LMG_quench.lyx

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\begin_body

\begin_layout Title
Quenches in the LMG model
\end_layout

\begin_layout Section
LMG Model: Details for implementing numerically
\end_layout

\begin_layout Standard
We start with the following Hamiltonian
\begin_inset Formula 
\begin{equation}
H=-\frac{J}{N}\sum_{i<j}\gamma_{x}\sigma_{i}^{x}\sigma_{j}^{x}+\gamma_{y}\sigma_{i}^{y}\sigma_{j}^{y}-\Gamma\sum_{i}\sigma_{i}^{z}
\end{equation}

\end_inset

A somewhat easier formula if we are more interested in the ferromagnetic
 phase is, to make the following rotation of axis about 
\begin_inset Formula $y$
\end_inset

 : 
\begin_inset Formula $z\rightarrow x$
\end_inset

, 
\begin_inset Formula $x\rightarrow-z$
\end_inset


\begin_inset Formula 
\begin{align}
H & =-\frac{J}{N}\sum_{i<j}\gamma_{z}\sigma_{i}^{z}\sigma_{j}^{z}+\gamma_{y}\sigma_{i}^{y}\sigma_{j}^{y}-\Gamma\sum_{i}\sigma_{i}^{x}\\
 & =+\frac{J}{2N}\left(\sum_{i}\gamma_{z}\left(\sigma_{i}^{z}\right)^{2}+\gamma_{y}\left(\sigma_{i}^{y}\right)^{2}\right)-\frac{J}{2N}\left(\sum_{ij}\left(\gamma_{z}\sigma_{i}^{z}\sigma_{j}^{z}+\gamma_{y}\sigma_{i}^{y}\sigma_{j}^{y}\right)\right)-\Gamma\sum_{i}\sigma_{i}^{x}\\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{J}{2N}\sum_{i,j}\left(\gamma_{z}\sigma_{i}^{z}\sigma_{j}^{z}+\gamma_{y}\sigma_{i}^{y}\sigma_{j}^{y}\right)-\Gamma\sum_{i}\sigma_{i}^{x}
\end{align}

\end_inset

Now defining the total spin variables,
\begin_inset Formula $S_{z}=\frac{1}{2}\sum_{i}\sigma_{i}^{z}$
\end_inset

, 
\begin_inset Formula $S_{x}=\frac{1}{2}\sum_{i}\sigma_{i}^{x}$
\end_inset

, 
\begin_inset Formula $S_{y}=\frac{1}{2}\sum_{i}\sigma_{i}^{y}$
\end_inset

, and 
\begin_inset Formula $S^{2}=S_{x}^{2}+S_{y}^{2}+S_{z}^{2}$
\end_inset

 ,
\begin_inset Formula $S_{\pm}=\left(S_{x}\pm iS_{y}\right)$
\end_inset

 we obtain,
\begin_inset Formula 
\begin{align*}
H & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{z}^{2}-\frac{2J\gamma_{y}}{N}S_{y}^{2}-2\Gamma S_{x}\\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{z}^{2}+\frac{J\gamma_{y}}{2N}\left(S_{+}-S_{-}\right)^{2}-2\Gamma S_{x}\\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{z}^{2}-\frac{J\gamma_{y}}{2N}\left(S_{+}S_{-}+S_{-}S_{+}\right)+\frac{J\gamma_{y}}{2N}\left(S_{+}^{2}+S_{-}^{2}\right)-2\Gamma S_{x}\\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{z}^{2}-\frac{J\gamma_{y}}{N}\left(S^{2}-S_{z}^{2}\right)+\frac{J\gamma_{y}}{2N}\left(S_{+}^{2}+S_{-}^{2}\right)-\Gamma\left(S_{+}+S_{-}\right)\\
\Rightarrow H & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\left(1-\frac{2S^{2}}{N}\right)\right)-\frac{J\left(2\gamma_{z}-\gamma_{y}\right)}{N}S_{z}^{2}+\frac{J\gamma_{y}}{2N}\left(S_{+}^{2}+S_{-}^{2}\right)-\Gamma\left(S_{+}+S_{-}\right)
\end{align*}

\end_inset

We know that the eigenstates are just total angular momentum states given
 by 
\begin_inset Formula $|S,M\rangle$
\end_inset

 with 
\begin_inset Formula $M\in\{-S,\cdots S\}$
\end_inset

 and 
\begin_inset Formula $S\in\left\{ 0\cdots\frac{N}{2}\right\} $
\end_inset

.
 Since different 
\begin_inset Formula $S$
\end_inset

-sectors do not talk to each other, we will focus on the largest 
\begin_inset Formula $S$
\end_inset

 given by 
\begin_inset Formula $S=\frac{N}{2}$
\end_inset

.
 This means 
\begin_inset Formula $S^{2}\rightarrow\frac{N}{2}\left(\frac{N}{2}+1\right)$
\end_inset

.
 Now in this sector we have,
\begin_inset Formula 
\begin{equation}
H=\frac{J}{2}\left(\gamma_{z}-\frac{\gamma_{y}N}{2}\right)-\frac{J\left(2\gamma_{z}-\gamma_{y}\right)}{N}S_{z}^{2}+\frac{J\gamma_{y}}{2N}\left(S_{+}^{2}+S_{-}^{2}\right)-\Gamma\left(S_{+}+S_{-}\right)
\end{equation}

\end_inset

Now, we need to find the action of the different spin operators on the states
\begin_inset Formula 
\begin{align}
S_{z}|S,M\rangle & =M|S,M\rangle\\
S^{2}\left|S,M\rangle\right| & =S\left(S+1\right)|S,M\rangle\\
S_{+}|S,M\rangle & =\sqrt{S\left(S+1\right)-M\left(M+1\right)}|S,M+1\rangle\\
S_{-}|S,M\rangle & =\sqrt{S\left(S+1\right)-M\left(M-1\right)}|S,M-1\rangle\\
S_{+}S_{-}|S,M\rangle & =\left(S^{2}-S_{z}^{2}-S_{z}\right)|S,M\rangle=\left[S\left(S+1\right)-M\left(M-1\right)\right]|S,M\rangle\\
S_{-}S_{+}|S,M\rangle & =\left[S\left(S+1\right)-M\left(M+1\right)\right]|S,M\rangle
\end{align}

\end_inset

Now we are ready to write down the matrix elements,
\begin_inset Formula 
\begin{align}
\left\langle S=\frac{N}{2},M\right|H\left|S=\frac{N}{2},M^{\prime}\right\rangle  & =\delta_{MM^{\prime}}\left[\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\left(1-\frac{2S\left(S+1\right)}{N}\right)\right)-\frac{J\left(2\gamma_{z}-\gamma_{y}\right)}{N}M^{2}\right]+\nonumber \\
 & \ \ \ \delta_{MM^{\prime}-2}\left[\frac{J\gamma_{y}}{2N}\sqrt{\left(S\left(S+1\right)-\left(M+2\right)\left(M+1\right)\right)\left(S\left(S+1\right)-M\left(M+1\right)\right)}\right]+\nonumber \\
 & \ \ \ \delta_{MM^{\prime}+2}\left[\frac{J\gamma_{y}}{2N}\sqrt{\left(S\left(S+1\right)-\left(M-2\right)\left(M-1\right)\right)\left(S\left(S+1\right)-M\left(M-1\right)\right)}\right]+\nonumber \\
 & \ \ \ \delta_{MM^{\prime}-1}\left[-\Gamma\sqrt{\left(S\left(S+1\right)-M\left(M+1\right)\right)}\right]+\nonumber \\
 & \ \ \ \delta_{MM^{\prime}+1}\left[-\Gamma\sqrt{\left(S\left(S+1\right)-M\left(M-1\right)\right)}\right]
\end{align}

\end_inset


\end_layout

\begin_layout Subsection
Ground state critical point
\end_layout

\begin_layout Standard
Let us examine the Hamiltonian first,
\begin_inset Formula 
\begin{align}
H & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{z}^{2}-\frac{2J\gamma_{y}}{N}S_{y}^{2}-2\Gamma S_{x}\nonumber \\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J}{N}\left[\left(\sqrt{\gamma_{z}}S_{z}\right)^{2}+\left(\sqrt{\gamma_{y}}S_{y}\right)^{2}\right]-2\Gamma S_{x}\nonumber \\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J}{N}\left[\left(\sqrt{\gamma_{z}}S_{z}+i\sqrt{\gamma_{y}}S_{y}\right)\left(\sqrt{\gamma_{z}}S_{z}-i\sqrt{\gamma_{y}}S_{y}\right)-i\sqrt{\gamma_{z}\gamma_{y}}\left(S_{y}S_{z}-S_{z}S_{y}\right)\right]-2\Gamma S_{x}\nonumber \\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J}{N}\left[\left(\sqrt{\gamma_{z}}S_{z}+i\sqrt{\gamma_{y}}S_{y}\right)\left(\sqrt{\gamma_{z}}S_{z}-i\sqrt{\gamma_{y}}S_{y}\right)-i\sqrt{\gamma_{z}\gamma_{y}}\left(iS_{x}\right)\right]-2\Gamma S_{x}\nonumber \\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J}{N}\left[\left(\sqrt{\gamma_{z}}S_{z}+i\sqrt{\gamma_{y}}S_{y}\right)\left(\sqrt{\gamma_{z}}S_{z}-i\sqrt{\gamma_{y}}S_{y}\right)\right]-\left(2\Gamma+\frac{2J}{N}\sqrt{\gamma_{z}\gamma_{y}}\right)S_{x}\nonumber \\
\Rightarrow H & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\sqrt{\gamma_{z}+\gamma_{y}}}{N}S_{\phi}^{\dagger}S_{\phi}-\left(2\Gamma+\frac{2J}{N}\sqrt{\gamma_{z}\gamma_{y}}\right)S_{x}
\end{align}

\end_inset

where, 
\begin_inset Formula $S_{\phi}=\frac{1}{\sqrt{\gamma_{z}+\gamma_{y}}}\left(\sqrt{\gamma_{z}}S_{z}-i\sqrt{\gamma_{y}}S_{y}\right)=\left|\cos\phi\right|S_{z}-i\left|\sin\phi\right|S_{y}$
\end_inset

 (when 
\begin_inset Formula $\gamma_{z}=\gamma\cos\phi$
\end_inset

, 
\begin_inset Formula $\gamma_{y}=\gamma\sin\phi$
\end_inset

).
 The critical point from a Holstein Primakoff approximation is 
\begin_inset Formula 
\[
\Gamma_{c}=J\max\left(\gamma_{y},\gamma_{z}\right)
\]

\end_inset

The order parameter to calculate is 
\begin_inset Formula $\left\langle S_{y}^{2}+S_{z}^{2}\right\rangle $
\end_inset


\end_layout

\begin_layout Subsection
Calculating Magnetization
\end_layout

\begin_layout Subsubsection
\begin_inset Formula $z$
\end_inset

-direction
\end_layout

\begin_layout Standard
Total magnetization, accounting for lack of symmetry breaking in finite
 size systems is
\begin_inset Formula 
\[
m_{z}^{2}=\left(\text{Magnetization}\right)^{2}=\left\langle \left(\frac{1}{N}\sum_{i}\sigma_{i}^{z}\right)^{2}\right\rangle =\left\langle \frac{4}{N^{2}}\left(S^{z}\right)^{2}\right\rangle =\frac{4}{N^{2}}\left(\sum_{M=-S}^{S}\left|\alpha_{M}\right|^{2}M^{2}\right)
\]

