Fotoc complexity in the Lipkin–Meshkov–Glick model and its variant

Abstract

We study fidelity out-of-time-order correlators (FOTOCs) in the Lipkin–Meshkov–Glick (LMG) model and one of its variants dubbed as the extended LMG model. We demonstrate that these exhibit distinctive behaviour at quantum phase transitions (QPTs) in both the ground and the excited states. We show that the dynamics of the FOTOC have different behaviour in the symmetric and broken phases and as one approaches QPT. Rescaling the FOTOC operator with time, we establish that for small times the rescaled operator is identical to the Loschmidt echo. We also compute the Nielsen complexity of the FOTOC operator for both models and apply this operator on the ground and excited states to obtain the quasi-scrambled states. The FOTOC operator introduces a small perturbation on the original ground and excited states. For this perturbed state, we compute the quantum information metric to first and second order in perturbation, in the thermodynamic limit. We find that the associated Ricci scalar does not show any divergence or discontinuity at QPTs in the LMG model. Instead, the amplitude of oscillations is relatively lower in the symmetric phase than in the broken phase. However, in the case of extended LMG model, the Ricci scalar diverges at the QPT from the broken phase side, in contrast to the zeroth order result. Finally, we comment upon the Fubini-Study complexity in the extended LMG model.

Graphical abstract

Appendices

Appendix A

In this section, we present a detailed calculation of the NC for LMG model following [49]. We apply Nielsen’s geometric approach to find the optimal circuit. The first step is to parametrize the unitary as a path-ordered exponential

$$\begin{aligned} U(\tau )=\overleftarrow{\mathcal {P}}\exp \left( \int _{0}^{\tau } d\tau ^{\prime }H(\tau ^{\prime })\right) , \end{aligned}$$

(A1)

with $H(\tau ^{\prime })=\sum _{I\in \{1,3\}}Y^{I}(\tau ^{\prime })M_{I}$, $Y^{I}(\tau ^{\prime })$ are control functions and specifies a particular circuit in the space of unitaries, and $M_{I}$ are the three-dimensional representation of the Heisenberg group generators

$$\begin{aligned} M_{1}= \begin{bmatrix} 0&{}1&{}0\\ 0&{}0&{}0\\ 0&{}0&{}0\\ \end{bmatrix} ,\quad M_{2}= \begin{bmatrix} 0&{}0&{}0\\ 0&{}0&{}1\\ 0&{}0&{}0\\ \end{bmatrix} ,\quad M_{3}= \begin{bmatrix} 0&{}0&{}1\\ 0&{}0&{}0\\ 0&{}0&{}0\\ \end{bmatrix}, \end{aligned}$$

(A2)

satisfying the Heisenberg algebra. The explicit form of the functions $Y^{I}(\tau )$ can be evaluated by using Eq. (A1) and the expressions of $M_{I}$,

$$\begin{aligned} Y^{I}(\tau )=-\text {Tr}\left[ (\partial _{\tau }U(\tau )) U^{-1}(\tau )M_{I}^{T}\right] .~ \end{aligned}$$

(A3)

By following the usual procedure [58], we can define the cost functional for various paths

$$\begin{aligned} \mathcal {D}[U(\tau )]=\int _{0}^{1}d\tau ^{\prime }\sum _{I}|Y^{I}(\tau ^{\prime })|^{2},~ \end{aligned}$$

(A4)

where we followed $\kappa =2$ cost functions. Note that the $\kappa =2$ cost function and the $F_{2}$ cost function will provide the exact same extremal trajectories or optimal circuits [63]. The minimal value of the above cost functional will give the required complexity, and it can be obtained by evaluating it on the geodesics of unitary space with the boundary conditions

$$\begin{aligned} \tau =0,\,U(\tau =0)=\mathbb {1},\quad \tau =1,\,U(\tau =1)=\hat{W}_{ls,lb}(t).~ \end{aligned}$$

(A5)

