I'm considering the evolution of a two-level quantum system given by

$$i\begin{pmatrix}\dot{c}_1\\ \dot{c}_2\end{pmatrix}=\begin{pmatrix} \frac{1}{2}E & \Delta(t)\\\Delta(t) & -\frac{1}{2}E\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}.$$

If $\,\Delta(t)\,$ is periodic, the model gives Rabi oscillations. But now I'm considering the function $\,\Delta(t)\,$ to be a real Gaussian white noise satisfying


In terms of the standard Wiener process $W_t$, the Schrodinger equation becomes

\begin{align} idc_1&=\frac{1}{2}Ec_1dt+\Gamma c_2dW_t,\\ idc_2&=\Gamma c_1dW_t-\frac{1}{2}Ec_2dt, \end{align}

where both $c_1dW_t$ and $c_2dW_t$ are in the Ito sense, meaning that $c_1$ and $c_2$ are at $t$ and $dW_t$ is during $[t,t+dt]$. I then go to the interaction picture


so that $\tilde{c}_1$ and $\tilde{c}_2$ do not evolve when $\,\Gamma=0$. In general, they satisfy \begin{align} id\tilde{c}_1&=\Gamma\tilde{c}_2e^{iEt}dW_t,\\ id\tilde{c}_2&=\Gamma\tilde{c}_1e^{-iEt}dW_t. \end{align}

Now how do I proceed from here to express $\tilde{c}_1$ and $\tilde{c}_2$ in terms of some integral of $W_t$? Also I'm wondering whether I get decoherence or still just Rabi oscillation because the system selectively absorbs frequency $E\,$ from the noise $W_t$.

  • $\begingroup$ It's not homework... $\endgroup$
    – Zhuoran He
    Nov 24, 2017 at 21:32
  • $\begingroup$ Fine. Make it seem easier so people can solve it.. $\endgroup$
    – Zhuoran He
    Nov 24, 2017 at 22:40

1 Answer 1


To solve these equations

\begin{align} id\tilde{c}_1&=\Gamma\tilde{c}_2e^{iEt}dW_t,\\ id\tilde{c}_2&=\Gamma\tilde{c}_1e^{-iEt}dW_t. \end{align}

let's look if we can find a pair of functions $F_1(t,x)$ and $F_2(t,x)$ such that $\tilde{c}_1=F_1(t,W_t)$ and $\tilde{c}_2=F_2(t,W_t)$. Plugging these in the formulas we have

\begin{align} idF_1(t,W_t)&=\Gamma F_2(t,W_t) e^{iEt}dW_t,\\ idF_2(t,W_t)&=\Gamma F_1(t,W_t) e^{-iEt}dW_t. \end{align}

We now apply Itô calculus to the differentials

\begin{align} i[F_{1t}(t,W_t)dt+F_{1x}(t,W_t)dW_t+1/2F_{1xx}(t,W_t)dt]&=\Gamma F_2(t,W_t) e^{iEt}dW_t,\\ i[F_{2t}(t,W_t)dt+F_{2x}(t,W_t)dW_t+1/2F_{2xx}(t,W_t)dt]&=\Gamma F_1(t,W_t) e^{-iEt}dW_t.\end{align}

In these the letters in the index notation are short hand for partial derivatives w.r.t. the corresponding variables. We hence obtain the following partial differential equations

\begin{align} F_{1t}(t,x)+1/2F_{1xx}(t,x) & = 0, \\ F_{2t}(t,x)+1/2F_{2xx}(t,x) & = 0, \\ iF_{1x}(t,x)& = \Gamma F_2(t,x) e^{iEt},\\ iF_{2x}(t,x)& = \Gamma F_1(t,x) e^{-iEt}. \end{align}

Taking the derivative to $x$ of the third equation and multiplying by $i$ we get

$$-F_{1xx}(t,x) = \Gamma iF_{2x}(t,x) e^{iEt}=\Gamma^2 F_1(t,x)$$

where in the last step I substitute in the fourth equation. Finally substituting this into the first equation we obtain


which is an easy first order linear partial differential equation in $t$. The solution is

$$F_1(t,x)=A(x)e^{\Gamma^2 t/2} \; .$$

With $A(x)$ some function to determine from the other equations. For instance, filling the result for $F_1$ in the first formula, we get the following differential equation for $A(x)$:

$$A_{xx}(x)+\Gamma^2 A(x)=0$$

the solutions of which are

$$A(x)=Be^{i\Gamma x}+De^{-i\Gamma x}$$

and thus

$$F_1(t,x)=(Be^{i\Gamma x}+De^{-i\Gamma x})e^{\Gamma^2 t/2}$$

and likewise

$$F_2(t,x)=(-Be^{i\Gamma x}+De^{-i\Gamma x})e^{(\Gamma^2/2-iE) t} \; .$$


\begin{align} \tilde{c}_1 & =(Be^{i\Gamma W_t}+De^{-i\Gamma W_t})e^{\Gamma^2 t/2} \; , \\ \tilde{c}_2 & =(-Be^{i\Gamma W_t}+De^{-i\Gamma W_t})e^{(\Gamma^2/2-iE) t} \; . \end{align}

  • $\begingroup$ I think one can show $|\tilde{c}_1|^2+|\tilde{c}_2|^2=\mathrm{const}$, so the probability amplitudes should not exponentially grow. But I'll try the idea of $\tilde{c}_i=F_i(t,W_t), i=1,2$. $\endgroup$
    – Zhuoran He
    Nov 24, 2017 at 22:42
  • $\begingroup$ I found that strange as well, but there might be a sign error somewhere. Also, I just noticed a discrepancy in your formulation where the change to the interaction picture has a $1/2$ in the exponents, but those don't show up in your equations below. $\endgroup$ Nov 24, 2017 at 22:44
  • $\begingroup$ I noticed one error in my computation. I forgot a $1/2$ factor myself. I'm going to fix it. $\endgroup$ Nov 24, 2017 at 22:47
  • $\begingroup$ Actually $\tilde{c}_i=F_i(t,W_t)$ is a very strong assumption. It assumes $\tilde{c}_i$ does not depend on earlier $W_{t'}$ with $t'<t$. If you plug in your solution to the equation and find it actually works, then it's fine. Otherwise the solution should be of some integral form. $\endgroup$
    – Zhuoran He
    Nov 25, 2017 at 0:03
  • $\begingroup$ That's a good point, but I don't think it is the issue here. $\endgroup$ Nov 25, 2017 at 7:46

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