Abstract
In this work, we propose the dual-code quantum computation model—a fault-tolerant quantum computation scheme which alternates between two different quantum error-correction codes. Since the chosen two codes have different sets of transversal gates, we can implement a universal set of gates transversally, thereby reducing the overall cost. We use code teleportation to convert between quantum states in different codes. The overall cost is decreased if code teleportation requires fewer resources than the fault-tolerant implementation of the non-transversal gate in a specific code. To analyze the cost reduction, we investigate two cases with different base codes, namely the Steane and Bacon-Shor codes. For the Steane code, neither the proposed dual-code model nor another variation of it achieves any cost reduction since the conventional approach is simple. For the Bacon-Shor code, the three proposed variations of the dual-code model reduce the overall cost. However, as the encoding level increases, the cost reduction decreases and becomes negative. Therefore, the proposed dual-code model is advantageous only when the encoding level is low and the cost of the non-transversal gate is relatively high.
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Notes
Note that to distinguish the encoded (or logical) gate from the physical gate, we use \(\overline{\mathtt{G}}\) to represent encoded gates and G for physical gates. Likewise, we use \(|\overline{\psi }\rangle \) for the encoded state and \(|\psi \rangle \) for the physical state.
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Appendices
Appendix 1: Fault-tolerant implementation of some elementary modules
1.1 Fault-tolerant preparation of \(|\overline{\mathtt{0}}\rangle \) and \(|\overline{\mathtt{+}}\rangle \)
There are two ways to prepare encoded computational basis states such as \(|\overline{\mathtt{0}}\rangle \) and \(|\overline{\mathtt{+}}\rangle \) for a stabilizer code. One method is to apply a sequence of stabilizer measurements and a logical \(\overline{\mathtt{Mz}}\) or \(\overline{\mathtt{Mx}}\) measurement for \(|\overline{\mathtt{0}}\rangle \) and \(|\overline{\mathtt{+}}\rangle \), respectively [36], Sect. 10.5.8]. The second method is to use an encoding circuit with the standard form of the generator matrix [15, 44]. Since the above teleportation method can be applied to arbitrary state encoding, it can be applied for these computational basis states. However, these two methods are generally used for computational basis states. Investigations demonstrate that these two methods have similar fault-tolerance [11, 52]. Therefore, either of the two methods can be applied for fault-tolerant encoding of logical qubits. In this work, we choose the encoding circuit method since it appears to have lower cost than the stabilizer measurement method. Figure 6a, b, and c shows the circuit for encoding computational basis states on the Steane, Bacon-Shor, and Reed-Muller codes, respectively. Since the encoding circuits are not fault-tolerant for the Steane and the Reed-Muller codes, we need to verify the logical state [39] as shown in Fig. 7. Note that since the target logical initial states are \(|\overline{\mathtt{0}}/\overline{\mathtt{1}}\rangle \) or \(|\overline{\mathtt{+}}/\overline{\mathtt{-}}\rangle \), the circuits shown in Fig. 7 can be augmented with Fig. 6, except (b), since it is already fault-tolerant. Based on this, we summarize the average cost of preparing encoded computational basis states for each code in Tables 7, 8, and 9. All cost evaluation tables are shown in the “Appendix” unless otherwise noted.
1.2 Fault-tolerant preparation of the \({\mathtt{CAT}}_n\) and Shor states
To prepare the \({\mathtt{CAT}}_n=\frac{1}{\sqrt{2}}(|\mathtt{0}\rangle ^{\otimes n} + |\mathtt{1}\rangle ^{\otimes n})\) state, we prepare \(n\) physical qubits which are all initialized to the physical \(|\mathtt{0}\rangle \) state. Then, we apply a physical H to the first qubit. After that, we apply \(n-1\) CNOTs using the first qubit as the control and the others as the targets. This is shown in the left part of Fig. 8. Since we cannot guarantee the fault-tolerance of this circuit, we have to verify \({\mathtt{CAT}}_n\) by checking the parity of all of the physical qubits. To do this, we prepare \(n\) physical qubits which are all initialized to \(|\mathtt{0}\rangle \). For the first additional qubit, we apply two CNOTs from the first and the second CAT qubits. Similarly, we apply two CNOTs sequentially for other \(n-1\) additional qubits. For the last additional qubit, we apply two CNOTs from the first and the last CAT qubit to the additional qubits. Finally, we measure all \(n\) additional qubits. If any of measurements is equal to one, we infer that an error occurs in the \({\mathtt{CAT}}_\mathtt{n}\) state. Therefore, we must repeat this entire procedure until we get all outputs equal to zero. This is shown in the right part of Fig. 8. The average cost is shown in Table 10.