We first represent the FOTOC operator in Heisenberg group generators as

$$\begin{aligned} \hat{W}_{ls}(t)= & {} e^{\varepsilon \left( \cos \left( \omega _{ls}t\right) M_{1} +\sqrt{\frac{h-\Upsilon }{h-1}}\sin \left( \omega _{ls}t\right) M_{2}\right) },\nonumber \\ \hat{W}_{lb}(t)= & {} e^{\varepsilon \left( \cos \left( \omega _{lb}t\right) M_{1}-\sqrt{\frac{1-\Upsilon }{1-h^{2}}}\sin \left( \omega _{lb}t\right) M_{2}\right) }, \end{aligned}$$

(A6)

which is solved as,

$$\begin{aligned} \hat{W}_{ls}(t)= & {} \begin{bmatrix} 1&{}\varepsilon \cos \left( \omega _{ls}t\right) &{}\frac{\varepsilon ^{2}}{4}\sqrt{\frac{h-\Upsilon }{h-1}} \sin \left( 2\omega _{ls}t\right) \\ 0&{}1&{}\varepsilon \sqrt{\frac{h-\Upsilon }{h-1}} \sin \left( \omega _{ls}t\right) \\ 0&{}0&{}1\\ \end{bmatrix},\nonumber \\ \hat{W}_{lb}(t)= & {} \begin{bmatrix} 1&{}\varepsilon \cos \left( \omega _{lb}t\right) &{} -\frac{\varepsilon ^{2}}{4}\sqrt{\frac{1-\Upsilon }{1-h^{2}}}\sin \left( 2\omega _{lb}t\right) \\ 0&{}1&{}-\varepsilon \sqrt{\frac{1-\Upsilon }{1-h^{2}}} \sin \left( \omega _{lb}t\right) \\ 0&{}0&{}1\\ \end{bmatrix}.\nonumber \\ \end{aligned}$$

(A7)

In general, we can parametrize an element of Heisenberg group by $U(\tau )$ where,

$$\begin{aligned} U(\tau )=\begin{bmatrix} 1&{}x_{1}(\tau )&{}x_{3}(\tau )\\ 0&{}1&{}x_{2}(\tau )\\ 0&{}0&{}1\\ \end{bmatrix}, \end{aligned}$$

(A8)

then the Eq. (A4) can be written in the form:

$$\begin{aligned} \mathcal {D}[U(\tau )]= & {} \int _{0}^{1}d\tau \Bigg ((1+x_{2}^{2})\left( \frac{dx_{1}}{d\tau }\right) ^{2}+ \left( \frac{dx_{2}}{d\tau }\right) ^{2}+\nonumber \\{} & {} \left( \frac{dx_{3}}{d\tau }\right) ^{2}-2x_{2} \left( \frac{dx_{1}}{d\tau }\right) \left( \frac{dx_{3}}{d\tau }\right) \Bigg ). \end{aligned}$$

(A9)

The geodesic equations corresponding to the above metric are

$$\begin{aligned} \dot{x}_{2}(x_{2}\dot{x}_{1}-\dot{x}_{3})+\ddot{x}_{1}=0,{} & {} \dot{x}_{1}(-x_{2}\dot{x}_{1}- \dot{x}_{3})+\ddot{x}_{2}=0,\nonumber \\ \dot{x}_{2}((x_{2}^{2}-1)\dot{x}_{1}-x_{2}\dot{x}_{3})+\ddot{x}_{3}= & {} 0, \end{aligned}$$

(A10)

where the dot represents the derivative with respect to $\tau$, and is solved using above boundary conditions for the symmetric phase to get

$$\begin{aligned} x_{1}(\tau )= & {} \varepsilon \cos \left( \omega _{ls}t\right) \tau ,\quad x_{2}(\tau )=\varepsilon \sqrt{\frac{h-\Upsilon }{h-1}} \sin \left( \omega _{ls}t\right) \tau ,\nonumber \\ x_{3}(\tau )= & {} \frac{\varepsilon ^{2}}{4} \sqrt{\frac{h-\Upsilon }{h-1}}\sin \left( 2\omega _{ls}t\right) \tau ^{2}. \end{aligned}$$

(A11)

Using the above solution in Eq. (A9), the final expression for the NC in the symmetric phase is

$$\begin{aligned} \mathcal {C}_{ls}(t)=\varepsilon ^{2}\left( \cos ^{2} \left( \omega _{ls}t\right) +\frac{h-\Upsilon }{h-1} \sin ^{2}\left( \omega _{ls}t\right) \right) . \end{aligned}$$

