I have a Hamiltonian of the form $H = H_0(t) + V(t)$ and wish to go into the interaction picture with $H_0(t)$. Normally this would be done by defining an evolution operator $U(t) = e^{-iH_0t/\hbar}$ and calculating $H_{interaction} = U ^{\dagger}V U $

However, I am having trouble exponentiating $H_0$, since it contains harmonic oscillator raising and lowering operators, $a$ and $a^{\dagger}$. Also, $H_0$ is time-dependent, which adds difficulty.

Is there a way to find $H_{interaction}$ without exponentiating $H_0$ ? Possibly a perturbative method?

Wikipedia mentions: 'It is possible to obtain the interaction picture for a time-dependent Hamiltonian $H_{0,S}(t)$ as well, but the exponentials need to be replaced by the unitary propagator for the evolution generated by $H_{0,S}(t)$, or more explicitly with a time-ordered exponential integral' on the interaction picture page but doesn't explain how one would go about doing this. Is this the Dyson series? Thanks.

  • 1
    $\begingroup$ You appear to have your notation backward. Normally, $H_0$ is the operator you exponentiate, and $V$ is the "small" operator. See: Wikipedia's interaction picture article. $\endgroup$ Jan 23, 2018 at 21:09
  • $\begingroup$ Hi, thanks for spotting that. I've now changed the question so that it uses standard notation and added some more information that I found on the wikipedia page. The main point of the question still remains, though. $\endgroup$
    – asph
    Jan 24, 2018 at 12:16

1 Answer 1


You can't get around doing something equivalent to exponentiating the operator. One way to do something equivalent is to solve the equations of motion that come from the commutation relations with $H_0(t)$. For example, when $H_0$ is the simple harmonic oscillator the equations of motion for $x$ and $p$ are \begin{align} \dot{x} & = i[H_0,x] = \frac{p}{m} \\ \dot{p} & = i[H_0,p] = -\omega^2 x, \end{align} with the initial condition that $x(0) = x_S$ and $p(0)=p_S$. That system of first order ordinary differential equations is solved by \begin{align} x(t) & = x_S \cos(\omega t) + p_S \sin(\omega t) \\ p(t) & = -x_S \sin(\omega t) + p_S \cos(\omega t). \end{align} Because $x(t) = \mathrm{e}^{-iH_0t}x_S\mathrm{e}^{iH_0t}$ (signs may be flipped), any reasonably smooth $V(x_S,p_S,t)$ will obey \begin{align} \mathrm{e}^{-iH_0t}V(x_S,p_S,t)\mathrm{e}^{iH_0t} = V(x(t),p(t),t). \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.