# 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. – Cosmas Zachos 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. – Cosmas Zachos 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 $U_F$.