Experimental realization of controlled quantum teleportation of arbitrary qubit states via cluster states

Controlled quantum teleportation involves a third party as a controller for the teleportation of state. Here, we present the novel protocols for controlling teleportation of the arbitrary two-qubit and three-qubit states through five-qubit and seven-qubit cluster states respectively. In these schemes, Alice sends the arbitrary qubit states to the remote receiver Bob through the cluster states as quantum channels under the control of Charlie. Bob can recover the mentioned states by making appropriate unitary operations, and we point out that the efficiency in our schemes is 100%. In the process of our analysis, we find the classical communication cost in our protocols is remarkably reduced when compared to the previous protocols. We perform the experimental realization of the above protocols on “IBM 16 Melbourne” quantum computer and “IBM quantum simulator” and we calculate the fidelity. We also examine the security analysis against Charlie, and these schemes which we considered here are secure against Charlie’s attacks.


theoretical and experimental realization of our protocols
There are many schemes for the controlled quantum teleportation 81,82 . In the following sub-sections, we discuss the controlled quantum teleportation of arbitrary qubit states via cluster states for two cases. Case I: scheme for the controlled quantum teleportation of an arbitrary two-qubit state using a five-qubit cluster state. Case II: scheme for the controlled quantum teleportation of an arbitrary three-qubit state using a seven-qubit cluster state.
In our protocols, the cluster state has been remotely prepared at the Alice place, where she performs all the necessary unitary operations including the deferred measurement 83 as shown in Figs. 1 and 2. After performing all the operations, Alice sends the respective qubits to the respective parties. Then Alice immediate measures her qubits in computational basis, which destroys her qubits, making her incapable of any further communication 55 . Now, after receiving qubits from Alice, Charlie measures his qubits in |±� basis ( |±� = (|0� ± |1�)/ √ 2 ). If the measurement outcome is |−� , then Charlie sends a classical bit of information to Bob within a certain time period. After receiving Charlie's classical information in a certain time period, Bob gets that Charlie's measurement outcome is |−� , and he has to perform a set of unitary operations on his qubits. If the measurement outcome is |+� , then Charlie does not need to send any classical information to Bob. After waiting for a certain time period and have not received any classical information, Bob understands that Charlie's measurement outcome is |+� and he has to perform another set of unitary operations on his qubits. The unitary operations, which Bob performs are discussed in the later sub-sections. By the above analysis, we conclude that the classical communication is taking place between Charlie and Bob, only when Charlie's measurement outcome is |−� . Hence, the average classical communication cost 84 necessary for our protocols is 0.5 bit. In Table 1, we compared the classical communication cost of our protocols with the results of the other protocols.
Scheme for controlled quantum teleportation of an arbitrary two-qubit state using a five-qubit cluster state. We consider a five-qubit cluster state which we used as a quantum channel to teleport a two- www.nature.com/scientificreports/ qubit state. The cluster state is shared between three parties Alice, Bob and Charlie which are far apart from each other. Alice shares two qubits, Charlie shares one qubit, and Bob shares the remaining two qubits of the fivequbit entangled cluster state. The average classical information shared between Charlie and Bob is 0.5 bit, which depends upon the measurement outcome of Charlie. Now, Bob takes the information of the classical channel into account and decides which set of unitary operations he has to perform on his qubits. Here, if Charlie wants to cheat and sends the wrong information to Bob through the classical channel, then after Bob's unitary operation and consulting with Alice, Bob finds out that Charlie cheated. Thus, even if Charlie wants to, he cannot cheat without getting caught, and this is the beauty of quantum communication 23 . The five-qubit cluster state from state |00000� 12345 is generated by the following circuit as shown in Fig. 3, is used here as a quantum channel for the quantum communication between Alice, Bob and Charlie. Given as In this scheme, we wish to teleport any two-qubit state |ψ� ab = (α|00� + β|01� + γ |10� + δ|11�) ab through the five-qubit cluster state (1), where |α| 2 + |β| 2 + |γ | 2 + |δ| 2 = 1 . The qubits 1 and 3 belong to Alice, the qubit 4 belongs to Charlie, and the qubits 2 and 5 belong to Bob. Now, the joint state of the arbitrary two-qubit state and the five-qubit cluster state can be written as |ψ� ab12345 = |ψ� ab ⊗ |C 5 � 12345 .
