How do irrational numbers give incommensurate potential periods?

[Luqman Saleem]

I am trying to understand Aubry-Andre model. It has the following form
$$H=∑_n c^†_nc_{n+1}+H.C.+V∑_n cos(2πβn)c^†_nc_n$$
This reference (at the 3rd page) says that if $\beta$ is irrational (rational) then the period of potential is quasi-periodic incommensurate (periodic commensurate) with underlying lattice period.
Question 1: What does incommensurate potential mean here?
Question 2: How does irrational $\beta$ guarantee that potential is quasi-period incommensurate with underlying lattice?
Furthermore, this reference says that with irrational $\beta$ (they are taking the inverse of Golden Mean i.e. $(\sqrt{5}−1)/2)$ to avoid the unwanted boundary effects, we have to take the system of a size of any number from Fibonacci series.
Question 3: How does the system of a size of any Fibonacci series' number avoid unwanted boundary effects?

 

[DrClaude]

Question 1: What does incommensurate potential mean here?
Question 2: How does irrational $\beta$ guarantee that potential is quasi-period incommensurate with underlying lattice?

Two periodic phenomena are incommensurate if there is no period with which the combined phenomena can repeat.

To answer question 2, think about how you would calculate the repeating pattern of two oscillations with periods $a$ and $b$.

Question 3: How does the system of a size of any Fibonacci series' number avoid unwanted boundary effects?

I don't know about that. Can you point out more specifically where this is written?

 

[TeethWhitener]

I imagine the answer to the third question has something to do with the fact that the nth Fibonacci number can be written in closed form as:
$$F_n = \frac{\varphi^n -\frac{1}{\varphi^n}}{\varphi -\frac{1}{\varphi}}$$
where $$\varphi=\frac{1+\sqrt{5}}{2}$$
the golden mean.

 

[Luqman Saleem]

About question 2:
If I remember correctly, to find the period of the sum of two periodic functions (with period $a$ and $b$) we try to find integers $n$ and $m$ such that $n a = m b = k$. If we can find $n,m$ then $k$ will be the period of the resultant. And in the case under study, the period of lattice point is $1$ and the period of potential is $1/\beta$. If $\beta$ is irrational then there will be no integers that can satisfy the above equation, which means we will get incommensurate potential. Am I right?

I don't know about that. Can you point out more specifically where this is written?

Here is the paragraph from that article:

An important aspect of the Aubry–Andr´e model is the incommensurability of the periodic potential in above Hamiltonian, with respect to the underlying lattice that is guaranteed by the choice of β as an irrational number. Numerical studies, however, will in general have to be carried out on a finite lattice with periodic boundary conditions imposed to avoid undesirable boundary effects. In this case, the periodic potential is no longer truly incommensurate and caution has to be exercised when increasing the system size. The inverse of the golden mean is a convenient choice because the convergents of its continued fraction representation are given by ratios of successive Fibonacci numbers defined by the recursion relation $F_{n+1} = F_n + F_{n−1}$ with $F_0 = 0$ and $F_1 = 1$. Therefore, if the system size is chosen as a Fibonacci number $F_i$, the period β in Hamiltonian can be approximated by $F_{i−1}/F_i$, which yields the inverse of the golden mean in the limit of large system sizes.

 

[DrClaude]

If $\beta$ is irrational then there will be no integers that can satisfy the above equation, which means we will get incommensurate potential. Am I right?

Correct.

Therefore, if the system size is chosen as a Fibonacci number $F_i$, the period β in Hamiltonian can be approximated by $F_{i−1}/F_i$, which yields the inverse of the golden mean in the limit of large system sizes.

The answer is there. Getting the period to be an irrational number is experimentally hard, so using Fibonacci numbers allows one to be guaranteed to have a good enough approximation over the size of the lattice.

 

[Luqman Saleem]

Correct.The answer is there. Getting the period to be an irrational number is experimentally hard, so using Fibonacci numbers allows one to be guaranteed to have a good enough approximation over the size of the lattice.

Thank you so much. I get the answer. I have also checked it by plotting the potential on MATLAB. Thanks again