$10$ points$)($Can a function be evaluated at two places with a single quantumquery?$)$Here we consider the problem where we have a query oracle for a function f : $\{0,1\}\{0,1\}$ and the goal is to obtain information about both f $(0)$ and f $(1)$ with a single query.We assume that the query oracle is in the usual form of a unitary operator Uf that, forall a$,$ b $\{0,1\},$ maps $|$a $|$b to $|$a $|$b f $($a$).$ For simplicity, we consider methods thatemploy only two qubits in all and are expressible by a circuit of the form$|0$ V Uf W$|0$ where V and W are two$-$qubit unitaries and the gates labelled are measurements in thestandard basis. Therefore, it can be assumed that the input state to the query $($i$.$e$.,$ rightafter V is applied$)$ is a two$-$qubit state of the form $00|00+01|01+10|10+11|11,$where $|00|2+|01|2+|10|2+|11|2=1\mathrm{.}1$We will talk about what normal matrix means later in the course, but you dont need to worry aboutits definition now. The matrices in parts $($a$)$ and $($b$)$ are normal.For each of the four functions of the form f : $\{0,1\}\{0,1\},$ give thequantum state right after the query has been performed $($i$.$e$.,$ right before W isapplied$).($b$)(2$ points$)$ If there is a measurement procedure that perfectly distinguishes betweenthe four states in part $($a$)$ then they must be mutually orthogonal. Show that, fora measurement to be able to perfectly determine the value of f $(0),$ it must be thecase that $10=11.($Hint: think of the orthogonality relationships that need tohold.$)($c$)(2$ points$)$ Show that, if the states are such that f $(0)$ can be determined perfectlyfrom them, then f $(1)$ cannot be determined with probability better than $1/2($whichis no better than random guessing$).($Hint: You may use the result in part $($b$)$ forthis.$)($d$)(4$ points$)$ Optional for bonus credit for all students: The above analysis isrestricted to methods that use two qubits. Show that, for all m $2,$ any strategythat uses m qubits $($V and W are m$-$qubit unitaries and the query gate Uf actson the last two qubits$)$ and determines f $(0)$ perfectly cannot determine f $(1)$ withprobability better than $1/2$