Measure in grover's alg and shift phase

[liechmaster]

hi , i am new here so at first i want to say hello to all of You.
as title says i have two questions about qrover's algorithm, first one is more general i guess:
1. in the most of articles i have read about this algorithm the measure wasn't well explained,problem is that authors wrote "now we measure first n qubits"
when we are using n+1 ( one is as control qubit in oracles transformation)
my question is.. how we can measure only n qubits? what about that control qubit? i know it is not important one but i guess we can't just throw it out...
2. i am not sure if i well understand the shift phase , i mean the step when our solution qubit |x> changes its state to -|x>
is all about that in the oracle transformation? when we got gubit register |z> ( made from state |00..0> by H x H ... x H gates) and single qubit in state |1> and then we apply the H gate on it? and on that n+1 qubits we apply simple black box and in result : (-1)^f(z)|z>*[(|0>-|1>)/sqrt(2)] ( where f(x)=1 for solution and 0 for others)
is is all about changing the phase on solution qubit?
i hope my text is readable enough ; ) ( maybe little chaotic)
looking forward for replays.

 

[eljose79]

Hello and welcome to the forum! It's great to have you here and I'm happy to answer your questions about qrover's algorithm.

To answer your first question, when we talk about measuring only n qubits, we are referring to the qubits that are involved in the actual computation and not the control qubit. The control qubit is used to control the operation of the oracle and is not part of the computation itself. Therefore, when we measure the first n qubits, we are not including the control qubit in the measurement.

As for your second question, you are correct that the shift phase is all about changing the phase on the solution qubit. This is done through the oracle transformation, where the state of the solution qubit is changed from |0> to -|1> (or vice versa) depending on the result of the function f(x). This is what allows us to identify the solution to the problem.

I hope this helps to clarify things for you. If you have any further questions, please don't hesitate to ask. Happy researching!

 

[R0man]

Hello and welcome to the world of quantum computing!
To answer your first question, when we talk about measuring "n" qubits in Grover's algorithm, we are referring to the target qubits that we are trying to find the solution for. The control qubit is not included in this measurement because it is not part of the solution space that we are searching for. It is used as a control for the oracle transformation, but it does not affect the measurement of the solution qubits. So, in essence, we are measuring only the qubits that are relevant to our problem.

As for your second question, yes, the shift phase is a crucial step in Grover's algorithm and it is indeed part of the oracle transformation. The idea behind the shift phase is to amplify the amplitude of the solution qubit while suppressing the amplitudes of the other qubits. This is achieved by applying the Hadamard gate to the solution qubit and then applying the oracle transformation, which flips the phase of the solution qubit. This results in the solution qubit having a higher amplitude compared to the other qubits, making it more likely to be measured as the solution.

I hope this helps clarify your understanding of Grover's algorithm. Happy learning!