\end_inset

where the ground state is given as 
\begin_inset Formula $|GS\rangle=\sum_{M}\alpha_{M}|S,M\rangle$
\end_inset

.
\end_layout

\begin_layout Standard
Let us discuss a couple of known properties of the ground state.
 The finite size dependence of the magnetization dependes on whether the
 system is in the critical point or deep in the disordered phase,
\begin_inset Formula 
\begin{align}
\left\langle m^{2}\right\rangle  & \sim N^{-2/3}\ \text{at the QCP,}\Gamma=J\\
\left\langle m^{2}\right\rangle  & \sim N^{-1}\ \text{disordered phase, }\Gamma\gg J\\
\left\langle m^{2}\right\rangle _{{\rm th}} & \sim N^{-1/2}\ \text{at a thermal CP}
\end{align}

\end_inset

where note that the last line is a thermal expectation value at the critical
 point.
 We have verified the first two properties in the numerically obtained magnetiza
tion.
\end_layout

\begin_layout Subsubsection
\begin_inset Formula $y$
\end_inset

-direction
\end_layout

\begin_layout Standard
The y-direction magnetization in the ground state 
\begin_inset Formula $|GS\rangle=\sum_{M}\alpha_{M}|S,M\rangle$
\end_inset

is defined as,
\begin_inset Formula 
\begin{align}
m_{y}^{2} & =\left\langle \left(\frac{1}{N}\sum_{i}\sigma_{i}^{y}\right)^{2}\right\rangle =\left\langle \frac{4}{N^{2}}\left(S^{y}\right)^{2}\right\rangle =\frac{1}{N^{2}}\left\langle \left(S_{+}-S_{-}\right)^{2}\right\rangle \nonumber \\
 & =\frac{1}{N^{2}}\left\langle \left(S_{+}\right)^{2}+\left(S_{-}\right)^{2}-\left(S_{+}S_{-}+S_{-}S_{+}\right)\right\rangle \nonumber \\
 & =\frac{1}{N^{2}}\left\langle \left(S_{+}\right)^{2}+\left(S_{-}\right)^{2}-2\left(S^{2}-S_{z}^{2}\right)\right\rangle \nonumber \\
 & =-\frac{2}{N^{2}}\left(S\left(S+1\right)-\sum_{M=-S}^{S}\left|\alpha_{M}\right|^{2}M^{2}\right)+\frac{1}{N^{2}}\left\langle \left(S_{+}\right)^{2}+\left(S_{-}\right)^{2}\right\rangle \nonumber \\
 & =-\frac{2}{N^{2}}\left(S\left(S+1\right)-\sum_{M=-S}^{S}\left|\alpha_{M}\right|^{2}M^{2}\right)\nonumber \\
 & +\frac{1}{N^{2}}\left(\sum_{M=-S}^{S}\alpha_{M+2}^{*}\alpha_{M}\sqrt{\left(S\left(S+1\right)-\left(M+2\right)\left(M+1\right)\right)\left(S\left(S+1\right)-M\left(M+1\right)\right)}\right.\nonumber \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \left.+\alpha_{M-2}^{*}\alpha_{M}\sqrt{\left(S\left(S+1\right)-\left(M-2\right)\left(M-1\right)\right)\left(S\left(S+1\right)-M\left(M-1\right)\right)}\right)
\end{align}

\end_inset


\end_layout

\begin_layout Subsubsection
'
\begin_inset Formula $\phi$
\end_inset

'-direction
\end_layout

\begin_layout Standard
We calculate the expectation value of a general operator :
\begin_inset Formula $S_{\phi}=\left(A_{z}S_{z}+A_{y}S_{y}\right)=\left(\frac{1}{2}\left(A_{s}\sum_{i}\sigma_{i}^{z}+A_{y}\sum_{i}\sigma_{i}^{y}\right)\right)$
\end_inset

.
 In the ground state 
\begin_inset Formula $|GS\rangle=\sum_{M}\alpha_{M}|S,M\rangle$
\end_inset

,
\begin_inset Formula 
\begin{align}
m_{\phi}^{2} & =\left\langle \frac{4}{N^{2}}S_{\phi}^{\dagger}S_{\phi}\right\rangle \nonumber \\
 & =\frac{4}{N^{2}}\left|\sum_{M}\left(A_{z}S_{z}+A_{y}S_{y}\right)\alpha_{M}|S,M\rangle\right|^{2}\nonumber \\
 & =\frac{4}{N^{2}}\left|\sum_{M}\alpha_{M}A_{z}M+\sum_{M}\alpha_{M}\left(A_{y}\frac{1}{2i}\left(S_{+}-S_{-}\right)\right)|S,M\rangle\right|^{2}\nonumber \\
 & =\frac{4}{N^{2}}\left|\sum_{M}\alpha_{M}A_{z}M|S,M\rangle\right.+\nonumber \\
 & \ \ \ \ \ \ \ \ \ \ \left.+\frac{A_{y}}{2i}\sum_{M}\alpha_{M}\left(\sqrt{S\left(S+1\right)-M\left(M+1\right)}|S,M+1\rangle-\sqrt{S\left(S+1\right)-M\left(M-1\right)}|S,M-1\rangle\right)\right|^{2}
\end{align}

\end_inset

Therefore we have the following definition for 
\begin_inset Formula $\left|m_{\phi}\right|^{2}=\frac{4}{N^{2}}\sum_{M=-S}^{S}\left|A_{M}\right|^{2}$
\end_inset


\begin_inset Formula 
\[
A_{M}=\begin{cases}
A_{z}S\alpha_{S}+\frac{A_{y}}{2i}\alpha_{S-1}\sqrt{2S} & M=S\\
-A_{z}S\alpha_{-S}-\frac{A_{y}}{2i}\alpha_{-S+1}\sqrt{2S} & M=-S\\
\alpha_{M}A_{z}M+\frac{A_{y}}{2i}\left(\alpha_{M-1}\sqrt{S\left(S+1\right)-M\left(M-1\right)}-\alpha_{M+1}\sqrt{S\left(S+1\right)-M\left(M+1\right)}\right) & \forall M\neq S,-S
\end{cases}
\]

\end_inset


\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout

\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
(a) Magnetization as a function of magnetic field.
 (b) Phase diagram as a function of 
\begin_inset Formula $\gamma_{z}$
\end_inset

 and 
\begin_inset Formula $\gamma_{y}$
\end_inset

.
 
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Properties at Finite Temperature
\end_layout

\begin_layout Standard
Let us calculate the properties at finite temperature of the LMG Model.
 We will use the standard mean-field picture to calculate the phase transition.
 We have the Hamiltonian,
\begin_inset Formula 
\begin{align}
H & =-\frac{J}{N}\sum_{i<j}\gamma_{z}\sigma_{i}^{z}\sigma_{j}^{z}+\gamma_{y}\sigma_{i}^{y}\sigma_{j}^{y}-\Gamma\sum_{i}\sigma_{i}^{x}\\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{z}^{2}-\frac{2J\gamma_{y}}{N}S_{y}^{2}-2\Gamma S_{x}
\end{align}

\end_inset

In this section, it is much more convenient to go back to the original represent
ation, 
\begin_inset Formula 
\begin{align}
H^{\prime} & =-\frac{J}{N}\sum_{i<j}\gamma_{x}\sigma_{i}^{x}\sigma_{j}^{x}+\gamma_{y}\sigma_{i}^{y}\sigma_{j}^{y}-\Gamma\sum_{i}\sigma_{i}^{z}\\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{x}^{2}-\frac{2J\gamma_{y}}{N}S_{y}^{2}-2\Gamma S_{z}
\end{align}

\end_inset

Using a mean-field ansatz, 
\begin_inset Formula $m_{a}=\frac{1}{N}\sum_{i}\left\langle \sigma_{i}^{a}\right\rangle $
\end_inset

, we have 
\begin_inset Formula 
\begin{align*}
H & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\sum_{i}\left[\frac{J\gamma_{x}}{2}m_{x}\sigma_{i}^{x}+\frac{J\gamma_{y}}{2}m_{y}\sigma_{i}^{y}+\Gamma\sigma_{i}^{z}\right]
\end{align*}

\end_inset

We drop the term proportional to identity.
 The partition function is
\begin_inset Formula 
\begin{align*}
Z & =\text{Tr}\left\{ \prod_{i}\exp\left[\beta\left(\frac{J\gamma_{x}}{2}m_{x}\sigma_{i}^{x}+\frac{J\gamma_{y}}{2}m_{y}\sigma_{i}^{y}+\Gamma\sigma_{i}^{z}\right)\right]\right\} \\
 & =\text{Tr}\left\{ \prod_{i}\exp\left[\left(\frac{\beta J\gamma_{x}m_{x}}{2},\frac{\beta J\gamma_{y}m_{y}}{2},\beta\Gamma\right)\cdot\boldsymbol{\sigma}_{i}\right]\right\} =\text{Tr}\left\{ \prod_{i}\exp\left[\boldsymbol{K}\cdot\boldsymbol{\sigma}_{i}\right]\right\} \\
 & =\text{Tr}\bigotimes_{i}\left[\cosh K+\hat{K}.\boldsymbol{\sigma}_{i}\sinh K\right]\text{, with }K=\beta\sqrt{\frac{1}{4}J^{2}\left(\gamma_{x}^{2}m_{x}^{2}+\gamma_{y}^{2}m_{y}^{2}\right)+\Gamma^{2}}\\
Z & =\left(2\cosh K\right)^{N}
\end{align*}

\end_inset

We have the self consistency equations,
\begin_inset Formula 
\begin{align*}
m_{x} & =\frac{\left(2\cosh K\right)^{N-1}\hat{K}_{x}2\sinh K}{\left(2\cosh K\right)^{N}}=\frac{\beta J\gamma_{x}m_{x}}{2K}\tanh K\\
m_{y} & =\frac{\beta J\gamma_{y}m_{y}}{2K}\tanh K
\end{align*}

\end_inset

To obtain the phase boundary, we take the limit 
\begin_inset Formula $m_{x},m_{y}\rightarrow0$
\end_inset


\begin_inset Formula 
\begin{align*}
\frac{2\Gamma}{J\gamma_{x,y}} & =\tanh K=\tanh\beta\Gamma\\
\Rightarrow & \frac{\frac{2\Gamma}{J\gamma_{x,y}}+1}{-\frac{2\Gamma}{J\gamma_{x,y}}+1}=e^{2\beta\Gamma}\\
\Rightarrow\beta & =\frac{1}{2\Gamma}\log\left(\frac{J\gamma_{x,y}+2\Gamma}{J\gamma_{x,y}-2\Gamma}\right)///check\ \ Calculation
\end{align*}

\end_inset

So basically we take the larger two 
\begin_inset Formula $\gamma_{x},\gamma_{y}$
\end_inset

 to obtain 
\begin_inset Formula $\Gamma$
\end_inset

.
\end_layout

\begin_layout Standard
To calculate numerically, we do the following :
\end_layout