(1)  www.nature.com/scientificreports/ Briefly, in this process of teleportation. Alice remotely prepared the cluster state at her place, and then she performs controlled-NOT (CNOT) gate on qubits (a, 1) and qubits (b, 3). Where the qubits a and b work as controlling qubits and qubits 1 and 3 work as target qubits (we use the abbreviation (x, y) where the qubit x works as controlling qubit and the qubit y works as target qubit, henceforth). After then, Alice performs Hadamard gate on her qubits a and b, and then she performs the deferred measurement on her qubits as discussed earlier.
Then Alice sends the respective qubits to Charlie and Bob. If the controller Charlie agrees to help the communication between Alice and Bob, Charlie has to perform a single-qubit measurement in |±� basis on his qubit 4. Finally, Bob can obtain the unknown state by performing the appropriate set of unitary transformations on his qubits 2 and 5. Indeed, Fig. 4 shows the generalized equivalent circuit for teleporting two-qubit state by using a five-qubit cluster state.
Circuit decomposition. The arbitrary two-qubit state which Alice wishes to teleport is given as |ψ� ab = (α|00� + β|01� + γ |10� + δ|11�) ab . The five-qubit cluster state which Alice used as a quantum channel for teleportation of two-qubit state is given as |C 5 � 12345 in Eq. (1). Now, Alice implements arbitrary two-qubit to her share of entangled qubits in following ways: First, Alice performs CNOT gate on qubits (a, 1) and qubits (b, 3), and the joint state |ψ� ab12345 is changed to where Next, Alice applies Hadamard gate on her qubits a and b and the new state is given as After expanding and rearranging the above equation, we have |ψ ′′ � ab12345 = 1/4[|0000� ab13 ⊗ |ψ 1 � + |0001� ab13 ⊗ |ψ 2 � + |0010� ab13 ⊗ |ψ 3 � + |0011� ab13 ⊗ |ψ 4 � + |0100� ab13 ⊗ |ψ 5 � + |0101� ab13 ⊗ |ψ 6 � + |0110� ab13 ⊗ |ψ 7 � + |0111� ab13 ⊗ |ψ 8 � + |1000� ab13 ⊗ |ψ 9 � + |1001� ab13 ⊗ |ψ 10 � + |1010� ab13 ⊗ |ψ 11 � + |1011� ab13 ⊗ |ψ 12 � + |1100� ab13 ⊗ |ψ 13 � + |1101� ab13 ⊗ |ψ 14 � + |1110� ab13 ⊗ |ψ 15 � + |1111� ab13 ⊗ |ψ 16  www.nature.com/scientificreports/ To make the teleportation successful and also to reduce the classical communication cost, we use the deferred measurement 83 . After using the deferred measurement, the average classical communication cost required for quantum teleportation is 0.5 bit. In the deferred measurement, CNOT gate is applied on qubits (1, 2), (3,5), (3,4), (1,4) and controlled-Z (CZ) gate is applied on qubits (a, 2) and (b, 5). After applying the deferred measurements, the state (5) becomes, Alice measured her sets of qubits in the computational basis, then the whole state is collapsed to |φ� 254 = (α|000� + β|011� + γ |101� + δ|110�) 254 and this can be written in the form as Charlie measures his qubit in |+� 4 or |−� 4 basis, and when his measured state is |−� 4 , then Charlie sends his qubit information in one bit to Bob within a certain time period through a secure classical channel. After receiving the information from Charlie, Bob gets that Charlie's qubit is in |−� 4 state, and he has to apply a phase-change unitary transformation on his qubits. If Charlie measurement outcome is |+� 4 , then he does not have to send any classical information