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I'm curious about the difference in physical interpretation between open and periodic boundary conditions (OBC and PBC) although they are identical in the thermodynamic limit. For simplicity, let's consider a 1-d free fermion field theory, $$ H=-\frac{1}{2}\sum_{j}^{N} (c_j^\dagger c_{j+1}+\text{h.c.}) $$ where the lattice number is $N$ and adding $c_1^\dagger c_{N}+\text{h.c.}$ for PBC. We often perform a Fourier transform to find the ground state energy and you can see both have the same dispersion relation $\cos k$ but the wavenumber is different \begin{align*} k=\begin{cases} 2n\pi/N & (n=-N/2,\cdots N/2-1, \text{PBC}) \\ n\pi/(N+1) & (n=1,\cdots N, \text{OBC}) \end{cases} \end{align*}

The ground state energy energy $E_0$ corresponds to half-filled case and both produce the same form and value in the thermodynamic limit. \begin{align*} \frac{E_0}{N}= \begin{cases} -\frac{1}{N}\sum _{k=-N/4}^{N/4-1}\cos k \to -\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}dk \cos k = -\frac{1}{\pi}\int_{0}^{\pi/2}dk \cos k =-\frac{1}{\pi} & \text{PBC} \\ -\frac{1}{N}\sum _{k=1}^{N/2}\cos k \to -\frac{1}{\pi}\int_{0}^{\pi/2}dk \cos k =-\frac{1}{\pi} & \text{OBC} \end{cases} \end{align*} Quantum mechanically, in the case of PBC, left-movers and right-movers are occupied up to the Fermi surface but the OBC seems to only occupy one side. Even if they match at the thermodynamic limit, is it okay to consider them physically equivalent?

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    $\begingroup$ Can you define OPC and PBC in mathematical terms, please? $\endgroup$
    – Tobias Fünke
    Commented Apr 16 at 7:09
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    $\begingroup$ In OBC you do not have "left-movers" and "right-movers". They are not "seems to only occupy one side". Instead, to not just exit the region they are confined to, the only states that are allowed are specific linear combinations of "left-movers" and "right-movers" that reduces the possible states to exactly half, thus accounting for the factor of 2 change in how they are labelled. There is quite a lot of literature going into showing that Born-von-Karmon PBC always works in the thermodynamic limit, because we care that these convenient fictions won't cause trouble. $\endgroup$
    – naturallyInconsistent
    Commented Apr 16 at 7:15
  • $\begingroup$ @ Tobias Fünke The conditions are just distinguished by whether both ends are connected. $\endgroup$
    – Kitchen
    Commented Apr 16 at 9:05
  • $\begingroup$ Yes of course, but it might not be clear for everyone and often terminology differs in different fields. As a general point: Be as clear and precise as possible. In this case, it really is straightforward to spell it out without any additional work. $\endgroup$
    – Tobias Fünke
    Commented Apr 16 at 9:17

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