\begin_layout Enumerate
Diagonalize 
\begin_inset Formula $H$
\end_inset

 in each spin sector, and obtain energies and eigenvectors : 
\begin_inset Formula $E_{n}(s=0\cdots S=N/2)$
\end_inset

 and 
\begin_inset Formula $|\Phi_{n}\rangle$
\end_inset


\begin_inset Newline newline
\end_inset


\begin_inset Formula 
\[
H=\oplus\sum_{n=-S}^{+S}E_{n}(S)|\Phi_{n}\rangle\langle\Phi_{n}|
\]

\end_inset


\end_layout

\begin_layout Enumerate
Obtain the magnetization of each eigenstate 
\begin_inset Formula $M_{\Phi_{n}}^{2}$
\end_inset

.
\begin_inset Newline newline
\end_inset


\end_layout

\begin_layout Enumerate
The average magnetization is 
\begin_inset Formula 
\begin{align}
\left\langle M\right\rangle  & =\frac{\sum_{s}D\left(S\right)\sum_{n}M_{\Phi_{n}}e^{-\beta E_{n}(s)}}{\sum_{s}D\left(S\right)\sum_{n}e^{-\beta E_{n}(s)}}\\
 & =\frac{\sum_{s}D\left(S\right)e^{-\beta E_{\text{min}}(s)}\sum_{n}M_{\Phi_{n}}e^{-\beta\left(E_{n}(s)-E_{\text{min}}(s)\right)}}{\sum_{s}D\left(S\right)e^{-\beta E_{\text{min}}(s)}\sum_{n}e^{-\beta\left(E_{n}(s))-E_{\text{min}}(s)\right)}}\\
 & =\frac{\sum_{s}D\left(S\right)e^{-\beta\left(E_{\text{min}}(s)-E_{\text{min}}^{{\rm tot}}\right)}\sum_{n}M_{\Phi_{n}}e^{-\beta\left(E_{n}(s)-E_{\text{min}}(s)\right)}}{\sum_{s}D\left(S\right)e^{-\beta\left(E_{\text{min}}(s)-E_{\text{min}}^{{\rm tot}}\right)}\sum_{n}e^{-\beta\left(E_{n}(s))-E_{\text{min}}(s)\right)}}
\end{align}

\end_inset

where we do the last two steps are taken to avoid having to take very large
 exponentials, which often happens because the minimum energy is often negative.
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout

\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
(a) Finite temperature phase diagram vs mean field solution.
 (b) Scaling with system size, of different quantities 
\begin_inset Formula $\left\langle S_{z}^{2}\right\rangle _{\beta}$
\end_inset

, 
\begin_inset Formula $\left\langle S_{y}^{2}\right\rangle _{\beta}$
\end_inset

 and 
\begin_inset Formula $\left\langle S_{z}^{2}+S_{y}^{2}\right\rangle _{\beta}$
\end_inset

 as a function of system size 
\begin_inset Formula $L$
\end_inset

.
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Unequal time correlation function 
\end_layout

\begin_layout Subsubsection
\begin_inset Formula $S_{\phi}$
\end_inset

 correlation 
\begin_inset Formula $\left\langle S_{\phi}\left(t_{2}\right)S_{\phi}\left(t_{1}\right)\right\rangle _{0}$
\end_inset


\end_layout

\begin_layout Standard
We note that 
\begin_inset Formula $S_{\phi}=A_{z}S_{z}+A_{y}S_{y}$
\end_inset

.
 We have, in the initial state, 
\begin_inset Formula $|\psi_{0}\rangle$
\end_inset


\begin_inset Formula 
\begin{align*}
\left\langle S_{\phi}\left(t_{1}\right)S_{\phi}\left(t_{2}\right)\right\rangle _{0} & =\left\langle \psi_{0}\left|S_{\phi}\left(t_{1}\right)S_{\phi}\left(t_{2}\right)\right|\psi_{0}\right\rangle \\
 & =\left\langle \psi_{0}\left|e^{iHt_{1}}S_{\phi}e^{iH\left(t_{2}-t_{1}\right)}S_{\phi}\left(t\right)e^{-iHt_{2}}\right|\psi_{0}\right\rangle 
\end{align*}

\end_inset

Let us obtain the action of 
\begin_inset Formula $S_{\phi}$
\end_inset

 on an arbitrary state 
\begin_inset Formula $|\Psi\rangle=\sum_{M}\alpha_{M}|S,M\rangle$
\end_inset

,
\begin_inset Formula 
\begin{align*}
S_{\phi}|\Psi\rangle & =\sum_{M}\left(A_{z}S_{z}+A_{y}S_{y}\right)\alpha_{M}|S,M\rangle\\
 & =\sum_{M}\alpha_{M}A_{z}M|S,M\rangle+\sum_{M}\alpha_{M}\left(A_{y}\frac{1}{2i}\left(S_{+}-S_{-}\right)\right)|S,M\rangle\\
 & =\sum_{M}\alpha_{M}A_{z}M|S,M\rangle+\frac{A_{y}}{2i}\sum_{M}\alpha_{M}\left(\sqrt{S\left(S+1\right)-M\left(M+1\right)}|S,M+1\rangle-\sqrt{S\left(S+1\right)-M\left(M-1\right)}|S,M-1\rangle\right)
\end{align*}

\end_inset

Therefore, 
\begin_inset Formula $S_{\phi}|\Psi\rangle=\sum_{M}B_{M}|S,M\rangle$
\end_inset

 is given as,
\begin_inset Formula 
\[
B_{M}=\begin{cases}
A_{z}S\alpha_{S}+\frac{A_{y}}{2i}\alpha_{S-1}\sqrt{2S} & M=S\\
-A_{z}S\alpha_{-S}-\frac{A_{y}}{2i}\alpha_{-S+1}\sqrt{2S} & M=-S\\
\alpha_{M}A_{z}M+\frac{A_{y}}{2i}\left(\alpha_{M-1}\sqrt{S\left(S+1\right)-M\left(M-1\right)}-\alpha_{M+1}\sqrt{S\left(S+1\right)-M\left(M+1\right)}\right) & \forall M\neq S,-S
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Finite temperature 
\begin_inset Formula $S_{\phi}$
\end_inset

 Correlation : 
\begin_inset Formula $\left\langle S_{\phi}\left(t_{2}\right)S_{\phi}\left(t_{1}\right)\right\rangle _{\beta}$
\end_inset


\end_layout

\begin_layout Standard
We have the correlation defined as follows,
\begin_inset Formula 
\begin{align*}
\left\langle S_{\phi}\left(t_{2}\right)S_{\phi}\left(t_{1}\right)\right\rangle _{\beta} & =\frac{\sum_{s}D\left(S\right)\sum_{n}\left\langle S_{\phi}\left(t_{2}\right)S_{\phi}\left(t_{1}\right)\right\rangle _{\Phi_{n}}e^{-\beta E_{n}(s)}}{\sum_{s}D\left(S\right)\sum_{n}e^{-\beta E_{n}(s)}}\\
 & =\frac{\sum_{s}D\left(S\right)e^{-\beta\left(E_{\text{min}}(s)-E_{\text{min}}^{{\rm tot}}\right)}\sum_{n}\left\langle S_{\phi}\left(t_{2}\right)S_{\phi}\left(t_{1}\right)\right\rangle _{\Phi_{n}}e^{-\beta\left(E_{n}(s)-E_{\text{min}}(s)\right)}}{\sum_{s}D\left(S\right)e^{-\beta\left(E_{\text{min}}(s)-E_{\text{min}}^{{\rm tot}}\right)}\sum_{n}e^{-\beta\left(E_{n}(s))-E_{\text{min}}(s)\right)}}
\end{align*}

\end_inset

We can thus use the above formulations to find the finite temperature expectatio
n value.
\begin_inset Formula 
\begin{align*}
\left\langle \left[S_{z}\left(t\right),S_{z}\left(t^{\prime}\right)\right]_{\pm}\right\rangle _{0} & =\left\langle S_{z}\left(t\right)S_{z}\left(t^{\prime}\right)\pm S_{z}\left(t^{\prime}\right)S_{z}\left(t\right)\right\rangle _{0}\\
 & =\sum_{M,M\prime}\langle S,M^{\prime}|\alpha_{M^{\prime}}^{*}\left(e^{iHt}S_{z}e^{iH\left(t^{\prime}-t\right)}S_{z}e^{-iHt^{\prime}}\pm e^{iHt^{\prime}}S_{z}e^{iH\left(t-t^{\prime}\right)}S_{z}e^{-iHt}\right)\alpha_{M}|S,M\rangle\\
 & =\sum_{M,M^{\prime},M^{\prime\prime}}\alpha_{M^{\prime}}^{*}\alpha_{M}\langle S,M^{\prime}|\left(e^{iHt}S_{z}e^{iH\left(t^{\prime}-t\right)}S_{z}e^{-iHt^{\prime}}\pm e^{iHt^{\prime}}S_{z}e^{iH\left(t-t^{\prime}\right)}S_{z}e^{-iHt}\right)|S,M\rangle
\end{align*}

\end_inset

Compared to 
\begin_inset Formula $\left\langle S_{z}^{2}\left(t\right)\right\rangle $
\end_inset

,
\begin_inset Formula 
\begin{align*}
\left\langle S_{z}^{2}\left(t\right)\right\rangle  & =\left\langle e^{iHt}S_{z}^{2}e^{-iHt}\right\rangle _{0}\\
 & =S_{z}^{2}e^{-iHt}\sum_{M}\alpha_{M}|S,M\rangle
\end{align*}

\end_inset


\end_layout

\begin_layout Section
Entanglement Entropy
\end_layout

\begin_layout Standard
Let us discuss the numerical method to compute the entanglement entropy.
 
\end_layout

\begin_layout Subsection
Bipartition
\end_layout

\begin_layout Standard
First let us consider a bipartition, 
\begin_inset Formula $A$
\end_inset

 to the left of 
\begin_inset Formula $B$
\end_inset

.
 Rewriting, 
\begin_inset Formula $S_{z}=S_{zA}+S_{zB}$
\end_inset

.We have,
\begin_inset Formula 
\begin{align*}
H & =-\frac{J}{N}\sum_{i<j}\gamma_{z}\sigma_{i}^{z}\sigma_{j}^{z}+\gamma_{y}\sigma_{i}^{y}\sigma_{j}^{y}-\Gamma\sum_{i}\sigma_{i}^{x}\\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{z}^{2}-\frac{2J\gamma_{y}}{N}S_{y}^{2}-2\Gamma S_{x}\\
\Rightarrow H & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\right)-\frac{2J\gamma_{z}}{N}S_{zA}^{2}-\frac{2J\gamma_{y}}{N}S_{yA}^{2}-2\Gamma S_{xA}\\
 & -\frac{2J\gamma_{z}}{N}S_{zB}^{2}-\frac{2J\gamma_{y}}{N}S_{yB}^{2}-2\Gamma S_{xB}-\frac{4J\gamma_{z}}{N}S_{zA}S_{zB}-\frac{4J\gamma_{y}}{N}S_{yA}S_{yB}
\end{align*}

\end_inset

Now, we can redefine the Hamiltonian in the basis of 
\begin_inset Formula $|S_{A},M_{A};S_{B},M_{B}\rangle$
\end_inset

.
 To do so we need to use the definitions of 
\begin_inset Formula $S_{\pm}=S_{\pm A}+S_{\pm B}$
\end_inset

