5
$\text{Tr}(AB)$ is always real and non-negative if $A,B$ are positive semi-definite hermitian matrices.
To see this note that $A = UDU^\dagger$, for some unitary $U$ and diagonal matrix $D$ with $d_{ii} \ge 0$.
Then $\text{Tr}(AB) = \text{Tr}(UDU^\dagger B) = \text{Tr}(DU^\dagger B U).$
But $B^\prime = U^\dagger B U$ is also positive semi-definite and ...
1
In the first circuit you only use 2 measurement gates on the first and second qubits. This means that the amount of outputs will behave similarly to a circuit that has only 2 qubits. If we were to list out all possible outputs, it would look like this: 0000, 0001, 0010, 0011. This is because the third and fourth qubits will always be 0, so you only have 2 ...
1
The key to figuring out the probability of any measurement result is Born's rule, which says that if you have a state $\left|\psi\right\rangle$ the probability of observing measurement outcome $\left|\phi\right\rangle$ is given by
$$
\begin{align}
\Pr(\phi | \psi) = \left|\left\langle \phi | \psi \right\rangle \right|^2.
\end{align}
$$
In the example you ...
Only top voted, non community-wiki answers of a minimum length are eligible
