Lecture 5: Phase space and decay rates
9
Lorentz invariant phase space
Particle physics calculations are occupied with transitions from one state to another (e.g. production cross sections, decay
rates). Transitions to a final state | f i from an initial state | ii are calculated from Fermi’s Golden Rule:
2
Γ f i = ~W f i = 2π T f i ρ(E f )
| {z } |{z}
dynamics
where:
(1)
kinematics
• Γ f i is the decay width to the final state in question;
• W f i is its transition rate: the number of transitions per unit time;
• T f i is the matrix element describing the dynamics (fundamental coupling, spin statistics...) of the transition:
T f i = hφ f | V | φi i;
(2)
where φi and φ f are the initial and final wavefunctions and describe m and n particles in some potential V.
• ρ(E f ) is the number of states available per unit energy in the final state: the phase-space factor.
To study the fundamental process T f i we need to be able to calculate the phase space factor ρ(E f ).
Dirac δ-functions
Dirac δ-functions prove to be very useful tool in relativistic mechanics as they can be used to neatly encode
conversation of energy and momentum. e.g. in a decay of particle i to particles 1 and 2, one can write:
Z
δ(Ei − E1 − E2 ) dE
Z
and
δ( p~i − p~1 − p~2 ) d3 p
Important identities relating to these functions are:
Z
Z
+∞
+∞
δ(x − a) dx
=
1
f (x)δ(x − a) dx
=
f (a)
δ( f (x))
=
−1
d f δ(x − a)
dx a
−∞
−∞
1
(3)
Phase space
This is the number of states available per unit of energy in the final state. Let’s first imagine a cube of sides L containing
one of the final-state particles with quantised momentum, p = ~k (though take ~ = 1) and k = n 2π/L , n ∈ I.
2πn x
L
Lp x
nx =
2π
px =
or,
2πny
L
Lpy
ny =
2π
2πnz
L
Lpz
nz =
2π
py =
pz =
(4)
Each p x , py , pz momentum state resides in the elemental volume (2π/L)3 = (2π)3 /V in momentum space, so that:
total phase space =
(2π)3
N1
V
where N1 is the number of momentum states available to one particle. Now let’s normalise to one particle per spatial
elemental volume and rewrite “total phase space” more mathematically as the integral over all momenta:
N1
=
,
Z
1
d p x d py d pz
(2π)3
V
1
(2π)3
=
V
Z
d3p
When we scale up to n particles, and because total momentum conservation in the transition will constrain the nth particle,
the number of available states is that of n − 1 free particles:
Nn−1 =
1
(2π)3(n−1)
Z Y
n−1
d3p j
j=1
From which the density of states, the number states per unit energy, is:
ρ(E) =
dNn−1
d
1
=
3(n−1)
dE
dE
(2π)
Z Y
n−1
d3 p j
(5)
j=1
Let’s re-express using Dirac δ-functions for the momentum conversation:


n−1
X



~
~p j 
~pn − P −
=
0
j=1



n−1
X





~
~p j  =
d pn δ  ~pn − P −
j=1


Z
n
X


similarly :
dE δ E −
E j  =
Z
3
1
(6)
1
(7)
j=1
ρ(E) =
=

 Z


Z Y
Z
n−1
n
n
X
X




d
1
3
3
~−
~p j  × dE δ E −
d p j × d pn δ P
E j 
3(n−1)
dE
(2π)
j=1
j=1
j=1

 

Z Y
n
n
n
3
X  
X 
d p j 
~−
~p j  δ E −
δ P
E j 
(2π)3
3
(2π)
j=1
j=1
j=1
2
(8)
Imposing Lorentz invariance
Let’s go back to our particle with wavefunction φ and energy E in volume V.
Z
| φ |2 dV = 1
This normalisation implies density = 1/V. At relativistic speeds, Lorentz contraction in direction of motion will increase
the particle density by γ/V. The phase space derivation we have used so far is manifestly not Lorentz invariant.
√
So require a wavefunction ψ that has normalisation ∝ γ. The usual convention is to 2E particles/unit-volume:
Z
| ψ |2 dV
=
2E
(9)
1
ψ
√
2E
φ =
So redefine the matrix element to accommodate the Lorentz invariant wavefunctions:
| T f i |2 =
m
n
Y
1 Y 1
| M f i |2
2E
2E
i
j
i=1
j=1
where
M f i = h ψ f | V f i | ψi i,
(10)
where there are m particles in the initial state.
So the Golden Rule becomes:


 
Z
n
n
m
n
X
X
Y
Y


 
d3 p j
1
2




~



~
E
−
E
δ
δ
P
−
p
Γ f i = (2π)
| Mfi |



j
j

 
3 (2E ) 
2E
(2π)
i
j
j=1
j=1
i=1
j=1
|
{z
}
Lorentz invariant phase space
4
Note: The transition rate is inversely proportional to energy of the initial state.
• This is expected in the case of a decaying particle whose decay rate is slowed by time dilation.
• The 1/(2Ei ) terms drop out the integral because the integration is over the final state momenta.
Checking the Lorentz invariance of the Lorentz invariant phase space
Let’s check this by, arbitrarily, choosing boost in z
p0z = γ(pz − βE)
E 0 = γ(E − βpz )
Differentiate E with respect to pz :
dE
d
=
d pz d pz