We must apply bitwise H gates since this CAT state should be used for logical ZZ measurements. This state is referred to as the Shor state [41]. Since the Shor state contains only the even parity qubits, it can used for measuring the parity between two encoded qubits, working as a logical ZZ measurement.
1.3 Fault-tolerant logical measurement and syndrome measurement
We apply the encoded measurement circuit as shown in Fig. 9 to measure an encoded qubit. Basically, we have to measure the encoded qubit using the CAT state. In the figure, M=X in \(Control\_M\) refers to \(\overline{\mathtt{Mx}}\); M=Z refers to \(\overline{\mathtt{Mz}}\). For fault-tolerant measurement, we measure three times and decide the final measurement output by the majority value. The overall cost of performing fault-tolerant encoded measurement is shown in Table 11.
Most schemes for syndrome measurement, \(\mathtt{S}_{i,k}\), where \(i\) is the index of the stabilizer generator and \(k\) is the number of physical qubits for a stabilizer generator, are equivalent to fault-tolerant encoded measurement, except that each syndrome must verify \(k\) physical qubits. Based on this, the average cost of syndrome measurement is shown in Table 12.
Appendix 2: Schemes
1.1 Single-code models
In this subsection, we describe the details of each scheme. The cost of each approach is shown in the following subsections.
1.1.1 Steane \(\overline{\mathtt{T}}\) gate
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1.
Fault-tolerantly prepare an encoded \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state.
The circuit for this is shown in Fig. 3b. The necessary circuit for TXT measurement is shown in Fig. 3c. Although the TXT measurement can prepare an encoded \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state, it is not fault-tolerant. Therefore, it should be verified twice using the error detection method. The state is considered high fidelity only when it passes the error detection twice.
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2.
Apply the \(\overline{\mathtt{CNOT}}\) gate between the encoded source state and the \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state.
Since \(\overline{\mathtt{CNOT}}\) is transversal, it is fault-tolerant.
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3.
Measure the source qubit in the Z basis.
To achieve fault-tolerance, we apply the fault-tolerant measurement method shown in “Fault-tolerant logical measurement and syndrome measurement.”
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4.
Apply \(\overline{\mathtt{S}}\) and \(\overline{\mathtt{X}}\) if the measurement is equal to one.
These operations are conditional on the measurement output. Since \(\overline{\mathtt{S}}\) and \(\overline{\mathtt{X}}\) are transversal in the Steane code, it is also fault-tolerant.
1.1.2 Bacon-Shor \(\overline{\mathtt{T}}\) gate
Figure 10a explains \(\overline{\mathtt{S}}\) in the Bacon-Shor code by using \(|\overline{\mathtt{+i}}\rangle \).
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1.
Fault-tolerantly prepare the \(|\overline{\mathtt{+i}}\rangle \) state.
The circuit for this is shown in Fig. 10b, c. More specifically, Fig. 10b explains how to distill two low-fidelity states into a high-fidelity state by using the TWIRL operation as shown in Fig. 10c. We can distill a high-fidelity quantum state by checking whether or not two quantum states are equal. Note that since the verification operation can also detect the existence of errors in the operation, it is fault-tolerant.
-
2.
Apply \(\overline{\mathtt{CNOT}}\) and \(\overline{\mathtt{CZ}}\) gates between the source and the \(|\overline{\mathtt{+i}}\rangle \) state.
