Quantum criticality from Fisher information

Abstract

Quantum phase transition is primarily characterized by a qualitative sudden change in the ground state of a quantum system when an external or internal parameter of the Hamiltonian is continuously varied. Investigating quantum criticality using information-theoretic methods has generated fruitful results. Quantum correlations and fidelity have been exploited to characterize the quantum critical phenomena. In this work, we employ quantum Fisher information to study quantum criticality. The singular or extremal point of the quantum Fisher information is adopted as the estimated thermal critical point. By a significant model constructed in Quan et al. (Phys Rev Lett 96: 140604, 2006), the effectiveness of this method is illustrated explicitly.

Appendix

Here, we present detailed derivation of Eq. (2). For the state $\rho _S(t)$ in Eq. (1), the explicit expression of the symmetric logarithmic derivatives $L_\lambda$ determined by

$$\begin{aligned} \partial _\lambda \rho _S(t)=\frac{1}{2} \bigl (\rho _S(t)L_{\lambda }+L_{\lambda }\rho _S(t) \bigr ) \end{aligned}$$

is given by

$$\begin{aligned} L_\lambda =\frac{1}{|f(\lambda )|^2-1}\left( \begin{array}{cc} |c_e|^2m(\lambda ) &{} \ \ c_gc_e^*n(\lambda )\\ c_g^*c_en^*(\lambda ) &{} \ \ |c_g|^2m(\lambda )\\ \end{array} \right) \end{aligned}$$

with

$$\begin{aligned} f(\lambda )= & {} \langle \varphi _e(t)|\varphi _g(t)\rangle ,\\ m(\lambda )= & {} f(\lambda )\dot{f}^*(\lambda )+f^*(\lambda )\dot{f}(\lambda ),\\ n(\lambda )= & {} 2\dot{f}(\lambda )(|f(\lambda )|^2-1)-m(\lambda )f(\lambda ). \end{aligned}$$

The quantum Fisher information $F_\lambda (t)$ can be evaluated as

$$\begin{aligned} F_\lambda (t)=\mathrm{tr}\rho _S(t) L_\lambda ^2=|c_g|^2|c_e|^2\biggl (4|\dot{f}(\lambda )|^2-\frac{\dot{L}(\lambda )^2}{L(\lambda )-1}\biggr ), \end{aligned}$$

where $L(\lambda )=|f(\lambda )|^2$. It remains to evaluate explicitly $f(\lambda ),$ which is the main difficult task.

Following the method in Ref. [24], the Hamiltonian $H(\lambda ,\delta )$ can be rewritten in a diagonal form as

$$\begin{aligned} H(\lambda ,\delta )=\sum _k\varepsilon _e^k(A_k^\dagger A_k-\mathbf {1}/2) \end{aligned}$$

with $\varepsilon _e^k=2J\sqrt{1+(\lambda +\delta )^2-2(\lambda +\delta )\cos (ka)}$ being the single quasiexcitation energy and

$$\begin{aligned} \begin{aligned} A_k=\sum _l\frac{\mathrm {e}^{-i{kal}}}{\sqrt{N}}\prod _{s<l}\sigma _s^x(u_e^k\sigma _l^{[+]}-iv_e^k\sigma _l^{[-]}) \end{aligned} \end{aligned}$$

being the normal mode operators which satisfy the canonical Fermion anticommunication relations. Here, $u_e^k=\cos (\theta _e^k/2)$ and $v_e^k=\sin (\theta _e^k/2)$ with

$$\begin{aligned} \theta _e^k=\arctan \frac{\sin (ka)}{\lambda +\delta -\cos (ka)}. \end{aligned}$$

The operator $\sigma _l^{[\pm ]}=(-\sigma _l^z\pm i\sigma _l^y)/2$ is defined by the Pauli matrices acting on the lth site of the spin chain.

