Graph states and entanglement

Graph state17 is a generalization of cluster state introduced in 2001,14 which is the resource state of one-way quantum computing15 and quantum error correction.16 GHZ state is an example of graph state and has been demonstrated in superconducting qubit system.18 However, GHZ state is fragile. Some other graph states are very robust to local operations, such as local measurements and noises. In order to disentangle the cluster state of N qubits, N/2 local measurements are needed.14 Because of this nice feature, we decide to generate and detect linear cluster states in the IBM cloud service ibmqx5.

X, Y, Z denote the Pauli operators. An undirected graph G(V, E) includes a set of vertices V and a set of edges E. A graph state that correspond to an undirected graph G(V, E) is a \(\left| V \right|\)-qubit state that has the form

$$\left| G \right\rangle = \mathop {\prod}\limits_{\left( {a,b} \right) \in E} {\kern 1pt} U_{ab}\left| + \right\rangle ^{ \otimes V},$$ (1)

where U ab is a control-Z operator acting on qubits a and b,19 and

$$\left| \pm \right\rangle = \frac{1}{{\sqrt 2 }}\left( {\left| 0 \right\rangle \pm \left| 1 \right\rangle } \right)$$ (2)

are eigenvectors of the X operator.

An equivalent definition, the graph state that corresponds to G(V, E), is the unique common eigenvector (of eigenvalue 1) of the set of independent commuting operators:

$$K_a = X^aZ^{N_a} = X^a\mathop {\prod}\limits_{b \in N_a} {\kern 1pt} Z^b,$$ (3)

where the eigenvalues to K a are +1 for all a ∈ V, and N a denotes the set of neighbor vertices of a in G.19 As implied by the first definition, a n-qubit graph state can be prepared by the following steps.

1. Initialize the state to \(\left| + \right\rangle ^{ \otimes n}\) by applying n Hadamard gates to \(\left| 0 \right\rangle ^{ \otimes n}\); 2. For every (a, b) ∈ E, apply a control-Z gate on qubits a and b; the order can be arbitrary.

Entanglement of general mixed states was discussed by Werner in 1989.20 Since then, many entanglement criteria were proposed; among them the widely used ones include the partial transpose criterion13,21,22 and the symmetric extension criterion.23

A bipartite state ρ AB on the Hilbert space \({\cal H} = {\cal H}_A \otimes {\cal H}_B\) is said to be separable if ρ AB can be written as

$$\rho _{AB} = \mathop {\sum}\limits_i {\kern 1pt} p_i\rho _A^i \otimes \rho _B^i,$$ (4)

where \(\rho _A^i\) and \(\rho _B^i\) are quantum states of the system A and B, respectively, with p i ≥ 0 and \(\mathop {\sum}

olimits_i {\kern 1pt} p_i = 1\). Otherwise ρ AB is entangled. For a state ρ of a many-body system, for any fixed bipartition AB of the system, if ρ is entangled with respect to the partition AB, then the entanglement of the many-body state ρ can also be examined via its subsystems. That is, if the subsystems are all entangled, the whole system must be also entangled.

To be more concrete, consider a 4-qubit subsystem ρ A,B,C,D in an n-qubit system. Suppose that we perform two local operations O A and O D on qubit A and D respectively, and then obtain the reduced density matrix of qubit B and C by tracing out qubit A and D. The reduced density matrix for qubits B and C reads

$$\rho _{B,C}^\prime = tr_{A,D}\left( {\frac{{O_AO_D\rho _{A,B,C,D}O_D^\dagger O_A^\dagger }}{{tr\left( {O_AO_D\rho _{A,B,C,D}O_D^\dagger O_A^\dagger } \right)}}} \right).$$ (5)

The entanglement of \(\rho _{B,C}^\prime\) can be determined by using entanglement monotones such as negativity and concurrence, which, in the 2-qubit case, has non-zero values if and only if the system is entangled.13,22 If \(\rho _{B,C}^\prime\) is entangled, we can conclude that in the original system, there could not exist a separation with qubit B and C on different sides. In other words, if the original system is biseparable with respect to a fixed partition, the qubit B and C must be on the same side. Otherwise, we will be able to create entanglement between the two separable parties with only local operations, which is not possible.13

For an n-qubit system {q 1 , q 2 , …, q n }, if we can show that among the n-qubit pairs (q 1 , q 2 ), …, (q n−1 , q n ), (q n , q 1 ), n − 1 of them must be on the same side in a separation, then we may conclude that there is no possible separation, and that the system is a n-qubit entangled state (meaning that the state is not biseparable with respect to (w.r.t.) a fixed partition, and that it involves all qubits). The (minimal) number of circuit configurations needed in this approach is 34(n − 1), which grows linear with respect to n. This method is far more efficient compared to a full n-qubit tomography, which requires exponential number of configurations.