These gates complete the entire procedure. Since \(\overline{\mathtt{CNOT}}\) and \(\overline{\mathtt{CZ}}\) are transversal, it is fault-tolerant.
The \(\overline{\mathtt{T}}\) gate can be implemented in a very similar way to the \(\overline{\mathtt{S}}\) gate. The overall circuit is shown in Fig. 11a. Note that the circuit shown in Fig. 11a differs from that in Fig. 3a only in the preparation of the \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state. Details follow.
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1.
Fault-tolerantly prepare the \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state.
The circuit for this is shown in Fig. 11b, c. Figure 11b shows how to distill several low-fidelity states into a high-fidelity state. Note that this is very different from the Steane TXT measurement. Figure 11c shows how to increase the fidelity of each encoded state using the TWIRL operation. Note that this circuit uses the \(\overline{\mathtt{S}}\) gate. Since the distillation protocol only succeeds when all measurements satisfy a predefined condition, it can generate an encoded \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state fault-tolerantly.
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2.
Apply \(\overline{\mathtt{CNOT}}\) gate between the source and \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state.
Since \(\overline{\mathtt{CNOT}}\) is transversal, it is fault-tolerant.
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3.
Measure the source qubit in the Z basis.
To achieve fault-tolerance, we apply the method shown in “Fault-tolerant logical measurement and syndrome measurement.”
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4.
Apply \(\overline{\mathtt{S}}\) and \(\overline{\mathtt{X}}\) if the measurement outcome is equal to one.
These operations are conditional on the measurement output. For the \(\overline{\mathtt{S}}\) gate, we apply the above fault-tolerant \(\overline{\mathtt{S}}\) gate operation. In contrast, \(\overline{\mathtt{X}}\) is transversal. Therefore, this step is also fault-tolerant.
1.2 Dual-code models
1.2.1 Steane \(\overline{\mathtt{T}}\) gate into Reed-Muller \(\overline{\mathtt{T}}\) gate using code teleporation
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1.
Conversion from the Steane state to the Reed-Muller state
The basic idea is to use the teleportation technique. Since teleportation uses a common entanglement between two encoded qubits, we should also prepare this entanglement. In this case, since the source and target qubits use different error-correction codes, the necessary entanglement should be prepared as follows.
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(a)
Fault-tolerant preparation of the \(|\overline{\mathtt{+}}\rangle \) state based on the Steane code
We must prepare a Steane \(|\overline{\mathtt{+}}\rangle \) state fault-tolerantly. Figure 6a shows the corresponding circuit. Note that the general method for encoding the \(|\overline{\mathtt{0}}\rangle \) or \(|\overline{\mathtt{+}}\rangle \) states on the stabilizer codes is proposed in [15]. Since this circuit is not fault-tolerant, we must apply a general method to render it fault-tolerant as shown in Fig. 7.
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(b)
Fault-tolerant preparation of the \(|\overline{\mathtt{+}}\rangle \) state on the Reed-Muller code
We must prepare a Reed-Muller \(|\overline{\mathtt{+}}\rangle \) state fault-tolerantly. Figure 6c shows how to encode the \(|\overline{\mathtt{+}}\rangle \) state with the Reed-Muller code [15]. To render it fault-tolerant, we apply the same method as used for the Steane case as shown in Fig. 7. Note that one way to prepare the \(|\overline{\mathtt{+}}\rangle \) state is to prepare the \(|\overline{\mathtt{0}}\rangle \) state and then apply the \(\overline{\mathtt{H}}\) gate. However, since the \(\overline{\mathtt{H}}\) gate cannot be implemented in a transversal way with the Reed-Muller code, we use a method as shown in [15].
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(c)
Fault-tolerant preparation of the \({\mathtt{CAT}}_{22}\) State
We must prepare a physical CAT state with 7 + 15 = 22 qubits since the Steane and the Reed-Muller codes require 7 and 15 physical qubits, respectively. A fault-tolerant way to prepare the \({\mathtt{CAT}}_{22}\) state is explained in “Fault-tolerant preparation of the \({\mathtt{CAT}}_n\) and Shor states ” of Appendix 1.