Similarly (or by putting $\delta =0$), the Hamiltonian $H(\lambda ,0)$ can be diagonalized as

$$\begin{aligned} H(\lambda ,0)=\sum _k\varepsilon _g^k(B_k^\dagger B_k-\mathbf {1}/2) \end{aligned}$$

with $\varepsilon _g^k=2J\sqrt{1+\lambda ^2-2\lambda \cos (ka)}$ and

$$\begin{aligned} \begin{aligned} B_k=\sum _l\frac{\mathrm {e}^{-i{kal}}}{\sqrt{N}}\prod _{s<l}\sigma _s^x(u_g^k\sigma _l^{[+]}-iv_g^k\sigma _l^{[-]}). \end{aligned} \end{aligned}$$

Here, $u_g^k=\cos (\theta _g^k/2)$, $v_g^k=\sin (\theta _g^k/2)$, and $\theta _g^k=\arctan \frac{\sin (ka)}{\lambda -\cos (ka)}.$

The Fermionic quasiexcitation operators $B_k$ can be verified to be related with the operators $A_k$ by the following Bogoliubov transformation

$$\begin{aligned} B_{\pm k}=\cos (\alpha _k)A_{\pm k}-i\sin (\alpha _k) A_{\mp k}^\dagger , \end{aligned}$$

where $\alpha _k=(\theta _g^k-\theta _e^k)/2$.

Suppose the Ising spin chain in a transverse field depicted by $H( \lambda ,0)$ is initially in the ground state $|\varphi (0)\rangle =|G\rangle$ which can be rewritten as

$$\begin{aligned} |G\rangle =\prod _{k>0}\bigl (\cos (\alpha _k)-i\sin (\alpha _k)A_k^\dagger A_{-k}^\dagger \bigr )|E\rangle , \end{aligned}$$

where $|E\rangle$ is the ground state of $H(\lambda ,\delta )$, then the expression of $f(\lambda )$ can be calculated as

$$\begin{aligned} f(\lambda )= & {} \langle \varphi _e(t)|\varphi _g(t)\rangle =\langle G|e^{iH(\lambda ,\delta )t}e^{-iH(\lambda ,0)t}|G\rangle \\= & {} e^{i\sum _k(\varepsilon _g^k-\varepsilon _e^k)t/2} \langle G|e^{i\sum _k\varepsilon _e^ktA_k^\dagger A_k}e^{-i\sum _k\varepsilon _g^ktB_k^\dagger B_k}|G\rangle . \end{aligned}$$

Since $B_k|G\rangle =0$, $[A_k^\dagger A_k, A_j^\dagger A_j]=0$, and $(A_k^\dagger A_k)^n=A_k^\dagger A_k$ for any positive integer n, the above expression turns out to be

$$\begin{aligned} f(\lambda )= & {} \mathrm{e}^{i\sum _k(\varepsilon _g^k-\varepsilon _e^k)t/2}\langle G|\prod _k \mathrm{e}^{i\varepsilon _e^ktA_k^\dagger A_k}|G\rangle \\= & {} \mathrm{e}^{i\sum _k(\varepsilon _g^k-\varepsilon _e^k)t/2}\langle E|\prod _{m>0} \bigl (\cos (\alpha _m)-i\sin (\alpha _m) A_{-m}A_{m}\bigr )\\&\times \prod _k \mathrm{e}^{i\varepsilon _e^ktA_k^\dagger A_k}\prod _{n>0}\bigl (\cos (\alpha _n)+i\sin (\alpha _n) A_n^\dagger A_{-n}^\dagger \bigr )|E\rangle . \end{aligned}$$

By virtue of the equations

$$\begin{aligned} \prod _k\mathrm{e}^{i\varepsilon _e^ktA_k^\dagger A_k}= & {} \mathrm{e}^{i\varepsilon _e^0tA_0^\dagger A_0}\prod _{l>0}\mathrm{e}^{i\varepsilon _e^{-l}tA_{-l}^\dagger A_{-l}}e^{i\varepsilon _e^ltA_l^\dagger A_l}\\= & {} \mathrm{e}^{i\varepsilon _e^0tA_0^\dagger A_0}\prod _{l>0}\mathrm{e}^{i\varepsilon _e^{l}tA_{-l}^\dagger A_{-l}}e^{i\varepsilon _e^ltA_l^\dagger A_l}, \end{aligned}$$