 12
 X

2
2


p
+
m
i



− 21
 X

pz
pz 
p2i + m2  =
E
i=xyz
=
i=xyz
Differentiate pz 0 with respect to pz and substitute in the above.
d pz 0
d pz
d pz 0
∴
E0
=
=
d
dE
γ(pz − βE) = γ 1 − β
d pz
d pz
d pz
E
3
!
pz γ(E − βpz ) E0
= γ 1−β
=
=
E
E
E
(11)
10
Decay rates
In a 2-body decay, the number of particles in the initial state is 1 and 2 in the final state: m = 1, n = 2. Then, as the final
~ = 0 , E i = mi .
state integral is Lorentz invariant, we are free to choose any frame! Choose CM frame: P
Γfi
(2π)4
2mi
=
Z
| M f i |2
d3 p1
d3 p2
δ ~p1 + ~p2 δ [mi − E1 − E2 ]
(2π)3 (2E1 ) (2π)3 (2E2 )
Gather constants and use one of the δ functions to remove one of the integrals:
Γfi
1
32π2 mi
=
Let’s switch to a polar coordinate system, d3 p
Γfi
=
7→
1
32π2 mi
Z
| M f i |2
d3 p1
δ [mi − E1 − E2 ]
E1 E2
p21 d p1 sin θ dθ dφ
Z
| M f i |2
7→
p21 d p1 dΩ:
p21 d p1 dΩ
δ [mi − E1 − E2 ]
E1 E2
Observing that E22 = m22 + p21 because p2 = −p1 , eq. 12 can be expressed as:
Γfi
=
where g(p1 ) =
and f (p1 ) =
Z
1
| M f i |2 g(p1 ) δ( f (p1 )) d p1 dΩ
32π2 mi
p21 /(E1 E2 )
q
q
mi − (m21 + p21 ) − (m22 + p21 )
Using eq. 3,
d f −1
δ(p1 − p∗ )
δ( f (p1 )) = d p 1 p∗
where p∗ denotes the value of p1 that conserves momentum, g(p1 ) can be easily integrated:
Z
Z
d f −1
g(p1 ) δ( f (p1 )) d p1 = g(p1 ) δ(p1 − p∗ ) d p1
d p1 p∗
|
{z
}
g(p∗ )
d f −1 (p∗ )2
= d p1 p∗ E1 E2
The inverted modulus term is:
p1
p1
p1
p1
E1 + E2
= −q
− q
=−
−
= −p1
E1 E2
E1 E2
m21 + p21
m22 + p21
df
d p1
−1
d f d p1 ∗
p
1 E1 E2
p∗ E1 + E2
=
Giving:
Γfi
=
1
E1 E2
(p∗ )2
32π2 mi p∗ (E1 + E2 ) E1 E2
4
Z
| M f i |2 dΩ
(12)
And as E1 + E2 = mi by energy conservation:
Γfi =
|p∗ |
32π2 m2i
Z
| M f i |2 dΩ
(13)
Finally p∗ can be obtained from the original condition on the δ-function, f (p1 ) = 0:
mi
=
mi
=
q
q
m21 + p∗2 + m22 + p∗2
q
E1 + m22 + E12 − m21
2
=
E12 + m22 − m21
−2E1 mi
=
−m2i + m22 − m21
(2mi )2 p∗2
=
p∗
=
(m2i + m21 − m22 )2 − 4m2i m21
q
1
[m2i − (m1 + m2 )2 ][m2i − (m1 − m2 )2 ]
2mi
(mi − E1 )
(14)
Example: W+ → e+νe
The [leading order] matrix element for this decay, as calculated by the SM is
h
i2
|M f i |2 = mW g 12 (1 − cos θ)
where mW = 80.4 GeV is the mass of the W boson and g is the weak coupling constant which is predicted to be gSM = 0.65.
1
1
The angular dependence; this is from the rotation matrix d−1,1
= 12 (1 − cos θ). For completeness, note that d1,1
does not
feature because unlike the photon, weak bosons couple to only one helicity amplitude.
Eq. 14 simplified when one realised that the mass of both decay products are negligible compared to the large W mass,
p∗
=
p∗
≈
1
2mW
mW
2
q
[m2W − (me + mν )2 ][m2W − (me − mν )2 ]
So using the main result from Eq. 13,
ΓW+→e+νe
=
=
=
=
Z h
i2
mW /2
mW g 21 (1 − cos θ) dΩ
2
2
32π mW
Z
mW g 2
(1 − cos θ)2 dΩ
256π2
Z 1
mW g 2
(−2π)
(1 − cos θ)2 d(cos θ)
256π2
−1
i1
mW g 2 h 1
− 3 (1 − cos θ)3
128π |
{z cos θ=−1
}
8/3
=
2
mW g
48π
(15)
The partial width of the electronic W decay was measured at LEP from total width and the branching fraction to be
0.228 GeV. Hence the weak coupling constant is obtained (gmeas = 0.65) just from the knowledge of two-body phasespace.
5
Phase space of 3-body decays
Though the calculation of the decay rate is more involved than in the two body case, an interesting observation can be
made about the phase space for three-body decays. For this purpose we consider the infinitesimal of the 3-body phase
space, dρ3 (i.e. before multiplying by the matrix element) and strip out the constant factors:
dρ3 ∝
d3 p1 d3 p2 d3 p3
δ ~p1 + ~p2 + ~p3 δ (E1 + E2 + E3 − M)
E1 E2 E3
(16)
• As in the 2-body case, one of the integrals over d3 pi is removed with the momentum δ-function.
dρ3 ∝
d 3 p1 d 3 p2
δ (E1 + E2 + E3 − M)
E1 E2 E3
(17)
• Change to spherical co-ordinates:
=
d3 p1 d3 p2
d p1 p1 dθ1 p1 sin θ1 dφ1 d p2 p2 dθ2 p2 sin θ2 dφ2
(18)
• Redefine the solid angle: dΩ1 dΩ2 = sin θ1 dθ1 sin θ2 dθ2 dφ1 dφ2 = sin θ1 dθ1 sin θ12 dθ12 dφ1 dφ2 :
=
d3 p1 d3 p2
p21 p22 d p1 d p2 sin θ1 dθ1 sin θ12 dθ12 dφ1 dφ2
(19)
And simplify noting that one part of the integration is trivial:
Z
3
3
d p1 d p2
=
Z
2
8π
p21 p22 d p1 d p2 sin θ12 dθ12
(20)
• Consider the (momentum)2 of the back-to-back systems, p3 vs. (p1 + p2 ):
p23
=
(p1 + p2 )2 = p21 + p21 + 2p1 p2 cos θ12
(21)
• For a given | p~1 | and | p~2 |, p3 depends only on θ12 :
2p3 d p3
=
sin θ12 dθ12
∝
−2p1 p2 sin θ12 dθ12
p3
d p3
p1 p2
(22)
Putting this decomposition into the infinitesimal phase space expression:
dρ3
∝
∝
1
p3
p2 p2 d p1 d p2
d p3 δ (E1 + E2 + E3 − M)
E1 E2 E3 1 2
p1 p2
p1 p2 p3
d p1 d p2 d p3 δ (E1 + E2 + E3 − M)
E1 E2 E3
(23)
• Perform a variable change: pi d pi = Ei dEi and remove one term using the δ-function.
(((
dρ3 ∝ dE1 dE
δ (E
+(
E2(
+(
E3 − M)
2 dE 3 (
(1(
6
(24)
We start by writing the Lorentz sum of m1 and m2 explicitly in terms of E3 and p3 :
2
M12
And identify m23 :
2
M12
=
(E1 + E2 )2 − (p1 + p2 )2
=
(M − E3 )2 − p23
=
M 2 − 2ME3 + E32 − p23
=
M 2 + m23 − 2ME3
(25)
And differentiating:
2
d(M12
) ∝
dE3
2
d(M23
) ∝
dE1
(26)
And by a similar reasoning:
Finally we plug these proportionality relationships back into Eq. 24 to obtain
2
2
dρ3 ∝ d(M12
) d(M13
)
(27)
Phase space available in a 3-body decay is proportional to the product of the invariant mass (squared) of decay products
(1 + 2) and (1 + 3) (or (2 + 3)). So what?
2
2
2
• phase space is uniform across the 2D distribution of M12
vs. M13
(or vs. M23
).
• the “dynamics”, like angular momentum dependence in the matrix element show-up as structure on the 2D plane.
Such distributions are called Dalitz plots after their first proponent, Prof. Dick Dalitz (Oxford).
The following data is from the Babar experiment at Stanford, CA. It is the Dalitz plot of the decay,
D →K π π
0
± ∓ 0
 
