 2
2]
|dddi
q = −1
∆− (1232)
[1 1
3]
|uusi
q = +1
Σ+ (1385)
[1 2
3]
|udsi
q=0
Σ0 (1385)
[2 2
3]
|ddsi
q = −1
Σ− (1385)
[1 3
3]
|ussi
q=0
Ξ0 (1530)
[2 3
3]
|dssi
q = −1
Ξ− (1530)
[3 3
3]
|sssi
q = −1
Ω− (1672)
Note that the masses – and in particular the mass differences – already set some of scales
for the quark model. For example, taking the quarks uud in an s-state with J = 1/2 (the
proton) and changing only the spin coupling so that J = 3/2 (the ∆+ ) results in a mass
increase of about 300 MeV. Also, just from the above table, we can see that the u and d
quarks must be very similar in mass (switching one to another has little effect), but every
time we replace a u or a d with a strange quark, the mass increases by about 150 MeV.
Much of what we “know” about the individual quark properties we have inferred (and not
directly measured) from such mass splittings.
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