# What are the allowed wavenumbers in the finite size system?

Usually, we introduce wavenumber $$\textbf{q}$$ by Fourier transform, for example, an operator $$A_{\textbf{q}}=1/\sqrt{N}*\sum_{i}e^{i \textbf{q}\cdot \textbf{r}_{i}}A_{i}$$, where $$N$$ is number of sites, $$i$$ denotes the site index. For a lattice or finite system with regular discrete $$i$$ (e.g. finite (spin) chain), $$\textbf{r}_i$$ takes integer number, then $$A_\textbf{q}=A_{\textbf{q}+2\pi\hat{1}}$$. So $$q$$ always takes value from $$(0,2\pi)$$.

From solid state physics textbook, $$i$$ component of $$\textbf{q}$$ takes quantized values as $$2\pi n/N_i$$ once we used period boundary condition, here $$i$$ denotes $$x, y, z$$. With period boundary condition, we have translation symmetry and $$\textbf{q}$$ is proportional to momentum.

But what if in finite size system, what are the allowed values of $$q$$? For example:

1) a finite chain, when the length $$L$$ takes finite value, such as 50.

2) a two-leg ladder, suppose there is translation symmetry along the leg direction $$x$$, then we have $$q_x=2\pi n/N_x$$, but does it make sense to define $$q_y$$? (On the other hand, the Brillouin zone is one dimension defined by period along $$x$$ direction.)

This ladder case can also be extended to bilayer case, such as bilayer square lattice. Then does it make sense to define $$q_z$$ along inter-layer direction?

If this $$q_y$$ in ladder can be defined, how to understand it? It is not the momentum anymore.

You already have your answer in the question. You said $$0. Plus $$q=n\frac{2\pi}{N}$$ (here you assumed a distance 1 between your spins otherwise it would be $$n\frac{2\pi}{Na}=n\frac{2\pi}{L}$$ where $$a$$ is the distance between spins). Since n is a positive integer, $$q<2\pi$$ induces $$n\leq N-1$$.

# edit :

Thanks to your comment I think I get what you ask. The problem is simply that to work with fourier transform you need a function defined on an infinite number of site. Otherwise, you won't get the inverse fourier transform. I'll stick to a continuous function $$f(x)$$ and its fourier transform $$F(q)$$ here for clarity.

\begin{align} F(q) &= \int\limits_{-\infty}^{+\infty}dx \, f(x) e^{-iqx} \\ &= \int\limits_{-\infty}^{+\infty}dx \, e^{-iqx} \int \frac{dk}{2\pi}F(k) e^{+ikx} \\ &= \int \frac{dk}{2\pi}F(k) 2\pi \delta(q-k)=F(q) \end{align}

I needed to use $$\int\limits_{-\infty}^{+\infty}dx \, e^{ix(k-q)}=2\pi\delta(q-k)$$. But if you want to do so on a finite domain you'll get $$\int\limits_{-L/2}^{+L/2}dx \, e^{ix(k-q)}=L\text{sinc}(\frac{qL}{2})$$ which gives back the delta function you need only for $$L\rightarrow{} \infty$$. Saying that your function is periodic is a way to define it on an infinite domain. You could also say that your function is $$0$$ everywhere outside the domain of interest.

• But this discrete q is obtained by assuming periodic boundary condition (for me, Born Karman boundary condition makes sense when N is large enough, and we only focus on the bulk). Now, N=2, there is only boundary, so does periodic boundary condition make sense in this case? Or there is other argument to obtain q here. – ZJX Jan 9 '19 at 15:03

In a coordinate where you have open boundary conditions, the problem is equivalent to a more familiar one: that of a particle in a box. Which wave numbers are allowed there? Well, it's precisely the $$k=\frac{n\pi}{L}$$ mentioned in @E. Bellec's answer. Multiply by $$\hbar$$ and you have a physical momentum. It might not have all the properties of the crystal momentum along another coordinate with periodic boundary conditions, but it certainly is a momentum.

Now, taking your ladder as an example - yes, it makes sense to define $$k_y$$. Similarly for the out-of-plane component $$k_z$$ in the bilayer example. However, for these cases, the gap between the two associated energy levels is usually quite large (since L is small), and the higher level is often considered to be "frozen out". Hence, a spin ladder can often be treated as a quasi-1D problem, and a bilayer as a quasi-2D problem.