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\heading{Entanglement with an accelerated observer} 
\auname{Judith Kupferman}


\section{Basic situation}

We work in 1+1 dimensions: one space dimension and time. 

Alice and Bob stand in the laboratory, and share an EPR state:\begin{equation}
\psi_{AB}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle _{A}\left|0\right\rangle _{B}+\left|1\right\rangle _{A}\left|1\right\rangle _{B}\right).\end{equation}
What does this mean: $\left|0\right\rangle $ means there is no probability
to observe a particle. $\left|1\right\rangle $ means there is a probability
to observe one particle. 

We work with momentum, so that $\left|1\right\rangle $ means probability
to observe one particle with momentum $k$ somewhere in all of space.
(we could write $\left|k\right\rangle $ but we will assume only one
possible momentum, for simplicity.)


\subsection{The formalism}

We work in Fock space. This means that we have creation operators
so that \begin{equation}
a^{\dagger}\left|0\right\rangle =\left|1\right\rangle ,\: a\left|0\right\rangle =0.\end{equation}
To create two particles (with the same momentum) we use the operator
twice:\begin{equation}
a^{\dagger}a^{\dagger}\left|0\right\rangle =a^{\dagger}\left|1\right\rangle =2\end{equation}
and to creat $n$ particles we do it $n$ times:\begin{equation}
\left(a^{\dagger}\right)^{n}\left|0\right\rangle \sim\left|n\right\rangle .\label{eq:n state}\end{equation}
Proportional (not equal) because\begin{equation}
a^{\dagger}\left|n\right\rangle =\sqrt{n+1}\left|n+1\right\rangle \rightarrow\left(a^{\dagger}\right)^{n}\left|0\right\rangle =\sqrt{n!}\left|n\right\rangle .\end{equation}
That means we have $n$ particles with the given momentum, which could
be found somewhere in all of space.

In Alice and Bob's EPR state, it means they share an amplitude that
either they will find no particles, or each will find one particle
somewhere in space.You could say they each hold a detector which can
register whether or not there's a particle somewhere in space with
the given momentum. Either both detectors will click, or neither of
them will click.


\subsection{Entanglement}

While Alice and Bob stand in the lab, we check entanglement. Tracing
out Alice, the reduced density matrix is\begin{eqnarray}
\rho_{B} & = & \left\langle 0_{A}\right.\left|\psi_{AB}\right\rangle \left\langle \psi_{AB}\right|\left.0_{A}\right\rangle +\left\langle 1_{A}\right.\left|\psi_{AB}\right\rangle \left\langle \psi_{AB}\right|\left.1_{A}\right\rangle \nonumber \\
 & = & \frac{1}{2}\left(\left|0_{B}\right\rangle \left\langle 0_{B}\right|+\left|1_{B}\right\rangle \left\langle 1_{B}\right|\right)\label{eq:bob on land partial trace}\end{eqnarray}
that is, $\frac{1}{2}$ of the 2 dimensional identity matrix. Using
Von Neuman entropy, \begin{equation}
S_{VN}=-Tr\left(\rho_{B}log\rho_{B}\right)=-\sum_{i=1}^{dim}\lambda_{i}log_{2}\lambda_{i}\end{equation}
where $\lambda_{i}$ are eigenvalues of the reduced density matrix,
we get $S_{VN}=-2(\frac{1}{2}log_{2}\left[\frac{1}{2}\right])=log_{2}2=1$
and see that the state is maximally entangled.

The question: if Bob gets into a car and drives off with constant
acceleration - will he still find that the state is maximally entangled?
That either both detectors will click, or neither will, with equal
amplitude?


\section{Bob accelerates}

Bob gets into a car and speed up with constant acceleration $a$ .
This means he is now in an accelerated coordinate system ({}``Rindler
space''). 


