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
.