AdS/QCD分析：半轻度B→K1衰变与非轻度B→K1γ衰减
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"这篇论文是关于在AdS/QCD对应关系中分析半轻子和非轻子B介子衰变的。研究集中在轴向矢量介子K1(1270)和K1(1400)上，这两个介子被视为|3P1⟩和|1P1⟩态的混合，混合角度θ约为(-34±13)°。作者计算了这两种轴向矢量介子的光前分布幅度（LFDA）和衰变常数，并基于此探讨了B → K1的半轻子衰变。"
在AdS/QCD（Anti-de Sitter/Quantum Chromodynamics）对应理论中，该研究详细分析了B介子到轴向矢量介子K1的半轻子$$ B \rightarrow K_1 \ell^+ \ell^- $$ 和非轻子$$ B \rightarrow K_1 \gamma $$ 衰变过程。K1(1270)和K1(1400)被处理为混合态，它们由两个不同量子态的组合构成：|3P1⟩和|1P1⟩，混合角度θ被估计为(-34±13)°。这个混合模型对于理解和预测这些衰变的性质至关重要。
LFDA（Light-Front Distribution Amplitudes）是量子色动力学（QCD）中用于描述粒子内部结构的重要工具。在这项研究中，LFDA被用于计算B介子到K1介子的过渡形状因子，这对于理解半轻子衰变过程是必要的。形状因子与衰变常数一起，用于确定$$ B \rightarrow K_1(1270,1400) \ell^+ \ell^- $$ 跃迁的差异分支比，其中$$ \ell $$ 表示轻子，如μ子或τ子。这些衰变的过程随着四动量传递平方$$ q^2 $$ 的变化被描绘出来，这提供了衰变动力学的详细视图。
此外，论文还估算了这些半轻子衰变的分支比值，以及非轻子衰变$$ B \rightarrow K_1(1270,1400) \gamma $$ 的分支比。这些计算结果对于实验观测和理论预测的比较，以及对基本物理参数的限制具有重要意义。研究的结果可能对未来的高能物理实验，如大型强子对撞机（LHC）上的B介子衰变实验提供理论依据。
这项工作深入研究了AdS/QCD框架下的B介子衰变，通过精确计算LFDA和衰变常数，对B介子到K1的半轻子和非轻子衰变进行了详尽的理论分析，这些结果有助于进一步理解强相互作用下基本粒子的行为和相互作用。
Eur. Phys. J. C (2018) 78 :805 Page 3 of 15 805
sitions, with respect to q
2
, are compared in the AdS/QCD
correspondence and 2HDM.
2 Distribution amplitudes and decay constants in
AdS/QCD
The physical states of K
1
(1270) and K
1
(1400) mesons are
considered as a mixture of two |
3
P
1
and |
1
P
1
states and can
be parameterized in terms of a mixing angle θ
K
, as follows:
|K
1
(1270)=sin θ
K
|
3
P
1
+cos θ
K
|
1
P
1
,
|K
1
(1400)=cos θ
K
|
3
P
1
−sin θ
K
|
1
P
1
, (1)
where |
3
P
1
≡|K
1A
and |
1
P
1
≡|K
1B
have differ-
ent masses and decay constants. Also, the mixing angle θ
K
can be determined by the experimental data. There are var-
ious approaches to estimate the mixing angle. The result
35
◦
≤|θ
K
|≤55
◦
was found in Ref. [46], while two possible
solutions were obtained as |θ
K
|≈33
◦
∨57
◦
in Ref. [47] and
as |θ
K
|≈37
◦
∨58
◦
in Ref. [48]. A new window for the value
of θ
K
is estimated from the results of B → K
1
(1270)γ and
τ → K
1
(1270)ν
τ
data as [49]
θ
K
=−(34 ±13)
◦
. (2)
Sofar this value is used in Refs. [1,2,4,13,15,17]. In this
study, we also use the result of θ
K
=−(34 ±13)
◦
.