\subsection{\label{sub:Coordinates:}Coordinates:}

We first find his trajectory. This is done in the {}``co-moving frame''
where Bob is instantaneously at rest, so that his path can be expressed
in the lab coordinates.

Bob moves at uniform acceleration of magnitude $\alpha$ in the x
direction. His path is
\begin{eqnarray}
x^{\mu}(\tau) & = & (t\left(\tau\right),x\left(\tau\right))\nonumber \\
t\left(\tau\right) & = & \frac{1}{\alpha}sinh(\alpha\tau),\: x(\tau)=\frac{1}{\alpha}coshc(\alpha\tau)\label{eq:bob's trajectory}\end{eqnarray}
To see why, differentiate $x^{\mu}$ twice, and you will get the constant
acceleration.\begin{eqnarray}
u^{\mu}(\tau) & = & \frac{dx^{\mu}}{d\tau}=\left(cosh(\alpha\tau),sinh(\alpha\tau)\right)\nonumber \\
a^{\mu}(\tau) & = & \frac{du^{\mu}}{d\tau}=\left(\alpha sinh\left(\alpha\tau\right),\alpha cosh\left(\alpha\tau\right)\right)\nonumber \\
\sqrt{a^{\mu}a_{\mu}} & = & \sqrt{g_{\mu\mu}a^{\mu}a^{\mu}}=\sqrt{\alpha^{2}\left(-sinh^{2}\left(\alpha\tau\right)+cosh^{2}\left(\alpha\tau\right)\right)}=\alpha.\end{eqnarray}
You also see that $x^{2}-t^{2}=\frac{1}{a^{2}}$ , a hyperbola. Figure
1 shows Bob's world line in Minkowsky space, assuming he's going in
the positive $x$ direction. If he were going in the $-x$ direction
the graph would be a mirror image on the left of the t axis.


%
\begin{figure}[H]
\caption{\label{fig:rindler}\protect\includegraphics[scale=0.65]{rindler}}

\end{figure}


Bob doesn't measure things in the lab frame but in his own {}``proper''
frame. We call Bob's coordinates $\eta$ (his time) and $\xi$ (his
space coordinate). The derivation is at the end, now we just write
down the relation between $t,x$ and $\eta,\xi:$\begin{eqnarray}
t & = & \frac{1}{\alpha}e^{\alpha\xi}sinh\left(\alpha\eta\right)\nonumber \\
x & = & \frac{1}{\alpha}e^{\alpha\xi}cosh\left(\alpha\eta\right)\nonumber \\
ds^{2} & = & e^{2\alpha\xi}\left(-d\eta^{2}+d\xi^{2}\right)\end{eqnarray}
This is known as the Rindler coordinate system. Note that time and
space coordinates mix: Minkowsky time contains Rindler space and time,
so does Minkowsky space, and vice versa. 

Also note that Rindler space only covers part of Minkowsky space because
the speed of light is finite. So there is a part of space from whch
no signal can reach Bob. 


\subsection{Bob's state}

Since his space and time coordinates are both different, his {}``measuring
ruler'' for finding particles is different. Details are in Appendix
A. For now I will just say what the difference is.

There are 2 important differences: First, the ground state in the
lab now seems to Bob to have particles in it:\begin{equation}
\left|0\right\rangle _{lab}\rightarrow\left|n\right\rangle _{Bob}.\label{eq:bob vacuum}\end{equation}
Second, the {}``size'' of the particles is different. This is because
the time and space coordinates have changed, so the momentum has changed
accordingly. Therefore we have a different creation operator:\begin{eqnarray}
b^{\dagger}\left|0\right\rangle _{Bob} & = & \left|1\right\rangle _{Bob}\end{eqnarray}
and $b^{\dagger}\neq a^{\dagger}$. 

So even without Alice, if Bob has his own state (for instance he holds
a hydrogen atom), if it looked like a ground state to him in the lab,
it will now look like an excited state. And if it looked like an excited
state in the lab, it looks like a \emph{different }excited state in
the car (more excited, and having a different amount of energy between
each level than he had in the lab.)


