Download Elementary Particles A Homework 2

Elementary Particles A Homework 2
Eric Cotner
October 7, 2014
Problem 1
Consider the Ds+ meson, with quark content cs̄.
a) Suggest a candidate strong decay for the Ds+ , and calculate the minimum mass
the Ds+ must have for this decay to be allowed.
My best guess at an allowed strong decay would be if uū pair were created by a hard gluon and
the resulting two mesons would be less total mass than the Ds+ . The resulting process would be
Ds+ (cs̄) → K + (us̄) D0 (cū).
Ds+
s̄
c
s̄
u
K+
ū
c
D0
Figure 1: Speculative Ds+ → K + D0 decay
The masses of the K + and D0 are 493 MeV and 1864 MeV, respectively, which sets the lower
bound on the Ds+ of 2357 MeV.
b) The mass of the Ds+ is 1970 MeV, so the above decay is not allowed. The
¯ has been observed though. Explain why this can’t
decay Ds+ (cs̄) → φ(ss̄) π + (ud)
be a strong decay and draw the diagram for it.
The decay Ds+ → φ π + cannot be strong because there is flavor violation. The original meson has
strangeness S = −1 and charm C = +1, while the resulting state has strangeness S = 0 and charm
C = 0. Therefore the decay has to be weak.
1
Ds+ s̄
c
s̄
s
φ
u
d¯
π+
W
Figure 2: Allowed Ds+ → φ π + decay
Problem 2
Consider the electromagnetic decay π 0 → γ γ (the way the pion decays 98.8% of the time).
a) Draw a diagram for the decay
γ
π0
u
ū
γ
Figure 3: π 0 → γ γ decay
b) Consider a different decay mode, π 0 → γ e+ e− . Explain roughly why this mode
has a ratio of occurrence to the first mode of 1:80.
The diagram for this decay is
γ
π0
u
ū
e+
e−
Figure 4: π 0 → γ e+ e− decay
To get a rough estimate of the branching ratio, we can compare the amplitudes of the two
processes. The first one has two vertexes, so it has an amplitude of M1 ∼ α, while the second has
three vertexes, so M2 ∼ α3/2 . The branching ratio is then roughly
B(π 0 → γ e+ e− )
|M2 |2
1
=
∼α≈
0
2
B(π → γ γ)
|M1 |
137
which is within a factor of two of 1/80, as expected.
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Problem 3
Scan the list of baryons containing one or more s quarks (both spin- 12 and spin- 23 ) in the particle
table inside the front cover of the book, and note that there are three regimes of lifetimes: 1023 s,
1020 s, and 1010 s.
a) Show how our simple estimates of relative rates for the three forces can very
roughly account for such a range of lifetimes.
We can use our knowledge of estimating the values of Feynman diagrams to aid us in calculating
order of magnitude estimates for particle decays. We know that the lifetime of a particle is inversely
proportional to the transition probability, which is proportional to the square of the probability
amplitude (by Fermi’s Golden Rule):
τ ∼ 1/Γ ∼ 1/|M|2
So if we can estimate the magnitude of the probability amplitude for various processes, we can
estimate the decay lifetime.
For electromagnetic decays, the diagram usually consists of one or two vertices. Internal photon
propagators don’t really affect the order of the diagram because they are massless (∼ 1/q 2 ). Because
of this there is usually little momentum transfer, so the mass of the decay products doesn’t really
matter. So most diagrams have amplitude M ∼ αe = 1/137.
For strong decays, we use the same logic, so that most diagrams have amplitude M ∼ αs ∼ 1.
For weak decays, we always have at least two vertices (because there can’t be external W or Z
particles), and usually one internal boson propagator. Because the W and Z are massive, the
amplitude is suppressed by their masses (∼ 1/m2W ) and because of significant momentum transfer,
is amplified by the mass of the decay products (usually the mass of a µ or π) because of the larger
phase space available. Therefore, most diagrams have order M ∼ αW (mµ /mW )2 ∼ 2 × 10−8 .
