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.