The twist-2 and twist-3 DAs for K
1
mesons are given in
termsoftheDAsofK
1A
and K
1B
states, as [3]:
K
1
(u) =C
1
f
K
1A
m
K
1A
f
K
1
m
K
1
K
1A
(u)+C
2
f
K
1B
m
K
1B
f
K
1
m
K
1
K
1B
(u),
⊥
K
1
(u) = C
1
f
⊥
K
1A
f
⊥
K
1
⊥
K
1A
(u) + C
2
f
⊥
K
1B
f
⊥
K
1
⊥
K
1B
(u),
g
(a,v)
⊥, K
1
(u) = C
1
f
K
1A
m
K
1A
f
K
1
m
K
1
g
(a,v)
⊥, K
1A
(u)
+C
2
f
K
1B
m
K
1B
f
K
1
m
K
1
g
(a,v)
⊥, K
1B
(u),
h
(t,p)
, K
1
(u) = C
1
f
⊥
K
1A
m
2
K
1A
f
⊥
K
1
m
2
K
1
h
(t,p)
, K
1A
(u)
+C
2
f
⊥
K
1B
m
2
K
1B
f
⊥
K
1
m
2
K
1
h
(t,p)
, K
1B
(u), (3)
where (C
1
, C
2
) = (sin θ
K
, cos θ
K
) for K
1
(1270) meson, and
(C
1
, C
2
) = (cos θ
K
, −sin θ
K
) for K
1
(1400). Here,
,⊥
are
the twist-2 DAs while, g
(a,v)
⊥
and h
(t, p)
are of twist-3. In this
phrases, u refer to the momentum fraction carried by the
quark in K
1
. In addition, f
K
1
and f
⊥
K
1
are decay constants,
written in terms of f
K
1A(1B)
and f
⊥
K
1A(1B)
as
f
K
1
= C
1
m
K
1A
m
K
1
(1270)
f
K
1A
+ C
2
m
K
1B
m
K
1
(1270)
a
,K
1B
0
f
K
1B
,
f
⊥
K
1
= C
1
a
⊥,K
1A
0
f
⊥
K
1A
+ C
2
f
⊥
K
1B
, (4)
where a
⊥,K
1A
0
and a
,K
1B
0
are G-parity invariant Gegenbauer
moments for K
1A
and K
1B
states which have been estimated
in Ref. [3].
First, we aim to calculate the twist-2 and twist-3 DAs for
K
1
mesons in the AdS/QCD correspondence. According to
Eq. (3), we need to investigate the DAs for two states K
1A
and
K
1B
in terms of the holographic LFWFs. In order to consider
the twist-2 and twist-3 DAs, the matrix elements of K
1A
and
K
1B
states should be considered. For instance, the following
two-particle matrix elements of state K
1A
in the light-front
coordinate, x
μ
= (x
+
, x
−
, x
⊥
), at equal light-front time x
+
,
up to twist-3 DAs are written as:
0|¯u(0)γ
μ
γ
5
s(x
−
)|K
1A
( p,λ)
= f
K
1A
m
K
1A
1
0
du e
−iup
+
x
−
p
μ
ε
λ
.x
p
+
x
−
K
1A
(u,μ)
+(ε
μ
λ
− p
μ
ε
λ
.x
p
+
x
−
)g
(a)
⊥,K
1A
+···
, (5)
0|¯u(0)[γ
μ
,γ
ν
]γ
5
s(x
−
)|K
1A
( p,λ)
= 2 f
⊥
K
1A
1
0
du e
−iup
+
x
−
[ε
μ
λ
, p
ν
]
⊥
K
1A
(u,μ)
+m
2
K
1A
ε
λ
.x
( p
+
x
−
)
2
[p
μ
, x
ν
]h
(t)
,K
1A
+···
, (6)
0|¯u(0)γ
μ
s(x
−
)|K
1A
( p,λ)
=−f
K
1A
m
K
1A
(
μ
νρσ
ε
ν
λ
p
ρ
x
σ
)
1
0
du e
−iup
+
x
−
g
(v)
⊥,K
1A
(u,μ)
4
,
(7)
0|¯u(0)γ
5
s(x
−
)|K
1A
( p,λ)
= if
⊥
K
1A
m
2
K
1A
(ε
λ
.x )
1
0
du e
−iup
+
x
−
h
( p)
,K
1A
(u,μ)
2
,
(8)
where γ
μ
= (γ
+
,γ
−
,γ
1
,γ
2
).The“···” describes the con-
tributions coming from higher twist DAs. In these relations,
p
+
is the “plus” component of the four-momentum of the
K
1A
state given by p
μ
=
p
+
,
m
2
K
1A
p
+
, 0
⊥
. The polariza-
tion vectors ε
λ
(λ = L, T ) for K
1A
state are chosen as
ε
L
=
p
+
m
K
1A
,
m
K
1A
p
+
, 0
⊥
, and ε
T (±)
=
1
√
2
(
0, 0, 1, ±i
)
.
Taking λ = L and μ =+in Eq. (5), in addition, the scalar
product of Eq. (6)in(ε
∗
T
)
μ
, we obtain:
0|¯u(0)γ
+
γ
5
s(x
−
)|K
1A
( p, L)
= f
K
1A
m
K
1A
1
0
du e
−iup
+
x
−
p
+
(
ε
L
.x
p
+
x
−
)
K
1A
(u,μ)
,
(9)
0|¯u(0)[γ.ε
∗
T
,γ
+
]γ
5
s(x
−
)|K
1A
( p, ±)
= 2 f
⊥
K
1A
1
0
du e
−iup
+
x
−
p
+
(ε
1
T
∓ iε
2
T
)
⊥
K
1A
(u,μ)
,
(10)
where γ.ε
∗
T
is placed instead of γ
1
∓iγ
2
. Applying the Fourier
transform of the above matrix elements with respect to the
longitudinal distance x
−
, the twist-2 DAs are given by:
123
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