\subsection{Translation of lab state into Bob's coordinates. }

Now it becomes more complicated yet.

In the lab we had two possibilities, $\left|0\right\rangle _{lab}$
and $\left|1\right\rangle _{lab}$. They refer to the possibility
to find particles anywhere in space. But Bob has to describe this
(all space) in two parts: the part to which he has access, (the right
side on the drawing) and the part to which he does not. So actually
eq.\ref{eq:bob vacuum} should be written \[
\left|0\right\rangle _{lab}\rightarrow\left|\psi_{I}\right\rangle \left|\psi_{II}\right\rangle _{Bob}\]
where $I,II$ refer to the left and right halves of Minkowsky space,
and we have to separate the parts which Bob can or cannot know about.
We expand the lab state in Bob's basis of $\left|n_{I}\right\rangle \left|n_{II}\right\rangle _{Bob}$.
It's a vacuum in the lab, but Bob sees it as a superposition of excited
states in his basis. (remember, an excited state in Bob's basis isn't
the same as an excited state in the lab.)

It can be shown \cite{milburn} that \begin{eqnarray}
\left|0\right\rangle _{lab} & = & \frac{1}{cosh(r)}\sum_{n=0}^{\infty}tanh^{n}\left(r\right)\left|n_{I}\right\rangle \left|n_{II}\right\rangle _{Bob}\nonumber \\
cosh\left(r\right) & = & \frac{1}{\sqrt{1-exp\left(-2\pi\Omega\right)}}\\
\Omega & \equiv & \frac{|k|}{a}\end{eqnarray}
Cosh{[}r{]} rises monotonically from 0, while Tanh{[}r{]} has a maximum
of 1. A plot of $r$ as a function of the acceleration, for constant
$k$, shows that $r$ rises as $a$ increases:%
\begin{figure}[H]
\caption{\protect\includegraphics[scale=0.5]{arccosh_r}}



\end{figure}
 For small acceleration, $cosh(r)\approx1,$ as acceleration increases
$cosh(r)$ increases

We say Bob is in the right half, labeled $I$. Then Bob's excited
state will be\begin{equation}
\left|1\right\rangle _{lab}=\frac{1}{cosh(r)}\sum_{n=0}^{\infty}tanh^{n}\left(r\right)\sqrt{n+1}\left|n+1_{I}\right\rangle \left|n_{II}\right\rangle _{Bob}.\end{equation}
We put the excited state in Bob's part of space because that describes
his amplitude to detect a photon. %
\footnote{Actually the reason is that Alice and Bob have the same time orientation.
What Alice considers a particle, Bob also considers a particle, whereas
if Bob were in the other part of space he'd see it as an antiparticle. %
} But since we are translating the lab state into Bob's coordinates,
we have to write$\left|n_{II}\right\rangle $ as well, because the
lab state related to ALL of space, not just Bob's part of it.

