Problem:
Show that
\(\sqrt{X} \sqrt{X} = I\)
Solution:
First note that
$$
(1+i)^2 + (1-i)^2 \eql 1 + i^2 + 2i + 1 + i^ - 2i \eql 0
$$
and
$$
(1+i)(1-i) + (1+i)(1-i) \eql 1^2 + 1^2 + 1^2 + 1^2 \eql 4
$$
Next,
$$
\sqrt{X} \: \sqrt{X}
\eql
\mat{ \frac{1+i}{2} & \frac{1-i}{2}\\
\frac{1-i}{2} & \frac{1+i}{2}}
\mat{ \frac{1+i}{2} & \frac{1-i}{2}\\
\frac{1-i}{2} & \frac{1+i}{2}}
\eql
\frac{1}{4}
\mat{ 1+i & 1-i\\
1-i & 1+i}
\mat{ 1+i & 1-i\\
1-i & 1+i}
\frac{1}{4}
\mat{0 & 4\\ 4 & 0}
\eql
\mat{0 & 1\\ 1 & 0}
\eql X
$$
Problem:
Consider the matrix
$$
A \eql \mat{a^* & b^*\\
-b & a}
$$
Show that \(A\) is unitary. Then, show how the numbers
\(a,b\) can be selected so that \(A\kt{0} = \ksi\)
for any arbitary \(\ksi = \alpha\kt{0} + \beta\kt{1}\).
That is, \(A\) can be used to generate any desired
state from \(\kt{0}\).
Solution:
First,
$$
A^\dagger \eql
\mat{a & -b^*\\
b & a^*}
$$
(The diagonal entries conjugate. The others reflect about the
diagonal and conjugate.) Thus,
$$
A^\dagger A \eql
\mat{a & -b^*\\
b & a^*}
\mat{a^* & b^*\\
-b & a}
\eql
\mat{aa^* + bb^* & ab^* - ab^*\\
a^*b - a^*b & bb^* + aa^*}
\eql
\mat{1 & 0\\ 0 & 1}
$$
because \(aa^* + bb^* = |a|^2 + |b|^2 = 1\). The squared magnitudes
add to 1 because the columns are orthonormal (hence, unit-length).
Next,
$$
A \kt{0}
\eql
\mat{a^* & b^*\\
-b & a}
\vectwo{1}{0}
\eql
\vectwo{a^*}{-b}
$$
Thus, by equating \(\alpha = a^*, \beta = -b\), we can generate
any vector.