Consider the Baker-Hausdorff formula for two operators $a_1$ and $iHt$: $$e^{iHt}a_1 e^{-iHt} =a_1+[iHt,a_1]+\frac{1}{2!}[iHt,[iHt,a_1]]+\frac{1}{3!}[iHt,[iHt,[iHt,a_1]]]+....,$$ where $[A,B]=AB-BA$. In the case of my problem, $H=\omega(a_1^\dagger a _1+a_2^\dagger a _2)+J(a_1^\dagger a _2+a_1a_2^\dagger)+\eta a_1^\dagger$. Thus one can write $[iHt,a_1]=\alpha\, a_1+\beta \,a_2+\gamma,$ and $[iHt,a_2]=\alpha\, a_2+\beta \,a_1,$ where the new parameters are either imaginary $(\alpha, \beta)$ or complex $(\gamma)$. So, I need to find the following: $$ a_1\,+(\alpha a_1+\beta a_2+\gamma)+\frac{1}{2!}\Big\{\alpha\Big(\alpha a_1+\beta a_2+\gamma\Big)+\beta\Big(\alpha\, a_2+\beta \,a_1\Big)\Big\}+\frac{1}{3!}\Big\{\alpha\Big(\alpha (\alpha\, a_1+\beta \,a_2+\gamma)+\beta (\alpha\, a_2+\beta \,a_1)+\gamma\Big)+\beta\Big(\alpha\, (\alpha\, a_2+\beta \,a_1)+\beta \,(\alpha\, a_1+\beta \,a_2+\gamma)\Big)\Big\}+....$$

In other words, in each new term, one should replace $a_1$ by $(\alpha\, a_2+\beta \,a_1+\gamma)$ and $a_2$ with $(\alpha\, a_2+\beta \,a_1)$, and repeat this procedure $N$ times, and finally calculate the limit when $N\rightarrow \infty$.

Does anyone know a systematic way to find a closed formula for the answer. The answer is given in equation 6 of enter link description here, and I'm trying to derive it. There might be a better approach there, but it doesn't come to my mind now!

  • $\begingroup$ Are you sure that that hamiltonian is meant to include unbalanced non-hermitian terms? That's unlikely to come up from realistic physical scenarios, but it is likely to make your life miserable. $\endgroup$ – Emilio Pisanty Mar 4 at 7:11
  • $\begingroup$ Quick Hint : Differentiate with 't' (aka equation of motion trick works perfect for any quadratic exponents of canonical operators). $\endgroup$ – Sunyam Mar 4 at 8:31
  • $\begingroup$ @Emiliobut Pisanty Yes, $H$ has another term like $-\eta^* a_1$ but since it commutes with $a_1$, I did not write it. By the way, today we Do have non-Hermitian Hamiltonians with real spectra. Google "non-Hermitian quantum mechanics". $\endgroup$ – Saeid Mar 4 at 16:25
  • $\begingroup$ @Sunyam That does not help me. I want to calculate the expression and get a closed formula, as is given in equation 6 of the above reference. $\endgroup$ – Saeid Mar 4 at 16:28
  • $\begingroup$ @Saeid That trick i mentioned is the quickest way to arrive at a closed expression. $\endgroup$ – Sunyam Mar 4 at 16:33

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