In the lab the density matrix for Alice and Bob's state was\begin{eqnarray*}
\left|\psi_{AB}\right\rangle \left\langle \psi_{AB}\right| & = & \frac{1}{2}\left[\left|0_{A}\right\rangle \left|0_{B}\right\rangle +\left|1_{A}\right\rangle \left|1_{B}\right\rangle \right]\left[\left\langle 0_{A}\right|\left\langle 0_{B}\right|+\left\langle 1_{A}\right|\left\langle 1_{B}\right|\right]\\
 & = & \frac{1}{2}\left[\left|0_{A}\right\rangle \left|0_{B}\right\rangle \left\langle 0_{A}\right|\left\langle 0_{B}\right|+\left|0_{A}\right\rangle \left|0_{B}\right\rangle \left\langle 1_{A}\right|\left\langle 1_{B}\right|+\left|1_{A}\right\rangle \left|1_{B}\right\rangle \left\langle 0_{A}\right|\left\langle 0_{B}\right|+\left|1_{A}\right\rangle \left|1_{B}\right\rangle \left\langle 1_{A}\right|\left\langle 1_{B}\right|\right].\end{eqnarray*}
We now rewrite this in terms of lab states for Alice and car states
for Bob, \[
\left|m_{A}\right\rangle \left|n_{B}\right\rangle \]
Using eq.\ref{eq:bob vacuum} we might think\begin{equation}
\left|0_{A}\right\rangle \left|0_{B}\right\rangle =\left|0_{A}\right\rangle \frac{1}{cosh(r)}\sum_{n=0}^{\infty}tanh^{n}\left(r\right)\left|n_{I}\right\rangle \left|n_{II}\right\rangle _{Bob}.\end{equation}
But this is a problem because Bob has no access to area $II$ . So
he can't write his state in terms of $\left|n_{I}\right\rangle \left|n_{II}\right\rangle $.
He's in a mixed state where he lacks information about area $II$,
and therefore has to trace out that area and write a density matrix.
The resulting density matrix for Alice and Bob will have components
such as\[
\left|0_{A}\right\rangle \left|n_{I,B}\right\rangle \left\langle 0_{A}\right|\left\langle n_{I,B}\right|\]
and so on. (without $n_{II}$). From now on we will write this as
$\left|0,n\right\rangle \left\langle 0,n\right|$ for simplicity (without
the subscripts), where the first part refers to Alice, in the lab,
and the second to Bob, in the half of space which he has access to,
after tracing out $n_{II}$: \begin{eqnarray}
\rho_{AB} & = & \frac{1}{2cosh^{2}r}\sum_{n=0}^{\infty}\left(tanh\left(r\right)\right)^{2n}\rho_{n}\nonumber \\
\rho_{n} & = & \left|0,n\right\rangle \left\langle 0,n\right|+\frac{\sqrt{n+1}}{cosh\left(r\right)}\left|0,n\right\rangle \left\langle 1,n+1\right|+\nonumber \\
 &  & \frac{\sqrt{n+1}}{cosh\left(r\right)}\left|1,n+1\right\rangle \left\langle 0,n\right|+\frac{n+1}{cosh^{2}\left(r\right)}\left|1,n+1\right\rangle \left\langle 1,n+1\right|.\end{eqnarray}



\section{Entanglement}

Here we can't use Von Neumann entropy because Bob is in a mixed state
even before tracing out Alice. For a mixed state in general\begin{equation}
\rho_{AB}=\sum_{ijkl}W_{ijkl}\left|i\right\rangle \left|j\right\rangle \left\langle k\right|\left\langle l\right|\end{equation}
you don't have a Schmidt decomposition. Instead we use the partial
transpose criterion for entanglement (valid for bipartite state).
Given a composite state $\rho_{AB},$ if it is a product state\[
\rho_{AB}=\sum_{i,j}A_{ij}\rho_{A}^{i}\otimes\rho_{B}^{j}\]
it will decompose into positive density matrices even if you transpose
components of the subsystems. If it's entangled you can't decompose
it. So if at least one eigenvalue of the partially transposed density
matrix is negative, the density matrix is entangled. (Further details
in the Appendix.)

After doing partial transpose, we look at the eigenvalues in the $n,n+1$
block of the transposed matrix (since $n$ goes to infinity, we have
to choose one block):

\begin{eqnarray}
\lambda_{\pm}^{n} & = & \frac{tanh^{2n}r}{4cosh^{2}r}\left[\left(\frac{n}{sinh^{2}r}+tanh^{2}r\right)\pm\sqrt{Z_{n}}\right]\\
Z_{n} & = & \left(\frac{n}{sinh^{2}r}+tanh^{2}r\right)^{2}+\frac{4}{cosh^{2}r}.\end{eqnarray}
If the acceleration is finite ($r<\infty$) one eigenvalue is always
negative. So the state is always entangled. The question is: is it
maximally entangled as before? 

