# 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?

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 $H=\begin{pmatrix} -\frac{\omega _0}{2} & \frac{1}{2} V \\ \frac{1}{2} V & \frac{\omega _0}{2} -\omega_D \end{pmatrix}$.
• Sorry, why is $\omega_a+\omega_D=\omega_b$? and which delta function is a solution to my equation above, is it $\delta(\omega)$ or $\delta(\omega-\omega_D)$or $\delta(\omega-\omega_0)$? – AimForClarity Sep 4 '13 at 19:10
• I would discourage you from working in that Fourier representation (it only obscures the fact that the components are simply oscillating exponentials). If you must know that, then solve that last eigensystem for $\omega_a$ and $\omega_b$, and Fourier transform the initial Ansatz for $c_a(t)$ and $c_b(t)$. – Emilio Pisanty Sep 4 '13 at 19:25
• Now that I think about it, I can cancel out $c_a$ in $c_a(\omega )=\frac{V^2 c_a(\omega )}{(2 \omega )^2+\omega _0^2}$ and solve for omega, which will give me exactly the correct rabi frequency: $\Omega = \sqrt{\left(\frac{V}{2}\right)^2-\left(\frac{\omega _0}{2}\right){}^2}$. This tells me that the equation will hold for any non-0 or non-infinite $c_a$ and $c_b$ only at $\omega = \pm \Omega$, and everywhere else $c_a$ and $c_b$ should be 0 (or Infinity, but that we leave out as unphysical). – AimForClarity Sep 4 '13 at 19:44