Sinusoidaly Driven Two-Level System (TLS)

Question

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?

Answer 1

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.