,
\begin_inset Formula 
\begin{align*}
H & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\left(1-\frac{2S^{2}}{N}\right)\right)-\frac{J\left(2\gamma_{z}-\gamma_{y}\right)}{N}S_{z}^{2}+\frac{J\gamma_{y}}{2N}\left(S_{+}^{2}+S_{-}^{2}\right)-\Gamma\left(S_{+}+S_{-}\right)\\
 & =\frac{J}{2}\left(\gamma_{z}+\gamma_{y}\left(1-\frac{2S^{2}}{N}\right)\right)-\frac{J\left(2\gamma_{z}-\gamma_{y}\right)}{N}\left(S_{zA}^{2}+S_{zB}^{2}+2S_{zA}S_{zB}\right)\\
 & \ \ +\frac{J\gamma_{y}}{2N}\left(S_{+A}^{2}+S_{+B}^{2}+2S_{+A}S_{+B}\right)+\frac{J\gamma_{y}}{2N}\left(S_{-A}^{2}+S_{-B}^{2}+2S_{-A}S_{-B}\right)\\
 & \ \ -\Gamma\left(S_{+A}+S_{+A}+S_{-A}+S_{-B}\right)
\end{align*}

\end_inset

Now the matrix has dimensions 
\begin_inset Formula $(2S_{A}+1)(2S_{B}+1)\times(2S_{A}+1)(2S_{B}+1)$
\end_inset

 with both 
\begin_inset Formula $S_{A}$
\end_inset

 and 
\begin_inset Formula $S_{B}$
\end_inset


\begin_inset Formula 
\[
|S=N/2,S\rangle=|S_{A},S_{A};S-S_{A},S-S_{A}\rangle
\]

\end_inset


\end_layout

\begin_layout Section
Quenching the magnetic field: Holstien-Primakoff Approximation
\end_layout

\begin_layout Subsection
Case with 
\begin_inset Formula $\gamma_{z}=1$
\end_inset

 and 
\begin_inset Formula $\gamma_{y}=0$
\end_inset

.
 [Numeric: Lmg_sim_v0.ipynb]
\end_layout

\begin_layout Standard
In the following I translate the Holstein Primakoff to my language (from
 Mohammad's).
 Consider the Hamilotnian 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
H=-\frac{2J}{N}(S_{{\rm tot}}^{z})^{2}-2\Gamma S_{{\rm tot}}^{x}
\end{equation}

\end_inset

Using the Holstein Primakoff transformation, we have
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
S_{{\rm tot}}^{x} & =\frac{N}{2}-a^{\dagger}a\\
S_{{\rm tot}}^{z} & =\frac{1}{2}\sqrt{N}(a+a^{\dagger})
\end{align}

\end_inset

the Hamiltonian turns into
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
H & =2\Gamma a^{\dagger}a-\frac{J}{2}(a+a^{\dagger})^{2}-\Gamma N\\
 & =2J[-\frac{1}{4}(a+a^{\dagger})^{2}+ga^{\dagger}a]
\end{align}

\end_inset

where 
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\begin_inset Formula $g\equiv\frac{\Gamma}{J}$
\end_inset

.
 It is more convenient to define 
\begin_inset Quotes eld
\end_inset

position
\begin_inset Quotes erd
\end_inset

 and 
\begin_inset Quotes eld
\end_inset

momentum
\begin_inset Quotes erd
\end_inset

 operators as
\begin_inset Formula 
\begin{align}
a & =\frac{1}{\sqrt{2}}(x+ip)\\
a^{\dagger} & =\frac{1}{\sqrt{2}}(x-ip)
\end{align}

\end_inset


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In this representation, the Hamiltonian takes the form
\begin_inset Formula 
\begin{align}
H & =2J\left[-\frac{1}{2}x^{2}+\frac{1}{2}g(x^{2}+p^{2})\right]=2J\left[\frac{1}{2}gp^{2}+\frac{1}{2}(g-1)x^{2}\right]\\
 & =2\Gamma\left[\frac{1}{2}p^{2}+\frac{1}{2}\frac{g-1}{g}x^{2}\right]
\end{align}

\end_inset

It is advantageous to redefine position and momentum variables as 
\begin_inset Formula $x\to\sqrt{2\Gamma}\tilde{x}$
\end_inset

 and 
\begin_inset Formula $p\to\tilde{p}/\sqrt{2\Gamma}$
\end_inset

 which satisfies their commutation relation, but leads to a standard form
 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
H_{0}=\frac{1}{2}\tilde{p}^{2}+\frac{1}{2}\omega^{2}\tilde{x}^{2}
\end{equation}

\end_inset

where 
\begin_inset Formula $\omega^{2}\equiv4\Gamma^{2}(g-1)/g$
\end_inset

.
 
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Now consider a quench of the form 
\begin_inset Formula $g_{0}\to g$
\end_inset

.
 Inital state 
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has equal distribution of energy in momentum and position terms of hamornic
 oscillator, 
\begin_inset Formula $\langle\tilde{p}^{2}/2\rangle=\omega_{0}/4$
\end_inset

 and 
\begin_inset Formula $\langle\omega_{0}^{2}\tilde{x}^{2}/2\rangle_{0}=\omega_{0}/4$
\end_inset

, which yields 
\begin_inset Formula $\langle\tilde{p}^{2}\rangle_{0}=\omega_{0}/2$
\end_inset

 and 
\begin_inset Formula $\langle\tilde{x}^{2}\rangle_{0}=1/(2\omega_{0})$
\end_inset

.
 This means we have in the initial state
\begin_inset Formula 
\begin{equation}
\langle p^{2}\rangle_{0}=\frac{\omega_{0}}{4\Gamma_{0}}\text{, and }\langle x^{2}\rangle_{0}=\frac{\Gamma_{0}}{\omega_{0}},\text{with \omega_{0}^{2}\equiv\ensuremath{\frac{4\Gamma_{0}^{2}(g_{0}-1)}{g_{0}}}}
\end{equation}

\end_inset

Now, evolving under the quench Hamiltonian, which is a harmonic oscillator,
 we have the operator equation of motion after the quench is simply 
\begin_inset Formula $\ddot{\tilde{x}}+\omega^{2}\tilde{x}=0$
\end_inset

 which admits a general solution of the form 
\begin_inset Formula $\tilde{x}(t)=\tilde{x}\left(0\right)\cos\omega t+\frac{1}{\omega}\tilde{p}\left(0\right)\sin\omega t.$
\end_inset

, which gives us,
\begin_inset Formula 
\begin{align}
x(t) & =\left(x\left(0\right)\cos\omega t+\frac{2\Gamma}{\omega}p\left(0\right)\sin\omega t.\right)\\
p\left(t\right) & =\left(p\left(0\right)\cos\omega t-2\Gamma\omega x\left(0\right)\sin\omega t.\right)\\
\langle x(t)^{2}\rangle_{0} & =\left(\cos^{2}\omega t\langle x^{2}\rangle_{0}+\frac{4\Gamma^{2}}{\omega^{2}}\sin^{2}\omega t\ \langle p^{2}\rangle_{0}\right)
\end{align}

\end_inset

where 
\begin_inset Formula $\left\langle \cdots\right\rangle _{0}$
\end_inset

 is the expectation value in the groundstate of the prequench Hamiltonian
 and we have used, 
\begin_inset Formula $\langle xp+px\rangle_{0}=\langle a^{2}+(a^{\dagger})^{2}\rangle_{0}=0$
\end_inset

.
 Simple algebra then gives
\begin_inset Formula 
\begin{align}
\langle x(t)^{2}\rangle & =\frac{1+\cos2\omega t}{2}\frac{\Gamma_{0}}{\omega_{0}}+\frac{1-\cos2\omega t}{4}\frac{\omega_{0}2\Gamma^{2}}{\omega^{2}\Gamma_{0}}\\
 & =\frac{1}{4}\left(\frac{1}{\tilde{\omega}_{0}}+\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)+\frac{1}{4}\left(\frac{1}{\tilde{\omega}_{0}}-\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)\cos2\omega t
\end{align}

\end_inset

where we have redefined variables, 
\begin_inset Formula $\tilde{\omega}^{2}=\frac{\omega^{2}}{4\Gamma^{2}}=\frac{(g-1)}{g}$
\end_inset

 and likewise for 
\begin_inset Formula $\tilde{\omega}_{0}$
\end_inset

.
 Next we relate the expectation value of 
\begin_inset Formula $S_{z}^{2}$
\end_inset

to that 
\begin_inset Formula $x^{2}$
\end_inset

.
 With 
\begin_inset Formula $S_{{\rm tot}}^{z}=\sqrt{\frac{N}{2}}x$
\end_inset

 we have
\begin_inset Formula 
\begin{align}
\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle & =\frac{N}{2}\langle x(t)^{2}\rangle\label{eq: 1}\\
 & =\frac{N}{8}\left[\left(\frac{1}{\tilde{\omega}_{0}}+\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)+\left(\frac{1}{\tilde{\omega}_{0}}-\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)\cos2\omega t\right]\label{eq:Analytical_time_dependence}
\end{align}

\end_inset

The combination of the last two equations gives the desrired result.
 In the limit that 
\begin_inset Formula $\omega\ll\omega_{0}$
\end_inset

we find the average correlation to be 
\begin_inset Formula 
\begin{equation}
\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}=\frac{N}{2}\frac{1}{4}\left(\frac{1}{\tilde{\omega}_{0}}+\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)\approx\frac{N}{8}\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}=\frac{N\Gamma^{2}}{8\Gamma_{0}}\frac{\omega_{0}}{\omega^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
In Fig.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:Sz2-dynamics"
plural "false"
caps "false"
noprefix "false"

\end_inset

, we compare the time dynamics for a fixed system size (
\begin_inset Formula $L=100$
\end_inset

) and compare it to the above analytical formulae.
 If we look at the system size dependence, the Holstien Primakoff approximation
 should get better as system size increases, regardless of how close to
 the critical point it is.
 This is seen in 
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
(a) 
\begin_inset Graphics
	filename plots/Sz2t_[0_0.2_20]_from_L_100,S_50.0,J_1,Γ_4,γ_0toL_100,S_50.0,J_1,Γ_2,γ_0.pdf
	lyxscale 50
	width 30line%

\end_inset

(b) 
\begin_inset Graphics
	filename plots/Sz2t_[0_0.2_20]_from_L_100,S_50.0,J_1,Γ_4,γ_0toL_100,S_50.0,J_1,Γ_1.5,γ_0.pdf
	lyxscale 50
	width 30line%

\end_inset

(c) 
\begin_inset Graphics
	filename plots/Sz2t_[0_0.2_20]_from_L_100,S_50.0,J_1,Γ_4,γ_0toL_100,S_50.0,J_1,Γ_1.02,γ_0.pdf
	lyxscale 50
	width 30line%

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Dynamics of 
\begin_inset Formula $\left\langle S_{z}^{2}\right\rangle $
\end_inset

 following a quench of the magnetic field, 
\begin_inset Formula $\Gamma_{0}\rightarrow\Gamma$
\end_inset

, comparing numerics with the predicted analytics [See Eq.
 