Assuming the scale of strong decays is τs ∼ 10−23 s, the timescale of electromagnetic and weak
decays should be
τe ∼ |Ms |2 /|Me |2 τs ∼ 10−19 s
τW ∼ |Ms |2 /|MW |2 τs ∼ 10−8 s
Comparing to the real decay timescales of τe ∼ 10−20 and τW ∼ 10−10 , these estimates are fairly
close in terms of order of magnitude.
b) Different decay paths
Consider the Λ, Σ0 , Σ0∗ baryons, with quark content uds. They have different masses (1115, 1192
and 1385 MeV, respectively), and different decay products. Explain how the different masses allow
different principle decay mechanisms.
The first decay, Λ → p π − , has the smallest invariant mass, and the difference in mass between the
Λ and proton is 117 MeV. Therefore the decay does not have much free phase space to decay to.
It must wait a lengthy time for the weak force to mediate the decay.
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Λ
u
d
s
u
d
u
p
d
ū
π−
Figure 5: Λ → p π − decay via intermediate W
The next smallest invariant mass decay, Σ0 → Λ γ, has the possibility of decaying through the
electromagnetic force. This is a much quicker decay, and therefore proceeds very rapidly, despite
the fact that the difference in mass between the Σ0 and Λ is only 77 MeV and therefore has less
phase space to decay to.
u
Σ0 d
s
u
d
s
Λ
γ
Figure 6: Σ0 → Λ γ decay via photon emission
The final decay, with the largest invariant mass, is of the Σ0∗ → Σ0 π 0 . Because the difference
in mass between the Σ0∗ and Σ0 is 193 MeV, it has the possibility of decaying to a pion through
the strong force. It can also decay to a pion and the Λ as well, giving it even more phase space and
a shorter lifetime.
u
Σ0∗ d
s
u
d
s
Σ0
u
ū
π0
Figure 7: Σ0∗ → Σ0 π 0 decay via pion production
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Problem 4
a) Draw the diagram for µ+ → e+ ν̄e νµ , which in the standard model has a branching ratio of almost 100%.
ν̄µ
µ+
νe
W+
e+
Figure 8: Standard Model µ+ → e+ ν̄e νµ decay
b) Calculate the branching ratio of the process µ+ → e+ e+ e− assuming the process
is mediated by a new boson H, with a mass mH and a vertex factor gH .
The diagram for the 3 electron decay is given by
e+
µ+
e+
H
e−
Figure 9: Non-Standard Model µ+ → e+ e+ e− decay
In fig. 8, the order of magnitude of the scattering amplitude is M1 ∼ g 2 /m2W times some phase
2 /m2 times some phase
space factor. In fig. 9, the order of magnitude of the amplitude is M2 ∼ gH
H
space factor. The two phase space factors are approximately the same for each decay since the
mass difference between the decay products is of order of the mass of the muon in both cases. The
branching ratio is then approximately
Γ(µ → 3e)
|M2 |2 gH 4 mW 4
B(µ → 3e) =
∼
=
Γ(all)
|M1 |2 g mH
c) The experimental limit on this branching ratio is about 10−12 . What does this
say about the mass of the H if we assume |gH /g| ≈ 1?
If our branching ratio is less than 10−12 , this means
B(µ → 3e) =
mW
mH
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< 10−12 =⇒ mH > mW × 103 ∼ 80 TeV
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Problem 5
∗
(1232) and Y1∗ (1385)? Draw figure 1 and
a) What are the modern names of N3/2
put the quark content of all the particles.
The production and subsequent decay of the Ω− is shown in the following diagram:
Figure 10: Scattering event decay chain from discovery of Ω−
∗ (1232) and Y ∗ (1385) particles, they are now known by different names. The
As for the N3/2
1
∗ (1232) is now the ∆ baryon, a spin-3/2 particle composed of u and d quarks but with the spins
N3/2
all aligned (as opposed to the proton/neutron, where two spins are aligned). The Y1∗ (1385) is now
known as the Σ∗ , which an excited state of the Σ, which is composed of various combinations of u,
d and s quarks.
b) Draw the diagram for the K − p → Ω− K + K 0 reaction.
The Ω− production reaction is:
p
u
d
u
K − ū
s
u
s̄
K+
d
s̄
K0
s
s
s
Ω−
Figure 11: K − p → Ω− K + K 0 reaction
c) Assuming the target proton is at rest, calculate the smallest K − beam energy
for which this reaction is allowed.
