Sinusoidaly Driven Two-Level System (TLS) I'm trying to solve the driven Two-Level System (TLS or qubit) question using a Fourier transform of the Schrodinger equation (SHE), but I'm getting stuck on solving the equation. 
Given Hamiltonian
 $$H=\left(
\begin{array}{cc}
 -\frac{\omega _0}{2} & \frac{1}{2} V e^{i t \omega _D} \\
 \frac{1}{2} V e^{-i t \omega _D} & \frac{\omega _0}{2} \\
\end{array}
\right)$$
and plugging into SHE:
$$i\frac{ d }{\text{dt}}\left(
\begin{array}{c}
  c_a(t) \\
  c_b(t) \\
\end{array}
\right)=H \left(
\begin{array}{c}
  c_a(t) \\
  c_b(t) \\
\end{array}
\right)$$
I get a set of two coupled 1st Order differential equations which i then Fourier transform and use the derivative rule and the shift rule to get:
\left(
\begin{array}{c}
 -\omega  c_a(\omega )=\frac{1}{2} V c_b\left(\omega -\omega _D\right)-\frac{1}{2} \omega _0 c_a(\omega ) \\
 -\omega  c_b(\omega )=\frac{1}{2} V c_a\left(\omega +\omega _D\right)+\frac{1}{2} \omega _0 c_b(\omega ) \\
\end{array}
\right)
If I shift the $c_b$ equation:
$$-\omega  c_b\left(\omega -\omega _D\right)=\frac{1}{2} V c_a\left(\omega _D-\omega _D+\omega \right)++\frac{1}{2} \omega _0 c_b\left(\omega -\omega _D\right)$$
and solve for term i can plug back into the first equation:
$$c_b\left(\omega -\omega _D\right)=\frac{V c_a(\omega )}{2 \omega +\omega _0}$$
and solve for $c_a$:
$$c_a(\omega )=\frac{V^2 c_a(\omega )}{(2 \omega )^2+\omega _0^2} $$
So it looks like a lorentzian function with a width of the resonance frequency and centered at 0 multiplies $c_a$, but $c_a$ cancels unless it is 0? or some delta function?
This is where I do not understand how to proceed to solve the problem further?
Is part of the challenge that I am solving the problem without boundary conditions, just trying to find the steady state?
 A: The solution is to realize that the steady-state solution of a harmonically driven system must also oscillate harmonically. (As regards your solution, this means that spectrally the $c_i(\omega)$ are delta functions, which resolves your contradiction.)
Thus one usually begins by postulating the oscillatory Ansatz
$$c_a(t)=c_a e^{-i\omega_a t},\,\, c_b(t)=c_b e^{-i\omega_b t},$$
where the $c_a$ and $c_b$ are now constants. As an Ansatz this is harmless and if it turns out to not be a solution you can drop it (but as it happens it will). Your Schrödinger equation thus reads
$$
\begin{pmatrix}
 -\frac{\omega _0}{2} & \frac{1}{2} V e^{i t \omega _D} \\
 \frac{1}{2} V e^{-i t \omega _D} & \frac{\omega _0}{2} 
\end{pmatrix}
\begin{pmatrix}
  c_a(t) \\
  c_b(t) 
\end{pmatrix}
=
i\frac d{dt}
\begin{pmatrix}
  c_a(t) \\
  c_b(t) 
\end{pmatrix}
$$
so
$$
\begin{pmatrix}
 -\frac{\omega _0}{2} & \frac{1}{2} V e^{i t \omega _D} \\
 \frac{1}{2} V e^{-i t \omega _D} & \frac{\omega _0}{2} 
\end{pmatrix}
\begin{pmatrix}
  c_a  e^{-i\omega_a t}\\
  c_b  e^{-i\omega_b t} 
\end{pmatrix}
=
\begin{pmatrix}
 \omega_a c_a  e^{-i\omega_a t} \\
  \omega_b c_b  e^{-i\omega_b t} 
\end{pmatrix}
$$
or
$$
\left\{
\begin{array}{ccc}
 -\frac{\omega _0}{2}  c_a  e^{-i\omega_a t} 
     &+\frac{1}{2} V e^{i t \omega _D}c_b  e^{-i\omega_b t}  
 &= \omega_a c_a  e^{-i\omega_a t},
\\
 \frac{1}{2} V e^{-i t \omega _D} c_a  e^{-i\omega_a t}
    & + \frac{\omega _0}{2} c_b  e^{-i\omega_b t} 
 & = \omega_b c_b  e^{-i\omega_b t} .
\end{array}
\right.
$$
This needs you to set $\omega_a+\omega_D=\omega_b$, after which you can eliminate the time dependence. That leaves you with the simple linear system
$$
\left\{
\begin{array}{ccc}
 -\frac{\omega _0}{2}  c_a  
     +\frac{1}{2} V c_b 
 = \omega_a c_a  ,
\\
\phantom+ \frac{1}{2} V  c_a 
     + \frac{\omega _0}{2} c_b  
  = (\omega_a+\omega_D) c_b,
\end{array}
\right.
$$
which is an eigenvalue system for the hamiltonian 
$H=\begin{pmatrix} 
 -\frac{\omega _0}{2} & \frac{1}{2} V \\
\frac{1}{2} V  & \frac{\omega _0}{2} -\omega_D
\end{pmatrix}$.
Since you tagged this as homework, I'll leave the calculation here, as I'm sure you're better off calculating eigenvectors and eigenvalues on your own.
