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\begin{document}
\openup 1 \jot

\noindent\bf
Chapter 3 Qubits and Quantum Key Distributions

\rm
\bigskip
In this chapter, we take a close look at qubits, 
and show that it is tricky to obtain complete information 
of a qubit even if an infinite supply of the qubits are given. 
Furthermore, we prove the no-cloning theorem showing that
it is impossible to make copy of a qubit. Hence, one can
use quantum states to set up secure cryptology schemes. 
This is illustrated by the BB84 protocol.

\bigskip\noindent
\bf Qubits
\rm
\begin{itemize}
\item
Mathematically, qubit is a vector in $|x\ra = a|0\ra + b|1\ra = 
\begin{pmatrix}a \cr b \cr
\end{pmatrix} \in \IC^2$ with $|a|^2 + |b|^2$ realized by 
physical quantum states such as the vertically and horizontally 
polarized  photons, or spin 1/2 in NMR system.
\item
Note that measurement will give $|0\ra$ or $|1\ra$ 
even a qubit can 
assume infinitely many states.
The probability for the measurement on $|x\ra$  yielding 
$|0\ra$ is $\la x|(|0\ra \la 0|) |x\ra = |a|^2$.
\item
Even if we can get the information $|a|$ and $|b|$ 
by {\bf measuring} many
identical $|x\ra$ if it is available,
we cannot get complete
information of $|x\ra \la x|$. 

\item Using the measurable states $P_1 = |0\ra \la 0|, 
P_2 = |1\ra \ra 1|$ 
to get information of $\la x|P_1x\ra, \la x|P_2 x\ra$,
we have the  ``diagonal entries'' of $\rho$. 

\item In order to obtain
complete information of $|x\ra \la x|$, we may
apply unitary $U_1, \dots, U_r$
and measure the diagonal $U_j|x\ra \la x|U_j^\dag$
to access information of the off-diagonal entries.
Such study is known as  state tomography problem.
\end{itemize}

\medskip\noindent
{\bf Example}  
Suppose $\rho =\frac{1}{2} 
\begin{pmatrix} 
1+a & b-ic \cr b+ic & 1-a  \end{pmatrix}$ 
with $a^2+b^2+c^2 \le 1$.
If $U_1 = \frac{1}{\sqrt{2}}
\begin{pmatrix} 1 & 1 \cr 1 & -1 \cr\end{pmatrix},
U_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i \cr i & 1 \cr
\end{pmatrix}$.  Show that $U_1, U_2$ are unitary and 
the measurements of $U_1\rho U_1^\dag, U_2\rho U_2^\dag$
will yield information of $b, c$.


One may consider qutrits in $\IC^3$  and qudits in $\IC^n$.


\medskip\noindent
{\bf Theorem} (No cloning theorem) 
There is no unitary $U$ such that 
$U|\psi\ra |0\ra = |\psi\ra |\psi\ra$ for every 
$|\psi\ra$.

\it Proof. \rm If $U$ exists, then for any orthonormal 
pairs $|x\ra, |y\ra$ and 
$|z\ra = (|x\ra + |y\ra)/\sqrt 2$, we have

\medskip \centerline{
$U|x0\ra = |xx\ra$, 
\quad $U|y0\ra= |yy\ra$,}


\medskip
\ $U|z0\ra = |zz\ra 
= \frac{1}{2} (|xx\ra + |xy\ra + |yx\rangle+ yy\ra)$, 
\quad and \quad $\frac{1}{\sqrt 2}U(|x0\ra+|y0\ra)
= \frac{1}{\sqrt 2}(|xx\ra + |yy\ra)$,

\medskip\noindent
which is impossible. \qed


\medskip\noindent
\bf Bloch sphere and Bloch ball \rm

\noindent
Note that a mixed state (density matrix) 
$\rho \in M_2$
has the standard form 

\medskip
\hskip .5in  $\rho = \frac{1}{2}(I_2 + u\cdot \sigma)
= \frac{1}{2}(\sigma_0 + u_1 \sigma_1 + u_2 \sigma_2 
+ u_3 \sigma_3)$
\  with 
\ $|u| = \sqrt{u_1^2+u_2^2+u_3^2} \le 1$.