To find out just how entangled it is we sum over all the negative
eigenvalues and calculate the logarithmic negativity.

Logarithmic negativity is defined as follows, where $\rho^{pt}$ is
the partially transposed $\rho_{AB}$ density matrix:\begin{eqnarray}
LN\left(\rho_{AB}\right) & = & log_{2}\left\Vert \rho^{pt}\right\Vert \nonumber \\
\left\Vert \rho^{pt}\right\Vert  & = & \sqrt{\rho^{pt}\left(\rho^{pt}\right)^{T}}\end{eqnarray}
(trace norm of the density matrix). It relates to the negativity thus\begin{eqnarray*}
N & = & \sum_{i}\frac{|\lambda_{i}|-\lambda_{i}}{2}\\
LN & = & log_{2}\left[2N+1\right].\end{eqnarray*}
For instance, for an EPR state in the lab, as Alice and Bob had at
first, there is 1 negative eigenvalue in the partially transposed
matrix, so the negativity is 1, and the logarithmic negativity is
$log_{2}3\approx1.585.$ When Bob is accelerating this becomes a function
of the acceleration parameter r:\begin{eqnarray}
LN(\rho_{AB}) & = & log_{2}\left(\frac{1}{2cosh^{2}r}+S\right)\nonumber \\
S & = & \sum_{n=0}^{\infty}\frac{tanh^{2n}r}{2cosh^{2}r}\sqrt{\left(\frac{n}{sinh^{2}r}+tanh^{2}r\right)^{2}+\frac{4}{cosh^{2}r}}.\end{eqnarray}
For vanishing acceleration $r=0$, $N=1$ and entanglement is maximal
like before. For finite acceleration entanglement is smaller. Figure
2 is a graph of the logarithmic negativity as a function of the acceleration
parameter $r$:%
\begin{figure}[H]
\caption{\protect\includegraphics[scale=0.6]{negativity}}



\end{figure}
. We see that the degree of entanglement is less as the acceleration
grows. This means that entanglement is an observer dependent quantity.
The physical situation of the particles hasn't changed, but Bob sees
them as less entangled than Alice does: when we write them using Bob's
basis the entanglement shrinks as a function of the acceleration.

\appendix



\section{Unruh effect }


\subsection{Rindler coordinates}


\subsubsection{Comoving frame:}

There are 3 coordinate frames involved: the lab frame where Alice
stands, the co-moving frame which freezes Bob for an instant, and
the proper frame where Bob lives. (Of course for each $t$ there's
a different comoving frame). We write the lab coordinates as functions
of his proper time $\tau$ as follows. We know the following:\[
u^{\mu}\equiv\frac{dx^{\mu}}{d\tau},\: u_{\mu}u^{\mu}=1,\: a_{\mu}a^{\mu}=\left|a\right|^{2}\]
We assume $u^{0}>0$ and $du^{1}/d\tau>0$ (acceleration is in positive
x direction) and so, dropping the $y,z$ space dimensions we have
\begin{eqnarray*}
u^{\mu} & = & \left(u^{0},u^{1}\right),\: a^{\mu}=\frac{du^{\mu}}{d\tau}\end{eqnarray*}
\begin{eqnarray}
\left(u^{0}\right)^{2}+\left(u^{1}\right)^{2}=1, &  & \left(\frac{du}{d\tau}^{0}\right)^{2}+\left(\frac{du}{d\tau}^{1}\right)^{2}=\alpha^{2}.\nonumber \\
u^{0}=\sqrt{1+\left(u^{1}\right)^{2}}, &  & \frac{du^{1}}{d\tau}=\alpha\sqrt{1+\left(u^{1}\right)^{2}}\end{eqnarray}
Taking initial condition as $u^{1}(0)=0,$ integration gives\begin{eqnarray*}
u^{1} & = & sinh(a\tau)\\
u^{0} & = & cosh(a\tau).\end{eqnarray*}
Integrate again, taking $\tau(0)=0:$\begin{eqnarray}
x & = & \frac{1}{\alpha}cosh(a\tau)-\frac{1}{\alpha}+x_{0}\nonumber \\
t & = & \frac{1}{\alpha}sinh(a\tau).\label{eq:comoving}\end{eqnarray}
For convenience we choose initial conditions $x_{0}=\frac{1}{\alpha}$
to obtain the results given in Sec.\ref{sub:Coordinates:}.


