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.

  • $\begingroup$ 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. $\endgroup$ Apr 28, 2016 at 18:27
  • $\begingroup$ 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. $\endgroup$
    – march
    Apr 28, 2016 at 20:19

1 Answer 1


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 $c_2$.

Better way:

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 $c_+$: $$ {\ddot c}_+ - \frac{i}{\hbar}{\dot V}c_+ - \frac{i}{\hbar} V {\dot c}_+ + \frac{i}{\hbar} B {\dot c}_- = 0 \\ {\ddot c}_+ - \frac{i}{\hbar}{\dot V}c_+ - \frac{i}{\hbar} V {\dot c}_+ + \frac{i}{\hbar} V {\dot c}_+ +\frac{B^2 + V^2}{\hbar^2}c_+ = 0\\ {\ddot c}_+ + \left[\frac{B^2 + V^2}{\hbar^2} - \frac{i}{\hbar}{\dot V} \right]c_+ = 0 $$

  • $\begingroup$ Thank you..I was trying to do this since I posted this question yesterday. Now, things are clear. $\endgroup$ Apr 29, 2016 at 6:34
  • $\begingroup$ Welcome. Added a better form, but I have a nagging feeling that I once saw an exact solution for this problem, but can't remember where or what I'm missing right now. $\endgroup$
    – udrv
    Apr 29, 2016 at 7:55
  • $\begingroup$ There are exact solutions for very special $V(t)$s. $\endgroup$ Apr 29, 2016 at 8:40

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