to Bob. Whereas Bob after waiting for a certain time period and have not received any information from Charlie, he understands that Charlie's qubit is in |+� 4 state. And he has to apply an identity gate unitary transformation on his qubits (see Table 2). For example, let us say Alice measures her state and it comes out |0111� , then the whole state is get collapsed to Eq. (8). After receiving qubits from Alice, Charlie measures his qubit state and let us say that the measurement outcome is |−� 4 . Then Charlie sends a 1 bit of classical information to Bob. Then Bob understands that he has to perform a phase-change unitary transformation on his qubits, means he applies Z gate on both of his qubits 2 and 5 to get the state sent by Alice. So this protocol is deterministic, i.e. the probability of success is 100%. We perform the above experiment on IBM QE, where we compare the statistical data of the teleportation of two-qubit between "IBM qasm-simulator" (with 32 qubits) and the "IBM 16 Melbourne" (IBM 16 Melbourne is www.nature.com/scientificreports/ a real device with 15 qubits). Indeed, we send an arbitrary two-qubit state from Alice to Bob with the help of a controller Charlie. We also figure out the density matrix of both the cases and from density matrices, we evaluate the fidelity 90,91 of the circuit.
Scheme for controlled quantum teleportation of an arbitrary three-qubit state using a seven-qubit cluster state. In this case, we consider a seven-qubit cluster state which we used as a quantum channel for teleportation of three-qubit state. The cluster state is shared between as usual Alice (sender), Bob (receiver), and Charlie (controller) which are far apart from each other. More precisely, Alice and Bob each share three qubits and Charlie shares one qubit of the seven-qubit cluster state. The procedure is the same as for the previous scheme, i.e. Alice remotely prepared the cluster state at her place. After applying the required unitary operations between the cluster state and the three-qubit state, Alice sends the respective qubits to the respective parties. And after that, she immediately measured her qubits state and the whole state is get collapsed to Eq. (16). Next, Charlie measures his qubit in |+� 7 or |−� 7 basis, and when his measured state is |−� 7 , then Charlie sends his qubit information in one bit to Bob within a certain time period through a secure classical channel. After receiving the information from Charlie, Bob gets that Charlie's qubit is in |−� 7 state, and he has to apply a phase-change unitary transformation on qubit 6 followed by CNOT operation on qubits (2,4) and (4,6). If Charlie measurement outcome is |+� 7 , then he does not have to send any classical information to Bob. Whereas Bob after waiting for a certain time period and have not received any information from Charlie, he gets that Charlie's qubit is in |+� 7 state. And he has to apply an identity gate unitary transformation on qubit 6 followed by CNOT operation on qubits (2,4) and (4,6). Herein, the qubits 1, 3 and 5 belong to Alice, the qubits 2, 4 and 6 belong to Bob, and the qubit 7 belongs to Charlie. Indeed, Fig. 5 shows the generalized equivalent circuit for teleporting three-qubit state by using a seven-qubit cluster state.
