Interpretation of the 1D transverve field Ising model vacuum state in a spin-language

Question

The 1D transverse field Ising model, \begin{equation} H=-J\sum_{i}\sigma_i^z\sigma_{i+1}^z-h\sum_{i}\sigma^x_i, \end{equation} can be solved via the Jordan-Wigner (JW) transformation (for further reference about the explicit form of the JW transformation, one may look at the following link). Here, $\sigma^z,\sigma^x$ are the Pauli matrices and the sum is carried over an infinite chain (consider $J>0$). After the JW non-local mapping to spinless fermionic operators (and a Bogoliubov transformation), one obtains a quadratic Hamiltonian that is diagonal in momentum space : \begin{equation} H=\sum_k \epsilon(k,h)(a^\dagger_ka_k+1/2) \end{equation} Where we express $h$ in units of $J$ and $\epsilon(k,h)$ represents the dispersion relation of the excitations of the system and depends on the transverse field $h$ and wave-vector $k$. $a^{(\dagger)}_k$ is the usual annihilation (creation) operator, which annihilates (creates) a spinless fermion with momentum $k$. For $h<1$ this dispersion relation is gapped, meaning that there is a finite energy cost to create $a^\dagger_k$ excitation with respect to the vacuum ground-state $\left|~0\right>$.

Here is my question : How do you explicitly write down this $\left|~0\right>$ ground-state in terms of the original $\sigma^z$ basis ? Explicitly : For $h=0$, we know that the ground-state corresponds to all spins up or down, i.e. $\left|~0\right>$=$\left|~\cdots\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\cdots\right>$ or $\left|~0\right>$=$\left|~\cdots\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\cdots\right>$. Now, how do you find the explicity form of $\left|~0\right>$ for $h\neq0$ ?

Answer 1

After thinking about it I must say it is not as simple as I thought it would be. The JW transformation on the transverse Ising model contains quite a few subtleties.

So to proceed,

1) Take your ground state for ANY $h$ expressed in the spinless fermion language. I stress ANY because this condition is true always - it's not just for $h<1$. Now this is the vacuum, specified by $| 0 \rangle$ s.t. $a_k |0\rangle = 0$. This is a non-trivial condition written in the spin-language, i.e. we take the operators $a_k$, and do the following:

2) Apply the reverse Bogoliubov transform on $a_k$: $\{a_k\}\to \{b_k\}$.

3) Apply an inverse Fourier transform: $\{b_k\} \to \{b_i\}$

4) Apply the inverse Jordan Wigner transformation: $b_i = f(\sigma^x,\sigma^y,\sigma^z)$ .

All these transformations are invertible (see this pdf for JW and inverse JW transformation), which is why you can do that. So composing all the maps one can express $a_k = g(\sigma^x, \sigma^y, \sigma^z)$, where $g$ is the highly non-trivial function.

Then one has to find the kernel of $g(\sigma^x, \sigma^y, \sigma^z)$, i.e. $|\psi\rangle $ s.t. $g(\sigma^x, \sigma^y, \sigma^z) |\psi\rangle = 0$. $|\psi \rangle $ is the ground state written in the spin-basis.

You can write a program to do it for you symbolically, but for all your effort what you'll end up with is a highly non-local ground state in the spin-basis because of all the Jordan-Wigner strings.

Remark:

There are many subtleties associated with this transformation. What is very often not mentioned when one derives the spectrum $\epsilon(k,h)$ is that the JW transformation MUST be performed separately on states with different parity in the Hilbert space.

This is because the imposition of periodic boundary conditions in spin space implies the imposition of periodic boundary conditions for ODD number of fermions but anti-periodic boundary conditions for EVEN number of fermions. This affects the Fourier transform. In the calculation of any macroscopic quantity in the thermodynamic limit, there is no difference, and many books/resources just discard talking about the two cases. But this distinction must be made if one wants to be careful about it.

One question that arose to me when I was thinking about this problem was: hm, in one limit, $h\to\infty$, the ground states is unique, while in the other limit $h \to 0$ the ground states is 2-fold degenerate. Can I see that in the fermion language easily?

There are two resolutions I can think to the problem.

1) It could be that the ground state $|0\rangle_k$ for each $k$ is not unique. That is, instead of the irreducible 2-dimension representation of the fermionic CAR that we usually assume $a_k$ acts in, $a_k$ could be operators in a $2 \times d$ (reducible) dimensional representation, with $d$ 'ground states'.

2) The even and odd sectors of the full Hilbert space give rise to two conditions on the ground state: $a_k$ in the even sector gives a condition $g(\sigma^x, \sigma^y, \sigma^z) |\psi \rangle = 0$ while $a_k$ in the odd sector gives another condition $g'(\sigma^x, \sigma^y, \sigma^z)|\psi'\rangle$ = 0.

It could perhaps be the case that when $h\to \infty$, $|\psi\rangle = |\psi'\rangle$, while when $h \to 0$, $|\psi\rangle \neq |\psi'\rangle$.

It would seem more likely to me that 2) is the correct analysis, though it's going to be one tough assertion to prove.

Answer 2

I think the bottom line is that everyone believes that the inversion of the Jordan-Wigner operation must be able to produce the simple eigenstates that one obtains very easily at either $h \rightarrow 0$ or $h \rightarrow \infty$, but having gone very thoroughly through the literature, it appears that no one has explicitly computed this.

A priori there's no reason why there would be any inconsistency, and the method outlined in nervxxx's answer is certainly expected to work, but it has never been actually done before.