All Questions
1 question
7
votes
1
answer
1k
views
Would IBM's "compiler" turn my identity circuit into nothing?
If I were to create a circuit with the following gate:
$$\tag{1}R_\phi = \begin{bmatrix} 1 & 0 \\ 0 & e^{i \phi} \end{bmatrix},$$
with $\phi$ specified to be equal to 0, then the gate that I ...
