We introduced unitary operators as those that map the Hilbert space to itself (i.e., those that maintain the normalization constraint that we require of all vectors in the Hilbert space). However, we didn't show the converse, that there always exists a unitary operator that connects two vectors of the Hilbert space. For a Hilbert space \(\mathcal{H}\) of two-dimensional vectors, consider the vectors \(\vec{u}, \vec{v} \in \mathcal{H}\).

(a) We can express these vectors in a particular basis as

\[\vec{u}=\left(\begin{array}{c}e^{i \phi_{1}} \cos u \tag{3.155}\\e^{i \phi_{2}} \sin u\end{array}\right), \quad \vec{v}=\left(\begin{array}{c}e^{i \theta_{1}} \cos v \\e^{i \theta_{2}} \sin \end{array}\right)\]

for real parameters \(u, v, \phi_{1}, \phi_{2}, \theta_{1}, \theta_{2}\). Show that these vectors are normalized to live in the Hilbert space.

(b) In this basis, determine the unitary matrix \(\mathbb{U}\) such that

\[\begin{equation*}\vec{v}=\mathbb{U} \vec{u} . \tag{3.156}\end{equation*}\]

Write the matrix \(\mathbb{U}\) as

\[\mathbb{U}=\left(\begin{array}{ll}a & b \tag{3.157}\\c & d\end{array}\right)\]

and determine the constraints on \(a, b, c, d\) if \(\mathbb{U}\) is unitary and maps \(\vec{u}\) to \(\vec{v}\).

(c) Can the matrix \(\mathbb{U}\) be degenerate? That is, can its determinant ever be 0 ?