Help on the notation of the 2-bit operation: $\frac 12{(I+Z_1Z_0)}$
The operation $I$ on two bits can be written as the identity matrix,
$$
I = 
\begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 1 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 1 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;1
\end{pmatrix}\,.
$$

The operation $Z_0Z_1$ that performs $Z$ on both qubits can be written as 
$$
Z_0Z_1 = 
\begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; -1 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; -1 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp; 1
\end{pmatrix}\,.
$$

Now, the operation that you ask about, $\frac12(I+Z_0Z_1)$ is simply the operation described by 

$$
\frac12(I+Z_0Z_1) = 
\frac12 \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 1 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 1 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;1
\end{pmatrix} +
\frac12 \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; -1 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; -1 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp; 1
\end{pmatrix}
= 
\begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;1
\end{pmatrix}\,.
$$
This indeed performs as you describe. If the $1/2$ wasn't there, we would get that $|0\rangle$ becomes $2|0\rangle$. Note that his operator is not unitary (it returns "$0$" for $|01\rangle$ and $|10\rangle$, which means the qubit "disappears", this shouldn't be allowed by quantum mechanics.)