\subsubsection{Proper frame}

Lab coordinates are $x$. The coordinates above were for the comoving
frame, instantaneously coinciding with the accelerating system. We
want the coordinates seen by an inertial observer in the lab (or floating
in space as the ship passes). So we do a Lorentz transform from the
comoving frame to the lab frame. 

We call Bob's coordinates $\tau,\xi$ . Since he doesn't see himself
as moving, he is always at $\xi=0$. To work out the relationship
of $\tau,\xi$ to the lab $x,t$ we give him a stick to hold out,
so one end of it is in his hand at $\xi=0$ and the other end at some
particular $\xi.$ The stick moves with Bob, so the coordinates at
the end of the stick are $(\tau,\xi)$ at some time $\tau$ (ignoring
the y, z coordinates). The comoving frame is an inertial system moving
at $u^{\mu}=dx^{\mu}/d\tau$. So we can make an inverse Lorentz transformation
to relate the comoving system to the lab. Call the lab frame $s$,
and the moving frame $\tilde{s}$. Then $s^{\mu}=\Lambda_{\nu}^{\mu}\tilde{s}^{\nu}$:\begin{eqnarray*}
\left(\begin{array}{c}
s^{0}\\
s^{1}\end{array}\right) & = & \left(\begin{array}{cc}
\frac{1}{\sqrt{1-v^{2}}} & \frac{v}{\sqrt{1-v^{2}}}\\
\frac{v}{\sqrt{1-v^{2}}} & \frac{1}{\sqrt{1-v^{2}}}\end{array}\right)\left(\begin{array}{c}
\tilde{s}^{0}\\
\tilde{s}^{1}\end{array}\right)\\
 & = & \left(\begin{array}{cc}
u^{0} & u^{1}\\
u^{1} & u^{0}\end{array}\right)\left(\begin{array}{c}
\tilde{s}^{0}\\
\tilde{s}^{1}\end{array}\right)=\left(\begin{array}{c}
u^{1}\xi\\
u^{0}\xi\end{array}\right)\end{eqnarray*}
because $\tilde{s}^{0}=0,\, s^{1}=\xi.$ So the lab coordinates will
be\begin{eqnarray}
t\left(\tau,\xi\right) & =x^{0}\left(\tau\right)+s_{lab}^{0}= & x^{0}\left(\tau\right)+\frac{dx^{1}\left(\tau\right)}{d\tau}\xi\nonumber \\
x\left(\tau,\xi\right) & =x^{1}\left(\tau\right)+s_{lab}^{1}= & x^{1}\left(\tau\right)+\frac{dx^{0}\left(\tau\right)}{d\tau}\xi.\end{eqnarray}
Now we plug in $x^{\mu}(\tau)$ from eq.(\ref{eq:comoving}) and their
derivatives, with the initial conditions as above $x^{0}(0)=0,$ $x^{1}(0)=\frac{1}{\alpha}$,
and obtain\begin{eqnarray}
t\left(\tau,\xi\right) & = & \left(\frac{1}{\alpha}+\xi\right)sinh\left(\alpha\tau\right)\nonumber \\
x\left(\tau,\xi\right) & = & \left(\frac{1}{\alpha}+\xi\right)cosh\left(\alpha\tau\right)\end{eqnarray}
and the inverse coordinates are\begin{eqnarray}
\tau\left(t,x\right) & = & \frac{1}{2\alpha}ln\frac{x+t}{x-t}\nonumber \\
\xi\left(t,x\right) & = & -\frac{1}{\alpha}+\sqrt{x^{2}+t^{2}}.\end{eqnarray}
Limits are $\xi>-1/\alpha.$ 