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Analytical_time_dependence"
plural "false"
caps "false"
noprefix "false"

\end_inset

].for a given system size (
\begin_inset Formula $L=100$
\end_inset

).
 (a) quench deep in the disordered phase.
 In this case the Holstein Primakoff approximation should be a very good
 approximation and we see very good agreement.
 (b) and (c) As we move closer to the critical point the approximation gets
 worse with it performing quite badly close to the critical point.
\begin_inset CommandInset label
LatexCommand label
name "fig:Sz2-dynamics"

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
(a)
\begin_inset Graphics
	filename plots/Sz2t_[0_0.2_20]_from_L_100,S_50.0,J_1,Γ_4,γ_0toL_100,S_50.0,J_1,Γ_1.02,γ_0.pdf
	lyxscale 50
	width 30line%

\end_inset

(b)
\begin_inset Graphics
	filename plots/Sz2t_[0_0.2_20]_from_L_1000,S_500.0,J_1,Γ_4,γ_0toL_1000,S_500.0,J_1,Γ_1.02,γ_0.pdf
	lyxscale 50
	width 30line%

\end_inset

(c)
\begin_inset Graphics
	filename plots/Sz2t_[0_0.2_20]_from_L_2000,S_1000.0,J_1,Γ_4,γ_0toL_2000,S_1000.0,J_1,Γ_1.02,γ_0.pdf
	lyxscale 50
	width 30line%

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
System size dependence for a fixed quench field 
\begin_inset Formula $\Gamma=1.02$
\end_inset

.(a) (b) and (c) correspond to 
\begin_inset Formula $L=100$
\end_inset

, 
\begin_inset Formula $1000$
\end_inset

 and 
\begin_inset Formula $2000$
\end_inset

 respectively.
 Clearly the agreement improves with system size.
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout Standard
Let us calculate the expectation values of energy.
 
\begin_inset Formula 
\begin{align*}
\left\langle H\right\rangle _{0} & =2\Gamma\left[\frac{1}{2}\left\langle p^{2}\right\rangle _{0}+\frac{1}{2}\frac{\omega^{2}}{4\Gamma^{2}}\left\langle x^{2}\right\rangle _{0}\right]=\frac{\omega_{0}}{2}\left[\frac{\Gamma}{2\Gamma_{0}}+\frac{\omega^{2}}{2\omega_{0}^{2}}\frac{\Gamma_{0}}{\Gamma}\right]\\
\left\langle H_{0}\right\rangle _{0} & =2\Gamma_{0}\left[\frac{1}{2}\left\langle p^{2}\right\rangle _{0}+\frac{1}{2}\frac{\omega_{0}^{2}}{4\Gamma_{0}^{2}}\left\langle x^{2}\right\rangle _{0}\right]=\left[\frac{\omega_{0}}{4}+\frac{\omega_{0}}{4}\right]=\frac{\omega_{0}}{2}=\ensuremath{\frac{4\Gamma_{0}^{2}(g_{0}-1)}{g_{0}}}\\
\left\langle H\right\rangle _{0}-\left\langle H_{0}\right\rangle _{0} & =\frac{\omega_{0}}{2}\left[\left(\frac{\Gamma}{2\Gamma_{0}}-1\right)+\frac{\omega^{2}}{2\omega_{0}^{2}}\frac{\Gamma_{0}}{\Gamma}\right]\\
\left\langle H\right\rangle _{0}-\left\langle H\right\rangle _{GS} & =\frac{\omega_{0}}{2}\left[\frac{\Gamma}{2\Gamma_{0}}+\frac{\omega^{2}}{2\omega_{0}^{2}}\frac{\Gamma_{0}}{\Gamma}-\frac{\omega}{\omega_{0}}\right]
\end{align*}

\end_inset


\end_layout

\begin_layout Subsection*
B.
 Case with 
\begin_inset Formula $\gamma_{z}=1$
\end_inset

, 
\begin_inset Formula $\gamma_{y}=\gamma\ne0$
\end_inset


\end_layout

\begin_layout Standard
In the following, I (Mohammad) extend the previous analysis (correctly executed
 by Paraj) to the case of the XY model
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
H=-\frac{2J}{N}[(S_{{\rm tot}}^{z})^{2}+\gamma(S_{{\rm tot}}^{y})^{2}]-2\Gamma S_{{\rm tot}}^{x}
\end{equation}

\end_inset

Using the Holstein Primakoff transformation, we have (signs and factors
 of 
\begin_inset Formula $i$
\end_inset

 are chosen to ensure the standard commutation relations)
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
S_{{\rm tot}}^{x} & =\frac{N}{2}-a^{\dagger}a=\frac{N+1}{2}-(a^{\dagger}a+\frac{1}{2})=\frac{N+1}{2}-\frac{1}{2}(x^{2}+p^{2})\\
S_{{\rm tot}}^{z} & \approx\frac{1}{2}\sqrt{N}(a+a^{\dagger})=\sqrt{\frac{N}{2}}x\\
S_{{\rm tot}}^{y} & \approx\frac{i}{2}\sqrt{N}(a-a^{\dagger})=-\sqrt{\frac{N}{2}}p
\end{align}

\end_inset

where we have 
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defined 
\begin_inset Quotes eld
\end_inset

position
\begin_inset Quotes erd
\end_inset

 and 
\begin_inset Quotes eld
\end_inset

momentum
\begin_inset Quotes erd
\end_inset

 operators as
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In this representation 
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\begin_inset Formula 
\begin{align}
a & =\frac{1}{\sqrt{2}}(x+ip)\\
a^{\dagger} & =\frac{1}{\sqrt{2}}(x-ip)
\end{align}

\end_inset


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\uuline default
\uwave default
\noun default
\color inherit
With this, the Hamiltonian takes the form
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
H & =-\Gamma(N+1)+\Gamma(x^{2}+p^{2})-J(x^{2}+\gamma p^{2})\\
 & =(\Gamma-J)x^{2}+(\Gamma-J\gamma)p^{2}+\mbox{const}
\end{align}

\end_inset

It is advantageous to redefine position and momentum variables as 
\begin_inset Formula $x\to\sqrt{2(\Gamma-J\gamma)}\tilde{x}$
\end_inset

 and 
\begin_inset Formula $p\to\tilde{p}/\sqrt{2(\Gamma-J\gamma)}$
\end_inset

 which satisfies their commutation relation, but leads to a standard form
 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
H_{0}=\frac{1}{2}\tilde{p}^{2}+\frac{1}{2}\omega^{2}\tilde{x}^{2}
\end{equation}

\end_inset

where 
\begin_inset Formula $\omega^{2}\equiv4(\Gamma-J)(\Gamma-J\gamma)$
\end_inset

.
 
\family roman
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\noun off
\color none
Now consider a quench of the form 
\begin_inset Formula $(\gamma_{0},\Gamma_{0})\to(\gamma,\Gamma)$
\end_inset

.
 Inital state 
\family default
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\noun default
\color inherit
has equal distribution of energy in momentum and position terms of hamornic
 oscillator, 
\begin_inset Formula $\langle\tilde{p}^{2}/2\rangle=\omega_{0}/4$
\end_inset

 and 
\begin_inset Formula $\langle\omega_{0}^{2}\tilde{x}^{2}/2\rangle_{0}=\omega_{0}/4$
\end_inset

, which yields 
\begin_inset Formula $\langle\tilde{p}^{2}\rangle_{0}=\omega_{0}/2$
\end_inset

 and 
\begin_inset Formula $\langle\tilde{x}^{2}\rangle_{0}=1/(2\omega_{0})$
\end_inset

.
 This means we have in the initial state
\begin_inset Formula 
\begin{equation}
\langle p^{2}\rangle_{0}=\frac{\omega_{0}}{4(\Gamma_{0}-J\gamma_{0})}\text{=\ensuremath{\frac{1}{2}}\ensuremath{\sqrt{\frac{\Gamma_{0}-J}{\Gamma_{0}-J\gamma_{0}}}\equiv\frac{\tilde{\omega_{0}}}{2}}, and }\langle x^{2}\rangle_{0}=\frac{\Gamma_{0}-J\gamma_{0}}{\omega_{0}}=\ensuremath{\frac{1}{2}}\ensuremath{\sqrt{\frac{\Gamma_{0}-J\gamma_{0}}{\Gamma_{0}-J}}}\equiv\frac{1}{2\tilde{\omega_{0}}},
\end{equation}

\end_inset

where we have used 
\begin_inset Formula $\omega_{0}^{2}\equiv\ensuremath{4(\Gamma_{0}-J_{0})(\Gamma_{0}-J\gamma_{0})}$
\end_inset

 and defined 
\begin_inset Formula $\tilde{\omega}_{0}\equiv\sqrt{\frac{\Gamma_{0}-J}{\Gamma_{0}-J\gamma_{0}}}$
\end_inset

.
\end_layout

\begin_layout Standard
Now, evolving under the quench Hamiltonian, which is a harmonic oscillator,
 we have the operator equation of motion after the quench is simply 
\begin_inset Formula $\ddot{\tilde{x}}+\omega^{2}\tilde{x}=0$
\end_inset

 which admits a general solution of the form 
\begin_inset Formula $\tilde{x}(t)=\tilde{x}\left(0\right)\cos\omega t+\frac{1}{\omega}\tilde{p}\left(0\right)\sin\omega t$
\end_inset

 which gives us,
\begin_inset Formula 
\begin{align}
x(t) & =\left(x\left(0\right)\cos\omega t+\frac{2(\Gamma-J\gamma)}{\omega}p\left(0\right)\sin\omega t.\right)\\
 & =\left(x\left(0\right)\cos\omega t+\frac{1}{\tilde{\omega}}p\left(0\right)\sin\omega t.\right)\\
\langle x(t)^{2}\rangle_{0} & =\left(\cos^{2}\omega t\langle x^{2}\rangle_{0}+\frac{1}{\tilde{\omega}^{2}}\sin^{2}\omega t\ \langle p^{2}\rangle_{0}\right)
\end{align}

\end_inset

where 
\begin_inset Formula $\left\langle \cdots\right\rangle _{0}$
\end_inset

 is the expectation value in the groundstate of the prequench Hamiltonian
 and we have used, 
\begin_inset Formula $\langle xp+px\rangle_{0}=\langle a^{2}+(a^{\dagger})^{2}\rangle_{0}=0$
\end_inset

 and defined 
\begin_inset Formula $\tilde{\omega}\equiv\sqrt{\frac{\Gamma-J}{\Gamma-J\gamma}}$
\end_inset

.
 Simple algebra then gives
\begin_inset Formula 
\begin{align}
\langle x(t)^{2}\rangle & =\frac{1+\cos2\omega t}{2}\frac{1}{2\tilde{\omega}_{0}}+\frac{1-\cos2\omega t}{2}\frac{\tilde{\omega}_{0}}{\tilde{2\omega}^{2}}\\
 & =\frac{1}{4}\left(\frac{1}{\tilde{\omega}_{0}}+\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)+\frac{1}{4}\left(\frac{1}{\tilde{\omega}_{0}}-\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)\cos2\omega t
\end{align}

\end_inset

Next we relate the expectation value of 
\begin_inset Formula $S_{z}^{2}$
\end_inset

to that 
\begin_inset Formula $x^{2}$
\end_inset

.
 With 
\begin_inset Formula $S_{{\rm tot}}^{z}=\sqrt{\frac{N}{2}}x$
\end_inset

 we have
\begin_inset Formula 
\begin{align}
\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle & =\frac{N}{2}\langle x(t)^{2}\rangle\label{eq: 1-1}\\
 & =\frac{N}{8}\left[\left(\frac{1}{\tilde{\omega}_{0}}+\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)+\left(\frac{1}{\tilde{\omega}_{0}}-\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)\cos2\omega t\right]\label{eq:Analytical_time_dependence-1}
\end{align}

