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      \hbox to 5.78in { {\bf PHYS 7895: Gaussian Quantum Information } \hfill #2 }
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\begin{document}

\lecture{12}{}{Lecturer: Mark M.\ Wilde}{Kevin Valson Jacob}

This document is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 Unported License.

\section{Overview}

In the last lecture we derived the formulas for the R\'enyi entropies, purity, and the entropy of Gaussian states.

In this lecture we derive the formulas for various overlap measures of two Gaussian states such as Holevo fidelity, Uhlmann fidelity, Petz-R\'enyi relative entropy, and sandwiched R\'enyi relative entropy.


\section{Overlap formulas for Gaussian states}
In quantum information, we are often interested in finding out how close two states are. A simple overlap formula between two states $\rho$ and $\tau$ is $\mathrm{Tr}[\rho\tau]$. More generally,  we compute overlap formulas of the following kind:
\begin{align}
    F_H(\rho,\tau) &=\mathrm{Tr}\left[\sqrt{\rho}\sqrt{\tau}\right]^2 
\\
    F(\rho,\tau)&=\left\Vert\sqrt{\rho}\sqrt{\tau}\right\Vert_1^2=\mathrm{Tr}\left[\sqrt{\sqrt{\tau}\rho\sqrt{\tau}}\right]^2
\end{align}
$F_H$ represents the Holevo fidelity whereas $F$ represents the Uhlmann fidelity.
Generalizing the above, we are interested in R\'enyi overlaps of the following kind:
\begin{align}
    Q_\alpha(\rho,\tau)&=\mathrm{Tr}\left[\rho^\alpha\tau^{1-\alpha}\right], \\
    \widetilde{Q}_\alpha(\rho,\tau)&=\mathrm{Tr}\left[( \tau^\frac{1-\alpha}{2\alpha}\rho \tau^\frac{1-\alpha}{2\alpha})^\alpha\right]\\
    &=\mathrm{Tr}\left[\left(\rho^{\frac{1}{2}}\tau^\frac{1-\alpha}{\alpha}\rho^{\frac{1}{2}}\right)^\alpha\right],
\end{align}
where $\alpha \in (0,1) \cup (1,\infty)$. Here, $Q$ represents the Petz-R\'enyi relative entropy while $\widetilde{Q}$ represents the sandwiched R\'enyi relative entropy.
Note that $Q_{\alpha=\frac{1}{2}}(\rho,\tau)=\sqrt{F_H(\rho,\tau)}$ and $\widetilde{Q}_{\alpha=\frac{1}{2}}(\rho,\tau)=\sqrt{F(\rho,\tau)}$. The reason these overlap functions are interesting is because we can bound the operationally meaningful trace distance between two states as
\begin{align}
    F_H(\rho,\tau)&\leq F(\rho,\tau) , \\
    1-\sqrt{F(\rho,\tau)}\leq 1-\sqrt{F_H(\rho,\tau)}&\leq\frac{1}{2}\left\Vert\rho-\tau\right\Vert_1\leq\sqrt{1-F(\rho,\tau)}\leq\sqrt{1-F_H(\rho,\tau)}.
\end{align}
As we do not possess a general formula for the trace distance between Gaussian states, the above relation proves useful in bounding it.
For simplicity in calculating these expressions, we will restrict ourselves to consider only zero-mean states.
\subsection{Simple overlap of Gaussian states}
Let us first consider the simple overlap formula $\mathrm{Tr}[\rho\tau]$ for states
\begin{align}
    \rho&=\frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_\rho+i\Omega}{2}\right)}}\exp\left(-\frac{1}{2}\Hat{r}^T H_\rho\Hat{r}\right),\label{eq:rho}\\
      \tau&=\frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_\tau+i\Omega}{2}\right)}}\exp\left(-\frac{1}{2}\Hat{r}^T H_\tau\Hat{r}\right).
      \label{eq:tau}
\end{align}
Thus we obtain
\begin{equation}
    \mathrm{Tr}[\rho\tau]= \frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_\rho+i\Omega}{2}\right)\mathrm{Det}\left(\frac{\sigma_\tau+i\Omega}{2}\right)}}\mathrm{Tr}\left[e^{-\frac{1}{2}\Hat{r}^T H_\rho\Hat{r}}e^{-\frac{1}{2}\Hat{r}^T H_\tau\Hat{r}}\right]
\end{equation}
We now wish to simplify the RHS of the above expression involving the product of two quadratic exponentials. In order to do so, we note the general result that for complex symmetric matrices $H_1$ and $H_2$, there exists another complex symmetric matrix $H_3$ such that if $H_3$ satisfies the relation
\begin{equation}
    e^{-\frac{1}{2}\Hat{r}^T H_1\Hat{r}}e^{-\frac{1}{2}\Hat{r}^T H_2\Hat{r}}=e^{-\frac{1}{2}\Hat{r}^T H_3\Hat{r}},
\end{equation}
then it also satisfies
\begin{equation}
    e^{-i\Omega H_1}e^{-i\Omega H_2}=e^{-i\Omega H_3}.
\end{equation}
The latter relation is useful in finding a form for $H_3$. Inverting the expression, we obtain
\begin{equation}
    e^{i\Omega H_3}=e^{i\Omega H_2}e^{i\Omega H_1}
\end{equation}
For simplicity in notations, define
\begin{align}
    W_3 &= \left(I+e^{i\Omega H_3}\right) \left(I - e^{i\Omega H_3}\right)^{-1} ,\\
\sigma_3 &=-W_3i\Omega .
  \end{align}
The latter implies that 
\begin{equation}
    \sigma_3=\mathrm{coth}\left(\frac{i\Omega H_3}{2}\right)i\Omega.
\end{equation}