\subsubsection{Causality}

Note that a signal sent from the origin $(x=0,t=0)$ will reach coordinate
$\xi=-\frac{1}{\alpha}$. If we assume Bob to be at $\xi=0$ , it
will never reach him. For any fixed $t$ there is thus an entire region
inaccessible to Bob. 


\subsubsection{Redefinition of metric}

The metric now is\begin{equation}
ds^{2}=dt^{2}-dx^{2}=\left(1+a\xi\right)^{2}d\tau^{2}-d\xi^{2}.\end{equation}
However - For ease of calculations we want a metric that is conformally
flat (you can only do it in 2d). Define $\tilde{\xi}$ so \begin{eqnarray}
d\xi & = & \left(\frac{1}{\alpha}+\xi\right)d\tilde{\xi}\nonumber \\
\tilde{\xi} & = & \frac{1}{\alpha}ln\left(\frac{1}{\alpha}+\xi\right).\end{eqnarray}
 Since $\xi>-1/\alpha$, limits for $\tilde{\xi}$ are $\pm\infty.$
Rename the time coordinate $\eta$ to avoid confusion with $t$. The
coordinates are now\begin{eqnarray}
t & = & \frac{1}{\alpha}e^{\alpha\xi}sinh\left(\alpha\eta\right)\nonumber \\
x & = & \frac{1}{\alpha}e^{\alpha\xi}cosh\left(\alpha\eta\right)\end{eqnarray}
and the metric is \begin{equation}
ds^{2}=e^{2\alpha\xi}\left(d\eta^{2}-d\xi^{2}\right)\end{equation}
which is conformal to Minkowsky space.


\subsection{Operators}

Rindler space operators are developed using quantum field theory.
A field is written with both creation and destruction operators: $\psi(p)=$
$ae^{-ip_{\mu}x^{\mu}}+cc$. Since now we have different $x,t$ and
thus different momenta and frequency for the Rindler system, the field
operators are different from the Minkowsky operators, as follows

\begin{eqnarray}
\psi_{mink}(p) & \sim & ae^{ipx-iwt}+a^{\dagger}e^{-ipx+iwt},a_{mink}\left|0\right\rangle _{m}=0\nonumber \\
\psi_{Rind}(p) & \sim & be^{i\tilde{p}\xi-i\tilde{w}\eta}+b^{\dagger}e^{-i\tilde{p}\xi+i\tilde{w}\eta},b_{Rind}\left|0\right\rangle _{R}=0.\end{eqnarray}
Since x and t \emph{each }contain both $\xi$ and $\eta$ , therefore
$iwt\neq i\tilde{w}\eta$. Finding the relation between the two frequencies
is a complicated problem in quantum field theory. The main point is
that since the field operators $\psi$ are different, so are the creation
and destruction operators, and so $a\neq b$ .

