On Pauli matrices being a basis for unitary transformations... I am confused about the Pauli matrices.
If you denote the Pauli matices plus the identity matrix as:
$$
\sigma^0 = \left(\begin{matrix}1&0\\
0&1
\end{matrix}\right)
$$
$$
\sigma^1 = \left(\begin{matrix}
0&1\\
1&0
\end{matrix}\right)
$$
$$
\sigma^2 = \left(\begin{matrix}
0&-i\\
i&0
\end{matrix}\right)
$$
$$
\sigma^3 = \left(\begin{matrix}
1&0\\
0&-1
\end{matrix}\right)\;,
$$
then any complex 2x2 matrix can be written as
$$
M = \sum_{\mu=0}^3m_\mu \sigma^\mu
$$
$$
=\left(\begin{matrix}
m_0+m_3&m_1-im_2\\
m_1+im_2&m_0-m_3
\end{matrix}\right)\;.
$$
If you demand that $M$ is unitary, you must have
$$
M^\dagger M = 1 = \sum_{\mu\nu}m_\nu^*m_\mu\sigma^\nu\sigma^\mu\;.\tag{1}
$$
This gives the conditions
$$
|m_0|^2 + |\vec m|^2 = 1
$$
and
$$
m_0^*\vec m + m_0\vec m^* + i\vec m^*\times\vec m = 0
$$
for $M$ to be unitary.
For example, as a special case, I could take $\vec m$ real and $m_0$ imaginary with
$$
-m_0^2 + \vec m^2 = 1\;.
$$
or with $m_0\equiv i m_4$ with $m_4$ real
$$
m_4^2 + \vec m^2 = 1\tag{2}\;.
$$
... I am confused, because the unitary matrices do not form a vector space. Linear combination of unitary matrices is not unitary in general. What am I missing here?
What this means is that if $U$ and $V$ are unitary then they can be represented as
$$
U=\sum_{\mu}u_\mu \sigma^\mu\;,
$$
$$
V=\sum_{\mu}v_\mu \sigma^\mu\;.
$$
But, indeed, the sum $U+V$ will not necessarily be unitary. In fact, if we use the special representation of Eq. (2) above we can see that we need $u\cdot v=-1/2$ for the sum to be unitary.
