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 $h\neq0$ ?

  • $\begingroup$ The wave function can be explicitly written as Pfaffians in $\sigma^x$ basis, and then you can formally transform this wave function to $\sigma^z$ basis, and the result won't be explicit, though. $\endgroup$ Feb 1, 2014 at 18:42

2 Answers 2


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.


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.

  • $\begingroup$ I thought of this before and this is most likely the way to go. However, there was two things that I didn't understand how to do : First : how do you invert JW transformation (I have not seen a fermionic variables to spin variables in the references I looked at) ? Second :in our problem, the ground-state for $h<1$ is the vacuum state, i.e. you do not fill anything. Thus how do you find what this corresponds to ? I guess my question is, what is the vacuum state in terms of a spin basis (is it simply not possible to write down and one must use perturbation theory to understand what's going on) ? $\endgroup$ Jan 28, 2014 at 15:46
  • $\begingroup$ Also note that since your Hilbert space $2^N$ dimensional, a general state could have just as many terms. So even for small $N$, your ground state could look very terrible and have lots of terms. I say this since it's often a misconception when doing many spins that a state is decomposable or has a simple form where some spins are up and others are down. If you really are bent on it, I'd try working it out explicitly for some low dimensional cases like $N = 2$, $3$ etc, in which case you could just write out a matrix for the Hamiltonian and use a computer to find your ground state.... $\endgroup$ Jan 28, 2014 at 23:52
  • 1
    $\begingroup$ ... and if you get to that, do post it here as an answer, as I'd also be curious to see what it looks like. If you have any questions about this particular process, don't hesitate to ask. $\endgroup$ Jan 28, 2014 at 23:55
  • 1
    $\begingroup$ @VanillaSpinIce Check out this set of notes: michaelnielsen.org/blog/archive/notes/… equations 31-34 where the JW transformation and its inverse are presented. I have also updated my answer to address your second question. $\endgroup$
    – nervxxx
    Jan 29, 2014 at 5:03
  • $\begingroup$ I think your answer is everything I was looking for and is most likely the only way to go. Could you add the link of your last comment in your answer (I think it would just make it better) ? My initial motivation for this question was to try to understand the ground-state of this system via non-perturbative method (here we have the exact solution !). But as you have noted, the answer is something quite involved... $\endgroup$ Jan 29, 2014 at 14:52

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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