# Floquet quasienergy spectrum, continuous or discrete?

+ 3 like - 0 dislike
4452 views

I haven't got a feeling about Floquet quasienergy, although it is talked by many people these days.

Floquet theorem:

Consider a Hamiltonian which is time periodic $H(t)=H(t+\tau)$. The Floquet theorem says the solution to the Schrödinger equation will have the form

$$\psi(r,t)=e^{-i\varepsilon t}u(r,t)\ ,$$

where u(r,t) is a function periodic in time.

We can rewrite the Schrödinger equation as

$$\mathscr{H}u(r,t)=[H(t)-\mathrm{i}\hbar\frac{\partial}{\partial t}]u(r,t)=\varepsilon u(r,t)\ ,$$

the $\mathscr{H}$ can be thought as a Hermitian operator in the Hilbert space $\mathcal{R}+\mathcal{T}$, where $\mathcal{T}$ is a Hilbert space with all square integrable periodic functions with periodicity $\tau$. Then the above equation can be thought analogy to the stationary Schrödinger equation, with the real eigenvalue $\varepsilon$ defined as Floquet quasienergy.

My question is, since in stationary Schrödinger equation, we have continuous and discrete spectrum. How about floquet quasienergy?

Another thing is, is this a measurable quantity? If it is, in what sense it is measurable? (I mean, in the stationary case, the eigenenergy difference is a gauge invariant quantity, what about quasienergy?)

This post imported from StackExchange Physics at 2014-10-07 10:52 (UTC), posted by SE-user luming

edited Oct 7, 2014
What is $\mathcal{R}$? And what does "$+$" in $\mathcal{R}+\mathcal{T}$ mean?

This post imported from StackExchange Physics at 2014-10-07 10:52 (UTC), posted by SE-user huotuichang
$\mathcal{R}$ is a linear space consists of square-integrable functions of $\vec{r}$ . "+" is direct sum

This post imported from StackExchange Physics at 2014-10-07 10:52 (UTC), posted by SE-user luming
Evolution equations with time-dependent generators are difficult to treat in a rigorous way. One standard source is the book by Pazy. A reference that seems more tailored on your question is this book.

This post imported from StackExchange Physics at 2014-10-07 10:52 (UTC), posted by SE-user yuggib

There exists nothing like the "space of square integrable periodic functions with periodicity $\tau$". Thus the question does not make sense as it stands.

You should be more precise on that. Moreover the only "object" could make sense in place of that $+$ is the tensor product $\otimes$.

@ValterMoretti I mean, $\mathcal{T}$ is consists of all the functions $a(t)$ with periodicity $\tau$, and the inner product is defined as $(a,b)=\int_0^\tau a^*(t)b(t)\mathrm{d}t/\tau$. Does this make sense?

@lumining That is nothing but $L^2(0,\tau)$! periodicity does not play any role here. What I mean is that the considered space includes also functions of the type $f(t)=t^{-1/4}$ with $t\in [0, \tau]$. These have nothing to do with periodic functions.

@ValterMoretti The reason consider only the periodic functions is that with the definition of inner product, we can make $i\frac{\partial}{\partial t}$ a Hermitian operator. I am confused with the mathematical definition. The question I came from is from this paper, probably you can see section II-A,B if you have time.

I see. So the candidate space is ${\cal R} \otimes L^2(0,\tau)$. And the candidate self-adjonit operator is your ${\cal H}$ where $i \partial_t$ is understood as defined on the dense domain of smooth functions $\psi: [0,\tau] \to \mathbb C$ with $\psi(0)=\psi(\tau)$. With this domain it is essentially self-adjoint. The fact that $\cal H$ is (essentially) self adjoint, strongly depends on the properties  of the  map $\tau \mapsto H(\tau)$... In general it is false.

@ValterMoretti So do you have further comment on my question "discrete or continous"? If you still think this question is ill defined, please help me improve it.

Well, if the functions $u$ you consider are elements of ${\cal R}\otimes L^2(0,\tau)$ (and this can be checked case by case),  the numbers $\epsilon$ certainly belong to the point spectrum of ${\cal H}$ by definition. What is physically disputable is whether or not  these $u$ could have the meaning of "stationary" states in any sense. I do not think so...(What is the relevant time evolution with respect to they are stationary? Time evolution is already embodied in the definition of the Hilbert space...)

+ 0 like - 0 dislike

You can think of a Floquet energy in a similar way to a Bloch state. In the latter case, because space is periodic, the momentum states are repeated at every reciprocal lattice vector, $\textbf{G}$. For a Floquet state, because time is periodic, energy states are repeated every $n\hbar \omega$ where $n$ is an integer and $\hbar \omega$ depends on the time, $\tau$ (where $\tau$ in the experiment is the time between laser pulses).

Here is an image from the attached paper in case you cannot view it, but I highly recommend reading the below paper if you are interested in Floquet states. You can see (barely) in the image below that the Dirac Cone (which was chosen to be the system studied here for no particular reason), is repeated at several values of $\hbar \omega$ above and below the "actual" Dirac Cone at $n=0$. You can see the $n=1$, $n=2$, and $n=-1$ states pretty clearly. See the paper here:

https://www.sciencemag.org/content/342/6157/453?related-urls=yes&legid=sci;342/6157/453

This post imported from StackExchange Physics at 2014-10-07 10:52 (UTC), posted by SE-user Xcheckr
answered Oct 5, 2014 by (0 points)
I happened to see this paper before. So your answer seem to say quasi-energy is a gauge invariant in the same sense as Bloch energy. Is that what you mean? Also, what about my first question?

This post imported from StackExchange Physics at 2014-10-07 10:52 (UTC), posted by SE-user luming
@luming Yes, the quasi-energy must be gauge-invariant or it wouldn't be measurable! Also, perhaps I don't understand in what sense you mean discrete vs. continuous? It seems to me that the quasi-energies are not very different than Bloch states with different bands, except separated by constant energy steps.

This post imported from StackExchange Physics at 2014-10-07 10:52 (UTC), posted by SE-user Xcheckr

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.