Going to the interaction picture in the Jaynes–Cummings model In the Jaynes–Cummings model for a two level atom, the Hamiltonian for the atom is defined as (I let $\bar{h}=1$)
$$H_a=\omega_a\frac{\sigma_z}{2}$$
and the field Hamiltonian is
$$H_f=\omega_ca^{\dagger}a.$$
The interaction Hamiltonian is 
$$V=(a+a^{\dagger})(\sigma_++\sigma_-).$$
Here $\sigma_z=\vert e\rangle\langle e\vert-\vert g\rangle\langle g\vert$ is the is the atomic inversion operator, $\sigma_+=\vert e \rangle\langle g\vert$ and $\sigma_-=\sigma_+^{\dagger}$. These are the raising and lowering operators for the atom. $a$ and $a^{\dagger}$ are the bosonic annihilation and creation operator.
Now I want to go to the interaction picture 
$$V_I=U^{\dagger}(t)VU(t).$$
Where $U(t)=e^{-i(H_a+H_f)t}$ 
Since $V$ is made up by two factors, one concerning the field, the other the atom I presume I can write
$$V=e^{i(H_f)t}(a+a^{\dagger})e^{-i(H_f)t}e^{i(H_a)t}(\sigma_++\sigma_-)e^{-i(H_at}.$$
So I would need to calculate
$$e^{i(\omega_ca^{\dagger}a)t}(a^{\dagger})e^{-i(\omega_ca^{\dagger}a)t}$$
$$e^{i(\omega_ca^{\dagger}a)t}(a)e^{-i(\omega_ca^{\dagger}a)t}$$
$$e^{i(\omega_a\frac{\sigma_z}{2})t}(\sigma_+)e^{-i(\omega_a\frac{\sigma_z}{2})t}$$
$$e^{i(\omega_a\frac{\sigma_z}{2})t}(\sigma_-)e^{-i(\omega_a\frac{\sigma_z}{2})t}$$
My problem lies here, I do not how to proceed. 
I have thought of using something like $[a,U(a^{\dagger}a)]\vert n\rangle$ but I don't get anywhere from here. 
 A: $ \def\ee{\mathrm{e}}
\def\ii{\mathrm{i}}
\def\dd{\mathrm{d}}
$
There are many ways to perform this calculation. Perhaps the simplest is to consider the object $a(t) = \ee^{\ii \omega_c t a^\dagger a } a \ee^{-\ii \omega_c t a^\dagger a }$ as the solution of the differential equation $$\dot{a}(t) = -\ii\omega_c a(t).$$ 
You can show that $a(t)$ satisfies this equation by directly differentiating it with respect to time. (You will need to use the fact that $\frac{\dd}{\dd t} \ee^{A t} = A \ee^{A t} =\ee^{A t} A,$ for an arbitrary operator $A$.) Now, it is easy to check that the solution of the above equation is
$$ a(t) = \ee^{-\ii\omega_c t}a(0).$$
Note that this property follows directly from the fact that $a$ is a lowering operator, i.e. it satisfies the commutation relation
$$ [H_f, a] = -\omega_c a.$$
(To prove this you need the identity $[AB,C] = A[B,C] + [A,C]B$ for arbitrary operators $A$, $B$ and $C$.) An analogous solution holds for the atomic operators $\sigma^{\pm}$, which are lowering operators with respect to the Hamiltonian $H_a$. 
