# Tag Info

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To obtain the stabilisers of a graph state, from its adjacency matrix: Change all 1s to Zs Change all 0s to identity operators Put X operators on the diagonal Each row then represents a stabiliser of the graph state, and any nontrivial stabiliser is a product of one or more rows.

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Remember that a graph state is simply the $|+\rangle$ state on every qubit together with a bunch of controlled phases enacted between them. So, assuming you have a list of the probability amplitudes of your state, you first check that, if there are $n$ qubits, every amplitude is $\pm1/\sqrt{2^n}$. Once you have done this, you need to determine the pattern of ...

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The state of Equation * is local unitary equivalent to all of the four Bell states. That is, there exists a unitary matrix of tensor-product structure $V_a \otimes V_b$, such that $U_{ab}|+\rangle\otimes |+\rangle = V_a \otimes V_b \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right)$ In terms of its entanglement properties, Equation * is completely ...

4

Imgine that there is a bit string $\vec{k}\in\{0,1\}^n$. We use this to specify sites (bit value 1) where an error has occurred, and sites (bit value 0) where no error has occurred. The number of 1s in the bit string is $k$. The probability of this particular error arising is then $p^k(1-p)^{n-k}$ because there are $k$ sites with an error and $n-k$ sites ...

4

Felix Huber's answer is correct, but I thought I'd fill in a few more details. You've written the two-qubit state of ($\star$) as $(|0\rangle|+\rangle+|1\rangle|-\rangle)/\sqrt{2}$. Notice that if you apply a Hadamard to the second qubit, you change it to $(|0\rangle|0\rangle+|1\rangle|1\rangle)/\sqrt{2}$, a Bell state. The point is that from the ...

4

The number of Bell pairs required to construct a given graph state can easily be given an upper bound: $|V|-1$, where $V$ is the set of vertices. You do this simply by preparing the entire state at one site, and teleporting all the other qubits to the relevant party. I wonder if this is actually all there is to it? If we assume that the entire graph is ...

3

The idea of erasure being projection onto $|0\rangle$ is perhaps misleading in this context (my fault for mentioning it in a comment without having looked at the full details of what this specific paper did). This paper does not project the set of qubits $y$ onto the $|0\rangle$ state. Instead, they trace out those qubits. Perhaps the best way of writing ...

3

When running on real backends you can't get a statevector since you can't return the quantum state of the device, only get measurement outcomes. The get_statevector method on the Result class only works if the backend (in this case a simulator) returns a statevector after executing the circuit. That being said you can however try to use quantum tomography to ...

3

The first definition says the nodes start in the $|+\rangle$ state and then you apply a CZ for each edge. The $|+\rangle$ state's stabilizer is $X$. When you conjugate an $X$ observable by a CZ operation, you end up with $X \otimes Z$. When you conjugate a $Z$ observable by a CZ you just get the $Z$ unchanged. This means propagating the $X$ stabilizer ...

3

You can think of a cluster state as a graph state, where the graph's vertices are on some $d$-dimensional lattice (normally just $2$-dimensional). Each vertex represents a qubit in the $|+\rangle$ state, and two vertices that are next to each other on the lattice may be connected, which means that a $CZ$ or controlled-$Z$ gate has been applied to them. ...

3

If you start with one graph state, which is an eigenstate of stabilizer (each of which comprises an X tensored with a bunch of Zs), then the other eigenstates of those stabilizers are the original state acted on by Z rotations (take all possible combinations). To see this note the commutation and anticommutation relations between a tensor product of Zs and ...

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The vertices and edges are the number of qubits and connections in DWaves quantum processing unit topology. It is called Chimera Graph, see DWaves docs . The graph is not fully connected. Therefore an embedding of the qubo problem has to be done. It is done heuristically, see minorminer docs. As an example, if your problem is fully connected, the maximum ...

2

I'm not familiar with how graph states extend to qudits, so let me just answer for the specific case of qubits. Consider a graph $G$, and we create the corresponding graph state $|G\rangle$ by placing a qubit on every vertex in the $|+\rangle$ state, and applying a controlled-phase gate along every edge. Now, take a bipartition of $G$. On either side of ...

2

the measurement is carried out on each individual qubit and the measurement on one qubit will not change the state of another qubit This is an incorrect statement. If the state that you are measuring is entangled (which it very much is for the 2D cluster state), measuring the state of one qubit absolutely changes the state of another qubit. The trivial ...