Also, we note that
\begin{align}
        \sigma_1 & =\mathrm{coth}\left(\frac{i\Omega  H_1}{2}\right)i\Omega , \\
        \sigma_2&=\mathrm{coth}\left(\frac{i\Omega H_2}{2}\right)i\Omega.
\end{align}
Using these, we can arrive at the final form of $\sigma_3$ and $H_3$ as (see page 13, Ref. \cite{seshadreesan})
\begin{align}
    \sigma_3 & =-i\Omega +(\sigma_2+i\Omega)(\sigma_1+\sigma_2)^{-1}(\sigma_1+i\Omega), \\
    H_3 &= 2i\Omega \mathrm{arccoth}\left({\sigma_3i\Omega}\right),
\end{align}
from which we can obtain
\begin{equation}
    \sqrt{\mathrm{Det}\left(\frac{\sigma_3+i\Omega}{2}\right)}= \sqrt{\mathrm{Det}\left(\left(\frac{\sigma_2+i\Omega}{2}\right)\left(\frac{\sigma_1+\sigma_2}{2}\right)^{-1}\left(\frac{\sigma_1+i\Omega}{2}\right)\right)}
\end{equation}
Note that $\sigma_3$ is complex symmetric. 
It can be shown that (see Prop. 11, Ref. \cite{seshadreesan})
\begin{equation}
    \mathrm{Tr}\left[e^{-\frac{1}{2}\Hat{r}^T H_3\Hat{r}}\right]= \sqrt{\mathrm{Det}\left(\frac{\sigma_3+i\Omega}{2}\right)}
    \end{equation}
 Thus finally we can simplify the expression for the overlap as
 \begin{align}
    \mathrm{Tr}\left[\rho\tau\right] & =\frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_\rho+i\Omega}{2}\right)
    \mathrm{Det}\left(\frac{\sigma_\tau+i\Omega}{2}\right)}}\frac{\sqrt{\mathrm{Det}\left(\frac{\sigma_\rho+i\Omega}{2}\right)\mathrm{Det}\left(\frac{\sigma_\tau+i\Omega}{2}\right)}}{\sqrt{\mathrm{Det}\left(\frac{\sigma_\rho+\sigma_\tau}{2}\right)}} \\
     & = \frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_\rho+\sigma_\tau}{2}\right)}} \\
     & =   \frac{2^n}{\sqrt{\mathrm{Det}(\sigma_\rho+\sigma_\tau)}}
     \end{align}
We note that the overlap expression is not a function of the Hamiltonian matrix which implies that it is valid for pure (coherent) states also.

