3
$\begingroup$

In the study of spin chains with periodic boundary condition ($S_{N+1}=S_{1}$) when one applies Jordan-Wigner transformation to map the spin chain to spinless fermion chain, one needs to make sure in the mapping the periodic boundary condition for the spin chain is consistent. In this regard the precise statement for Transverse field Ising (TFI) model

To make the JW mapping exact, we supplement JW transformation with the following boundary conditions:

  1. the negative spin-inversion symmetry sector maps to the fermion Hamiltonian with periodic boundary conditions (PBC) and odd total number of fermions;
  2. the positive spin-inversion symmetry sector maps to the fermion Hamiltonian with anti-periodic boundary conditions (APBC) and even total number of fermions.

Anti-periodic boundary conditions differ from PBC by a negative sign attached to all coupling constants that cross a single, fixed lattice bond.

Where the spin Hamiltonian for TFI model is $$ H=\sum_{j=0}^{L-1}-J\sigma^z_{j}\sigma^z_{j+1}-h\sigma^x_j $$

Jordan Wigner transformations are given by \begin{align} c_i&=\frac{\sigma^x_i-i\sigma^y_i}{2}\prod_{j<i}\sigma^z_j \\ c_i^{\dagger}&=\frac{\sigma^x_i+i\sigma^y_i}{2}\prod_{j<i}\sigma^z_j \end{align}

And the mapped fermionic Hamiltonian is $$ H=\sum_{j=0}^{L-1}J(-c_j^{\dagger}c_{j+1}+c_jc_{j+1}^{\dagger})+J(-c_j^{\dagger}c_{j+1}^{\dagger}+c_jc_{j+1})+2h\left(n_j-\frac{1}{2}\right) $$

How the PBC and APBC come into play while JW transformation that is not clear to me. This concern appears mainly while creating a simulation, and most of the books overlook this.

The quote is taken from arXiv:1804.06782 sec 2.1.

$\endgroup$
  • 2
    $\begingroup$ A few comments: (1) Your Hamiltonian and JW-convention don't match, you need to swap X<->Z in one of them. (2) Do you understand how the JW-transformed Hamiltonian without periodic boundaries (your last equation) is obtained? (I am a bit doubtful in the light of comment 1). (3) Have you tried to apply the same to the term coupling sites L and 1? If yes, where did you get stuck? $\endgroup$ – Norbert Schuch Oct 6 at 10:20
  • $\begingroup$ I just realize the formulas are verbatim copied from the paper. Looks like a "do my work" question ... $\endgroup$ – Norbert Schuch Oct 6 at 10:22
3
$\begingroup$

It is important to note that there are two different boundary conditions, the first is the boundary condition for real spin model, the second is for fermion model. In fact, for real spin model with a certain boundary condition, there may be different corresponding boundary conditions for fermion operator. When we make a simulation in reality, we need to deal with this "boundary conditions problem" carefully.

Jordan-Wigher transformation

For a Ising model with spin-$\frac{1}{2}$, we can maop it into a spinless fermion model via Jordan-Wigher transformation: $$\sigma_i^z=1-c_i^\dagger c_i$$ which means we map the spin-up state to empty state, and map spin-down state to occupied state: $$|\uparrow\rangle \longmapsto |0\rangle\\|\downarrow\rangle \longmapsto |1\rangle$$ Naively speaking, we can unify the creation and annihilate operator for fermion with the ladder operator for spin: $c_i^\dagger\sim \sigma_i^-$, $c_i\sim \sigma_i^+$. However, the fermion operator follows the anti-communication relation and spin operator behaves like hard-core boson, which follows communication relation. To unify such two different communication relation, we need to introduce an additional string: $$\sigma_i^+=P_{i-1}c_i\\\sigma_i^-=P_{i-1}c_i^\dagger$$ where $P_{i}$ is the so-called string operator, which is defined as $$P_{i}\prod_{j=1}^i(1-c_j^\dagger c_j)$$. The effect of string operator is to measure the parity, even or odd, of the number of fermion which on the left-side of i-site.