Graph states on ibmqx5

ibmqx5 is a 16-qubit superconducting quantum processor. It allows independent single-qubit operations with fidelity >99% and control operations with fidelity 95–97% (see Fig. 1) marked as the edges in the connectivity map (see Fig. 2). That is, controlled NOT (CNOT) operations with qubit a as the control qubit and b as the target is allowed if and only if a → b is an edge in the map.

Fig. 1 Calibration parameters of ibmqx5, archived 10 January 2018 from ref.4 It should be noted that these parameters are updated on a daily basis Full size image

In our experiment, as shown in Fig. 3, the following five graph states are employed. The first state is a 8-qubit graph state involving qubits q5–q12 that corresponds to a ring of length 8; the second one is a 10-qubit state involving qubits q4–q13 corresponding to a ring of length 10; the third one involves qubits q3–q14 and corresponds to a ring of length 12; the fourth one involves qubits q2–q15 and corresponds to a ring of length 14; the fifth one involves all the 16 qubits. We employ these particular graph states based on the following considerations. First, these states are genuinely entangled and will remain entangled after tracing out a large number of qubits. Second, research has shown that one-dimensional (1D) cluster states are robust against decoherence, meaning that it would be more likely to find entanglement in a rather large graph state close to a 1D chain, compared to GHZ states and two-dimensional (2D) graph states.24 At last, even rings are two-edge colorable; as a result, on the 16-qubit ibmqx5, these “even-ring” states could be prepared using low-depth circuits (see Fig. 4).

Fig. 3 Graph states employed in this experiment. Colored lines illustrate the graph of the 8-qubit graph state (in red), 10-qubit graph state (orange), 12-qubit graph state (yellow), 14-qubit state (blue) and 16-qubit graph state (purple) Full size image

Fig. 4 a The quantum circuit for preparing a 8-qubit graph state implied by the definition of graph states. b The optimized circuit that suits ibmqx5’s connectivity Full size image

To prepare the desired graph state, we start from the circuit implied by the definition of graph states (see Fig. 4a). The control-Z gates are implemented using a CNOT gate and two Hadamard gates. We then optimize this circuit by adjusting the order of commuting gates and removing redundant Hadamard gates (see Fig. 4b). The circuit that we implemented are shown in Fig. 4b and Fig. 5a–d.

Fig. 5 The quantum circuit implemented on ibmqx5 for the preparation of a 10-qubit graph state, b 12-qubit graph state, c 14-qubit graph state and d 16-qubit graph state Full size image

Experimental results

For each n-qubit ring state, n partial tomographies are performed for every subsystem with 4 qubits that forms a chain in the ring. For example, for the 8-qubit graph state, the 8 subsystems are (q5, q6, q7, q8), (q6, q7, q8, q9), …, (q12, q5, q6, q7). For every state, 34n experimental configurations are used; 2048 measurements are taken under each configuration. The n 4-qubit reduced density matrices are obtained using the maximum likelihood method proposed by Smolin et al.25

Due to Eq. (3), for a ring graph state, each 4-qubit density matrix of neighboring four qubits, as illustrated in Fig. 6, is given by

$$\rho _{A,B,C,D} = \frac{1}{4}\left( {I + Z_AX_BZ_C} \right)\left( {I + Z_BX_CZ_D} \right).$$ (6)

Fig. 6 A 4-qubit subsystem that forms a chain Full size image

Then, for each 4-qubit density matrix, we apply the local operations \(O_A = {\textstyle{{Z_A + I} \over 2}}\) and \(O_D = {\textstyle{{Z_D + I} \over 2}}\) and calculate the negativity of the resulting 2-qubit subsystem. For instance, we may choose (q5, q6, q7, q8) as our subsystem; after applying O A and O D to q5 and q8 respectively, we will trace out q5 and q8, and measure the negativity of the remaining subsystem, (q6, q7). We choose \(O_A = {\textstyle{{Z_A + I} \over 2}}\) and \(O_D = {\textstyle{{Z_D + I} \over v}}\) for the following reason. If ρ is graph state, and the 4-qubit subsystem corresponds to 4 vertices that form a chain in the graph, then the resulting 2-qubit state is a maximally entangled state

$$\left| \phi \right\rangle = \frac{1}{{\sqrt 2 }}\left( {\left| 0 \right\rangle \left| + \right\rangle + \left| 1 \right\rangle \left| - \right\rangle } \right).$$ (7)

Therefore, for a state close to this graph state, we should expect the resulting 2-qubit state to have a negativity significantly greater than 0. The results are plotted in Fig. 7.