If the mean vectors of the states are represented by $\underline{r}_\rho$ and $\underline{r}_\tau$, then it can be shown that the overlap expression is 
\begin{equation}
    \mathrm{Tr}[\rho\tau] = \frac{2^n}{\sqrt{\mathrm{Det}(\sigma_\rho+\sigma_\tau)}} \mathrm{exp}[-\underline{\delta}^T(\sigma_\rho+\sigma_\tau)^{-1}\underline{\delta}]
\end{equation}  
where $\underline{\delta}=\underline{r}_\rho- \underline{r}_\tau$.

\subsection{Petz-R\'enyi relative entropy of Gaussian states}

Having computed an expression for the simple overlap formula, we now move on to compute the Petz-R\'enyi overlap of Gaussian states, defined as
\begin{equation}
     Q_\alpha(\rho,\tau)=\mathrm{Tr}\left[\rho^\alpha\tau^{1-\alpha}\right] .
\end{equation}
We will restrict ourselves to first consider the case when $\alpha \in (0,1)$. 
For notational simplicity, label the normalization of Gaussian states as 
\begin{equation}
   Z_\rho \equiv \sqrt{\mathrm{Det}\left(\frac{\sigma_\rho+i\Omega}{2}\right)},
\end{equation}
For states $\rho$ and $\tau$ as defined in \eqref{eq:rho} and \eqref{eq:tau}, 
the Petz-R\'enyi overlap is 
\begin{align}
     Q_\alpha(\rho,\tau)&=\frac{1}{(Z_\rho)^\alpha(Z_\tau)^{1-\alpha}} \mathrm{Tr}\left[e^{-\frac{1}{2}\Hat{r}^T \alpha H_\rho\Hat{r}}e^{-\frac{1}{2}\Hat{r}^T (1-\alpha)H_\tau\Hat{r}}\right] \\
         & = \frac{Z_{\rho(\alpha)}Z_{\tau(1-\alpha)}}{(Z_\rho)^\alpha(Z_\tau)^{1-\alpha}} \mathrm{Tr}\left[\frac{e^{-\frac{1}{2}\Hat{r}^T \alpha H_\rho\Hat{r}}}{Z_{\rho(\alpha)}}\frac{e^{-\frac{1}{2}\Hat{r}^T (1-\alpha)H_\tau\Hat{r}}}{Z_{\tau(1-\alpha)}}\right]
    \end{align}
    wherein we have used the fact that the Hamiltonian matrix of the exponent of a Gaussian state is the product of that exponent with the original Hamiltonian matrix. Now we can apply the simple overlap that we calculated earlier to obtain
    \begin{equation}
      Q_\alpha(\rho,\tau)=  \frac{Z_{\rho(\alpha)}Z_{\tau(1-\alpha)}}{(Z_\rho)^\alpha(Z_\tau)^{1-\alpha}}\frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_{\rho(\alpha)}+\sigma_{\tau(1-\alpha)}}{2}\right)}},
    \end{equation}
    where
    \begin{align}
        Z_{\rho(\alpha)} &= \sqrt{\mathrm{Det}\left(\frac{\sigma_{\rho(\alpha)}+i\Omega}{2}\right)},\\
        Z_{\tau(1-\alpha)} &= \sqrt{\mathrm{Det}\left(\frac{\sigma_{\tau(1-\alpha)}+i\Omega}{2}\right)},
    \end{align}
    and
    \begin{align}
        \sigma_{\rho(\alpha)} &=\frac{\left[I+(\sigma_\rho i \Omega)^{-1}\right]^\alpha+\left[I-(\sigma_\rho i \Omega)^{-1}\right]^\alpha}{\left[I+(\sigma_\rho i \Omega)^{-1}\right]^\alpha-\left[I-(\sigma_\rho i \Omega)^{-1}\right]^\alpha} i\Omega, \\
        \sigma_{\tau(1-\alpha)} &=\frac{\left[I+(\sigma_\tau i \Omega)^{-1}\right]^{1-\alpha}+\left[I-(\sigma_\tau i \Omega)^{-1}\right]^{1-\alpha}}{\left[I+(\sigma_\tau i \Omega)^{-1}\right]^{1-\alpha}-\left[I-(\sigma_\tau i \Omega)^{-1}\right]^{1-\alpha}} i\Omega.
    \end{align} 