The total 4-momentum of the incoming particles in the lab frame is given by
pbefore = (EK − + mp , pK − )
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and the total 4-momentum of the decay products in the CM frame is minimized by the energy
being the sum of the masses:
pafter = (mΩ− + mK + + mK 0 , 0)
We can solve for EK − by using the equation p2before = p2after :
p2before = p2after = (mΩ− + mK + + mK 0 )2 ≡ M 2
M 2 = (EK − + mp )2 − |pK − |2
2
2
2
2
= EK
− + mp + 2mp EK − − (EK − − mK − )
= 2mp EK − + m2p + m2K −
=⇒ EK − =
M 2 − m2K − − m2p
(mΩ− + mK + + mK 0 )2 − m2K − − m2p
=
2mp
2mp
This corresponds to a beam energy of of 3.2 GeV. The beam used was 5.0 GeV, just barely enough.
d) Draw a diagram to produce Ω− by proton-proton collision. Why do you think
a kaon beam was used?
Because the strong force cannot change flavor or violate strangeness, the only way we can produce
the Ω− is if hard gluons create 3 ss̄ pairs, creating an Ω− in addition to a proton and 3 kaons to soak
up the s̄ quarks and the constituents of the other proton via the process p p → Ω− p K + K + K 0 :
p
p
u
u
d
u
u
d
u
u
d
p
s̄
u
K+
s̄
u
K+
s̄
d
K0
s
s
s
Ω−
Figure 12: p p → Ω− p K + K + K 0 reaction
If we perform a similar calculation (as in part c) for this process, we see that the minimum
proton beam energy is
(mΩ + mp + 3mK )2 − 2m2p
Ep =
= 8.0 GeV
2mp
Perhaps a beam energy of this magnitude was difficult or just beyond reach at the time. It is
certainly more than the 5.0 GeV beam used in the experiment.
e) Draw quark line diagrams for the entire decay chain of the Ω− .
I will break the full decay down into a series of diagrams showing the subsequent decay chain. The
first decay is that of the Ω− to a Ξ0 and π − :
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s
Ω− s
s
W−
s
s
u
Ξ0
ū
d
π−
Figure 13: Ω− → Ξ0 π − decay
Note the π − is lost before subsequent decay. The next decay in the chain is Ξ0 → Λ0 π 0 :
Ξ0
u
s
s
W−
u
s
d
Λ0
ū
u
π0
Figure 14: Ξ0 → Λ0 π 0 decay
The next two diagrams show the decays Λ0 → p π − and π 0 → γ γ → e+ e+ e− e− . Note that the
decay of the photons into electrons is not kinematically allowed unless we consider the source pion
(since there is no center of momentum frame for a photon), which is not reflected in the paper.
e+
Λ0
u
d
s
W−
u
d
u
p
d
ū
π−
e−
u
ū
e+
e−
Figure 15: Λ0 → p π − and π 0 → γ γ → e+ e+ e− e− decays
Problem 6
a) Which of the following are (polar) vectors? Axial vectors? Scalars? Pseudoscalars?
i) S = spin momentum
Any type of angular momentum is an axial vector, which is even under a parity transformation.
ii) r = position in coordinate space
By definition, r is odd under a parity transformation, so it is a polar vector.
iii) p = three-momentum
This one is also odd under a parity transformation (reverses direction of motion), so is a polar
vector.
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iv) L = r × p
Because both r and p are odd under parity, L is even, so is an axial vector.
v) Helicity h = Ŝ · p̂
Because S is even and p is odd under parity, the helicity is odd, which makes is a pseudoscalar.
vi) Magnetic field B
Looking at how a magnetic field from a particle is defined (as a cross product of the displacement
vector r and the momentum p), the magnetic field is clearly even under parity, and is therefore an
axial vector.
vii) Magnetic moment µ
The magnetic moment of a particle is proportional to its spin, so that µ is also an axial vector.
viii) Energy of magnetic moment in an external field: −µ · B
Because both µ and B are odd under parity, the energy is even, and is therefore a true scalar.
ix) The triple product (p1 × p2 ) · S
The momenta are odd, and the spin is even under parity, so this object is even under parity. It is
therefor an axial vector.
x) p1 × (p2 × p3 )
All three momenta are parity odd, so the resulting object is parity odd. It is therefore a polar
vector.
b) Parity in Electrodynamics
i) Suppose the Lorentz force equation appeared as F = q(E + v × E). Why should you
be suspicious?
