3
$\begingroup$

Consider Kicked field Ising model Hamiltonian given as follows. (I am following the this paper, relevant calculations are in appendix A.)

$$ H_I=2 J \sum_k\left[\cos k\left(\hat{b}_k^{\dagger} \hat{b}_k-1 / 2\right)+\frac{\imath}{2}\left(\hat{b}_k^{\dagger} \hat{b}_{-k}^{\dagger}-\hat{b}_{-k} \hat{b}_k\right) \sin k\right] $$

and

$$ H_K=-2 b \sum_k\left(\hat{b}_k^{\dagger} \hat{b}_k-1 / 2\right). $$

This is obtained using Jordan-Wigner transformation. $\hat{b}_k$ are the Fermionic operator corresponding to momentum $k$.

Further using the transformation

$$ \hat{\eta}_k=\cos \vartheta_k \hat{b}_k-i \sin \vartheta_k \hat{b}_{-k}^{\dagger}, \quad \vartheta_k=k / 2+\pi / 2 $$

the Hamiltonian can be expressed as

$$ H_{\mathrm{I}}=-2 J \sum_k\left(\hat{\eta}_k^{\dagger} \hat{\eta}_k-1 / 2\right) $$

and

$$ H_{\mathrm{K}}=-2 b \sum_k\left[\cos 2 \vartheta_k\left(\hat{\eta}_k^{\dagger} \hat{\eta}_k-1 / 2\right)+\frac{i}{2} \sin 2 \vartheta_k\left(\hat{\eta}_k^{\dagger} \hat{\eta}_{-k}^{\dagger}-\hat{\eta}_{-k} \hat{\eta}_k\right)\right] . $$

Using that $\hat{b}_k$ and $\hat{\eta}_k$ are fermionic operators we obtain for the Ising and kick part of the Floquet operator

$$ \begin{aligned} U_I= exp(- i H_{I})= & \mathrm{e}^{2 i J \sum_k\left(\hat{\eta}_k^{\dagger} \hat{\eta}_k-1 / 2\right)}=\mathrm{e}^{-i N J} \prod_k\left[1+\left(\mathrm{e}^{2 i J}-1\right) \hat{\eta}_k^{\dagger} \hat{\eta}_k\right] \\ U_K= exp(- i H_{K})= & \mathrm{e}^{2 i b \sum_k\left(\hat{b}_k^{\dagger} \hat{b}_k-1 / 2\right)}=\mathrm{e}^{-i N b} \prod_k\left[1+\left(\mathrm{e}^{2 i b}-1\right) \hat{b}_k^{\dagger} \hat{b}_k\right] \\ = & \mathrm{e}^{-i N b} \prod_k\left[1+\left(\mathrm{e}^{2 i b}-1\right)\left(\cos ^2 \vartheta_k \hat{\eta}_k^{\dagger} \hat{\eta}_k+\sin ^2 \vartheta_k \hat{\eta}_{-k} \hat{\eta}_{-k}^{\dagger}\right.\right. \\ & \left.\left.+\frac{i}{2} \sin 2 \vartheta_k\left(\hat{\eta}_k^{\dagger} \hat{\eta}_{-k}^{\dagger}-\hat{\eta}_{-k} \hat{\eta}_k\right)\right)\right] \end{aligned} $$

All the above steps are clear to me and I was able to obtain all of them. After this the paper states " In the latter expression k is only coupled to itself and to −k, the Floquet operator thus splits into $4 \times 4$ subblocks ($k$ and $-k$ occupied, $k$ and $-k$ unoccupied, only $k$ occupied, only $-k$ occupied) that can be diagonalized analytically". It is this part at which I am stuck.

My approach was to consider the states corresponding to momentum $|k,-k \rangle$ and label them as $|0,0 \rangle ,|1,1 \rangle,|0,1 \rangle,|1,0 \rangle$ corresponding to the four cases mentioned above. Then all I need to know is how the operator $\eta_{k}$ and $\eta_{k}^{\dagger}$ (similarly for $-k$ ) acts on these states. This in principle should have given me the $4\times $ matrix, that I can diagonalize analytically and obtain the eigenvalues. However, it seems that I am missing something as this simple approach did not work. My eigenvalues did not match the eigenvalues of the paper which are given by following set