The seven-qubit cluster state from state |0000000� 1234567 is generated by the following circuit as shown in Fig. 6, is used as a quantum channel for teleportation of the three-qubit state reads   www.nature.com/scientificreports/ The joint state of the arbitrary three-qubit state and the seven-qubit cluster state can be written as |χ� abc1234567 = |χ� abc ⊗ |C 7 � 1234567 , i.e. Now, Alice implements arbitrary three-qubit to her share of entangled qubits in following ways: Step 1 Considering the initial joint state |χ� abc1234567 , Alice applies CNOT gate on her qubits (a, 1), (b, 3), and (c, 5). Therefore, the transformed state is given as Step 2 Alice applies Hadamard gate on her qubits a, b and c and after that, the deferred measurement is performed on the circuit in a systematic way. Now in the deferred measurement, CNOT gate and CZ gate applied on qubits, i.e. Alice performs CNOT gate on qubits (1, 2), (1,4), (1,6), (1,7), (3,4), (3,6), (  |χ l � abc135 ⊗ (α|0000� + β|0011� + γ |0111� + δ|0100� and this can be written in the following form Step 4 Now if Charlie's measurement outcome is |+� 7 , then Bob has to perform first an identity gate on qubit 6 and then CNOT gate on qubits (4, 6) and (2,4) to get the state |χ� abc as Alice wants to send. But if Charlie's measurement outcome is |−� 7 , then Bob has to perform first a phase-change unitary transformation on qubit 6 and then CNOT gate on qubits (4, 6) and (2,4) to get the initial state |χ� abc (see Table 3).
For example, let us say Alice measures her state and it comes out |χ 9 � . Now Charlie measured his state and if his measurement outcome is |−� 7 , then Bob understands that he has to perform a set of unitary transformation on his qubits to get the state, as discussed earlier. As a result, our protocol is deterministic, i.e. the probability of success achieves 100%.
We perform the above experiment in IBM QE, where we compare the statistical data of the teleportation of three-qubit state between "IBM qasm-simulator" and "IBM 16 Melbourne". In this experiment, we send an arbitrary three-qubit state from Alice to Bob with the help of a controller Charlie. We run the experiment on both the simulator and the real device and from there we figure out the density matrix of both the cases and from density matrices, we evaluate the fidelity 90,91 of the circuit.

Results
In this section, we present the experimental results for both the cases. Namely the teleportation of an arbitrary two-qubit state using a five-qubit cluster state and then teleportation of an arbitrary three-qubit state using a seven-qubit cluster state.
for teleportation of two-qubit state. First, we run our circuit on "IBM qasm simulator" (with 8192 shots for more accuracy and to reduce statistical errors) and on "IBM 16 Melbourne" (real device). Then we compare both results as seen in Fig. 8. We observe that there are some errors in the "IBM 16 Melbourne" results, and these are due to decoherence 66 , state preparation, and also due to the number of gates used in the circuit 92 as each gate inherently contains gate errors. Now, we check the fidelity of the circuit so that we can know how well the state is teleported. Fidelity is given by the following formula 90,91 here ρ T is the theoretical density matrix whereas, ρ E is the experimental density matrix, and for a positive semidefinite matrix M, √ M denotes its unique positive square root. The general equation of density matrix is given by Eq. (17). Now let us analyze a two-qubit state which is given by for this case ρ T is obtained as  www.nature.com/scientificreports/ Besides, the experimental density matrix (18) for two qubits is given by the following formula 67,93 here T j 1 j 2 is defined as T j 1 j 2 = S j 1 × S j 2 where the Stokes parameters are S 0 = P |0I� + P |1I� , S 1 = P |0X� − P |1X� , S 2 = P |0Y � − P |1Y � , S 3 = P |0Z� − P |1Z� . Here, P |0j 1 � represents the probability of the qubit to be found in |0� state when it is measured in j 1 basis. P |1j 1 � represents the probability of the qubit to be found in |1� state when it is measured in j 1 basis, and same thing with j 2 basis. So, for a two-qubit state (21), the experimental density matrix is calculated as what follows (see Fig. 9) www.nature.com/scientificreports/ Finally, the fidelity between the theoretical density matrix (22) and the experimental density matrix (24) is calculated to be F(ρ T , ρ E ) = 0.4919.
