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\title{Asymmetric delay attack on an entanglement-based bidirectional clock synchronization protocol -- Supplementary Material}

\author{Jianwei~Lee}
\affiliation{Centre for Quantum Technologies, National University of
  Singapore, 3 Science Drive 2, Singapore 117543, Singapore}

\author{Lijiong~Shen}
\affiliation{Centre for Quantum Technologies, National University of
  Singapore, 3 Science Drive 2, Singapore 117543, Singapore}
\affiliation{Department of Physics, National University of Singapore, 2
  Science Drive 3, Singapore 117551, Singapore}

\author{Alessandro~Cer\`{e}}
\affiliation{Centre for Quantum Technologies, National University of
  Singapore, 3 Science Drive 2, Singapore 117543, Singapore}

\author{James~Troupe}
\affiliation{Applied Research Laboratories, The University of Texas at Austin,
  Austin, Texas, USA}

\author{Antia~Lamas-Linares}
\affiliation{SpeQtral, 73 Science Park Drive, Singapore 118254, Singapore}
\affiliation{Centre for Quantum Technologies, National University of
  Singapore, 3 Science Drive 2, Singapore 117543, Singapore}

\author{Christian~Kurtsiefer}
\affiliation{Centre for Quantum Technologies, National University of
  Singapore, 3 Science Drive 2, Singapore 117543, Singapore}
\affiliation{Department of Physics, National University of Singapore, 2
  Science Drive 3, Singapore 117551, Singapore}


\email[]{christian.kurtsiefer@gmail.com}
\date{\today}

\maketitle

In this supplementary material, we show that when circulators rotate 
the polarization state of one of the photons in an entangled pair by 180$^{\text{o}}$,
the geometric phase imposed on the rotated photon does not produce a measurable change in polarization entanglement.

We first introduce the formalism to deal with the fact that points on the Poincaré sphere carry no phase information;
the beginning and end points of a cyclic evolution correspond on the same point on the sphere.

To reflect this property, we define a ``basis vector field'' $\ket{\tilde{\psi}(t)}$, 
such that 
\begin{equation*}
\ket{\tilde{\psi}(t)}=e^{-if(t)}\ket{\psi(t)} \quad\text{and}\quad \ket{\tilde{\psi}(\tau)}=\ket{\tilde{\psi}(0)}, 
\end{equation*}
where $f(t)$ is the phase of $\ket{\psi(t)}$ expressed in terms of its basis state $\ket{\tilde{\psi}(t)}$ on the Poincaré sphere~\cite{anandan1992geometric}. 

The change in $f$ comprises of two terms
\begin{equation}\label{eqn:full_phase_change}
\Delta f = \beta + \gamma, 
\end{equation}
where the geometric phase
\begin{equation}
  \beta = \int\displaylimits_0^\tau\bra{\tilde{\psi}(t)}i\frac{\diff }{\diff t}\ket{\tilde{\psi}(t)}
\end{equation}
is due to the evolution of the basis state along a curved geometry, 
and the dynamic phase
\begin{equation}\label{eqn:dp_raw}
  \gamma = - \int\displaylimits_0^\tau \bra{\psi(t)}i\frac{\diff}{\diff t}\ket{\psi(t)} \diff t
\end{equation} 
is due to the photon's dynamics through the rotation medium~\cite{Troupe:2018}.
\subsection{Geometric Phase}\label{sec:gp}
Berry showed that the geometric phase is proportional only to the solid angle
$\Omega$ subtended by the cyclic trajectory on the Poincaré
sphere~\cite{berry:87},
\begin{equation}
\beta = -\frac{1}{2} \Omega.
\end{equation}
Thus, a qubit in the initial state
\begin{equation} \label{eq:qubit}
\ket{\psi(t=0)} = e^{-i\phi}\cos(\theta/2)\ket{R}+ \sin(\theta/2)\ket{L}
\end{equation}
that underwent a 180$^{\text{o}}$ rotation in the plane of polarization ($\phi\rightarrow\phi+2\pi$)
will accumulate a geometric phase $\beta = -\pi(1-\cos\theta)$.

\subsection{Dynamic Phase}\label{sec:dp}
To evaluate the dynamic phase $\gamma$ accumulated by the photon at end of a Faraday Rotator of length $d$, we parameterize its expression in Eq.~\ref{eqn:dp_raw} in terms of the penetration depth $z$
\begin{align}\label{eqn:dp}
\gamma &= -\int\displaylimits_0^d \bra{\psi(z)}i\frac{\diff }{\diff z}\ket{\psi(z)} dz
&=-\int\displaylimits_0^d \bra{\psi(z)}\hat{N}\ket{\psi(z)} dz,
\end{align}
where 
\begin{equation}\label{eqn:operator_and_state}
\hat{N}
    = k\begin{pmatrix} 
      n_R & 0 \\
      0 & n_L
    \end{pmatrix} \quad\text{and}\quad 
\ket{\psi(z)} = \begin{pmatrix} 
e^{ikn_Rz}e^{-i\phi}\cos(\theta/2) \\
e^{ikn_Lz}sin(\theta/2)
\end{pmatrix}
\end{equation}
are expressed in the $\{\ket{R},\ket{L}\}$ basis, and $k =
\frac{2\pi}{\lambda}$ is the wave number of the photon mode in free space.

