# Exact functions of a single-iteration Grover Search Algorithm's operators

I'm doing a practice assignment where I'm asked to identify specific features of the Grover Search Algorithm's second operator (picture in post, further on "$$Us$$"), which mirrors the system relative to the $$|s\rangle$$ vector, defined as the equal-weighted sum of the input parameter's subspace basis vectors (definition below). Naturally, this only concerns the contents of a single iteration.

$$|s\rangle=H|0\rangle^{\otimes n}=\frac{1}{2^{n/2}}\sum_{x=0}^{2^n-1}|x\rangle.$$

The operator is defined by the following scheme, which I have, for reference, built in the Quirk QC simulator. It has the following proposed features (which I need to verify as true or false):

1. The Hadamard and NOT gates applied to the qubits 1 and 2 (further on $$|C_I\rangle$$, $$|C_{II}\rangle$$) in succession map the $$|s\rangle$$ vector onto the $$|11\rangle$$ vector.

2. The Hadamard and NOT gates applied to the qubits 1 and 2 (further on $$|C_I\rangle$$, $$|C_{II}\rangle$$) in succession map the $$|s\rangle$$ vector onto the $$|00\rangle$$ vector.

3. The CNOT gate (with $$C_I$$ and $$C_{II}$$ as controls) affects only the $$|11C_{III}\rangle$$ vector.

4. The CNOT gate multiplies the $$|11C_{III}\rangle$$ vector by $$-1$$ and does not affect other vectors.

5. The Hadamard, then the NOT gate applied to $$C_I$$ and $$C_{II}$$ maps $$|11\rangle$$ onto $$|s\rangle$$. The $$|s\rangle$$ component in the system vector is thus multiplied by $$-1$$ while components orthogonal to it are unaffected.

6. The NOT gate on $$C_{III}$$ multiplies the system vector by $$-1$$, thus the system vector is reflected relative to $$|s\rangle$$.

7. The NOT gate on the $$C_{III}$$ qubit is unnecessary.

My perspective:

For the first two points, I don't fully grasp the concept of the mapping described here. However, it should be noted that by definition, as far as I understand, the Hadamard transformation on either side of the scheme is to allow the "$$Us$$" operator to function as a reflection relative to the $$|0\rangle^{\otimes n}$$ vector.

For (3) and (4), while I have maintained the original wording, I understand the question misnames the CCNOT gate as CNOT - I'm not sure how important this distinction is, other than technically referring to two different types of gates. Both of these are true if only the controlled vector is affected, which I believe to be correct.

I believe (5) is incorrect as, by definition, "$$Us$$" is a reflection relative to $$|s\rangle$$. Because of this definition, I believe (6) is correct.

I believe (7) is incorrect because it changes the outcome of the "$$Us$$" operator.

My understanding of quantum computing is severely limited at the moment, so I'm hoping that I'll be able to gain some much-needed intuition by having my logic here fact-checked.

• Hi @Andre R. Welcome to QCSE. It looks like you put a lot of effort into building your question - I've converted it according to my understanding into Latex format for readability. I think this is a homework question, wherein you are given the circuit that you included and are asked some TRUE/FALSE questions about it. The effort you put in should be applauded, but it's a little confusing what's going on. For example, I assume the top two qubtits are what you are calling $C_{1}$ and $C_{II}$, while the bottom is $C_{III}$? – Mark S Oct 25 '19 at 12:32
• Also, unless specified otherwise, the qubits coming in to the circuit can be assumed to be $|0\rangle$. If so, it's not clear why you would Hadamard a qubit $|0\rangle$ and then negate them - the negation does not do anything, as the state $1/\sqrt{2}(|0\rangle+|1\rangle)$ is not affected by a NOT gate. – Mark S Oct 25 '19 at 12:34
• Further, in order for any of 4), 5), or 6) to make any sense, the gates should be a CZ (controlled rotation of $Z$) rather than a CNOT, correct? – Mark S Oct 25 '19 at 17:16