$$a_1(1260)$$-meson longitudinal twist-2 distribution amplitude and the $$D\rightarrow a_1(1260)\ell ^+\nu _\ell $$ decay processes
Authors: Hu Dan-Dan , Fu Hai-Bing , Zhong Tao , Wu Zai-Hui , Wu Xing-Gang 
Published: July 2022

In the paper, we investigate the moments $$\langle \xi _{2;a_1}^{\Vert ;n}\rangle $$

⟨

ξ

2
;

a
1

‖
;
n

⟩

of the axial-vector $$a_1(1260)$$

a
1

(
1260
)

-meson distribution amplitude by using the QCD sum rules approach under the background field theory. By considering the vacuum condensates up to dimension-six and the perturbative part up to next-to-leading order QCD corrections, its first five moments at an initial scale $$\mu _0=1~{\mathrm{GeV}}$$

μ
0

=
1

GeV

are $$\langle \xi _{2;a_1}^{\Vert ;2}\rangle |_{\mu _0} = 0.223 \pm 0.029$$

⟨

ξ

2
;

a
1

‖
;
2

⟩

|

μ
0

=
0.223
±
0.029

, $$\langle \xi _{2;a_1}^{\Vert ;4}\rangle |_{\mu _0} = 0.098 \pm 0.008$$

⟨

ξ

2
;

a
1

‖
;
4

⟩

|

μ
0

=
0.098
±
0.008

, $$\langle \xi _{2;a_1}^{\Vert ;6}\rangle |_{\mu _0} = 0.056 \pm 0.006$$

⟨

ξ

2
;

a
1

‖
;
6

⟩

|

μ
0

=
0.056
±
0.006

, $$\langle \xi _{2;a_1}^{\Vert ;8}\rangle |_{\mu _0} = 0.039 \pm 0.004$$

⟨

ξ

2
;

a
1

‖
;
8

⟩

|

μ
0

=
0.039
±
0.004

and $$\langle \xi _{2;a_1}^{\Vert ;10}\rangle |_{\mu _0} = 0.028 \pm 0.003$$

⟨

ξ

2
;

a
1

‖
;
10

⟩

|

μ
0

=
0.028
±
0.003

, respectively. We then construct a light-cone harmonic oscillator model for $$a_1(1260)$$

a
1

(
1260
)

-meson longitudinal twist-2 distribution amplitude $$\phi _{2;a_1}^{\Vert }(x,\mu )$$

ϕ

2
;

a
1

‖

(
x
,
μ
)

, whose model parameters are fitted by using the least squares method. As an application of $$\phi _{2;a_1}^{\Vert }(x,\mu )$$

ϕ

2
;

a
1

‖

(
x
,
μ
)

, we calculate the transition form factors (TFFs) of $$D\rightarrow a_1(1260)$$

D
→

a
1

(
1260
)

in large and intermediate momentum transfers by using the QCD light-cone sum rules approach. At the largest recoil point ($$q^2=0$$

q
2

=
0

), we obtain $$ A(0) = 0.130_{ - 0.013}^{ + 0.015}$$

A

(
0
)

=
0
.

130

-
0.013

+
0.015

, $$V_1(0) = 1.898_{-0.121}^{+0.128}$$

V
1

(
0
)

=
1
.

898

-
0.121

+
0.128

, $$V_2(0) = 0.228_{-0.021}^{ + 0.020}$$

V
2

(
0
)

=
0
.

228

-
0.021

+
0.020

, and $$V_0(0) = 0.217_{ - 0.025}^{ + 0.023}$$

V
0

(
0
)

=
0
.

217

-
0.025

+
0.023

. By applying the extrapolated TFFs to the semi-leptonic decay $$D^{0(+)} \rightarrow a_1^{-(0)}(1260)\ell ^+\nu _\ell $$

D

0
(
+
)

→

a
1

-
(
0
)

(
1260
)

ℓ
+

ν
ℓ

, we obtain $${\mathcal {B}}(D^0\rightarrow a_1^-(1260) e^+\nu _e) = (5.261_{-0.639}^{+0.745}) \times 10^{-5}$$

B

(

D
0

→

a
1
-

(
1260
)

e
+

ν
e

)

=

(
5
.

261

-
0.639

+
0.745

)

×

10

-
5

, $${\mathcal {B}}(D^+\rightarrow a_1^0(1260) e^+\nu _e) = (6.673_{-0.811}^{+0.947}) \times 10^{-5}$$

B

(

D
+

→

a
1
0

(
1260
)

e
+

ν
e

)

=

(
6
.

673

-
0.811

+
0.947

)

×

10

-
5

, $${\mathcal {B}}(D^0\rightarrow a_1^-(1260) \mu ^+ \nu _\mu )=(4.732_{-0.590}^{+0.685}) \times 10^{-5}$$

B

(

D
0

→

a
1
-

(
1260
)

μ
+

ν
μ

)

=

(
4
.

732

-
0.590

+
0.685

)

×

10

-
5

, $${\mathcal {B}}(D^+ \rightarrow a_1^0(1260) \mu ^+ \nu _\mu )=(6.002_{-0.748}^{+0.796}) \times 10^{-5}$$

B

(

D
+

→

a
1
0

(
1260
)

μ
+

ν
μ

)

=

(
6
.

002

-
0.748

+
0.796

)

×

10

-
5

.