\end_inset

The combination of the last two equations gives the desrired result.
 In the limit that 
\begin_inset Formula $\omega\ll\omega_{0}$
\end_inset

we find the average correlation to be 
\begin_inset Formula $\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}=\frac{N}{2}\frac{1}{4}\left(\frac{1}{\tilde{\omega}_{0}}+\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}\right)\approx\frac{N}{8}\frac{\tilde{\omega}_{0}}{\tilde{\omega}^{2}}$
\end_inset


\end_layout

\begin_layout Standard
For harmonic oscillator,
\begin_inset Formula $a|n\rangle=\sqrt{}$
\end_inset


\begin_inset Formula 
\begin{align*}
\left\langle x^{2}\right\rangle _{\beta} & =\frac{e^{-\beta\left(n\right)\omega}\left\langle n\left|x^{2}\right|n\right\rangle }{}\left(1-e^{-\beta\omega}\right)\\
 & =\sum_{n}\frac{e^{-\beta\left(n\right)\omega}\left\langle n\left|\left(aa^{\dagger}+a^{\dagger}a\right)\right|n\right\rangle }{2\omega}\left(1-e^{-\beta\omega}\right)\\
 & =\frac{\left(1-e^{-\beta\omega}\right)}{\omega e^{-\beta\omega/2}}\sum_{n}e^{-\beta\left(n+\frac{1}{2}\right)\omega}\left(n+\frac{1}{2}\right)\\
 & =\frac{\left(1-e^{-\beta\omega}\right)}{\omega e^{-\beta\omega/2}}\frac{\partial}{\partial\beta\omega}\left(\frac{1}{1-e^{-\beta\omega}}\right)=\frac{\left(1-e^{-\beta\omega}\right)}{\omega e^{-\beta\omega/2}}\left(\frac{1}{1-e^{-\beta\omega}}\right)^{2}e^{-\beta\omega}\\
 & =\frac{1}{\omega}\left(\frac{1}{e^{\beta\omega/2}-e^{-\beta\omega/2}}\right)=\frac{T}{\omega^{2}}
\end{align*}

\end_inset


\end_layout

\begin_layout Subsubsection
Different limits:
\end_layout

\begin_layout Standard
Let's set 
\begin_inset Formula $J=1$
\end_inset

for simplicity.
 Consider different situations (we only consider 
\begin_inset Formula $\gamma_{0},\gamma$
\end_inset

<1):
\end_layout

\begin_layout Standard
1.
 Near 
\series bold
quantum critical point (QCP) in equilibrium.
\series default

\begin_inset Formula $\Gamma_{0}=\Gamma\gtrapprox1$
\end_inset

 and 
\begin_inset Formula $\gamma_{0}=\gamma$
\end_inset

.
 In this case we have 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N\frac{1}{\sqrt{\Gamma-1}}
\]

\end_inset


\end_layout

\begin_layout Standard
At the critical point, we should have 
\begin_inset Formula $\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N^{4/3}.$
\end_inset


\end_layout

\begin_layout Standard
2.
 
\series bold
Quench from a noncritical point to QCP .
\series default

\begin_inset Formula $\Gamma_{0}\gg1\to\Gamma\gtrapprox1$
\end_inset

 and 
\begin_inset Formula $\gamma_{0}=\gamma$
\end_inset

.
 In this case we have 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N\frac{1}{\Gamma-1}
\]

\end_inset


\end_layout

\begin_layout Standard
It is expected that the quench to the critical point gives 
\begin_inset Formula $\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N^{3/2}$
\end_inset


\end_layout

\begin_layout Standard
3.
 
\series bold
Quench from QCP to QCP 
\series default

\begin_inset Formula $\Gamma_{0}=\Gamma\gtrapprox1$
\end_inset

and 
\begin_inset Formula $\gamma_{0}\to\gamma$
\end_inset

.
 In this case we have
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N\frac{1}{\sqrt{\Gamma-1}}
\]

\end_inset


\end_layout

\begin_layout Standard
The scaling is similar to the QCP in equilibrium.
 The 
\bar under
milder
\bar default
 divergence with 
\begin_inset Formula $\Gamma-1$
\end_inset

suggestes a milder divergence with the system size which may be identical
 to the QCP.
 The overall constant of proportionality will be however different.
 
\end_layout

\begin_layout Subsection
Case with 
\begin_inset Formula $\gamma_{z}$
\end_inset

, 
\begin_inset Formula $\gamma_{y}$
\end_inset

 arbitrary
\end_layout

\begin_layout Standard
In the following, let us extend the previous analysis to the case of the
 arbitrary 
\begin_inset Formula $\gamma_{z}$
\end_inset

 and 
\begin_inset Formula $\gamma_{y}$
\end_inset

model
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
H=-\frac{2J}{N}[\gamma_{z}(S_{{\rm tot}}^{z})^{2}+\gamma_{y}(S_{{\rm tot}}^{y})^{2}]-2\Gamma S_{{\rm tot}}^{x}
\end{equation}

\end_inset

Using the Holstein Primakoff transformation, we have (signs and factors
 of 
\begin_inset Formula $i$
\end_inset

 are chosen to ensure the standard commutation relations)
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
S_{{\rm tot}}^{x} & =\frac{N}{2}-a^{\dagger}a=\frac{N+1}{2}-(a^{\dagger}a+\frac{1}{2})=\frac{N+1}{2}-\frac{1}{2}(x^{2}+p^{2})\\
S_{{\rm tot}}^{z} & \approx\frac{1}{2}\sqrt{N}(a+a^{\dagger})=\sqrt{\frac{N}{2}}x\\
S_{{\rm tot}}^{y} & \approx\frac{i}{2}\sqrt{N}(a-a^{\dagger})=-\sqrt{\frac{N}{2}}p
\end{align}

\end_inset

where we have 
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defined 
\begin_inset Quotes eld
\end_inset

position
\begin_inset Quotes erd
\end_inset

 and 
\begin_inset Quotes eld
\end_inset

momentum
\begin_inset Quotes erd
\end_inset

 operators as
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In this representation 
\family roman
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\begin_inset Formula 
\begin{align}
a & =\frac{1}{\sqrt{2}}(x+ip)\\
a^{\dagger} & =\frac{1}{\sqrt{2}}(x-ip)
\end{align}

\end_inset


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\xout default
\uuline default
\uwave default
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\color inherit
With this, the Hamiltonian takes the form
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
H & =-\Gamma(N+1)+\Gamma(x^{2}+p^{2})-J(\gamma_{z}x^{2}+\gamma_{y}p^{2})\\
 & =(\Gamma-J\gamma_{z})x^{2}+(\Gamma-J\gamma_{y})p^{2}+\mbox{const}
\end{align}

\end_inset

It is advantageous to redefine position and momentum variables as 
\begin_inset Formula $x\to\sqrt{2(\Gamma-J\gamma_{y})}\tilde{x}$
\end_inset

 and 
\begin_inset Formula $p\to\tilde{p}/\sqrt{2(\Gamma-J\gamma_{y})}$
\end_inset

 which satisfies their commutation relation, but leads to a standard form
 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
H_{0}=\frac{1}{2}\tilde{p}^{2}+\frac{1}{2}\omega^{2}\tilde{x}^{2}
\end{equation}

\end_inset

where 
\begin_inset Formula $\omega^{2}\equiv4(\Gamma-J\gamma_{z})(\Gamma-J\gamma_{y})$
\end_inset

.
 So the critical point occurs at 
\begin_inset Formula $\Gamma=J\max\left(\gamma_{z},\gamma_{y}\right)$
\end_inset

.
 Note that, to do the above step we require 
\begin_inset Formula $\Gamma>J\max\left(\gamma_{y}\right)$
\end_inset

.
 
\end_layout

\begin_layout Standard

\family roman
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\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
Now consider a quench of the form 
\begin_inset Formula $(\gamma_{0z},\gamma_{0y},\Gamma_{0})\to(\gamma_{z},\gamma_{y},\Gamma)$
\end_inset

.
 Inital state 
\family default
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has equal distribution of energy in momentum and position terms of hamornic
 oscillator, 
\begin_inset Formula $\langle\tilde{p}^{2}/2\rangle=\omega_{0}/4$
\end_inset

 and 
\begin_inset Formula $\langle\omega_{0}^{2}\tilde{x}^{2}/2\rangle_{0}=\omega_{0}/4$
\end_inset

, which yields 
\begin_inset Formula $\langle\tilde{p}^{2}\rangle_{0}=\omega_{0}/2$
\end_inset

 and 
\begin_inset Formula $\langle\tilde{x}^{2}\rangle_{0}=1/(2\omega_{0})$
\end_inset

.
 This means we have in the initial state
\begin_inset Formula 
\begin{equation}
\langle p^{2}\rangle_{0}=\frac{\omega_{0}}{4(\Gamma_{0}-J\gamma_{0y})}\text{=\ensuremath{\frac{1}{2}}\ensuremath{\sqrt{\frac{\Gamma_{0}-J\gamma_{0z}}{\Gamma_{0}-J\gamma_{0y}}}\equiv\frac{\tilde{\omega_{0}}}{2}}, and }\langle x^{2}\rangle_{0}=\frac{\Gamma_{0}-J\gamma_{0y}}{\omega_{0}}=\ensuremath{\frac{1}{2}}\ensuremath{\sqrt{\frac{\Gamma_{0}-J\gamma_{0y}}{\Gamma_{0}-J\gamma_{0z}}}}\equiv\frac{1}{2\tilde{\omega_{0}}},
\end{equation}

\end_inset

where we have used 
\begin_inset Formula $\omega_{0}^{2}\equiv\ensuremath{4(\Gamma_{0}-J_{0})(\Gamma_{0}-J\gamma_{0})}$
\end_inset

 and defined 
\begin_inset Formula $\tilde{\omega}_{0}\equiv\sqrt{\frac{\Gamma_{0}-J\gamma_{0z}}{\Gamma_{0}-J\gamma_{0y}}}$
\end_inset

.
\end_layout

\begin_layout Standard
Now, evolving under the quench Hamiltonian, which is a harmonic oscillator,
 we have the operator equation of motion after the quench is simply 
\begin_inset Formula $\ddot{\tilde{x}}+\omega^{2}\tilde{x}=0$
\end_inset

 which admits a general solution of the form 
\begin_inset Formula $\tilde{x}(t)=\tilde{x}\left(0\right)\cos\omega t+\frac{1}{\omega}\tilde{p}\left(0\right)\sin\omega t$
\end_inset