(A12)

Similarly, the expression of the NC in the broken phase is

$$\begin{aligned} \mathcal {C}_{lb}(t)=\varepsilon ^{2}\left( \cos ^{2} \left( \omega _{lb}t\right) +\frac{1-\Upsilon }{1-h^{2}} \sin ^{2}\left( \omega _{lb}t\right) \right) . \end{aligned}$$

(A13)

Appendix B

In this Appendix, we will give the proof of Eq. (31) and show the detailed calculation of metric components of the LMG model. From Eq. (26),

$$\begin{aligned} {|{\Psi (t)}\rangle }_{ls,lb}= & {} e^{i\varepsilon \hat{Q}_{ls,lb}(t)}{|{0}\rangle }_{ls,lb},\nonumber \\= & {} e^{i\varepsilon \mathcal {A}_{ls,lb}(t)\gamma _{ls,lb}^{\dagger }} \,e^{i\varepsilon \mathcal {A}_{ls,lb}^{*}(t)\gamma _{ls,lb}}\times \,\nonumber \\{} & {} e^{-\frac{1}{2}([i\varepsilon \mathcal {A}_{ls,lb}(t)\gamma _{ls,lb}^{\dagger },\, i\varepsilon \mathcal {A}_{ls,lb}^{*}(t)\gamma _{ls,lb}])}{|{0}\rangle }_{ls,lb}.\nonumber \\ \end{aligned}$$

(B1)

The last step is written by using the Baker-Campbell-Hausdorff theorem: $e^{\hat{A}+\hat{B}}=e^{\hat{A}}e^{\hat{B}}e^{-\frac{1}{2}[\hat{A},\,\hat{B}]}$, and $[\hat{A},\,[\hat{A},\,\hat{B}]]=[\hat{B},\,[\hat{A},\,\hat{B}]]=0$ Then,

$$\begin{aligned} {|{\Psi (t)}\rangle }_{ls,lb}= & {} e^{-\frac{\varepsilon ^{2}}{2}|\mathcal {A}_{ls,lb}(t) |^{2}}e^{i\varepsilon \mathcal {A}_{ls,lb}(t) \gamma _{ls,lb}^{\dagger }}{|{0}\rangle }_{ls,lb},\nonumber \\= & {} \sum \limits _{m=0}^{\infty }e^{ -\frac{\varepsilon ^{2}}{2}|\mathcal {A}_{ls,lb}(t)|^{2}} \frac{\left( i\varepsilon \mathcal {A}_{ls,lb}(t)\right) ^{m}}{\sqrt{m!}}{|{m}\rangle }_{ls,lb}.\nonumber \\ \end{aligned}$$

(B2)

This completes the proof of Eq. (31). Next, we turn to the metric components of the LMG model. For simplicity, we only derive the $g_{hh}$ component in the symmetric phase using real part of Eq. (28),

$$\begin{aligned} g_{hh}= & {} \text {Re}\big [_{ls}\!\!{\langle {\partial _{h} \Psi (t)|\partial _{h}\Psi (t)}\rangle }_{ls} - \nonumber \\{} & {} _{ls}{\langle {\partial _{h}\Psi (t)|\Psi (t)}\rangle }_{ls} {\langle {\Psi (t)|\partial _{h}\Psi (t)}\rangle }_{ls}\big ], \end{aligned}$$

(B3)

and other components follow from the similar calculations. By taking the derivative of Eq. (33) with respect to h, we have up to $\mathcal {O}(\varepsilon ^{2})$

