# Help on the notation of the 2-bit operation: $\frac 12{(I+Z_1Z_0)}$

This was adapted from Quantum Mechanichs for Computer Scientists. On this equation, we have $I$ representing the identity operator, with $Z$ being defined for 1 bit as:

$$Z|0⟩ = |0⟩, \quad Z|1⟩ = -|1⟩\,.$$

The operation $\frac 12{(I+Z_1Z_0)}$ acts as the identity for the two-bit states $|00⟩$ and $|11⟩$ while returning $0$ for the states $|01\rangle$ or $|10\rangle$.

I'm assuming the operator addition is the linear algebra definition, but I can't understand the $\frac 12$ fraction and its effect on the operation.

Could you point me to what concept I am missing, so I can derive the resulting states of this operation?

The operation $I$ on two bits can be written as the identity matrix, $$I = \begin{pmatrix} 1 & 0 & 0 &0 \\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 &0 \\ 0 & 0 & 0 &1 \end{pmatrix}\,.$$
The operation $Z_0Z_1$ that performs $Z$ on both qubits can be written as $$Z_0Z_1 = \begin{pmatrix} 1 & 0 & 0 &0 \\ 0 & -1 & 0 &0 \\ 0 & 0 & -1 &0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\,.$$
Now, the operation that you ask about, $\frac12(I+Z_0Z_1)$ is simply the operation described by
$$\frac12(I+Z_0Z_1) = \frac12 \begin{pmatrix} 1 & 0 & 0 &0 \\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 &0 \\ 0 & 0 & 0 &1 \end{pmatrix} + \frac12 \begin{pmatrix} 1 & 0 & 0 &0 \\ 0 & -1 & 0 &0 \\ 0 & 0 & -1 &0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 &0 \\ 0 & 0 & 0 &0 \\ 0 & 0 & 0 &0 \\ 0 & 0 & 0 &1 \end{pmatrix}\,.$$ 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.)