, , 
\begin_inset Formula $\tilde{p}(t)=-\omega\tilde{x}\left(0\right)\sin\omega t+\tilde{p}\left(0\right)\cos\omega t$
\end_inset

 which gives us,
\begin_inset Formula 
\begin{align}
x(t) & =\left(x\left(0\right)\cos\omega t+\frac{2(\Gamma-J\gamma_{y})}{\omega}p\left(0\right)\sin\omega t.\right)\\
 & =\left(x\left(0\right)\cos\omega t+\frac{1}{\tilde{\omega}}p\left(0\right)\sin\omega t.\right)\\
\langle x(t)^{2}\rangle_{0} & =\left(\cos^{2}\omega t\langle x^{2}\rangle_{0}+\frac{1}{\tilde{\omega}^{2}}\sin^{2}\omega t\ \langle p^{2}\rangle_{0}\right)\\
\langle p(t)^{2}\rangle_{0} & =\left(\tilde{\omega}^{2}\sin^{2}\omega t\langle x^{2}\rangle_{0}+\cos^{2}\omega t\ \langle p^{2}\rangle_{0}\right)
\end{align}

\end_inset

where 
\begin_inset Formula $\left\langle \cdots\right\rangle _{0}$
\end_inset

 is the expectation value in the groundstate of the prequench Hamiltonian
 and we have used, 
\begin_inset Formula $\langle xp+px\rangle_{0}=\langle a^{2}+(a^{\dagger})^{2}\rangle_{0}=0$
\end_inset

 and defined 
\begin_inset Formula $\tilde{\omega}\equiv\sqrt{\frac{\Gamma-J\gamma_{z}}{\Gamma-J\gamma_{y}}}$
\end_inset

.
 Simple algebra then gives
\begin_inset Formula 
\begin{align}
\langle x(t)^{2}\rangle & =\frac{1+\cos2\omega t}{2}\frac{1}{2\tilde{\omega}_{0}}+\frac{1-\cos2\omega t}{2}\frac{\tilde{\omega}_{0}}{2\tilde{\omega}^{2}}\\
 & =\frac{1}{4\tilde{\omega}_{0}}\left[\left(1+\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)+\left(1-\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)\cos2\omega t\right]\label{eq: 78}\\
\langle p(t)^{2}\rangle & =\frac{1-\cos2\omega t}{2}\frac{\tilde{\omega}^{2}}{2\tilde{\omega}_{0}}+\frac{1+\cos2\omega t}{2}\frac{\tilde{\omega}_{0}}{2}\\
 & =\frac{\tilde{\omega}_{0}}{4}\left[\left(1+\frac{\tilde{\omega}^{2}}{\tilde{\omega}_{0}^{2}}\right)+\left(1-\frac{\tilde{\omega}^{2}}{\tilde{\omega}_{0}^{2}}\right)\cos2\omega t\right]\\
 & =\frac{\tilde{\omega}^{2}}{4\tilde{\omega}_{0}}\left[\left(1+\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)-\left(1-\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)\cos2\omega t\right]
\end{align}

\end_inset

Next we relate the expectation value of 
\begin_inset Formula $S_{z}^{2}+S_{y}^{2}$
\end_inset

 to that 
\begin_inset Formula $x^{2}+p^{2}$
\end_inset

.
 With 
\begin_inset Formula $S_{{\rm tot}}^{z}=\sqrt{\frac{N}{2}}x$
\end_inset

 we have
\begin_inset Formula 
\begin{align}
\langle S_{z}^{2}(t)+S_{y}^{2}(t)\rangle & =\frac{N}{2}\langle x(t)^{2}+p(t)^{2}\rangle\label{eq: 1-1-1}\\
 & =\frac{N}{8\tilde{\omega}_{0}}\left[\left(1+\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)\left(1+\tilde{\omega}^{2}\right)+\left(1-\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)\left(1-\tilde{\omega}^{2}\right)\cos2\omega t\right]\label{eq:Analytical_time_dependence-1-1}
\end{align}

\end_inset

The combination of the last two equations gives the desrired result.
 
\end_layout

\begin_layout Standard
Let us now calculate the energy in the quench
\begin_inset Formula 
\begin{align*}
\left\langle H\right\rangle _{0} & =2(\Gamma-J\gamma_{y})\left[\frac{1}{2}\left\langle p^{2}\right\rangle _{0}+\frac{1}{2}\right]=(\Gamma-J\gamma_{y})\frac{\tilde{\omega}^{2}}{2\tilde{\omega}_{0}}\left(1+\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)\\
\left\langle H_{0}\right\rangle _{0} & =2\Gamma_{0}\left[\frac{1}{2}\left\langle p^{2}\right\rangle _{0}+\frac{1}{2}\frac{\omega_{0}^{2}}{4\Gamma_{0}^{2}}\left\langle x^{2}\right\rangle _{0}\right]=\left[\frac{\omega_{0}}{4}+\frac{\omega_{0}}{4}\right]=\frac{\omega_{0}}{2}=\ensuremath{\frac{4\Gamma_{0}^{2}(g_{0}-1)}{g_{0}}}\\
\left\langle H\right\rangle _{0}-\left\langle H_{0}\right\rangle _{0} & =\frac{\omega_{0}}{2}\left[\left(\frac{\Gamma}{2\Gamma_{0}}-1\right)+\frac{\omega^{2}}{2\omega_{0}^{2}}\frac{\Gamma_{0}}{\Gamma}\right]\\
\left\langle H\right\rangle _{0}-\left\langle H\right\rangle _{GS} & =\frac{\omega_{0}}{2}\left[\frac{\Gamma}{2\Gamma_{0}}+\frac{\omega^{2}}{2\omega_{0}^{2}}\frac{\Gamma_{0}}{\Gamma}-\frac{\omega}{\omega_{0}}\right]
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Correlation functions are obtained in a straightforward way from correlators
 of 
\begin_inset Formula $x(t)$
\end_inset

 at different times.
 Simple algebra yields 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
\frac{1}{2}\langle x(t)x(t')+x(t')x(t)\rangle & =\cos\omega t\cos\omega t'\langle x^{2}\rangle_{0}+\frac{1}{\tilde{\omega}^{2}}\sin\omega t\sin\omega t'\ \langle p^{2}\rangle_{0}\\
 & =\frac{1}{2}(\cos\omega(t-t')+\cos\omega(t+t'))\langle x^{2}\rangle_{0}+\frac{1}{\tilde{\omega}^{2}}\frac{1}{2}(\cos\omega(t-t')-\cos\omega(t+t'))\langle p^{2}\rangle_{0}\\
 & =\frac{1}{2}\cos\omega(t-t')(\langle x^{2}\rangle_{0}+\frac{1}{\tilde{\omega}^{2}}\langle p^{2}\rangle_{0})+\frac{1}{2}\cos\omega(t+t')(\langle x^{2}\rangle_{0}-\frac{1}{\tilde{\omega}^{2}}\langle p^{2}\rangle_{0})\\
 & \approx\frac{1}{2}\cos\omega(t-t')(\langle x^{2}\rangle_{0}+\frac{1}{\tilde{\omega}^{2}}\langle p^{2}\rangle_{0})+\mbox{highly oscillating terms}
\end{align}

\end_inset

where in the last line, I dropped the terms which depend on 
\begin_inset Formula $t+t'$
\end_inset

 as they are highly oscillatory in time (
\bar under
we should confirm if this is justified in the numerics
\bar default
).
 Therefore, the same algebra that leads to Eq.
 (
\begin_inset Formula $\text{\ref{eq: 78}}$
\end_inset

) now leads to 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
\frac{1}{2}\langle x(t)x(t')+x(t')x(t)\rangle & =\frac{1}{4\tilde{\omega}_{0}}\left[\left(1+\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)\cos\omega(t-t')+\left(1-\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)\cos\omega(t+t')\right]\\
 & \frac{1}{4\tilde{\omega}_{0}}\left(1+\frac{\tilde{\omega}_{0}^{2}}{\tilde{\omega}^{2}}\right)\cos\omega(t-t')+\mbox{highly oscillating terms}
\end{align}

\end_inset

One can find the commutator in a similar fashion as
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\langle[x(t),x(t')]\rangle=-\frac{1}{\omega}(\cos\omega t\sin\omega t'-\sin\omega t\cos\omega t')=\frac{1}{\omega}\sin\omega(t-t')
\end{equation}

\end_inset

The commutator is rather trivial in the sense that it does not depend on
 the initial state, and is just identical to the commutator of a free harmonic
 oscillator with natural frequency 
\begin_inset Formula $\omega.$
\end_inset


\end_layout

\begin_layout Subsubsection
Different limits:
\end_layout

\begin_layout Standard
Let's set 
\begin_inset Formula $J=1$
\end_inset

for simplicity.
 Consider different situations (we only consider 
\begin_inset Formula $\gamma_{0},\gamma$
\end_inset

<1):
\end_layout

\begin_layout Standard
1.
 Near 
\series bold
quantum critical point (QCP) in equilibrium.
\series default

\begin_inset Formula $\Gamma_{0}=\Gamma\gtrapprox1$
\end_inset

 and 
\begin_inset Formula $\gamma_{0}=\gamma$
\end_inset

.
 In this case we have 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N\frac{1}{\sqrt{\Gamma-1}}
\]

\end_inset


\end_layout

\begin_layout Standard
At the critical point, we should have 
\begin_inset Formula $\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N^{4/3}.$
\end_inset


\end_layout

\begin_layout Standard
2.
 
\series bold
Quench from a noncritical point to QCP .
\series default

\begin_inset Formula $\Gamma_{0}\gg1\to\Gamma\gtrapprox1$
\end_inset

 and 
\begin_inset Formula $\gamma_{0}=\gamma$
\end_inset

.
 In this case we have 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N\frac{1}{\Gamma-1}
\]

\end_inset


\end_layout

\begin_layout Standard
It is expected that the quench to the critical point gives 
\begin_inset Formula $\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N^{3/2}$
\end_inset


\end_layout

\begin_layout Standard
3.
 
\series bold
Quench from QCP to QCP 
\series default

\begin_inset Formula $\Gamma_{0}=\Gamma\gtrapprox1$
\end_inset

and 
\begin_inset Formula $\gamma_{0}\to\gamma$
\end_inset

.
 In this case we have
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\overline{\langle(S_{{\rm tot}}^{z}(t))^{2}\rangle}\sim N\frac{1}{\sqrt{\Gamma-1}}
\]

\end_inset


\end_layout

\begin_layout Standard
The scaling is similar to the QCP in equilibrium.
 The 
\bar under
milder
\bar default
 divergence with 
\begin_inset Formula $\Gamma-1$
\end_inset

suggestes a milder divergence with the system size which may be identical
 to the QCP.
 The overall constant of proportionality will be however different.
 
