Question

Quantum Operator • Here is the quantum circuit that implementation of the function f: gives a unitary x}ly) "3 x) ly+f(x)) — \x> — lyf(x)> Binary addition (addition modulo 2) 1000> = 10> 1001> = 11> (100> = 11> ||101> = 10>

Problem regarding the Deutsh algorithm for Quantum computing.

Answer 1

It is given in the question that this problem can be solved by a trial and error method.

The possible values of f(x) is 0 and 1

For f(x) = 0,

The LHS is

10602-11 +0>

From the tabular column, this is

000> -10 >= 0 > -1 >

on the RHS,

(-1)^f(x) = -1^0 = 1

So,

the RHS becomes

RHS=1*(0 > -1>) = 0 > -1 >= LHS

For f(x) = 1,

The LHS is

101>-|11>

From the tabular column, this is

000> -10 >= 1 >-0>

on the RHS,

(-1)^f(x) = -1^1 = -1

So,

the RHS becomes

RHS = -1*(1 >-0>) = 0 > -1 >= LHS

Since LHS = RHS for all possible values of f(x), the equation is true