\medskip\noindent
Here $(\sigma_1, \sigma_2, \sigma_3) = 
(\sigma_x, \sigma_y, \sigma_3)$

\begin{itemize}
\item 
The eigenvalues of $\rho$ are $\frac{1}{2}(1\pm |u|)$.
\item $\rho$ is a pure state if and only if 
$|u| = 1$. In such a case 
$$u = 
\hat \bn(\theta,\phi) = 
(\sin\theta \cos \phi, \sin \theta \sin \phi, 
\cos\theta)$$
and
$$\rho = \frac{1}{2}(\sigma_0 + \hat \bn \cdot {\sigma})
= \frac{1}{2}(I_2 + P_1-P_2),$$
where  
$P_1 = |x\ra \la x|$ with 
$|x\ra = |\psi(\theta,\phi)\ra = 
\cos (\theta/2) |0\ra +  e^{i\phi} \sin(\theta/2)|1\ra = 
\begin{pmatrix}
\cos(\theta/2) \cr e^{i\phi} \sin(\theta/2)\cr\end{pmatrix}$.

Clearly, $\rho |x\ra = |x\ra$.
Also, $(\hat \bn \cdot \sigma)|x\ra = 
(2\rho - I)|x\ra = |x\ra$.
\end{itemize}

\iffalse
Let ${\bf \sigma} = (\sigma_x, \sigma_y, \sigma_z)$.
If $|x\ra = \cos (\theta/2) |0\ra +  e^{i\phi} \sin(\theta/2)|1\ra = 
\pmatrix{\cos(\theta/2) \cr e^{i\phi} \sin(\theta/2)\cr}$, 
then 
\begin{eqnarray*}|x\ra \la x| 
&=& 
\pmatrix{
\cos^2 (\theta/2) & e^{-i\phi}\sin(\theta/2)\cos(\theta/2)\cr
e^{i\phi}\sin(\theta/2)\cos(\theta/2) & \sin^2(\theta/2)\cr}
={1\over 2} \pmatrix{1 + \cos \theta & e^{-i\phi} \sin \theta \cr
e^{i\phi} \sin \theta & 1 - \cos \theta \cr}\\
&=&
{1\over 2} \pmatrix{1 + \cos \theta & \cos\phi \sin \theta - i \sin\phi \sin \theta \cr
\cos\phi \sin \theta + i\sin\phi\sin\theta & 1 - \cos \theta \cr} \cr
&=& {1\over 2} 
(\sigma_0 +  \sin\theta\cos\phi\sigma_x + \sin\theta\sin \phi \sigma_y + \cos\theta \sigma_z).
\end{eqnarray*}


\noindent
If 
$$\hat \bn(\theta,\phi) = (\sin\theta \cos \phi, \sin \theta \sin \phi, 
\cos\theta)^t,$$ 
then 
$$\hat \bn(\theta,\phi)\cdot {\bf \sigma} |x\ra = 
(2 |x\ra \la x| - I_2)|x\ra = |x\ra.$$
So, every state vector $|x\ra$ corresponds to a vector $\hat 
\bn(\theta,\phi)$ on the surface of the unit sphere, called the
Bloch sphere in this context.

\fi

\medskip\noindent
{\bf Remark} Note that for $\rho = \frac{1}{2}(\sigma_0 + n_x 
\sigma_x + n_y \sigma_y + n_z\sigma_z) \in D_2$, if $U$ is unitary 
such that
$$\rho \mapsto U\rho U^\dag = 
 \frac{1}{2}(\sigma_0 + \tilde n_x 
\sigma_x + \tilde n_y \sigma_y + \tilde n_z\sigma_z)$$
then there is an orthogonal matrix $T_U$ with determinant
1 such that
$(\tilde n_x, \tilde n_y, \tilde n_z)^t = 
T_u(n_x, n_y, n_z)^t$.


\bigskip\noindent
{\bf Multi-qubit systems and entangled states}


Given $n$ qubits $|x_1\ra, \dots, |x_n\ra$, we can consider the
tensor product $|x_1 \ra \otimes \cdots \otimes |x_n\ra \in \IC^N$
with $N = 2^n$. Most state vectors 
$$\sum_{i_k = 0, 1} a_{i_1 \cdots i_n} |x_{i_1}\ra \otimes
\cdots \otimes |x_{i_n}\ra \in \IC^N$$ 
are entangled state vectors, which are not of the tensor form.

\medskip\noindent
{\bf Notation} We often assume $|x_j\ra \in\{|0\ra, |1\ra\}$,
and regard $|x\ra = |x_{i_1} \cdots x_{i_n}\rangle= 
|q_{n-1} \cdots q_0\ra$ as a 
binary number, and $|\psi\ra = \sum_{i_k = 0,1}
a_{i_1 \cdots i_n} |x_{i_1} \cdots x_{i_n}\ra$.

\medskip\noindent
{\bf Example} $|x\ra = \frac{1}{2}
\sum_{i,j \in \{0,1\}} |ij\rangle 
= \frac{1}{2} (|00\ra + |01\ra + |10\ra + |11\ra)
= \frac{1}{2} \sum_{x=0}^3 |x\ra$.

