Deutsch's Problem (Part 5)
Thursday, June 3, 2010 at 12:00PM Today, we will explore the oracle Uf. Yesterday, we saw that the first two Hadamard gates produce two qubits that equally weight all four binary states in superposition (two of them with negative coefficients). I am now going to drop the normalization constant to keep the equations a bit simpler. So, we can describe the input to the oracle as the state
|00>-|01>+|10>-|11> .
The oracle has two qubit inputs, |x> and |y>, and two qubit outputs, |x> and |y+f(x)>, where f(x) is one of the four secret mappings we previously discussed. Given the input from the output of the two Hadamards, we can write the oracle output as the sum of these four factors:

Adding the above four terms together and simplifying yields

And tomorrow, we will see what happens when we put this through the final Hadamard gate.