For simplicity, we will rotate the axis: $$\sigma_z \longmapsto \sigma_x\\\sigma_x \longmapsto -\sigma_z$$

Open boundary condition(OBC)

For the real spin system, i.e. Ising model, with open boundary condition: $$H_I=-J\sum_{i=1}^N g\sigma_i^x-J\sum_i^{N-1}\sigma_i^z\sigma_{i+1}^z$$ we can re-express it via fermion operator: $$H_I=-Jg\sum_{i=1}^N (1-2c_i^\dagger c_i)-J\sum_i^{N-1}P_{i-1}(c_i^\dagger+c_i)P_i(c_{i+1}^\dagger+c_{i+1}) \\=-Jg\sum_{i=1}^N (1-2c_i^\dagger c_i)-J\sum_i^{N-1}P_{i-1}(c_i^\dagger+c_i)P_{i-1}(1-2n_i)(c_{i+1}^\dagger+c_{i+1}) \\=-Jg\sum_{i=1}^N (1-2c_i^\dagger c_i)-J\sum_i^{N-1}(c_i^\dagger+c_i)(1-2n_i)(c_{i+1}^\dagger+c_{i+1}) \\=-Jg\sum_{i=1}^N (1-2c_i^\dagger c_i)-J\sum_i^{N-1}(c_i^\dagger-c_i)(c_{i+1}^\dagger+c_{i+1})$$

We can find that the OBC of real spin model will be consistent with the OBC of spinless fermion model after Jordan-Wigher transformation.

Periodic boundary condition(PBC)

For the real spin system, i.e. Ising model, with periodic boundary condition: $$\sigma_{N+1}=\sigma_1$$ there is an additional term: $$-J\sigma_N^z\sigma_{N+1}^z \\=-J\sigma_N^z\sigma_{1}^z \\=-JP_{N-1}(c_N^{\dagger}+c_N)(c_1^{\dagger}+c_1) \\=-JP_{N}(c_N^{\dagger}-c_N)(c_1^{\dagger}+c_1)$$ the $P_N$ operator measure the parity of the number of fermion in the whole system. It is important to note that for the whole system with odd fermion number, $P_N=1$, we can unify this additional term to the normal expression of fermion model via $c^\dagger_{N+1}=c_i^\dagger$, which means the PBC for fermion model. On the other hands, for the whole system with even fermion number, $P_N=-1$, we can unify this additional term to the normal expression of fermion model via $c^\dagger_{N+1}=-c_i^\dagger$, which means the APBC for fermion model. As the result, for the real spin system with periodic boundary condition, the corresponding periodic boundary condition of fermion model has two situation: - PBC: if total number of fermion is odd, the spin-less fermion follows PBC. The effect of PBC will constrain the value of momentum. Namely, after Fourier transformation, $c_{k}=\frac{1}{\sqrt{N}} \sum_{j=1}^{N} c_{j} \exp (-i k j)$, where $k=2 \pi n / N$, the PBC restricts $n$ can only be integer. - APBC: if total number of fermion is even, the spin-less fermion follows APBC. which restricts $n$ can only be half-integer.

As the result, the spin-less fermion Hamiltonian can be written in the momentum space: $$H=J \sum_{k}\left[2(g-\cos k) c_{k}^{\dagger} c_{k}+i \sin k\left(c_{-k}^{\dagger} c_{k}^{\dagger}+c_{-k} c_{k}\right)-g\right]$$ where the value of $k$ depends on the boundary condition.

After Bogoliubov transformation: $$\gamma_{k}=u_{k} c_{k}-i v_{k} c_{-k}^{\dagger}$$ we can obtain the final diagonal Hamiltonian: $$H_{I}=\sum_{k} \varepsilon_{k}\left(\gamma_{k}^{\dagger} \gamma_{k}-1 / 2\right)$$ where the dispersion is: $$\varepsilon_{k}=2 J\left(1+g^{2}-2 g \cos k\right)^{1 / 2}$$

Unifying boundary condition

Now, there will emerge a contradiction. For a real spin-$\frac{1}{2}$ model with PBC, assuming there are $N$ sites, the number of state is $2^N$. However, for the spin-less fermion with certain boundary condition, $k$ have $N$ values, thus, there are totally $2^N$ states for a certain boundary condition. In the other words, if we consider two boundary conditions, i.e. PBC and APBC for fermion model, there are $2\times2^N$ states, which means half of them are redundant.

In fact, not all the states which calculated from fermion model for a certain boundary condition are physical, we only need to take the states with correct fermion number. For example, when we choose $k=2 \pi n / N, n\in Z$, which satisfies the PBC, the number of total fermion should be odd, thus, we need to leave out all the states with even fermion number. As the result, for each boundary condition, we just need to take half of them.

Till now, we have discussed how to deal with boundary condition problem for real spin model with PBC. For real spin model with APBC, the methods is similar.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.