Decoupling coupled differential equations in dynamically coupled two state system

Consider the following dynamically coupled two state hamiltonian, $$H=-B\sigma_z-V(t)\sigma_x.$$Taking the eigenfunctions of $\sigma_z$ ($|+>$ and $|- >$) as basis vectors, we have the wave function to be $$\Phi=c_ 1|+>+ c_2|->$$ and we get coupled differential equations for the time evolution of these two coefficients.

$$\left[ \begin{array}{c} \frac{dc_1}{dt} \\ \frac{dc_2}{dt} \end{array} \right] = \begin{bmatrix} -B & -V(t) \\ -V(t) & B \end{bmatrix} \times \left[ \begin{array}{c} c_1 \\ c_2 \end{array} \right]$$

To decouple the equations I tried diagonalyzing the Hamiltonian involved. But, then the eigenvectors themselves involve time dependence due to $V(t)$ and thus, i'm not able to decouple the differential equations. So, is there any other method do it? Any hints are welcome.

• I will add the steps that I did soon. I'm not used to math TEX commands and it takes me time to type it out. Apr 28, 2016 at 18:27
• Being able to decouple the equations depends on the form of $V(t)$. If $V(t)$ is a simple complex exponential (often what happens in the context of the rotating wave approximation), then it can be done; if it is a cosine or a sine, it cannot (this is called the Rabi model, I believe). There are examples on this site: see here and here. Apr 28, 2016 at 20:19

The system can be separated, but not necessarily in nice form. For instance, the time derivative of the first eq. reads $$i\hbar {\ddot c}_1 = - B {\dot c}_1 - {\dot V}c_2 - V {\dot c}_2$$ Now remove $c_2$ using again the first eq., $$c_2 = -\frac{i\hbar}{V} {\dot c}_1 - \frac{B}{V} c_1$$ and ${\dot c}_2$ using the second eq., ${\dot c_2} = \frac{i}{\hbar}Vc_1 - \frac{i}{\hbar}B c_2$: $$i\hbar {\ddot c}_1 = - B {\dot c_1} + i\hbar \frac{d\ln V}{dt} {\dot c}_1 + B \frac{d\ln V}{dt} c_1 - \frac{i}{\hbar}V^2c_1 + \frac{i}{\hbar} BV\left(-\frac{i\hbar}{V} {\dot c}_1 - \frac{B}{V} c_1\right) = 0$$ Simplify, rearrange, and obtain $${\ddot c}_1 - \frac{d\ln V}{dt} {\dot c}_1 + \left[\frac{i}{\hbar}B\frac{d\ln V}{dt} + \frac{B^2 + V^2}{\hbar^2} \right]c_1 = 0$$ Similarly for $c_2$.
Change from $c_1$, $c_2$ to $$c_+ = c_2 + c_1\\ c_- = c_2 - c_1$$ such that the system becomes $$i\hbar {\dot c}_+ = -V(t) c_+ + B c_-\\ i\hbar {\dot c}_- = B c_+ + V(t) c_-\\$$ Applying the same elimination procedure for $c_-$, this time using $$c_- = \frac{i\hbar}{B}{\dot c}_+ + \frac{V}{B}c_+\\ {\dot c_-} = - \frac{i}{\hbar}Bc_+ - \frac{i}{\hbar}Vc_- = - \frac{i}{\hbar}Bc_+ - \frac{i}{\hbar}V\left[\frac{i\hbar}{B}{\dot c}_+ + \frac{V}{B}c_+\right] = \frac{V}{B}{\dot c}_+ - \frac{i}{\hbar}\frac{B^2 + V^2}{B}c_+$$ yields a much simpler looking eq. for $c_+$: $${\ddot c}_+ - \frac{i}{\hbar}{\dot V}c_+ - \frac{i}{\hbar} V {\dot c}_+ + \frac{i}{\hbar} B {\dot c}_- = 0 \\ {\ddot c}_+ - \frac{i}{\hbar}{\dot V}c_+ - \frac{i}{\hbar} V {\dot c}_+ + \frac{i}{\hbar} V {\dot c}_+ +\frac{B^2 + V^2}{\hbar^2}c_+ = 0\\ {\ddot c}_+ + \left[\frac{B^2 + V^2}{\hbar^2} - \frac{i}{\hbar}{\dot V} \right]c_+ = 0$$
• There are exact solutions for very special $V(t)$s. Apr 29, 2016 at 8:40