Angular Distribution of Two-Body Decay
Let's begin with a general amplitude for the two-body decay of a state with angular momentum quantum numbers J ,m . Specifically, we want to know the amplitude of this state having daughter 1 with trajectory
Ω
=
(
ϕ
,
θ
)
{\displaystyle \Omega =(\phi ,\theta )}
.
We can also describe the angular momentum between the daughters as being L and spin sum as s . Alternatively, we will label the daughters as having helicities of
λ
1
{\displaystyle \lambda _{1}}
and
λ
2
{\displaystyle \lambda _{2}}
- projections on the direction of decay (specified by daughter 1)
⟨
Ω
λ
1
λ
2
|
U
|
J
m
⟩
=
∑
L
,
S
⟨
Ω
λ
1
λ
2
|
J
m
λ
1
λ
2
⟩
⟨
J
m
λ
1
λ
2
|
J
m
L
S
⟩
⟨
J
m
L
S
|
U
|
J
m
⟩
{\displaystyle \langle \Omega \lambda _{1}\lambda _{2}|U|Jm\rangle =\sum _{L,S}\langle \Omega \lambda _{1}\lambda _{2}|Jm\lambda _{1}\lambda _{2}\rangle \langle Jm\lambda _{1}\lambda _{2}|JmLS\rangle \langle JmLS|U|Jm\rangle }
simple insertion of complete sets of states for recoupling
=
∑
L
,
S
[
2
J
+
1
4
π
D
m
λ
J
∗
(
Ω
,
0
)
]
[
2
L
+
1
2
J
+
1
(
L
S
J
0
λ
λ
)
(
S
1
S
2
S
λ
1
−
λ
2
λ
)
]
a
L
S
J
{\displaystyle =\sum _{L,S}\left[{\sqrt {\frac {2J+1}{4\pi }}}D_{m\lambda }^{J*}(\Omega ,0)\right]\left[{\sqrt {\frac {2L+1}{2J+1}}}\left({\begin{array}{cc|c}L&S&J\\0&\lambda &\lambda \end{array}}\right)\left({\begin{array}{cc|c}S_{1}&S_{2}&S\\\lambda _{1}&-\lambda _{2}&\lambda \end{array}}\right)\right]a_{LS}^{J}}
Substitution of each bra-ket with their respective formulae.
λ
=
λ
1
−
λ
2
{\displaystyle \lambda =\lambda _{1}-\lambda _{2}}
Note that in the event of one daughter being spin-less, the second
Clebsch-Gordan coefficient is 1
Isospin Projections
One must also take into account the various ways isospin of daughters can add up to the isospin quantum numbers of the parent, requiring a term:
C
a
,
b
=
(
I
a
I
b
I
I
z
a
I
z
b
I
z
a
+
I
z
b
)
{\displaystyle C^{a,b}=\left({\begin{array}{cc|c}I^{a}&I^{b}&I\\I_{z}^{a}&I_{z}^{b}&I_{z}^{a}+I_{z}^{b}\end{array}}\right)}
where a=1 and b=2 , referring to the daughter number. Because an even-symmetric angular wave function (i.e. L=0,2... ) imply that 180 degree rotation is equivalent to reversal of daughter identities (a,b becoming b,a ) one must write down the symmetrized expression:
C
(
L
)
=
1
2
[
C
a
,
b
+
(
−
1
)
L
C
b
,
a
]
{\displaystyle C(L)={\frac {1}{\sqrt {2}}}\left[C^{a,b}+(-1)^{L}C^{b,a}\right]}
⟨
Ω
X
0
λ
b
1
|
U
X
|
J
X
m
X
⟩
=
∑
L
X
[
2
J
X
+
1
4
π
D
m
X
λ
b
1
J
X
∗
(
Ω
X
,
0
)
]
[
2
L
X
+
1
2
J
X
+
1
(
L
X
1
J
X
0
λ
b
1
λ
b
1
)
]
a
L
X
J
X
{\displaystyle \langle \Omega _{X}0\lambda _{b_{1}}|U_{X}|J_{X}m_{X}\rangle =\sum _{L_{X}}\left[{\sqrt {\frac {2J_{X}+1}{4\pi }}}D_{m_{X}\lambda _{b_{1}}}^{J_{X}*}(\Omega _{X},0)\right]\left[{\sqrt {\frac {2L_{X}+1}{2J_{X}+1}}}\left({\begin{array}{cc|c}L_{X}&1&J_{X}\\0&\lambda _{b_{1}}&\lambda _{b_{1}}\end{array}}\right)\right]a_{L_{X}}^{J_{X}}}