where we have used $\varepsilon _e^{-l}=\varepsilon _e^{l}$ in the second equality,

$$\begin{aligned} \mathrm{e}^{i\varepsilon _e^ktA_k^\dagger A_k}=\mathbf {1}+(\mathrm{e}^{i\varepsilon _e^kt}-1)A_k^\dagger A_k, \end{aligned}$$

and for all $m>l\ge 0$,

$$\begin{aligned}&[\cos (\alpha _m)-i\sin (\alpha _m) A_{-m}A_{m},A_{-l}^\dagger A_{-l}]=0,\\&[A_l^\dagger A_l,\cos (\alpha _m)+i\sin (\alpha _m) A_{m}^\dagger A_{-m}^\dagger ]=0,\\&[\cos (\alpha _m)-i\sin (\alpha _m) A_{-m}A_{m},\cos (\alpha _l)+i\sin (\alpha _l) A_{l}^\dagger A_{-l}^\dagger ]=0, \end{aligned}$$

the expression can be further simplified as

$$\begin{aligned} f(\lambda )= & {} \langle \varphi _e(t)|\varphi _g(t)\rangle \\= & {} \mathrm{e}^{i\sum _k(\varepsilon _g^k-\varepsilon _e^k)t/2}\langle E|\mathrm{e}^{i\varepsilon _e^0tA_0^\dagger A_0}\prod _{k>0}\bigl (\cos (\alpha _k)-i\sin (\alpha _k )A_{-k}A_{k}\bigr )\\&\times \mathrm{e}^{i\varepsilon _e^{k}tA_{-k}^\dagger A_{-k}} \mathrm{e}^{i\varepsilon _e^ktA_k^\dagger A_k}\bigl (\cos (\alpha _k)+i\sin (\alpha _k) A_k^\dagger A_{-k}^\dagger \bigr )|E\rangle . \end{aligned}$$

Finally, using

$$\begin{aligned}&\mathrm{e}^{i\varepsilon _e^{k}tA_{-k}^\dagger A_{-k}}\mathrm{e}^{i\varepsilon _e^ktA_k^\dagger A_k}A_k^\dagger A_{-k}^\dagger \\&\quad =\mathrm{e}^{i\varepsilon _e^ktA_{-k}^\dagger A_{-k}}(\mathbf {1}+(\mathrm{e}^{i\varepsilon _e^kt}-1)A_k^\dagger A_k)A_k^\dagger A_{-k}^\dagger \\&\quad =\mathrm{e}^{i\varepsilon _e^ktA_{-k}^\dagger A_{-k}}\mathrm{e}^{i\varepsilon _e^kt}A_k^\dagger A_{-k}^\dagger \nonumber \\&\quad =\mathrm{e}^{i\varepsilon _e^kt}(\mathbf {1}+(\mathrm{e}^{i\varepsilon _e^kt}-1)A_{-k}^\dagger A_{-k})A_k^\dagger A_{-k}^\dagger \\&\quad =\mathrm{e}^{2i\varepsilon _e^kt}A_k^\dagger A_{-k}^\dagger , \end{aligned}$$

and $A_k|E\rangle =0$ for all k, we get

$$\begin{aligned} f(\lambda )=\mathrm{e}^{i\sum _k(\varepsilon _g^k-\varepsilon _e^k)t/2}\prod _{k>0}\bigl (\cos ^2(\alpha _k)+\sin ^2(\alpha _k) \mathrm{e}^{2i\varepsilon _e^kt}\bigr ). \end{aligned}$$

Furthermore, it is worth mentioning that the explicit expression of Loschmidt echo $L(\lambda )$ can be directly derived from $f(\lambda )$. Specifically,

$$\begin{aligned} L(\lambda )=|f(\lambda )|^2=\prod _{k>0}\bigl (1-\sin ^2(2\alpha _k) \sin ^2(\varepsilon _e^kt)\bigr ), \end{aligned}$$

which is in accordance with the result in Ref. [24].