Since E is a polar vector and v is an axial vector, their cross product is also an axial vector.
Therefore, this version of the Lorentz force is adding a polar and axial vector together, in clear
violation of the fact that the force is a polar vector.
ii) Magnetic monopoles
If magnetic monopoles existed, Maxwell’s equations would have a lot more symmetry, and the
Lorentz force would become
F = qE + qv × B + qm B + qm v × E
If we consider the additional terms, we see that they must have the same parity as the original
ones, which means they must be polar vectors. The magnetic field is axial and the electric field
polar though, so in order for the combinations qm B and qm v × E to be parity odd, this demands
that qm be parity odd, so magnetic charge is pseudoscalar!
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c) Parity in Particle Decays
i) In the decay π 0 → µ+ νµ , the muons are found to have spins aligned preferentially in
the direction of their momentum. Explain why this violates parity.
The reason this violates parity comes down to the fact that under a parity transformation, the
helicity of the muon is reversed. If parity were conserved, we would expect both helicities with
equal probability since the parity-reversed scenario should be equally probable. However, since
more muons are found to have positive helicity, there is a clear asymmetry between the positive
and negative helicity muons, meaning parity-reversal symmetry has been violated.
ii) The muons were then brought to rest and it was observed that in the decay µ+ →
e+ ν̄µ νe , the momentum of the positron preferentially aligned with the spin of the
muon. Explain why this is further evidence of parity violation.
This is further evidence of parity violation because in the parity-reversed situation, the momentum
of the positrons is reversed, but the spin of the muon is not, so it would appear that the positrons’
momenta are preferentially anti-aligned with the muon spin. This further shows that the parity
symmetry of the decay is violated.
Problem 7
a) Calculate the probability to measure strangeness S = −1. What is the sum of
the probabilities for measuring both S = −1 and S = +1? What is this at times
long compared to the K1 lifetime but short compared to the K2 lifetime?
The given equation for the time-dependent wavefunction of the kaon system is
1 ψ(t) = √ |K1 i e−(im1 +Γ1 /2)t + |K2 i e−(im2 +Γ2 /2)t
2
The decomposition of the flavor eigenstates into the mass eigenstates is
1
|K 0 i = √ (|K1 i + |K2 i) (S = +1)
2
1
|K 0 i = √ (|K1 i − |K2 i) (S = −1)
2
Taking the inner product of the K 0 with ψ(t) and squaring, we find the probability of measuring
S = −1:
PS=−1 (t) = | hK 0 |ψ(t)i |2
2
1
−(im
+Γ
/2)t
−(im
+Γ
/2)t
1
1
2
2
= (hK1 | − hK2 |) |K1 i e
+ |K2 i e
2
2
1 = e−(im1 +Γ1 /2)t − e−(im2 +Γ2 /2)t 4
1 −Γ1 t
=
e
+ e−Γ2 t − 2e−(Γ1 +Γ2 )t/2 cos (∆mt)
4
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We see that this term oscillates while decaying away with respect to two different decay rates.
The sum of the two probabilities for S = +1 and S = −1 are
1 −Γ1 t
e
+ e−Γ2 t + 2e−(Γ1 +Γ2 )t/2 cos (∆mt)
4
1 −Γ1 t
+
e
+ e−Γ2 t − 2e−(Γ1 +Γ2 )t/2 cos (∆mt)
4
1 −Γ1 t
=
e
+ e−Γ2 t
2
PS=+1 (t) + PS=−1 (t) =
At timescales long compared to the K1 lifetime (Γ1 t 1) and short compared to the K2 lifetime
(Γ2 t 1), we can see that the first exponential is heavily damped, so that the effective probability
of measuring a kaon in the beam is 12 e−Γ2 t . This is because all the K1 mesons have decayed into
other particles, leaving only the K2 mesons.
b) Derive the mass splitting of the kaons.
