Hi,

Suppose I have a system of a N spin chain labeling each spin from 1 to N.

So, when I use open boundary conditions in code, does it mean the wave function vanishes at S=1 and S=N or at S=0 and S=N+1?

I believe here you are thinking of a wavefunction of a collection of spins as if it is a function of a single parameter S. So you are thinking the wavefunction is like this:

But really for N spins it is a function like this:

So one cannot really speak of the wavefunction vanishing “at” S=0 or “at” S=1 or “at” S=N because it does not have a value at a one-dimensional spatial position. It is a function that depends on the states of all of the spins.

Maybe you are really asking a different question? Perhaps your question is: what does using open boundaries do to the wavefunction of my system? What are the implications of using open boundaries in terms of the observable properties of the ground state?

Actually I am talking about position of spins in 1D lattice.

Do you mean a system of spins similar to a 1D Ising model or a Heisenberg model? Then my answer as given above would be correct and there is no way to answer the original question you asked, unfortunately.

To help understand this matter in more detail, I would suggest you set up a two-site system (two S=1/2’s) just on paper I mean, and then write the full wavefunction for these two spins when the ground state is a singlet, say. Then see if you can answer the question of whether it takes the value zero at the location 1. You will see that it is not a question that can be answered for a function like that. The wavefunction of these two spins is a function of two variables, and these two variables are the value of each spin. So its domain is \{\uparrow \uparrow, \uparrow \downarrow, \downarrow \uparrow, \downarrow \downarrow\} and it lives in a four-dimensional space, not a two-dimensional space (like \{1,2\}). More generally a wavefunction of N spins has a domain which is a 2^N dimensional space, the space of all spin configurations.

I hope that helps answer your question and that I’m not misunderstanding it!

Thank you. I got it.