\medskip\noindent
In quantum computing, we often implement quantum 
operation of the form: 
$$\sum_{x} |x\ra |0\ra \mapsto \sum_{x} |x\ra |f(x)\ra.$$
For example, if $f(0) = f(1) = 1$, we want 
$U$ such that 
$U|00\ra = |01\ra, U|10\ra = |11\ra$.
So, we may set $U|01\ra = |00\ra, U|11\ra = |10\ra$.
 

\medskip\noindent
{\bf Example} The Bell states
$$|\Phi^+\ra = {1\over \sqrt 2}(|00\ra + |11\ra), \quad
|\Phi^-\ra = {1\over \sqrt 2}(|00\ra - |11\ra), $$
$$|\Psi^+\ra = {1\over \sqrt 2}(|01\ra + |10\ra), \quad
|\Psi^-\ra = {1\over \sqrt 2}(|01\ra - |10\ra) $$
are entangled states and form an orthonormal basis for the two qubit systems.

\iffalse
\medskip\noindent
{\bf Remark} If we measure the first system, and do nothing about
the second one, then we use the observable 
$\diag(1,-1) \otimes I_2$. If the fire photon collapses to 
$|0\ra$, then the second one becomes $|0\ra$;
if the first phone collapses to $|1\ra$ then the second photon 
becomes $|1\ra$.
\fi

\medskip\noindent
{\bf Example} In the 3 qubit system, we have that GHZ state and 
W state:
$$|GHZ\ra = {1\over \sqrt 2} (|000\ra + |111\ra) 
\quad \hbox{ and } \quad 
|W\ra {1\over \sqrt 3} (|100\ra + |010\ra + |001\ra).$$

\medskip\noindent
{\bf Measurements}

For each outcome $m$, construct a measurement operator $M_m$ so that
the probability of obtaining outcome $m$ in the state $|x\ra$
is computed by
$$p(m) = \la x |M_m^\dag M_m| x\ra$$
and the state immediately after the measurement is 
$$|m\ra = {M_m |x\ra  \over \sqrt{p(m)}}.$$

\medskip\noindent
{\bf Example} Let $M = \{M_0, M_1\}$ with $M_0 = |0\ra \la 0|$ and $M_1 = |1\ra \la 1|$.
Then for $|x\ra = a|0\ra + b|1\ra$ with $a \ne 0$, 
$p(0) = |a|^2$, $M_0|x\ra = a|0\ra/|a|$, which is
the same as the vector state $|0\ra$.

\medskip
In general,
suppose an observable $M$ is given with measurement 
operators $M_m$. Then setting $P_i = M_i^{\dag}M_i$,
we require that $\sum_m P_m = I_n$.

If there are many copy of a state $|x\ra$, then the expected value
of $M$ is 
$$E(M) = \la M \ra = \sum_{m} mp(m) 
= \sum_{m} m \la x|P_m| x \ra = \la x|M|x\ra.$$
Here $M$ can be identified with $\sum_m mP_m$.

The standard derivation is 
$$\Delta(M) = \sqrt{\la (M-\la M\ra)^2} = \sqrt{\la M^2\ra - \la M\ra^2}.$$

The variance (square of standard deviation) is 
$$\la (M -  \la M \ra)^2 \ra = \la x |M^2| x\ra - \la x |M| x \ra^2.$$

\medskip\noindent
\bf Example \rm One can do measurement of the first qubit
for a state vector in a $n$ qubit system. For instance,
$$|x\ra = a|00\ra + b |01\ra + c|10\ra + d|11\ra, 
\quad |a|^2 + |b|^2 + |c|^2 + |d|^2 = 1.$$
We measure the first qubit with respect to the basis $\{|0\ra, |1\ra\}$.
Set
$$|x\ra = |0\ra (a|0\ra + b|1\ra) + |1\ra (c|0\ra + d|1\ra)
= u|0\ra ((a/u)|0\ra + (b/u)|1\ra) + v|1\ra ((c/v)|0\ra + (d/v) |1\ra),$$
where $u = \sqrt{|a|^2+|b|^2}$ and $v = \sqrt{|c|^2 + |d|^2}$.
Now,
$$M_0 = |0\ra \la 0| \otimes I_2, \quad M_1 = |1\ra \la 1| \otimes I_2.$$
Applying $M_0$ and $M_1$, we obtain $0$ with  probability $\la x|M_0|x\ra = u^2$
and 1 with probability $v^2$;
the state $|x\ra$ collapses to $|0\ra \otimes ((a/u)|0\ra + (b/u)|1\ra)$ 
and $|1\ra \otimes ((c/v)|0\ra + (d/v)|1\ra)$, respectively, upon measurement.
\newpage
\medskip\noindent
{\bf Einstein-Podolsky-Rosen (EPR) Paradox}

Consider the EPR state
$$|\Psi^-\ra = \frac{1}{\sqrt 2}(|01\ra - |10\ra).$$
Alice gets the first particle and Bob gets the second one.
When Alice measures, 
Bob's particle will change instantaneously to $|1\ra$
or $|0\ra$ depending on the measured outcome of Alice 
being $|0\ra$ or $|1\ra$.