$$ \Lambda(k)=\left\{\mu_{-}(k), 1,1, \mu_{+}(k)\right\}, $$

where $\mu_{ \pm}(k)=\alpha(k) \pm \sqrt{\beta(k)}$ with

$$ \begin{aligned} & \alpha(k)=\frac{1}{4} \mathrm{e}^{2 \mathrm{i}\left(J+b^x\right)}\left[\left(1+\cos 2 \vartheta_k\right)\left(1+\mathrm{e}^{-4 \mathrm{i}\left(J+b^x\right)}\right)+\left(1-\cos 2 \vartheta_k\right)\left(\mathrm{e}^{-4 \mathrm{i} J}+\mathrm{e}^{-4 \mathrm{i}^x}\right)\right] \\ & \beta(k)=\frac{\mathrm{e}^{4 \mathrm{i}\left(J+b^x\right)}}{16}\left(\left(1+\mathrm{e}^{-4 \mathrm{i} J}\right)\left(1+\mathrm{e}^{-4 \mathrm{i}^x}\right)+\left(1-\mathrm{e}^{-4 \mathrm{i} J}\right)\left(1-\mathrm{e}^{-4 \mathrm{i}^x}\right) \cos 2 \vartheta_k\right)^2-1 . \end{aligned} $$

Can someone help me to identify the problem and rectify it. For comparison, I obtain following $U_{I}$ and $U_{k}$ ( which are obviously wrong as the eigenvalues do not match). The basis I choose is $|0,0 \rangle ,|1,1 \rangle,|0,1 \rangle,|1,0 \rangle$ so that the matrix is block diagonal (because of Parity)

$U_{I}= \left( \begin{array}{cccc} e^{-2 i J} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & e^{-2 i J} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right), \quad \quad \quad U_{K}= \left( \begin{array}{cccc} 1+\left(-1+e^{2 i b}\right) \sin ^2(\nu ) & -\frac{1}{2} i \left(-1+e^{2 i b}\right) \sin (2 \nu ) & 0 & 0 \\ \frac{1}{2} i \left(-1+e^{2 i b}\right) \sin (2 \nu ) & 1+\left(-1+e^{2 i b}\right) \cos ^2(\nu ) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{2 i b} \\ \end{array} \right) $ $b= b^{x}$ and $\nu = \vartheta_k$

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The products $\prod\limits_k$ in the operators $U_I$ and $U_K$ consist of pairs of factors corresponding to pairs of opposite momenta $(k',-k')$: $$ \prod\limits_k A_k = A_0\left(\prod_{0<k<\pi}A_kA_{-k}\right)A_\pi $$ Accordingly, the part of the operator $U_I$ acting on the states $|0\rangle$, $\eta^\dagger_k\eta^\dagger_{-k}|0\rangle$, $\eta_k^\dagger|0\rangle$, $\eta_{-k}^\dagger|0\rangle$ is equal to $$ U_I^k = \left[ 1+(e^{2iJ}-1)\eta^\dagger_k\eta_k\right]\left[ 1+(e^{2iJ}-1)\eta^\dagger_{-k}\eta_{-k}\right]. $$ So I would say that the corresponding matrix is ​​as follows $$ U_I^k= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & e^{4iJ} & 0 & 0 \\ 0 & 0 & e^{2 i J} & 0 \\ 0 & 0 & 0 & e^{2 i J} \\ \end{array} \right). $$ The same applies to the $U_K$ operator.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you for the heads up (I will try it). I got confused by the notation. Maybe a silly question but isn't it weird notation that the product say $k$ lies between $0$ and $\pi$ (completely positive) but you have $A_{-k}$ inside it? I think I can alternatively take it from $-\pi/2$ to $\pi/2$ which rectifies the problem? $\endgroup$ Commented 2 days ago
  • $\begingroup$ @Erosannin In the article $k$ takes values ​​between $0$ and $2\pi$, see formula $(A.7)$. So what is $-k$? Since $k$ actually is quasi-momentum, it is defined up to the addition or subtraction of $2\pi$. Therefore, if $-k\in(0,2\pi)$, then $-k \equiv 2\pi-k$. Thus operation of changing sign of $k$ maps the interval $(0,\pi)$ to the interval $(\pi,2\pi)$. To account for each pair of momenta $k,-k$ only once, I wrote the first formula as is. $\endgroup$ Commented 2 days ago
  • 1
    $\begingroup$ Thanks a lot for clearing the doubt. I redid my calculations and the eigenvalues are in agreement now. $\endgroup$ Commented yesterday

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.