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I am looking for a detailed explanation of when and why we can neglect terms like $a_n a_{n+1}$ or $a_n^\dagger a_{n+1}^\dagger$ in Hamiltonians like $$ H = \omega \sum_n a^\dagger_n a_n + \lambda\sum_{n} \left(a_n a_{n+1} + a_n^\dagger a_{n+1}^\dagger + a^\dagger_n a_{n+1} +a^\dagger_{n+1}a_n \right) \approx \omega \sum_n a^\dagger_n a_n + \lambda\sum_{n} \left( a^\dagger_n a_{n+1} +a^\dagger_{n+1}a_n \right).$$ I've seen this arise in the context of quantum circuits and other models where one encounters interactions of the form $x_n x_{n+1}$ , $p_n p_{n+1}$, $x_n p_{n+1}$, etc. The typical explanation given is that "$\omega \gg \lambda$" or that "in the rotating wave approximation, terms like $a_n a_{n+1}$ oscillate fast and can be ignored".

In the rotating frame, I see that transformations like $$H(t) = \exp(-it\omega\sum_n a^\dagger_n a_n)H\exp(it\omega\sum_n a^\dagger_n a_n)$$ leave a fast oscillating factor in front of terms like $a_n a_{n+1}$ but why/when can it be disregarded?

I don't see how $\omega \gg \lambda$ is relevant for the approximation.


Edit 1, based on the comments:

  1. I will stress that I am looking for a way of justifying neglecting $\lambda \sum_n (a_n a_{n+1} + a_n^\dagger a_{n+1}^\dagger)$ but NOT the hopping term $\lambda \sum_n (a_n^\dagger a_{n+1} + a_n a_{n+1}^\dagger)$. I suspect this will likely involve talking about the spectrum/eigenvectors of $H$ in a particular limit.
  2. I have for instance seen this arise in the context of quantum electrical circuits. For instance, one can approximate a chain of capacitively/inductively coupled harmonic LC oscillators to a hopping chain. That is, I can show that the Hamiltonian of a chain of capacitively/inductively coupled harmonic LC oscillators is $$ H = \sum_n \alpha_n p_n^2 + \beta_n x_n^2 + \lambda_n x_n x_{n+1}.$$ Note that the coupling term $x_n x_{n+1} = (a^\dagger_n + a_n)(a^\dagger_{n+1} + a_{n+1})$ yields exactly my first Hamiltonian. However, following reference "Bloch oscillations and Wannier-Stark ladder in the coupled LC circuits (Bahmani 2020)" equation (12), one claims that if $\omega \gg \lambda$ then $$H \approx \sum_n \omega_n a^\dagger_n a_n + \lambda_n\sum_{n} \left( a^\dagger_n a_{n+1} +a^\dagger_{n+1}a_n \right),$$ if I understand the ref correctly.
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    $\begingroup$ Just to clarify where you are in your understanding: Are you familiar with the RWA in the context of two-level systems and how it works in that simplest case? I imagine it's the same here, and so that would be a first good case to look at. It's essentially a resonance condition. $\endgroup$
    – march
    Commented Oct 7 at 15:29
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    $\begingroup$ A dumb answer is that the Hamiltonian is schematically $\omega a^2 + \lambda a^2$, and the approximation $\omega \gg \lambda$ means you can ignore the second term. Are you looking for more than that, and if so can you be a bit more specific on where you get stuck? Like, "when is $\omega \gg \lambda$ true", or "why does the first term come with a factor of $\omega$ and the second a factor of $\lambda$", or "does this approximation of the Hamiltonian operator work for every state in the Hilbert space"? $\endgroup$
    – Andrew
    Commented Oct 7 at 16:58
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    $\begingroup$ You'll want to look at how the RWA works for two-level systems, because it illuminates both the physics and the mathematics of this approximation. You can start here. $\endgroup$
    – march
    Commented Oct 7 at 20:54
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    $\begingroup$ @RogerV. No worries! I left the clarification I was needing in your answer to the post you shared. PD: Second quantisation is what made me "understand" QM, so I guess everyone is different! $\endgroup$ Commented Oct 10 at 14:24
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    $\begingroup$ I had a new answer I was thinking of adding! Oh well, it doesn't quite fit in the other question's answers, just a different perspective. Think about the Heisenberg equations $dA/dt=i[H,A]/\hbar$ for the evolution of any operator $A$. In the interaction picture (rotating frame, not rotating wave approximation), the operator is actually $a e^{-i\omega_a t}$. If you integrate that part of the Hamiltonian in a classical way, you find $A(t)-A(0)=i[a b,A]\int_0^t e^{i(\omega_a+\omega_b)t} dt/\hbar$, ignoring temporal commutation. The integral gives $1/(\omega_a+\omega_b)$, which is small $\endgroup$ Commented Oct 12 at 16:46

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