 
 
 c 
 s 
u 
−
+
where D =   , K =   , π =   ,
u
u
d
0
The sharp boundary to the Dalitz plot is defined by the total mass-energy available in the centre-of-mass frame, i.e. the
mass of the decaying D0 meson. As we have just shown, If the three-body decay occurred with no angular preferences
coming from the matrix element, then the distribution of events across the Dalitz plot would be flat (i.e. a uniform colour
in the plot). Therefore one can analysis this decay by inspecting (e.g. performing a fit to) the structure across this plot.
For example, the three dominant spin-1 vector resonances show up as bands in the plot. K ∗− → K − π0 can be seen
as a vertical band, ρ+ → π+ π0 is a wide horizontal band and K ∗0 → K − π+ is evident as a diagonal band. Perhaps
most interestingly, because these intermediate resonances are decaying to the same, indistinguishable final state, their
amplitudes can interfere in the regions where the bands overlap.
7
102
BABAR
230 fb-1 preliminary
1.5
10
1
1
0.5
Wtd. Cnd./0.025 [GeV/c2 ]2
0
1
2
3
BABAR
15000
230 fb-1 preliminary
10000
5000
0
1
2
2
3
− 0
2 2
m (K π ) [GeV/c ]
8
Weighted Candidates
m2 (π + π 0 ) [GeV/c2 ]2
2