The Faraday Rotator is a birefringent medium whose refractive indices $n_{R,L}$ depend on the magnitude of an applied magnetic field $B$ in the direction of light propagation,
\begin{equation}\label{eqn:refractive_index}
  n_{R,L} = n_0\left(1\pm\frac{VB}{kn_0}\right),
\end{equation}
where $V$ is the Verdet constant and $n_0$ is the index of refraction in the absence of a magnetic field.

Substituting \ref{eqn:operator_and_state} into \ref{eqn:dp}, we obtain
\begin{equation}\label{eqn:dp_psi}
  \gamma = kn_0d + VBd\cos\theta,
\end{equation}
where the product $VBd$
can be shown~\cite{zak1991geometric} to be the anti-clockwise rotation angle for a linearly polarized input.

Consider an initial input state 
$\ket{\psi(\phi=0,\theta=0)}=\ket{H}$. 
For the evolution cycle ($\phi = 0\rightarrow2\pi$) considered earlier,
$\ket{H}\rightarrow\ket{-45}\rightarrow{\ket{V}}\rightarrow\ket{+45}\rightarrow\ket{H}$ corresponds to a \textit{clockwise} 180$^{\text{o}}$ in the plane-of-polarization. 
Thus, the the rotation must be realized by a medium whose product $VBd = -\pi$.
Consequently, the dynamic phase $\gamma=kn_0d - \pi\cos\theta$ for the state considered in Eq.~\ref{eq:qubit}.

\subsection{Overall Phase \& the Circulator Attack}
We have already shown that an initial state 
\begin{equation}\label{eq:psi}
\ket{\psi} = \cos(\theta/2)\ket{R} + \sin(\theta/2)\ket{L}, 
\end{equation}
will accumulate a geometric phase $\beta = -\pi(1-\cos\theta)$ and a dynamic phase $\gamma = kn_0d - \pi\cos\theta$, resulting in an overall phase $\phi=kn_0d-\pi$.
Repeating this procedure for the orthogonal state
\begin{equation}\label{eq:psi_perp}
\ket{\psi_\perp} = -\sin(\theta/2)\ket{R} + \cos(\theta/2)\ket{L}, 
\end{equation}
we obtain a geometric phase of $\beta'=-\beta = +\pi(1-\cos\theta)$ and a dynamic phase $\gamma' = kn_0d + \pi\cos\theta$, resulting in an overall phase $\phi'=kn_0d+\pi= \phi+2\pi$.  

Let the entangled pair initially be in the Bell state $\ket{\Psi^-} = \frac{1}{\sqrt{2}}\left( \ket{HV} - \ket{VH}\right)$.  With the first qubit Alice's photon and the second one Bob's photon.  We can re-write the Bell state in the basis defined by  Equations \ref{eq:psi} and \ref{eq:psi_perp},
\begin{align}
\ket{\Psi^-} & =\frac{1}{\sqrt{2}} \Big( \ket{HV} - \ket{VH} \Big) \nonumber \\
 &= \frac{i}{\sqrt{2}} \Big( \ket{\psi_\perp}_A\ket{\psi}_B - \ket{\psi}_A\ket{\psi_\perp}_B \Big). 
\end{align}
The state of the Bell pair after Bob's photon goes through Eve's circulator based attack, $\hat{U}_{Attack}$, is given by 
\begin{align}
\ket{\Psi^-} 
& \rightarrow 
\hat{U}_{Attack}\frac{i}{\sqrt{2}}\Big( \ket{\psi_\perp}_A\ket{\psi}_B - 
\ket{\psi}_A\ket{\psi_\perp}_B \Big) \nonumber\\
&= 
\frac{i}{\sqrt{2}} \Big( 
e^{i\phi}\ket{\psi_\perp}_A\ket{\psi}_B 
- e^{i\phi'}\ket{\psi}_A\ket{\psi_\perp}_B \Big)\\
&=
\frac{ie^{i\phi}}{\sqrt{2}} \Big( 
\ket{\psi_\perp}_A\ket{\psi}_B 
- e^{i2\pi}\ket{\psi}_A\ket{\psi_\perp}_B \Big)\nonumber\\ 
&=e^{i\phi}\ket{\Psi^-}=-e^{ikn_0d}\ket{\Psi^-} \nonumber\\
&\equiv -\ket{\Psi^-}.
\end{align}
We can see from this expression, 
that the initial Bell state remains unchanged from the introduction of the circulators, 
and is equivalent to the result obtained by direct calculation in Eq.~5 in
the main text. %\ref{eqn:direct_result}. 

Recent work assumed that the contribution from the dynamic phase was ``zero, or is known and compensated for'' and predicted instead that the circulators imparted a non-local geometric phase to produce a dramatic change~\cite{Troupe:2018}
\begin{align}
\ket{\Psi^-} 
& \rightarrow 
\hat{U}_{Attack}\frac{i}{\sqrt{2}}\Big( \ket{\psi_\perp}_A\ket{\psi}_B - 
\ket{\psi}_A\ket{\psi_\perp}_B \Big) \nonumber\\
&= 
\frac{i}{\sqrt{2}} \Big( 
e^{i\beta}\ket{\psi_\perp}_A\ket{\psi}_B 
- e^{-i\beta}\ket{\psi}_A\ket{\psi_\perp}_B \Big).
\end{align}
However, we note that the dynamic phase (Eq.~\ref{eqn:dp_psi}) is likewise non-local (due to its dependence on $\theta$) and combines with the geometric phase to produce no measurable net change in the state.
\end{document}

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