\end_layout

\begin_layout Section
Numeric results for Equilibrium state
\end_layout

\begin_layout Standard
In this section we tabulate exponents for different equilibrium states 
\end_layout

\begin_layout Standard
\begin_inset Float table
wide false
sideways false
status open

\begin_layout Plain Layout
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="6">
<features tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Ground State Hamiltonian
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\left\langle S_{z}^{2}\right\rangle _{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
 
\begin_inset Formula $\left\langle S_{y}^{2}\right\rangle _{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\left\langle S_{z}^{2}+S_{y}^{2}\right\rangle _{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\left\langle H_{0}\right\rangle _{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\text{S}=-\text{Tr}\left[\rho_{A}\log\rho_{A}\right]$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=4}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=1}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${J=1,\gamma_{z}=1,\gamma_{y}=0.5,\Gamma=1}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${J=1,\gamma_{z}=0,\gamma_{y}=1,\Gamma=1}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Tabulating the different exponents for the behavior of 
\begin_inset Formula $\left\langle S_{z}^{2}\right\rangle $
\end_inset

, 
\begin_inset Formula $\left\langle S_{y}^{2}\right\rangle $
\end_inset

 and 
\begin_inset Formula $\left\langle S_{z}^{2}+S_{y}^{2}\right\rangle $
\end_inset

 of different quenches.
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout Section
Numeric results for Different Quenches
\end_layout

\begin_layout Standard
In this section, we consider 4 different quenches 
\end_layout

\begin_layout Enumerate
\begin_inset Formula $H_{0}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=4}$
\end_inset

 to(a) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=2}$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $H_{0}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=4}$
\end_inset

 to(a) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=1}$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $H_{0}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=1}$
\end_inset

 to(a) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0.5,\Gamma=1}$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $H_{0}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=1}$
\end_inset

 to(a) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=0,\gamma_{y}=1,\Gamma=1}$
\end_inset


\end_layout

\begin_layout Subsection
Local Observables: 
\begin_inset Formula $\left\langle S_{z}^{2}\right\rangle $
\end_inset

, 
\begin_inset Formula $\left\langle S_{y}^{2}\right\rangle $
\end_inset

 and 
\begin_inset Formula $\left\langle S_{z}^{2}+S_{y}^{2}\right\rangle $
\end_inset

 
\end_layout

\begin_layout Standard
In this section we consider a quench between two critical points for the
 Hamiltonian.
\begin_inset Note Comment
status collapsed

\begin_layout Plain Layout
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
(a) (b)
\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Example phase diagram.
 (a) Fixed 
\begin_inset Formula $\Gamma=1$
\end_inset

 as a function of 
\begin_inset Formula $\gamma_{z}$
\end_inset

 and 
\begin_inset Formula $\gamma_{y}$
\end_inset

 (b) Magnetization as a function of 
\begin_inset Formula $\Gamma$
\end_inset

 for 
\begin_inset Formula $\gamma_{z}=1$
\end_inset

 and 
\begin_inset Formula $\gamma_{y}=0$
\end_inset

.
\end_layout

\end_inset


\end_layout

\end_inset


\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
(a) (b) (c)
\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
In this figure we show the finite size dependence in the ground state of
 
\begin_inset Formula $\left\langle S_{z}^{2}+S_{y}^{2}\right\rangle _{0}$
\end_inset

 for the Hamiltonians (a) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=0,\gamma_{y}=1,\Gamma=1}$
\end_inset

 (b) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0.5,\Gamma=1}$
\end_inset

, (c) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=0.5,\gamma_{y}=1,\Gamma=1}$
\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
(a) (b) (c)
\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
In this figure we show the time dependence of 
\begin_inset Formula $\left\langle S_{z}^{2}\left(t\right)+S_{y}^{2}\left(t\right)\right\rangle $
\end_inset

 for a particular system size 
\begin_inset Formula $L=1000$
\end_inset

, with the initial state as the ground state of 
\begin_inset Formula $H_{0}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=1}$
\end_inset

 to(a) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=0,\gamma_{y}=1,\Gamma=1}$
\end_inset

 (b) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0.5,\Gamma=1}$
\end_inset

, (c) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=0.5,\gamma_{y}=1,\Gamma=1}$
\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
(a) (b) (c)
\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
In this figure we show the finite size dependence long-time value/ time-averaged
 value 
\begin_inset Formula $\overline{\left\langle S_{z}^{2}\left(t\right)+S_{y}^{2}\left(t\right)\right\rangle }=\frac{1}{t}\int_{0}^{t}\left\langle S_{z}^{2}\left(t\right)+S_{y}^{2}\left(t\right)\right\rangle $
\end_inset

, with the initial state as the ground state of 
\begin_inset Formula $H_{0}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0,\Gamma=1}$
\end_inset

 quenched to (a) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=0,\gamma_{y}=1,\Gamma=1}$
\end_inset

 (b) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=1,\gamma_{y}=0.5,\Gamma=1}$
\end_inset

, (c) 
\begin_inset Formula $H_{f}\rightarrow{J=1,\gamma_{z}=0.5,\gamma_{y}=1,\Gamma=1}$
\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
(a) (b) (c)
\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
In this figure we show dependence of the correlator 
\begin_inset Formula $\left\langle S_{z}\left(t\right)\right\rangle $
\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Float table
wide false
sideways false
status open

\begin_layout Plain Layout
\begin_inset Tabular
<lyxtabular version="3" rows="6" columns="4">
<features tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Quench protocol
\end_layout

\end_inset
</cell>
<cell multirow="3" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\left\langle S_{z}^{2}\right\rangle _{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell multirow="3" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
 
\begin_inset Formula $\left\langle S_{y}^{2}\right\rangle _{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell multirow="3" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\left\langle S_{z}^{2}+S_{y}^{2}\right\rangle _{0}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $H_{0}\rightarrow{J_{0},\gamma_{z0},\gamma_{y0},\Gamma_{0}}$
\end_inset

to 
\begin_inset Formula $H_{f}\rightarrow{J_{f},\gamma_{zf},\gamma_{yf},\Gamma_{f}}$
\end_inset


\end_layout

\end_inset
</cell>
<cell multirow="4" alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell multirow="4" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell multirow="4" alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${1,1,0,4}\rightarrow{1,1,0,2}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${1,1,0,4}\rightarrow{1,1,0,1}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${1,1,0,1}\rightarrow{1,1,0.5,1}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${1,1,0,1}\rightarrow{1,0,1,1}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Tabulating the different exponents for the behavior of 
\begin_inset Formula $\left\langle S_{z}^{2}\right\rangle $
\end_inset

, 
\begin_inset Formula $\left\langle S_{y}^{2}\right\rangle $
\end_inset

 and 
\begin_inset Formula $\left\langle S_{z}^{2}+S_{y}^{2}\right\rangle $
\end_inset

 of different quenches.
\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Energy density 
\end_layout

\begin_layout Standard
In this section we tabulate the different exponents for different quenches.
\end_layout

\begin_layout Standard
\begin_inset Float table
wide false
sideways false
status open

\begin_layout Plain Layout
\begin_inset Tabular
<lyxtabular version="3" rows="6" columns="5">
<features tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Quench protocol
\end_layout

\end_inset
</cell>
<cell multirow="3" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
 
\begin_inset Formula $\left\langle H_{f}\right\rangle _{0}-\left\langle H_{0}\right\rangle _{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell multirow="3" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\left\langle H_{f}^{2}\right\rangle _{0}-\left\langle H_{f}\right\rangle _{0}^{2}$
\end_inset


\end_layout

\end_inset
</cell>
<cell multirow="3" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\left\langle H_{f}\right\rangle _{0}-\left\langle H_{f}\right\rangle _{GS}$
\end_inset


\end_layout

\end_inset
</cell>
<cell multirow="3" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\frac{\left\langle H_{f}\right\rangle _{0}-\left\langle H_{f}\right\rangle _{GS}}{\left\langle H_{f}\right\rangle _{FES}-\left\langle H_{f}\right\rangle _{GS}}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $H_{0}\rightarrow{J_{0},\gamma_{z0},\gamma_{y0},\Gamma_{0}}$
\end_inset

to 
\begin_inset Formula $H_{f}\rightarrow{J_{f},\gamma_{zf},\gamma_{yf},\Gamma_{f}}$
\end_inset


\end_layout

\end_inset
</cell>
<cell multirow="4" alignment="center" valignment="middle" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell multirow="4" alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell multirow="4" alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell multirow="4" alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${1,1,0,4}\rightarrow{1,1,0,2}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{1}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${1,1,0,4}\rightarrow{1,1,0,1}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{1}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0.1}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0.42}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${1,1,0,1}\rightarrow{1,1,0.5,1}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0.05}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{-0.64}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{-0.31}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0.01}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula ${1,1,0,1}\rightarrow{1,0,1,1}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0.34}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0.64}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0.34}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $L^{0.66}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Tabulating the different exponents for the behavior of energy density of
 different quenches.
\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Entanglement entropy 
\end_layout

\begin_layout Standard
In this section we tabulate the different exponents for different quenches.
\end_layout

\begin_layout Standard
\begin_inset Float table
wide false
sideways false
status open

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Tabulating the different exponents for the behavior of energy density of
 different quenches.
\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Effective temperature
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
N^{\beta-2\alpha}\begin{cases}
N^{0}=\omega^{0} & \text{Type-I}\\
N^{-1/3}=\omega^{1} & \text{Type-II}\\
N^{1/3}=\omega^{-2} & \text{Type-III}\\
\frac{\omega^{2}}{\omega_{0}}+\omega_{0} & \text{Deep Disordered}
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Section*
Future plans
\end_layout

\begin_layout Itemize
Fluctuation-dissipation relation is satisfied? Fluctuations are characterized
 by 
\begin_inset Formula $\langle S_{z}^{2}\rangle$
\end_inset

, while dissipation is given by 
\begin_inset Formula $[S_{z}(t),S_{z}(t^{\prime})]_{-}.$
\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $t,t^{\prime},\left|t-t^{\prime}\right|\rightarrow\infty$
\end_inset

, with 
\begin_inset Formula $\frac{1}{t}\left|t-t^{\prime}\right|,\frac{1}{t^{\prime}}\left|t-t^{\prime}\right|\rightarrow0$
\end_inset


\end_layout

\end_deeper
\begin_layout Itemize
Aging exponents: 
\end_layout

\begin_deeper
\begin_layout Itemize
Start from a nonzero magnetization (
\begin_inset Formula $\langle S_{z}\rangle\ne0$
\end_inset

) on the ordered side of the critical point (
\begin_inset Formula $\Gamma_{0}\lessapprox1$
\end_inset

 and 
\begin_inset Formula $\gamma_{0}$
\end_inset

 to 
\begin_inset Formula $\Gamma=1$
\end_inset

and 
\begin_inset Formula $\gamma\ne\gamma_{0}$
\end_inset

) near the critical point.
 The decay of the magnetization to zero as a function of time defines the
 aging behavior.
\end_layout

\begin_layout Itemize
Alternatively, we can take
\begin_inset Formula $\Gamma_{0}=\Gamma=1$
\end_inset

 and 
\begin_inset Formula $\gamma_{0}\to\gamma\ne\gamma_{0}$
\end_inset

and compute the response function 
\begin_inset Formula $\langle[S_{z}(t),S_{z}(t')]_{\pm}\rangle$
\end_inset

.
 Consider two different limits:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $|t-t'|\ll t,t'$
\end_inset

which gives the reponse function in the late-time steady state
\end_layout

\begin_layout Itemize
\begin_inset Formula $t'\ll t$
\end_inset

 which gives the response to an early perturbation and produces the aging
 exponent.
\end_layout

\end_deeper
\end_deeper
\begin_layout Itemize
OTOC: any nontrivial exponents?
\end_layout

\begin_layout Itemize
Coarsening
\end_layout

\begin_layout Section
Questions for Mohammad
\end_layout

\begin_layout Itemize
Is it necessary to restrict the symmetry sector.
 The ground state obtained numerically seem to be a random superposition
 of the symmetric states.
\end_layout

\begin_layout Itemize
\begin_inset Formula $U(1)\rightarrow\mathbb{Z}_{2}$
\end_inset

 How to handle this symmetry.
\end_layout

\end_body
\end_document