Because m1 and m2 are the eigenvalues of the kaon Hamiltonian, we can derive how to calculate it:
∆m = m1 − m2 = hK1 |H|K1 i − hK2 |H|K2 i
1
1 0
=
hK | + hK 0 | H |K 0 i + |K 0 i −
hK 0 | − hK 0 | H |K 0 i − |K 0 i
2
2
1 0
0
0
hK |H|K i + hK 0 |H|K 0 i + hK |H|K 0 i + hK 0 |H|K 0 i
=
2
1 0
−
hK |H|K 0 i + hK 0 |H|K 0 i − hK 0 |H|K 0 i − hK 0 |H|K 0 i
2
∗
= hK 0 |H|K 0 i + hK 0 |H|K 0 i = hK 0 |H|K 0 i + hK 0 |H|K 0 i
= 2Re hK 0 |H|K 0 i
c) Draw a box diagram for the above matrix element in the Standard Model.
What CKM matrix elements is this sensitive to?
The diagram for the matrix element hK 0 |H|K 0 i is that which evolves a |K 0 i state to a |K 0 i state:
qu
s
d
W
K0
d¯
K0
W
q̄u
s̄
Figure 16: K 0 → K 0 oscillation
where the qu denotes any up-type quark (u, c, t). This process is sensitive to the elements of the
CKM matrix which mix the s and d with the up-type quarks (the effects of the top quark are
probably washed out because of its mass), so the relevant parameters are Vud , Vus , Vcd and Vcs .
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Problem 8: B Meson Oscillation
Similar to the kaon system, the neutral B system also has mass eigenstates which do not match
exactly with the flavor eigenstates (there is some mixing).
a) Unlike K1 and K2 , the lifetimes of B1 and B2 are not noticeably different.
Why?
In the kaon system, the fact that the K1 is is CP even while the K2 is CP odd is the reason the
two have such different lifetimes. The K1 decays to two pions, so it does so quickly, but the K2
must decay to three pions because of CP. This decay takes much longer since the mass of the K2 is
just barely larger than the mass of three pions; the difference in phase space available to decay to
causes the difference in lifetimes. In the B system, the mass of the b quark is so large ( 4 GeV) and
the decay modes so many that the difference in phase space available to the B1 and B2 to decay
to is negligible. This results in the two mass eigenstates having very similar lifetimes.
b) At the Belle accelerator, 8 GeV electrons collide with 3.5 GeV positrons.
What special mass was this tuned to create?
Since the beam energies are in the GeV, they are much higher than the rest mass of the electron
(me /Ee 1), which means we can safely neglect the mass, and Ee ≈ |pe |. In the lab frame, the
total 4-momentum is
ptot = (Ee + Ep , pe + pp )
The invariant mass M of this collision (which is the mass of the particle produced) is then
M 2 = p2tot = (Ee + Ep )2 − (pe + pp )2
= (Ee2 + Ep2 + 2Ee Ep ) − (|pe |2 + |pp |2 + 2pe · pp )
≈ (Ee2 + Ep2 + 2Ee Ep ) − (Ee2 + Ep2 − 2Ee Ep )
= 4Ee Ep
p
Therefore, the mass if the particle created is M = 2 Ee Ep = 10.6 GeV.
c) The particle created was in fact the Υ(4S), with quark content bb̄ and mass
10.6 GeV. What was the speed of the Υ in the lab frame?
The magnitude of the total 3-momentum in the lab frame is
q
|ptot | = |pe + pp | = Ee2 + Ep2 − 2Ee Ep
The momentum and energy are conserved in the reaction so that the Υ will carry the total energy/momentum as well. From simple kinematics we recall that β = p/E, so that the speed is
then
q
Ee2 + Ep2 − 2Ee Ep
|p |
= 0.39
βΥ = tot =
Etot
Ee + Ep
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Show that this situation is ideally suited to study B mesons, both neutral and
charged, by drawing diagrams for the decay of the Υ(4S) to B mesons. Explain
why the Υ(3S) would not have been useful for these studies.
We can see that the Υ(4S) can decay strongly to either a B 0 B 0 or B + B − pair.
b̄
d
B0
Υ(4S) b̄
b
b̄
u
B+
ū
b
B−
Υ(4S) b̄
b
d¯
b
B0
Figure 17: Υ(4S) → B B̄ decays
According to the Particle Data Group, these decays make up 95% of the branching ratio for Υ(4S).
As for why the Υ(3S) isn’t a good candidate for these types of decays, we can look at their masses
(mΥ(3S) = 10.4 Gev, mB = 5.27 GeV) and see that the Υ(3S) does not have enough mass to create
two B mesons! According to the PDG, it decays to the Υ(2S) and pions most of the time.
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