###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](http://aip.scitation.org/doi/abs/10.1063/1.451710), 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 $(\lambda-n\omega)$.