\paragraph{Holevo Fidelity:}  
 
For $\alpha = \frac{1}{2}$ the above expression simplifies to 
\begin{align}
    Q_{\frac{1}{2}}(\rho,\tau) &=\sqrt{F_H(\rho,\tau)}=\mathrm{Tr}\left[\sqrt{\rho}\sqrt{\tau}\right] \\
    &=\frac{Z_{\rho(1/2)}Z_{\tau(1/2)}}{(Z_\rho)^\frac{1}{2}(Z_\tau)^\frac{1}{2}}\frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_{\rho(1/2)}+\sigma_{\tau(1/2)}}{2}\right)}}
\end{align}
where
\begin{align}
    \sigma_{\rho(1/2)} &= \left( \sqrt{I+(\sigma_\rho\Omega)^{-2}}+I\right)\sigma_\rho, \\
    \sigma_{\tau(1/2)} &= \left( \sqrt{I+(\sigma_\tau\Omega)^{-2}}+I\right)\sigma_\tau ,
\end{align}
and
  \begin{align}
        Z_{\rho(1/2)} &= \sqrt{\mathrm{Det}\left(\frac{\sigma_{\rho(1/2)}+i\Omega}{2}\right)},\\
        Z_{\tau(1/2)} &= \sqrt{\mathrm{Det}\left(\frac{\sigma_{\tau(1/2)}+i\Omega}{2}\right)}.
    \end{align}
    
Now we shall consider the case of $\alpha > 1$. This is interesting because we will have to deal with inverses of Gaussian states which are in general unbounded operators. However, we can find expressions for  overlaps. In order to derive such an expression, note that for $H_1,H_2 >0$ such that $\sigma_2>\sigma_1$, we have (see Ref. \cite{seshadreesan})
\begin{equation}
    \mathrm{Tr}\left[e^{-\Hat{r}^T H_1\Hat{r}}e^{-\Hat{r}^T (-H_2)\Hat{r}}\right] = \frac{\sqrt{\mathrm{Det}\left(\frac{\sigma_1+i\Omega}{2}\right)\mathrm{Det}\left(\frac{\sigma_2+i\Omega}{2}\right)}}{\sqrt{\mathrm{Det}\left(\frac{\sigma_2-\sigma_1}{2}\right)}} 
\end{equation}
Now we can consider 
\begin{align}
     Q_\alpha(\rho,\tau)&=\mathrm{Tr}\left[\rho^\alpha\tau^{1-\alpha}\right]  \\
     & = \frac{1}{(Z_\rho)^\alpha(Z_\tau)^{1-\alpha}} \mathrm{Tr}\left[e^{-\frac{1}{2}\Hat{r}^T \alpha H_\rho\Hat{r}}e^{-\frac{1}{2}\Hat{r}^T [-(\alpha-1)H_\tau]\Hat{r}}\right]
\end{align}
We can apply the above relation to obtain the following, when $\sigma_{\tau(\alpha-1)}>\sigma_{\rho(\alpha)}$
\begin{equation}
\mathrm{Tr}\left[\rho^\alpha\tau^{1-\alpha}\right] = \frac{Z_{\rho(\alpha)}Z_{\tau(\alpha-1)}}{(Z_\rho)^\alpha(Z_\tau)^{1-\alpha}}\frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_{\tau(\alpha-1)}-\sigma_{\rho(\alpha)}}{2}\right)}}
\end{equation}
where $Z_{\rho(\alpha)}$, $Z_{\tau(\alpha-1)}$, and  $\sigma_{\rho(\alpha)}$ and $\sigma_{\tau(1-\alpha)}$ are defined similarly. The above expression simplifies significantly for $\alpha=2$. In that case, for $\sigma_{\tau} > \sigma_{\rho(2)}$
\begin{equation}
    \mathrm{Tr}[\rho^2\tau^{-1}] = \frac{Z_{\rho(2)}(Z_{\tau})^2}{(Z_\rho)^2}\frac{1}{\sqrt{\mathrm{Det}\left(\frac{\sigma_{\tau}-\sigma_{\rho(2)}}{2}\right)}}
\end{equation}
where
\begin{equation}
    \sigma_{\rho(2)}=\frac{1}{2}\left(\sigma_\rho+\Omega\sigma_\rho^{-1}\Omega^T\right)
\end{equation}