$$\begin{aligned} \Big |\frac{\partial \Psi (t)}{\partial h}\Big \rangle _{ls}= & {} \frac{\partial }{\partial h}\left( \!1-\frac{\varepsilon ^{2}}{2}| \mathcal {A}_{ls}(t)|^{2}\!\right) \!\!{|{\psi _{0}}\rangle }_{ls} + i\varepsilon \frac{\partial \mathcal {A}_{ls}(t)}{\partial h} {|{\psi _{1}}\rangle }_{ls}\nonumber \\{} & {} +\left( \!1-\frac{\varepsilon ^{2}}{2}| \mathcal {A}_{ls}(t)|^{2}\!\right) \!\!\Big | \frac{\partial \psi _{0}}{\partial h}\Big \rangle _{ls} + i\varepsilon \mathcal {A}_{ls}(t)\Big |\frac{\partial \psi _{1}}{\partial h}\Big \rangle _{ls} \nonumber \\{} & {} - \frac{\varepsilon ^{2}}{\sqrt{2}}\frac{\partial \mathcal {A}^{2}_{ls}(t)}{\partial h}{|{\psi _{2}}\rangle }_{ls} - \frac{\varepsilon ^{2}}{\sqrt{2}}\mathcal {A}^{2}_{ls}(t)\Big |\frac{\partial \psi _{2}}{\partial h}\Big \rangle _{ls}, \end{aligned}$$

(B4)

with ${|{0}\rangle }\equiv {|{\psi _{0}}\rangle }$, ${|{1}\rangle }\equiv {|{\psi _{1}}\rangle }$, ${|{2}\rangle }\equiv {|{\psi _{2}}\rangle }$, etc. By doing calculations using Eq. (30), the only non-zero terms in the $g_{hh}$ component are

$$\begin{aligned} g_{hh}= & {} \left( 1-\varepsilon ^{2}|\mathcal {A}_{ls}(t)|^{2}\right) {\langle {\partial _{h}\psi _{0}|\partial _{h}\psi _{0}}\rangle } + \varepsilon ^{2} | \partial _{h}\mathcal {A}_{ls}(t)|^{2}\nonumber \\{} & {} -\frac{\varepsilon ^{2}}{\sqrt{2}}\left( \left( \partial _{h} \mathcal {A}^{2}_{ls}(t)\right) ^{*}{\langle {\psi _{2}|\partial _{h}\psi _{0}}\rangle } +\partial _{h}\mathcal {A}^{2}_{ls}(t){\langle {\partial _{h} \psi _{0}|\psi _{2}}\rangle }\right) \nonumber \\{} & {} +\varepsilon ^{2}|\mathcal {A}_{ls}(t)|^{2} {\langle {\partial _{h}\psi _{1}|\partial _{h}\psi _{1}}\rangle }, \end{aligned}$$

(B5)

with

$$\begin{aligned} {\langle {\partial _{h}\psi _{0}|\partial _{h}\psi _{0}}\rangle }= & {} \frac{\left( \Upsilon - 1\right) ^{2}}{32 (h-1)^{2}\left( h-\Upsilon \right) ^{2}},\nonumber \\ {\langle {\partial _{h}\psi _{0}|\psi _{2}}\rangle }= & {} {\langle {\psi _{2}|\partial _{h} \psi _{0}}\rangle }=-\frac{1}{\sqrt{2}}\frac{\sinh \theta _{ls}}{\omega _{ls}},\nonumber \\ {\langle {\partial _{h}\psi _{1}|\partial _{h}\psi _{1}}\rangle }= & {} \frac{3}{2}\frac{\sinh ^{2} \theta _{ls}}{\omega _{ls}^{2}}, \end{aligned}$$

(B6)

where $\omega _{ls}$ and $\theta _{ls}$ are given in sections 4 and 5, respectively. In the $\varepsilon \rightarrow 0$ limit, it reproduces the result obtained for $g_{hh}$ in [86]. Using similar mathematical steps as above, we can also compute other metric components which are $g_{h\Upsilon }$, and $g_{\Upsilon \Upsilon }$.