for teleportation of three-qubit state. Here we perform the same as we did for two-qubit teleportation, i.e. we run our experiment on "IBM qasm simulator" (with 8192 shots for more accuracy) and on "IBM 16 Melbourne" (real device). As we can see in Fig. 10, that the probability of getting each possible states for the three-qubit system is nearly the same in the case of "IBM qasm simulator". However, in the case of "IBM 16 Melbourne" we can observe that the probability of getting each possible states for the three-qubit system is different and this is due to the noise errors present in the quantum channel. These noise errors are due to decoherence 66 in the quantum channel, state preparation error, and gate errors. Indeed, all these factors play an important role in reducing the fidelity of the states. Now let us consider a three-qubit state as follows In the previous section, we presented the general equation of the density matrix in Eq. (17) and then we defined fidelity in Eq. (20). Thus, for this state (25) the theoretical density matrix is obtained as Besides, the experimental density matrix (18) for three-qubit is given by the following formula 67,93 where, T j 1 j 2 j 3 = S j 1 × S j 2 × S j 3 and the Stokes parameters are S 0 = P |0I� + P |1I� , S 1 = P |0X� − P |1X� , S 2 = P |0Y � − P |1Y � , S 3 = P |0Z� − P |1Z� . Here, P |0j 1 � represents the probability of the qubit to be found in |0� state when it is measured in j 1 basis. P |1j 1 � represents the probability of the qubit to be found in |1� state when it is measured in j 1 basis, and the same things with j 2 and j 3 bases. Hence, for the three-qubit state (25), the experimental density matrix is calculated as follows (see Fig. 11)  www.nature.com/scientificreports/ Thereby, the fidelity between ρ T and ρ E is calculated to be 0.4006.

Security analysis against charlie
We perform a security analysis against Charlie to check his honesty. In the security analysis protocol, Bob has to contact Alice through a classical channel, as well as he has to measure his state in the basis decided by Alice.
For realizing security analysis of our protocols, we have taken Bell state and GHZ state as a two-qubit state and three-qubit state respectively for teleportation.
for teleportation of two-qubit state. Now, Alice wants to share a Bell state as |ψ − � = 1 √ 2 (|01� − |10�) , and after performing all the unitary transformations in a systematic way as discussed before, Eq. (7) can be rewritten as In this case, Bob has to measure his state in Bell basis. Now, Alice measures her states and let us say she gets |0000� then the whole state is get collapsed to 1 for teleportation of three-qubit state. In this case, suppose Alice wants to share a GHZ state given as , and after performing all the unitary transformations as discussed above in a systematic way, Eq. (13) becomes In this case, Bob has to measure his state in GHZ basis 94,95 . Now, Alice measures her states and let us say she gets |χ 0 � then the whole state is get collapsed to 1 (−|011� − |110�) 246 ⊗ |−� 7 ) . Next, Charlie measures his state and let us say that the measurement outcome is |+� 7 . Now Charlie does not need to send classical information to Bob as discussed earlier. Suppose, Charlie is a dishonest person and he sends the one bit of classical information to Bob within a time period. As usual, after receiving the information, Bob gets that Charlie's qubit is in |−� 7 state. Then, Bob first performs the phase-change unitary transformation on his qubit 6 and then he applies CNOT gate on his qubits (4,6) and (2,4). After performing the unitary transformations, Bob measures his state in GHZ basis and he gets |Bob − state� = 1 √ 2 (−|010� − |101�) . Then Bob communicates with Alice and they both exchange their information about the state and thereby they know that Charlie cheated.

conclusions
In this work, we theoretically and experimentally demonstrate the teleportation of two-qubit and three-qubit states through five-qubit and seven-qubit cluster states respectively. As shown in the circuit diagrams, we successfully run those circuits on "IBM qasm simulator" as well as on "IBM 16 Melbourne" and report their probability distribution results. And we show that the teleportation of qubits through five-qubit and seven-qubit cluster states is possible. We also calculate the fidelity for both the cases and obtain a genuine fidelity over 40%. Remarkably, we also examine the security analysis against Charlie, and these schemes which we consider here are secure against Charlie's attacks. We show that the average classical communication cost of our protocols is 0.5 bit. As compared to other protocols, our protocols reduce substantially the classical communication cost. Hence, our protocols are economical and achievable.