2

It is incorrect to use modulo arithmetic in this context. Instead finite field arithmetic should be applied. In $\textrm{GF}(4) = \{0, 1, x, x^2\}$ where $x^2 = x + 1$ and conjugation of $a$ is defined as $\bar{a} = a^2$. Addition, multiplication and conjugation tables are then as follows: In this picture we have $0 \equiv 0$, $1 \equiv 1$, $2 \equiv x$, ...

2

I'm not sure I understand the question, since this seems quite straightforward. Graph states are Clifford states, so for a state on $n$ qubits, the set of stabilizers has $n$ generators looking like $$i^k P_1\otimes\ldots\otimes P_n$$ where $k\in\{0,1,2,3\}$ and the $P$'s are Pauli $X,Y,$ or $Z$ operators. (For graph states, these stabilizers look like $\... 2 The calculation of the resulting state of the described circuit: After applying the same circuit identities described in this answer (and here) to the connected question we will obtain a "simplified" circuit: Here are the links to the initial, first intermediate, second intermediate and final circuits presented in the Quirk. Whenever it was ... 2 Because$CNOT = I\otimes H \cdot CZ \cdot I\otimes H$as was mentioned here, and because$CZ(q_1, q_2) = CZ(q_2, q_1)$, we can rewrite the circuit in this way (by adding Hadamards as needed): The link to the circuit created with Quirk. As one can see from the circuit above, after the first Hadamard gate and two$CNOT$gates, we will have a GHZ state for the ... 2 We need to go throught the gates one by one to understand what's happening. We have to keep a few things in mind.$H|0\rangle=\frac{1}{\sqrt2}(|0\rangle + |1\rangle) = |+\rangle$and$H|1\rangle=\frac{1}{\sqrt2}(|0\rangle - |1\rangle) = |-\rangle$. In the Hadamard basis the$Z$gate acts like the$X$gate. Namely$Z|+\rangle=|-\rangle$and$Z|-\rangle=|+\...

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A partial explanation is motivated by the proof in Theorem 1 (bottom of page 2). Assuming two locally non-commuting stabilizing operators, using the Cauchy-Schwarz inequality, for pure product states it is shown that $$\langle S_{l}^{(GHZ_N)} + S_{m}^{(GHZ_N)} \rangle \leq 1.$$ But since we also assume these are stabilizer operators (GHZ eigenstate with ...

2

Absolutely! Think about the eigenstates of $H_{ab}$: $|11\rangle$ has eigenvalue 1, while $|00\rangle, |01\rangle$ and $|10\rangle$ have eigenvalue 0. This tells us that the time evolution must be $$U_{ab}(\varphi)=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{-i\varphi} \... 2 Sure, I would say this is pretty clearly a typo. They very helpfully write out what they mean immediately after that statement so you can compare your understanding! 1 The calculation in the paper is correct. One of the easiest ways to think about this is that you start all qubits in the |+\rangle state, so |+++\rangle=(|0\rangle+|1\rangle)|++\rangle)/\sqrt{2}. Now you've got to co controlled-phase between 1 and 2, and 1 and 3. Let's take qubit 1 to be the control both times (controlled-phase is symmetric, so it doesn'... 1 The distance of an error correcting code is the smallest number of single-qubit rotations that you have to apply to map one logical codeword into an orthogonal one. If I express it in terms of the stabilizers of the code, it's the smallest tensor product of single-qubit unitaries (usually Paulis) that commutes with all the generators. Let's see how this ... 1 About checking. I do not quite understand why you can't check how the GHZ-state was teleported in the most ordinary ways: in addition to measuring the state immediately after teleportation, as well as by YYX, YXY, XYY, XXX measurements, you can inverse your GHZ-state and make sure that all 0 are obtained, e.g. like as for this with the most usual GHZ-state ... 1 I think it is worth trying to understand the circuit that the authors really want to implement: Here they produce the Bell state they want to teleport onto the final two qubits, and cluster state (personally, I wouldn't call it a cluster state after they've added the two extra hadamards). Then they do two single-qubit teleportation protocols to make the ... 1 1) This equation is at best confusing, and at worst perhaps even straight-up wrong. Firstly, they use a different edge contraction notation as in the equations description and secondly, as you observe, they appear to have a Z operator acting upon qubit v^{\prime\prime} which is no longer a vertex of the edge-contracted state. Perhaps the Z-operators on ... 1 Yes for QUBO problems, vertices of the graph are qubits, and edges of the graphs are "couplers" which couple two qubits together. For example for the QUBO problem:$$\tag{1} b_1b_2 - 3b_1b_3 + b_3,  you need a machine with 3 qubits and 2 couplers (one between qubits 1 and 2, and one between qubits 1 and 3).

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