# Two possible expressions for Hamiltonian of quantum critical Ising chain

While reading an article I encountered an expression for Hamiltonian of so called "critical chain": $$H = \sum_{k} c^{\dagger}_k[\sigma_x \sin k +B(1-\cos k)\sigma_y]c_k \quad (1)$$

where $c_k$ is 2-component spinor, $\sigma_i$ - Pauli matrix, $k$ is momentum, B is some constant. I want to see whether this is just different expression of critical Ising chain. Its Hamiltonian is known to be given by following expression: $$H(\sigma) = -J \sum_{i=1,...N}\sigma_i^{z}\sigma_{i+1}^{z} - h\sum_{i} \sigma_i^{x} \quad(2)$$

First, I transform (1) to spacial representation by means of discrete Fourier transformations: $$c^{\dagger}_{k}=\frac{1}{\sqrt{N}} \sum_{r} e^{-ikr} c^{\dagger}_{r}; \quad \frac{1}{N} \sum_{k} e^{ik(n-n')}=\delta(n-n')$$

After some effort I obtain following expression: $$H =\sum_{r} c^{\dagger}_{[r+1]} \bigg(\frac{\sigma_x}{2i} - \frac{B}{2} \sigma_y \bigg)c_{[r]} - c^{\dagger}_{[r ]}\bigg(\frac{\sigma_x}{2i} + \frac{B}{2} \sigma_y \bigg)c_{[r+1]} +c^{\dagger}_{[r]} B \sigma_{y}c_{[r]}$$

Next step I'm trying to do is to rewrite $\sigma_i$ in terms of $c_{[i]},c^{\dagger}_{[i+1]}$. I have found in Wikipedia following expressions:

$$\sigma_z = \sum K_1 c_{[i]}^{\dagger}c_{[i]}$$ $$\sigma_i^x = Dc_{[i]}^{\dagger}c_{[i+1]} +D^{*} c_{[i]}^{\dagger}c_{[i-1]} + K c_{[i]}c_{[i+1]}+ K^{*} c_{[i]}^{\dagger}c_{[i+1]}^{\dagger}$$ where K,D are some constants.
However I still can't show that these two expressions for Hamiltonians are equivalent. Do they actually describe different systems? Or is there a better way to show their equivalence?

• I've also tried to use Jordan–Wigner transformation. It is able to rewrite $c[r]$ in terms of sigma-matrices living on given site [r]. Yet in this case I don't understand how to interpret sigma-matrices from brackets, which don't seem to live on any specific site. – Yaroslav Shustrov Mar 13 '17 at 10:26

So far I have two possible solutions to this problem(which seem to contradict to each other). I've decided to write an answer rather than edit question itself because:
1) I have found plenty of useful information about the subject. So it may also be useful to other people.
2)If I put this information to the question it'll become too large and hence unreadable.

So here it goes.
First strategy

Obtained result: According to this approach Hamiltonian (1) is equivalent to the following one: $$H =\sum_{r} \frac{1}{2}(\sigma^{y}_r\sigma^{x}_{r+1}+\sigma^{x}_r\sigma^{y}_{r+1}) \quad (3)$$

Used techniques and (possibly) tricky moments:
Firstly I've used definition of two-component spinor $c_k$, k-momentum found in the internet: \begin{align} c_k=\begin{pmatrix} a_{k} \\ a^{\dagger}_{-k} \end{pmatrix} \quad c^{\dagger}_{k}= (a^{\dagger}_k; a_{-k}) \end{align}

where $a_k$ is fermionic annihilation operator in momentum space representation.

Once above expression used we obtain following expression: $$H=a_{-k}a_{k}(\sin k+ Bi(1-\cos k))+a^{\dagger}_k a^{\dagger}_{-k}(\sin k- Bi(1-\cos k))$$ Then I use following discrete Fourier transformations: $$a_r=\frac{1}{\sqrt{N}}e^{ikr}a_k; \quad a_k=\frac{1}{\sqrt{N}}e^{-ikr}a_r; \quad \frac{1}{N}\sum_{k}e^{ik(n-n')}=\delta(n-n')$$ where $k = \frac{2\pi}{N}n;\quad n= - \frac{N}{2}+1, ... \frac{N}{2}.$ Substitution of these expressions into the Hamiltonian produces:

$$H =i \sum_r (a_{r+1} a_{r} + a^{\dagger}_{r+1} a^{\dagger}_r)$$ This expression looks strange to me, because it doesn't depend on B - original parameter of theory. Yet I have arrived to this expression in two different ways.

Now I use Jordan–Wigner transformation. It helps me to rewrite $a_r, a^{\dagger}_r$ operators in terms of Pauli matrices. Among other properties of sigma matrices I've used the fact that they commute at different sites. This is how I arrived to (3).

Second strategy
This one is mainly based on following article. They start from Hamiltonian (2) and after certain transformations(similar to ones described above) they arrive to following expression:

$$H=N-2\sum_{q}(1+\lambda \cos q)a^{\dagger}_q a_q - \lambda \sum_q (e^{-iq} a^{\dagger}_q a^{\dagger}_{-q } -e^{iq} a_{q}a_{-q})$$ where $\lambda$ is a constant defined by Ising model.

Now if I take B=1 in (1) I may arrive to following expression: $$H=\sum_{k} -ic_{-k}c_{k} e^{ik} + i c^{\dagger}_{k} c^{\dagger}_{-k} e^{-ik}$$ This expression resembles the previous one. This may hint to the fact that Hamiltonians (1) and (2) are the same one after all. Yet I can't reconcile this hyphotesis with expression (3). If one can see some mistakes in my reasoning or has useful links, please let me know.

• If $c_k$ is a two component spinor wouldn't $c_k=\begin{pmatrix} c_{k\alpha} \\ a^{\dagger}_{k\beta} \end{pmatrix}$ where $\alpha$ and $\beta$ are some indices such as spin up/down? Why did you set it to two operators with opposite $k$ signs? I am not too familiar with this subject so maybe incorrect. – honey.mustard Mar 23 '17 at 13:59
• Let me show explicitly how this expression came to life. We start from so called Nambu-spinor representation: \begin{align} c_r=\begin{pmatrix} a_{r} \\ a^{\dagger}_{r} \end{pmatrix} \end{align} where r denotes number of site. Now we use aforementioned Fourier transforms for $a_r, a^{\dagger}_r$. In particular $a^{\dagger}_r = \frac{1}{\sqrt{N}} e^{ikr} a^{\dagger}_{-k}$. Then we can see that \begin{align} c_r= \frac{1}{\sqrt{N}} e^{ikr} \begin{pmatrix} a_{k} \\ a^{\dagger}_{-k} \end{pmatrix} \end{align} where column now is nothing more than $\psi_k$ – Yaroslav Shustrov Mar 24 '17 at 13:11
• I guess in my notation $\alpha$ and $\beta$ are implied yet not written explicitly. – Yaroslav Shustrov Mar 24 '17 at 13:14