For example, we can set up the apparatus for the observable  
$H = \diag(1,-1) \otimes I_2 = 
|0\ra \la 0|\otimes I_2 -
|1\ra \la 1| \otimes I_2$.
If Alice sees the reading 1, then Bob's qubit is to $|1\ra$;
if Alice sees the reading $-1$, then Bob's qubit is $|0\ra$.

Alice cannot control her measurement
and hence the reading of Bob! So, it does not violate the special theory of relativity.
(It is impossible that information travels faster than light!)

However, they can measure their individual states around  
the same time, and decide to make a move according to 
$|01\ra$ or $|10\ra$ occur.

\medskip\noindent
{\bf 3.2 Quantum Key Distribution (BB84 protocol)}

Alice wants to send $n$ bits to Bob securely.
Eve is the notorious eavesdropper.

\medskip
Alice and Bob will choose two randomly chosen systems:


\medskip
(1) $0 \mapsto |e_1\ra$, \ $1 \mapsto |e_2\ra$.

\medskip
(2) $0 \mapsto (|e_1\ra + |e_2\ra)/\sqrt 2$, \ $1 \mapsto (|e_1\ra - |e_2\ra)/\sqrt 2$.


\begin{itemize}

\item Alice send $4n$ bits to Bob, each using the two systems randomly.


\item Announce the types of systems; discard about $2n$ bits with wrong channel match.

Probability for Alice and Bob use the same system: (1)(1), (2)(2), vs.
(1)(2), (2)(1). 

\item Alice tells Bob $n$ of the remaining bits to test whether they have not been
tempered.

Alice, Eve, Bob use channels

(1)(1)(1) \  (1)(1)(2) \ (1)(2)(1) \ (1)(2)(2) \ (2)(1)(1) \ (2)(1)(2) \ (2)(2)(1) \ (2)(2)(2)

\quad
  Y \hskip .6in N \hskip .6in Y  \hskip .6in N  \hskip .5in
  N \hskip .6in Y \hskip .6in N \hskip .6inY 

\quad
  G \hskip .6in * \hskip .5in $\pm$ G  \hskip .6in 
   * \hskip .5in * \hskip .5in $\pm$ G  \hskip .6in * \hskip .6in G 


\item If not, use the remaining $n$ bits as the private key. Otherwise, repeat the process.
\end{itemize}

\medskip 
\noindent
{\bf Open problems}
\begin{itemize}

\item It is easy to show that one can use the following observable
$$H \in S = \{E_{jj}: 1 \le j \le N\} \cup \{E_{rs} + E_{sr}: 1 \le r < s \le N\} 
\cup \{i(E_{rs}-E_{sr}): 1 \le r < s \le N\}$$
to determine a density matrix $\rho \in D_N$ using the measurements
$\tr (H\rho)$ with $H \in S$.

Since $\rho$ has trace one, we may remove $E_{NN}$ from $S$.

One can show that one need at least $N^2-1$ observable to determine $\rho$.

\item If $\rho \in D_N$ has rank one, one may use $3N-3$ observable to determine $\rho$
as follows. Determine $\tr E_{jj} \rho$ to determine which diagonal entry $r$ is nonzero,
and then use $\tr H\rho$ for $E_{rs}+E_{rs}, i(E_{rs}-E_{rs})$ for $s = 1, \dots, N$
with $r \ne s$.


\item One can show that for a density matrix $\rho \in D_N$, there are $N$ 
unitary matrices $U_1, \dots, U_N$ so that one can use
the diagonal entries of $\rho, U_1 \rho U_1^\dag, \dots, U_N\rho U_N^\dag$
can be used to determine $\rho$.
It is an interesting problem to construct simple 
unitary matrices $U_1, \dots, U_N$, which could be easily implemented,
to carry out the scheme.

\item If $N = 2^n$ is the previous part, there are 
$r = 3^n$ unitary matrices of the form $U_j = U_{j1} \otimes \cdots \otimes U_{jn}$
such that one can use the diagonal entries of 
$U_1 \rho U_1^\dag, \dots, \cdots U_r \rho U_r^\dag$ to determine $\rho$.
If is an open problem to show that $r = 3^n$ is the minimum number for such 
a tomography scheme.

\end{itemize}

\end{document}

