3
votes
Accepted
Equivalence of circuits in measurement-based QC
In step 2, you are doing two measurements, $X$ on qubit 1 and $X$ on qubit 2. To move to step 3, you need to bring these measurements earlier in the circuit, which means updating what they look like ...
- 55.7k
2
votes
Accepted
Problem with eigenvalue evaluation algorithm application on matrix $U$
TL;DR: Qubit order in the top register is reversed.
QFT qubit order in Quirk
Quirk's QFT gate treats the top qubit as the least significant and the bottom qubit as the most significant. Thus, if you ...
- 20.8k
2
votes
Equivalence of circuits in measurement-based QC
If $AU \equiv UB$ then $M_{A} \cdot U \equiv U \cdot M_B$.
Measuring $B$ before $U$ is equivalent to measuring $A$ after $U$ if $U$ conjugates $B$ into $A$.
The transition from step 2 to step 3 is ...
- 33.3k
2
votes
Accepted
Constructing a two 3-qubit state involving either X, Y or Z rotation gate
One way to achieve this is the following way.
Initialise 1 qubit in the $|0\rangle$-state and rotate along the $Y$-axis to
\begin{equation}
\frac{1}{2}|0\rangle+\frac{\sqrt{3}}{2}|1\rangle.
\end{...
- 179
2
votes
creating a generalised Bell state with gates
Starting from any of the four basis states, Hadamard + controlled-not will produce each of the four Bell states. So, expressed in terms of the gates you need, you just need to add $X$ to either qubit ...
- 55.7k
2
votes
Role of qubit registers in HHL circuit
community wiki
Your first bullet is probably correct, although I wouldn't refer to the one single qubit $|0\rangle$ at the top as storing eigenvalues of $A^{-1}$ - the eigenvalues of $A^{-1}$ are ...
2
votes
Does the $\text{CNOT}$ gate activate if the control qubit is $-|1⟩?$
Yes, it would. It is sometimes better to think in the vector-matrix representation than in the dirac notations.
$$|1\rangle = \begin{bmatrix}
0\\
1
\end{bmatrix}\,.$$
Now,
$$Z|1\rangle = -|1\rangle =...
- 1,376
2
votes
Does the $\text{CNOT}$ gate activate if the control qubit is $-|1⟩?$
Yes.
Think of the $-|1\rangle$ state as a superposition of basis states $|0\rangle$ and $|1\rangle$ (albeit a weird one). In this case, you can apply the general rule for the CNOT gate acting on ...
- 8,785
1
vote
Accepted
Custom gate labels in PennyLane
First, good news: Your approach to provide an id in order to impact the drawn circuit has been implemented very recently. So recent in fact, it is not in the newest ...
- 221
1
vote
Accepted
creating a generalised Bell state with gates
I think I nailed it.
The 4 Bell states are
$$\frac{1}{\sqrt{2}}\big
[|00\rangle + |11\rangle\big],\quad \frac{1}{\sqrt{2}}\big[|01\rangle + |10\rangle\big],\quad \frac{1}{\sqrt{2}}\big[|00\rangle - |...
- 510
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