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

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.