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\title{2-Qubits with spins}

At the beginning most things here can be done analytically. You can
still try to program them if you like.

Consider two spin $\frac{1}{2}$ particles $a$ and $b$ with Hamiltonian 

\begin{equation}
H=H_{B}+H_{J}+H_{By}\label{h}
\end{equation}


$H_{B}=-B\:(S_{az}+S_{bz})$

$H_{By}=-B_{y}S_{by}$

$H_{J}=JS_{az}S_{bz}$

($S_{az}$ is the $z$ component of the spin of particle $a$, etc.)

later, we will use the spin-rotation invariant interaction 

$H_{J}^{'}=J\vv S_{a}\cdot\vv S_{b}$

The two spins represent qubits with the rule

\begin{equation}
\ket 0=\ket{+\frac{1}{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ket 1=\ket{-\frac{1}{2}}\label{01}
\end{equation}
 

for both particles. We work with $\hbar=1$


\lyxframeend{}\underline{Realisation of C-phase gate}

We first want a process producing the so-called C-phase gate, which
maps

\begin{gather}
\ket{0\: x}\to e^{i\phi}\ket{0\: x}\nonumber \\
\ket{1\: x}\to e^{i\phi}(-1)^{x}\ket{0\: x}\label{cph}
\end{gather}


here $\ket{x\: y}$ is a two-qubit register in which $x\in(0,1)$
is the state of the first spin and $y\in(0,1)$ is the one of the
second spin. $\phi$ is an arbitrary common phase which is immaterial
for a quantum state.

The transformation \ref{cph} can be realized by time evolution of
the Hamiltonian \ref{h}, whereby $B$ is switched on for a time $t_{B}$
and $J$ for a time $t_{J}$ , while $B_{y}=0$. \textcolor{blue}{Find
$t_{B}$ and $t_{J}$}

Try the same using $H_{J}^{'}$ instead of $H_{J}$ (I don't know the
answer)


\lyxframeend{}\underline{Realisation of C-not gate}

Show that a C-not gate is realized by first carrying a rotation of
$\pi/2$ around the $y$ axis, which is produced by switching on $B_{y}$
for a certain time $t_{y}$, then carrying out the C-phase as above,
and then reversing the $B_{y}$ field for the same time $t_{y}$. 

The C-not maps

$\ket{0\: x}\to\ket{0\: x}\:\:\:\:\:\:\:\:\:\:\:\:\:\ket{1\: x}\to\ket{1\: NOT(x)}$


\lyxframeend{}\underline{Reduced density operator as a measure of entanglement}

Given a normalized two-spin state $\ket{\Psi}$, the reduced density
operator on the space of the second spin is given by

\[
\rho=Tr_{1}\ket{\Psi}\bra{\Psi}
\]


where $Tr_{1}$ is the trace over the degrees of freedom of the first
spin.

The von Neumann entropy is given by 

$S=-\mbox{tr}\rho\ln\rho$

For the following initial states, write $\rho$ as a $2\times2$ matrix
and evaluate the entropy before and after the c-not or c-phase transformation
and evaluate the entropy (be careful about normalisation!):

$\ket{00}$

$\ket{10}$

$\ket{00}+\ket{10}$

$\ket{00}+\ket{01}+\ket{10}+\ket{11}$


\lyxframeend{}\underline{Next step: effect of environment on two-qubit gate, Lindblad approach}

TODO


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