⟨
Ω
b
1
0
λ
ω
|
U
b
1
|
1
,
m
b
1
=
λ
b
1
⟩
=
∑
L
b
1
[
2
J
b
1
+
1
4
π
D
m
b
1
=
λ
b
1
λ
ω
1
∗
(
Ω
b
1
,
0
)
]
[
2
L
b
1
+
1
2
J
b
1
+
1
(
L
b
1
1
1
0
λ
ω
λ
ω
)
]
b
L
b
1
{\displaystyle \langle \Omega _{b_{1}}0\lambda _{\omega }|U_{b_{1}}|1,m_{b_{1}}=\lambda _{b_{1}}\rangle =\sum _{L_{b_{1}}}\left[{\sqrt {\frac {2J_{b_{1}}+1}{4\pi }}}D_{m_{b_{1}}=\lambda _{b_{1}}\lambda _{\omega }}^{1*}(\Omega _{b_{1}},0)\right]\left[{\sqrt {\frac {2L_{b_{1}}+1}{2J_{b_{1}}+1}}}\left({\begin{array}{cc|c}L_{b_{1}}&1&1\\0&\lambda _{\omega }&\lambda _{\omega }\end{array}}\right)\right]b_{L_{b_{1}}}}
⟨
Ω
ω
0
λ
ρ
|
U
ω
|
1
,
m
ω
=
λ
ω
⟩
=
∑
L
ω
J
ρ
[
2
J
ω
+
1
4
π
D
m
ω
=
λ
ω
λ
ρ
1
∗
(
Ω
ω
,
0
)
]
[
2
L
ω
+
1
2
J
ω
+
1
(
L
ω
1
1
0
λ
ρ
λ
ρ
)
]
c
L
ω
J
ρ
{\displaystyle \langle \Omega _{\omega }0\lambda _{\rho }|U_{\omega }|1,m_{\omega }=\lambda _{\omega }\rangle =\sum _{L_{\omega }J_{\rho }}\left[{\sqrt {\frac {2J_{\omega }+1}{4\pi }}}D_{m_{\omega }=\lambda _{\omega }\lambda _{\rho }}^{1*}(\Omega _{\omega },0)\right]\left[{\sqrt {\frac {2L_{\omega }+1}{2J_{\omega }+1}}}\left({\begin{array}{cc|c}L_{\omega }&1&1\\0&\lambda _{\rho }&\lambda _{\rho }\end{array}}\right)\right]c_{L_{\omega }J_{\rho }}}
⟨
Ω
ρ
0
λ
ρ
|
U
ρ
|
J
ρ
,
m
ρ
=
λ
ρ
⟩
=
∑
L
ρ
[
2
J
ρ
+
1
4
π
D
m
ρ
0
J
ρ
∗
(
Ω
ρ
,
0
)
]
[
2
L
ρ
+
1
2
J
ρ
+
1
(
L
ρ
0
J
ρ
0
0
0
)
]
d
L
ρ
=
∑
L
ρ
2
L
ρ
+
1
4
π
Y
m
ρ
J
ρ
∗
(
Ω
ρ
)
d
L
ρ
{\displaystyle \langle \Omega _{\rho }0\lambda _{\rho }|U_{\rho }|J_{\rho },m_{\rho }=\lambda _{\rho }\rangle =\sum _{L_{\rho }}\left[{\sqrt {\frac {2J_{\rho }+1}{4\pi }}}D_{m_{\rho }0}^{J_{\rho }*}(\Omega _{\rho },0)\right]\left[{\sqrt {\frac {2L_{\rho }+1}{2J_{\rho }+1}}}\left({\begin{array}{cc|c}L_{\rho }&0&J_{\rho }\\0&0&0\end{array}}\right)\right]d_{L_{\rho }}=\sum _{L_{\rho }}{\sqrt {\frac {2L_{\rho }+1}{4\pi }}}Y_{m_{\rho }}^{J_{\rho }*}(\Omega _{\rho })d_{L_{\rho }}}
A
J
X
=
∑
λ
b
1
,
λ
ω
,
λ
ρ
⟨
Ω
X
0
λ
b
1
|
U
|
J
X
m
X
⟩
C
X
(
L
X
)
k
L
X
⟨
Ω
b
1
0
λ
ω
|
U
|
1
,
m
b
1
=
λ
b
1
⟩
C
b
1
(
L
b
1
)
q
L
b
1
⟨
Ω
ω
0
λ
ρ
|
U
|
1
,
m
ω
=
λ
ω
⟩
C
ω
(
L
ω
)
u
L
ω
⟨
Ω
ρ
0
λ
ρ
|
U
|
J
ρ
,
m
ρ
=
λ
ρ
⟩
C
ρ
(
L
ρ
)
v
L
ρ
{\displaystyle A^{J_{X}}=\sum _{\lambda _{b_{1}},\lambda _{\omega },\lambda _{\rho }}\langle \Omega _{X}0\lambda _{b_{1}}|U|J_{X}m_{X}\rangle C_{X}(L_{X})k^{L_{X}}\langle \Omega _{b_{1}}0\lambda _{\omega }|U|1,m_{b_{1}}=\lambda _{b_{1}}\rangle C_{b_{1}}(L_{b_{1}})q^{L_{b_{1}}}\langle \Omega _{\omega }0\lambda _{\rho }|U|1,m_{\omega }=\lambda _{\omega }\rangle C_{\omega }(L_{\omega })u^{L_{\omega }}\langle \Omega _{\rho }0\lambda _{\rho }|U|J_{\rho },m_{\rho }=\lambda _{\rho }\rangle C_{\rho }(L_{\rho })v^{L_{\rho }}}