Appendix C

In this section, for the sake of completeness, we evaluate the metric components of the perturbed ground state. The Hamiltonian of the ground state takes the form:

$$\begin{aligned} \hat{H}_{g}\simeq -j\sqrt{1+\Omega _{x}^{2}}+\left( \frac{\sqrt{1+\Omega _{x}^{2}} +2\xi _{y}}{2}\right) \hat{P}^{2}+ \frac{\sqrt{1+\Omega _{x}^{2}}}{2}\hat{Q}^{2}, \end{aligned}$$

(C1)

which describes the harmonic oscillator with frequency $\omega _{g}=(1+\Omega _{x}^{2})^{1/4}\sqrt{\Gamma _{+}}$ and $\Gamma _{+}=\sqrt{1+\Omega _{x}^{2}}+2\xi _{y}$. The Bogoliubov transformation of the ground state is

$$\begin{aligned} \hat{Q}= & {} \left( \frac{\Gamma _{+}}{4\sqrt{1+\Omega _{x}^{2}}}\right) ^{1/4} \left( \gamma _{g}^{\dagger }+\gamma _{g}\right) ,\nonumber \\ \hat{P}= & {} i\left( \frac{\sqrt{1+\Omega _{x}^{2}}}{4\Gamma _{+}}\right) ^{1/4} \left( \gamma _{g}^{\dagger }-\gamma _{g}\right) , \end{aligned}$$

(C2)

with $\gamma _{g}{|{0}\rangle }_{g}=0$ and ${|{0}\rangle }_{g}$ is the Bogoliubov ground state. Following a similar analysis as for the excited state, we perturb the ground state as

$$\begin{aligned} {|{\Psi (t)}\rangle }_{g}=e^{i\hat{H}_{g}t}e^{i\varepsilon \hat{Q}}e^{-i\hat{H}_{g}t}{|{0}\rangle }_{g}\equiv e^{i\varepsilon \hat{Q}_{g}(t)}{|{0}\rangle }_{g}, \end{aligned}$$

(C3)

with

$$\begin{aligned} \hat{Q}_{g}=\hat{Q}\cos \omega _{g}t+\hat{P}\, \frac{\Gamma _{+}^{1/2}\sin \omega _{g}t}{(1+\Omega _{x}^{2})^{1/4}}, \end{aligned}$$

(C4)

and the metric components up to $\mathcal{O}(\varepsilon )$ turn out to be

$$\begin{aligned} g_{\Omega _{x}\Omega _{x}}= & {} {\frac{j}{2(1+\Omega _{x}^{2})^{7/4} \Gamma _{+}^{1/2}} + \frac{\xi _{y}^{2}\Omega _{x}^{2}}{8(1+\Omega _{x}^{2})^{2}\Gamma _{+}^{2}}} + \nonumber \\{} & {} \varepsilon \frac{\sqrt{j}\,\Omega _{x}t\left( 2\xi _{y}^{2}+3\xi _{y} \sqrt{1+\Omega _{x}^{2}}+1+\Omega _{x}^{2}\right) \cos \omega _{g}t}{(1+\Omega _{x}^{2})^{7/4}\Gamma _{+}^{3/2}}\nonumber \\{} & {} -\varepsilon \frac{\sqrt{j}\,\Omega _{x}\xi _{y}\sin \omega _{g}t}{(1+\Omega _{x}^{2})^{2}\Gamma _{+}},\nonumber \\ g_{\Omega _{x}\xi _{y}}= & {} {-\frac{\xi _{y}\Omega _{x}}{8(1+\Omega _{x}^{2})\Gamma _{+}^{2}}}+\varepsilon \frac{\sqrt{j}\,t\cos \omega _{g}t}{2(1+\Omega _{x}^{2}) ^{3/4}\Gamma _{+}^{1/2}}+\nonumber \\{} & {} \varepsilon \frac{\sqrt{j}\,\sin \omega _{g}t}{2(1+\Omega _{x}^{2})\Gamma _{+}},\nonumber \\ g_{\xi _{y}\xi _{y}}= & {} {\frac{1}{8\Gamma _{+}^{2}}},\quad g_{\Omega _{x}t}=\varepsilon \frac{\sqrt{j}\,\Gamma _{+}^{1/2}\cos \omega _{g}t}{2(1+\Omega _{x}^{2})^{3/4}},\nonumber \\ g_{\xi _{y}t}= & {} g_{tt}=0. \end{aligned}$$

(C5)

We have computed the Ricci scalar in the parameter space $(\Omega _{x},\xi _{y})$ for fixed t and find that $R\rightarrow -4$ as $\xi _y \rightarrow -\xi _{yc}$ similar to the situation in the symmetric phase.