\subsection{Sandwiched Petz-R\'enyi relative entropy}

Let's now consider the sandwiched Petz-R\'enyi relative entropy:
\begin{align}
    \widetilde{Q}_\alpha(\rho,\tau)&=\mathrm{Tr}\left[\left( \tau^\frac{1-\alpha}{2\alpha}\rho \tau^\frac{1-\alpha}{2\alpha}\right)^\alpha\right]\\
    &=\mathrm{Tr}\left[\left(\rho^{\frac{1}{2}}\tau^\frac{1-\alpha}{\alpha}\rho^{\frac{1}{2}}\right)^\alpha\right]
\end{align}
For states $\rho$ and $\tau$ as defined in Eqs. \eqref{eq:rho} and \eqref{eq:tau}, 
\begin{equation}
   \widetilde{Q}_\alpha(\rho,\tau)=  \frac{1}{(Z_\rho)^\alpha(Z_\tau)^{1-\alpha}} \mathrm{Tr}\left[ \left(e^{-\frac{1}{2}\Hat{r}^T \frac{1}{2} H_\rho\Hat{r}}e^{-\frac{1}{2}\Hat{r}^T [\beta H_\tau]\Hat{r}}   e^{-\frac{1}{2}\Hat{r}^T \frac{1}{2} H_\rho\Hat{r}    }\right)^\alpha\right],
\end{equation}
where $\beta = (1-\alpha)/\alpha$.
We use the fact that 
\begin{equation}
    e^{-\frac{1}{2}\Hat{r}^T \frac{1}{2} H_1\Hat{r}}e^{-\frac{1}{2}\Hat{r}^T  H_2 \Hat{r}}   e^{-\frac{1}{2}\Hat{r}^T \frac{1}{2} H_1\Hat{r} } = e^{-\frac{1}{2}\Hat{r}^T  H_3 \Hat{r}}  
\end{equation}
where (see Prop. 8, Ref. \cite{seshadreesan})
\begin{align}
    H_3 &= 2i\Omega\mathrm{arccoth}(\sigma_3i\Omega), \\
    \sigma_3 &=\sigma_1-\left(\sqrt{I+(\sigma_1\Omega)^{-2}}\right)\sigma_1(\sigma_1+\sigma_2)^{-1}\sigma_1 \left(\sqrt{I+(\Omega\sigma_1)^{-2}}\right).
\end{align}
Using this, we find that
\begin{equation}
 e^{-\frac{1}{2}\Hat{r}^T \frac{1}{2} H_\rho\Hat{r}}e^{-\frac{1}{2}\Hat{r}^T \beta H_\tau \Hat{r}}   e^{-\frac{1}{2}\Hat{r}^T \frac{1}{2} H_\rho \Hat{r} } = e^{-\frac{1}{2}\Hat{r}^T  H_\zeta \Hat{r}},   
\end{equation}
where
\begin{align}
    H_\zeta &= 2i\Omega\mathrm{arccoth}(\sigma_\zeta i\Omega), \\
    \sigma_\zeta &=\sigma_\rho-\left(\sqrt{I+(\sigma_\rho\Omega)^{-2}}\right)\sigma_\rho(\sigma_\rho+\sigma_{\tau(\beta)})^{-1}\sigma_\rho \left(\sqrt{I+(\Omega\sigma_\rho)^{-2}}\right), \\
    \sigma_{\tau(\beta)} &= \frac{\left[I+(\sigma_\tau i \Omega)^{-1}\right]^{\beta}+\left[I-(\sigma_\tau i \Omega)^{-1}\right]^{\beta}}{\left[I+(\sigma_\tau i \Omega)^{-1}\right]^{\beta}-\left[I-(\sigma_\tau i \Omega)^{-1}\right]^{\beta}} i\Omega.
\end{align}
Finally, we exponentiate this expression with $\alpha$ from the definition of sandwiched  Petz-R\'enyi overlap. We use the fact that this scales the resultant Hamiltonian matrix by a factor $\alpha$ so as to obtain
\begin{equation}
    \mathrm{Tr}\left[e^{-\frac{1}{2}\hat{r}^T\alpha H_\zeta\hat{r}}\right]=\sqrt{\mathrm{Det}\left(\frac{\sigma_{\zeta(\alpha)}+i\Omega}{2}\right)}
\end{equation}
where
\begin{equation}
     \sigma_{\zeta(\alpha)} = \frac{\left[I+(\sigma_\zeta i \Omega)^{-1}\right]^{\alpha}+\left[I-(\sigma_\zeta i \Omega)^{-1}\right]^{\alpha}}{\left[I+(\sigma_\zeta i \Omega)^{-1}\right]^{\alpha}-\left[I-(\sigma_\zeta i \Omega)^{-1}\right]^{\alpha}} i\Omega
\end{equation}
Thus we obtain our formula for the sandwiched  Petz-R\'enyi overlap as:
\begin{equation}
    \widetilde{Q}_\alpha(\rho,\tau)=  \frac{1}{(Z_\rho)^\alpha(Z_\tau)^{1-\alpha}}\sqrt{\mathrm{Det}\left(\frac{\sigma_{\zeta(\alpha)}+i\Omega}{2}\right)}
\end{equation}
Fidelity is a special case of sandwiched Petz-R\'enyi overlap. When $\alpha=\frac{1}{2}$,  we have
\begin{equation}
    F[\rho,\tau]=\left(\widetilde{Q}_{\alpha=\frac{1}{2}}(\rho,\tau)\right)^2
\end{equation}
\paragraph{Fidelity:}
If $\alpha=\frac{1}{2}$, we have $\beta =1$, and $\tau(\beta) = \tau$. Thus we have
\begin{align}
    \sigma_\zeta &=\sigma_\rho-\left(\sqrt{I+(\sigma_\rho\Omega)^{-2}}\right)\sigma_\rho(\sigma_\rho+\sigma_{\tau})^{-1}\sigma_\rho \left(\sqrt{I+(\sigma_\rho\Omega)^{-2}}\right), \\
    \sigma_{\zeta(\alpha=\frac{1}{2})} &= \left( \sqrt{I+(\sigma_\zeta\Omega)^{-2}}+I\right)\sigma_\zeta,
\end{align}
and thus the fidelity becomes
\begin{align}
     F[\rho,\tau]&=\mathrm{Tr}\left[\sqrt{\rho^\frac{1}{2}\tau\rho^\frac{1}{2}}\right]^2 \\ 
     &= \frac{\mathrm{Det}\left[\sigma_{\zeta(\frac{1}{2})}\right]}{Z_\rho Z_\tau}
\end{align}