To find the relation we expand the Rindler operator in terms of the
Minkowsky ones:\begin{equation}
b_{\tilde{w}}=\intop_{0}^{\infty}dw\left[\alpha_{\tilde{w}\omega}a_{\omega}-\beta_{\tilde{w}\omega}a_{\omega}^{\dagger}\right]\label{eq:b}\end{equation}
or in the discrete case this would be\begin{equation}
b_{i}=\sum_{j=0}^{\infty}\left[\alpha_{ij}a_{j}-\beta_{ij}a_{j}^{\dagger}\right].\end{equation}
We can work out the relation between $\alpha$ and $\beta$ . The
normalization condition is\begin{equation}
\intop_{0}^{\infty}dw\left(\alpha_{\tilde{w}\omega}\alpha_{\tilde{w}'\omega}^{*}-\beta_{\tilde{w}\omega}\beta_{\tilde{w}'\omega}^{*}\right)=\delta\left(\tilde{w}-\tilde{w}'\right).\end{equation}
The procedure is to write down $\psi$ in each set of coordinates
and then plug in eq.\ref{eq:b} and equate them. Then extract the
coordinates of the ladder operator $a$ from each side and equate
them. This is tricky because the measure of the integral includes
$\sqrt{g}$ from the Rindler metric, and it affects the Fourier transform
as well. (This is how the acceleration enters into the final result).
For details see, for instance, \cite{mukhanov},\cite{unruh}. You
wind up with \begin{equation}
\left|\alpha\right|^{2}=e^{2\pi\tilde{w}/a}\left|\beta\right|^{2}.\end{equation}
(The $a$ in the exponent refers to acceleration.) The number of particles
a Rindler observer sees in a Minkowsky vacuum is\begin{equation}
N={}_{M}\left\langle O\right|b^{\dagger}b\left|0\right\rangle _{M}=\int d\omega|\beta|^{2}=\frac{1}{e^{2\pi\tilde{w}/a}-1}\delta(0).\end{equation}
The delta function is the infinite volume of space, and diving by
that we get the particle density:\begin{equation}
n=\frac{N}{V}=\frac{1}{e^{2\pi\tilde{w}/a}-1}\end{equation}
This is just the Planck thermal distribution, with inverse temperature
$\beta=2\pi/a$ and so the Unruh temperature is \begin{equation}
T=\frac{a}{2\pi}.\end{equation}


\section{Partial transpose criterion}

Take the composite density matrix and transpose the components of
one of the subsystems. If the resulting matrix has positive eigenvalues,
the system is separable and was composed of non entangled systems.
To understand why, write out the matrix elements:\begin{equation}
\rho_{m\mu,n\nu}=\sum_{r}p_{r}(\rho_{A}^{r})_{mn}(\rho_{B}^{r})_{\mu\nu}\end{equation}
Latin indices refer to Alice, Greek indices to Bob. The partial transpose
is given by transposing only one of the subsystems. The entries of
the density matrix are then\begin{equation}
\rho_{m\mu,n\nu}^{PT}=(\rho_{A}^{T})_{m\mu,n\nu}=\rho_{n\mu,m\nu}\end{equation}
If the joint density matrix is separable, it's a tensor product of
two density matrices each non negative with unit trace: \[
\rho_{sep}^{PT}=\sum_{i}p_{i}(\vert A_{i}\rangle\langle A_{i}\vert)^{T}\otimes\vert B_{i}\rangle\langle B_{i}\vert=\sum_{i}p_{i}(\rho_{i}^{A})^{T}\otimes\rho_{i}^{B}\]
The transposed matrices $(\rho^{A})^{T}=(\rho^{A})^{*}$ are also
non-negative with unit trace, and so none of the eigenvalues of $\rho^{PT}$
is negative. If $\rho^{PT}$ has negative eigenvalues, it means it
was not derived from a separable product of two density matrices,
and so they were entangled. If the eigenvalues are positive, it has
been shown that the density matrix might be entangled but the entanglement
can't be distilled and used; this is called bound entanglement. 
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\bibitem{milburn}D.F.Walls and G.J.Milburn, Quantum Optics, Springer-Verlag,
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\bibitem{fulling}S.A. Fulling, Phys.Rev. D7, 2850 (1973)

\bibitem{mukhanov}V.F.Mukhanov and S.Winitzki, Introduction to Quantum
Effects in Gravity, Cambridge University Press, Cambridge (2007)

\bibitem{unruh}W.G.Unruh, Physical Review D 14:4,870 (1976)
\end{thebibliography}

\end{document}
