How the useful, unitary "Folding Transform" is applied to a Hamiltonian

Summary

Given a unitary transformation $$U_F(t)=\sum_n e^{in\omega t}\sum_{\lambda\in n}\left| \lambda, n\right\rangle \left\langle \lambda,n \right|$$ applied to a Hamiltonian $$H_0$$ (with $$H_0 \left|\lambda , n\right\rangle=\lambda\left|\lambda , n\right\rangle$$, and where $$(2n-1)\omega/2 < \lambda \leq (2n+1)\omega/2$$ defines $$n$$), how does one obtain the "$$(\lambda-n\omega)$$" in the result $$H_F=U_F(t)H_0U_F(t)^{-1}=\sum_n \sum_{\lambda \in n}(\lambda-n\omega)\left| \lambda, n\right\rangle \left\langle \lambda,n \right|$$?

Details

I'm trying to understand a "folding transform" as discussed by Maricq in his paper here, pages 5169-5170.

I'm leaving out extra bits of the Hamiltonian and other things, but the important features are these:

Setup

You have a Hamiltonian $$H_0$$, with eigenvalues $$\lambda$$. You're interested in how some value $$\omega$$ compares with the $$\lambda$$'s ($$\omega$$ is some driving frequency, say), so you break up your eigenvalues into regions indexed by $$n$$, with

$$(2n-1)\omega/2 < \lambda \leq (2n+1)\omega/2$$

and define your eigenvectors with this $$n$$ explicitly included: $$H_0 \left|\lambda , n\right\rangle=\lambda\left|\lambda , n\right\rangle$$. So, $$n$$ tells you what region your eigenvalue $$\lambda$$ is in.

Folding Transform

Now you define two operators: a projection operator $$Q_n=\sum_{\lambda\in n}\left| \lambda, n\right\rangle \left\langle \lambda,n \right|$$ and a transformation operator $$U_F(t)=\sum_n e^{in\omega t}Q_n$$.

The transformation operator $$U_F(t)$$ is applied to the Hamiltonian $$H_0$$ to produce a new "folded" Hamiltonian $$H_F$$ (presumably $$H_F(t)=U_F(t)H_0U_F(t)^{-1}$$, though this isn't made explicit), to produce the "folded" Hamiltonian:

$$H_F(t)=\sum_n \sum_{\lambda \in n}(\lambda-n\omega)\left| \lambda, n\right\rangle \left\langle \lambda,n \right|$$

Now the question: how did $$(\lambda-n\omega)$$ appear there? It seems to me it should just boil right back down to $$\lambda$$ again, not $$(\lambda-n\omega)$$.

• U is a Sylvester clock matrix; work out its commutation relations with the original hamiltonian. The are not null. Apr 18 '17 at 19:31
• @CosmasZachos I thought it would simplify to $\lambda$ for a different reason, namely that $\sum_{n_1,\lambda \in n_1, n_2, \lambda \in n_2}e^{i(n_1-n_2) \omega t} \left| \lambda, n_1 \right\rangle \left\langle \lambda, n_1 \right| H_0 \left| \lambda, n_2 \right\rangle \left\langle \lambda, n_2 \right| = \sum_{n,\lambda \in n} \lambda \left| \lambda, n \right\rangle \left\langle \lambda, n \right|$ because of the matrix elements, not the exponential
– JDR
Apr 18 '17 at 20:39
• I stand corrected. Their eqn (15) appears malformed, if one heeds the figure. One would expect the λs to now all be in the fundamental domain, n=1, or whatever. He appears to be transforming the propagators (3) with his operator, not the static hamiltonian you are writing here. Apr 18 '17 at 22:22

The error was in assuming $H_F=U_F H_0 U_F^{-1}$. As we ALL know, this isn't how to go to a transformed frame! See here for a good local explanation of the rotating frame.
When you do things properly (transform your basis eigenvectors and deduce the evolution and effective Hamiltonian from there --- chain rule!) you get exactly the form above by differentiating $U_F$.