Now we will find an expression for the sandwiched Petz-R\'enyi overlap when $\alpha >1$. For simplicity, define $\gamma=-\beta >0$. When $\sigma_{\tau(\gamma)}>\sigma_\rho$, we have
\begin{equation}
    \widetilde{Q}_\alpha(\rho,\tau)=  \frac{1}{(Z_\rho)^\alpha(Z_\tau)^{1-\alpha}}\sqrt{\mathrm{Det}\left(\frac{\sigma_{\zeta(\alpha)}+i\Omega}{2}\right)}
\end{equation}
where
\begin{align}
     \sigma_{\zeta(\alpha)} &= \frac{\left[I+(\sigma_\zeta i \Omega)^{-1}\right]^{\alpha}+\left[I-(\sigma_\zeta i \Omega)^{-1}\right]^{\alpha}}{\left[I+(\sigma_\zeta i \Omega)^{-1}\right]^{\alpha}-\left[I-(\sigma_\zeta i \Omega)^{-1}\right]^{\alpha}} i\Omega, \\
     \sigma_\zeta &=\sigma_\rho + \left(\sqrt{I+(\sigma_\rho\Omega)^{-2}}\right)\sigma_\rho(\sigma_{\tau(\gamma)-\sigma_\rho})^{-1}\sigma_\rho \left(\sqrt{I+(\sigma_\rho\Omega)^{-2}}\right), \\
    \sigma_{\tau(\gamma)} &= \frac{\left[I+(\sigma_\tau i \Omega)^{-1}\right]^{\gamma}+\left[I-(\sigma_\tau i \Omega)^{-1}\right]^{\gamma}}{\left[I+(\sigma_\tau i \Omega)^{-1}\right]^{\gamma}-\left[I-(\sigma_\tau i \Omega)^{-1}\right]^{\gamma}} i\Omega.
\end{align}


%\bibliography{mybib}
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\bibliographystyle{Alpha}
\bibitem{seshadreesan} K. P. Seshadreesan, L. Lami, and M. M. Wilde, {\it{J. Math. Phys.}} {\bf{59}},  072204 (2018).
\end{thebibliography}
\end{document}