Fig. 7 The result of a the 8-qubit graph state, b the 10-qubit graph state, c the 12-qubit graph state, d the 14-qubit graph state and e the 16-qubit graph state. The negativity of the final 2-qubit states are plotted. The 95% confidence intervals are estimated using bootstrapping techniques Full size image

For the 8-qubit graph state, the measured negativities are all significantly greater than 0. For the 10-qubit graph state, 9 out of 10 measured negativities are significantly greater than 0. Based on our argument above, both the 8-qubit state and 10-qubit state are fully entangled.

In the 12-qubit case, as shown in Fig. 7c, 10 out of 12 measured negativites are significantly non-zero. The two zeros come from (q9, q10) and (q14, q3) pairs. Therefore, there is only one possible separation, namely {q10, q11, q12, q13, q14} | {q3, q4, q5, q6, q7, q8, q9}. Should this be true, the reduced density matrix of qubits q8,q9,q10,q11 should also be separable with the separation {q8, q9} | {q10, q11}. In that case, its partial transpose with respect to qubit q8 and q9 must be positive. However, with respect to this partial transpose, ρ q8,q9,q10,q11 has negativity 0.0391 ± 0.0039 (standard deviation estimated via bootstrapping). Therefore, this possibility is ruled out with very high confidence. We can now conclude that the 12-qubit graph state is fully entangled.

In the 14-qubit case, as shown in Fig. 7d, 12 out of 14 measured negativites are significantly greater than 0. Here, we may apply the same trick again. The only possible separation is {q2, q3, q4, q5, q6, q7, q8, q9, q12, q13, q14} | {q10, q11}. In this case, subsystem {q8, q9, q10, q11} should have zero negativity with respect to the partial transpose on q8 and q9. However, the measured negativity is 0.0698 ± 0.0048 (standard deviation estimated via bootstrapping). Hence, this possibility is ruled out with very high confidence. We may conclude that this state is fully entangled.

In the 16-qubit case, as shown in Fig. 7e, 15 out of 16 measured negativites are significantly greater than 0. As argued above, this means that this state is not biseparable w.r.t. a fixed partition, thereby showing that all 16 qubits in ibmqx5 are in full entanglement.

It may be noted that the subsystem of qubits {q8, q9, q10, q11} yields close-to-zero negativity in 3 out of 4 experiments. This can be due to relatively high readout errors or gate errors involving these qubits, which is compatible with the measured parameters provided by IBM website4 (see Fig. 1). For instance, the CNOT gate between q10 and q11 has the largest error among all possible CNOT gates, while the readout errors of q10 and q11 are also above the average level (6.5%).

Further exploration of the 16-qubit state

The results above could be understood as an ability to generate localized entanglement on physically neighboring qubits.26 That is, neighboring qubits can be put into entanglement by performing ideal local operations on the 16-qubit state. Using the same data obtained above, we will show that localized entanglement on qubits with distance 2 and 3 could also be generated.

Suppose {E, A, B, C, D, F} is a 6-qubit subsystem that forms a chain. We first apply \(O_E = {\textstyle{{Z_E \pm I} \over 2}}\) and \(O_F = {\textstyle{{Z_F \pm I} \over 2}}\) on E and F respectively (four possibilities). On our data, this can effectively be done by first postselecting 0s on qubits E and F before calculating the tomography of {A, B, C, D}. Next, \(O_B = {\textstyle{{X_B + I} \over 2}}\) and \(O_D = {\textstyle{{Z_D + I} \over 2}}\) are performed (see Fig. 8). At last, B, E, F and D are traced out, while the negativity in subsystem {A, C} is calculated. If the 16-qubit state is perfect, this resulting system would be maximum entangled.

Fig. 8 Operations performed to produce entanglement on subsystem {A,C} Full size image

Based on data obtained in previous experiments, we have calculated the corresponding negativity for each 6-qubit subsystem and shown them in Table 1. Using this method, we have identified localized entanglement in 13 out of 16 pairs of qubits with distance 2.

Table 1 Negativities of qubits with distance 2 in the 16-qubit state Full size table

To generate localized entanglement on qubits with distance 3, we may apply the same O E and O F , and then apply \(O_B^\prime = {\textstyle{{X_B + I} \over 2}}\) and \(O_C^\prime = {\textstyle{{X_C + I} \over 2}}\) (see Fig. 9). The negativity of subsystem {A, D} would be calculated. Again, if the 16-qubit state is perfect, these two qubits would be maximum entangled; therefore, we should expect a non-zero negativity if the actual state is close to the theoretical one.

Fig. 9 Operations performed to produce entanglement on subsystem {A, D} Full size image

Among 16 pairs of qubits with distance 3, we have identified localized entanglement in 6 pairs of them. The results based on our data is presented in Table 2.