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(d)
Two \(\overline{\mathtt{CNOT}}\)s
After that, we apply \(\overline{\mathtt{CNOT}}\) between the Steane \(|\overline{\mathtt{+}}\rangle \) and 7 qubits of \({\mathtt{CAT}}_{22}\), and the Reed-Muller \(|\overline{\mathtt{+}}\rangle \) and the remaining 15 qubits of \({\mathtt{CAT}}_{22}\). Since these \(\overline{\mathtt{CNOT}}\)s are transversal, it is fault-tolerant.
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(e)
Measure \({\mathtt{CAT}}_{22}\) state
Next, we measure the \({\mathtt{CAT}}_{22}\) state for each code using the fault-tolerant measurement as explained in “Fault-tolerant logical measurement and syndrome measurement” of Appendix 1.
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(f)
Additional \(\overline{\mathtt{X}}\)
Based on the measurement output, we should apply an additional \(\overline{\mathtt{X}}\) on the Reed-Muller state. Since the \(\overline{\mathtt{X}}\) gate is transversal in the Reed-Muller code, it is also fault-tolerant.
After preparing an encoded entanglement between the two different codes, we must apply conventional teleportation. It applies an encoded Bell measurement on the Steane code side and additional \(\overline{\mathtt{X}}\) and \(\overline{\mathtt{Z}}\) gates on the Reed-Muller code side. Since the Bell measurement consists of \(\overline{\mathtt{CNOT}}\) and \(\overline{\mathtt{Mx}}\) and \(\overline{\mathtt{Mz}}\), they are all fault-tolerant. At the same time, \(\overline{\mathtt{X}}\) and \(\overline{\mathtt{Z}}\) on the Reed-Muller code are transversal, and hence all steps are fault-tolerant.
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(a)
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2.
Applying the Reed-Muller encoded \(\overline{\mathtt{T}}\) gate
The \(\overline{\mathtt{T}}\) gate in the Reed-Muller code is simply the application of physical T gates on the qubits transversally since \(\overline{\mathtt{T}}\) is transversal in the Reed-Muller code. Therefore, the \(\overline{\mathtt{T}}\) gate in the Reed-Muller code is also fault-tolerant.
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3.
Conversion from the Reed-Muller state into the Steane state
In this case, most of the circuits are very similar to the circuit for the conversion from the Steane state to the Reed-Muller state, except that the source and target states are interchanged.
1.2.2 Bacon-Shor \(\overline{\mathtt{T}}\) gate into Reed-Muller \(\overline{\mathtt{T}}\) gate using code teleportation
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1.
Conversion from the Bacon-Shor state to the Reed-Muller state
This method is the same as the conversion from the Steane state into the Reed-Muller state. The circuit to prepare a Bacon-Shor encoded \(|\overline{\mathtt{+}}\rangle \) state is shown in Fig. 6b. Note that this circuit is fault-tolerant.
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2.
Applying the Reed-Muller \(\overline{\mathtt{T}}\) gate
In this case, we can implement the \(\overline{\mathtt{T}}\) gate by applying 15 physical T gates for each physical qubit since \(\overline{\mathtt{T}}\) is transversal in the Reed-Muller code.
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3.
Conversion from the Reed-Muller state to the Bacon-Shor state
This conversion is almost same as the conversion from the Bacon-Shor state to the Reed-Muller state.
1.2.3 Bacon-Shor \(\overline{\mathtt{T}}\) gate with Reed-Muller \(\overline{\mathtt{T}}\) state
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1.
Prepare \(\overline{T}|\overline{+}\rangle \) state based on the Reed-Muller code
Since the encoded \(\overline{T}\) gate is transversal in the Reed-Muller code, this process is fault-tolerant.
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2.
Teleport this state into the Bacon-Shor encoded state
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3.
Apply Fig. 11a
This process realizes the encoded \(\overline{T}\) gate in the Bacon-Shor code.
Appendix 3: Cost analysis
1.1 Single-code models
In this section, we describe the details of the cost of each approach. The necessary tables for every component are shown in “Appendix 3.”
1.1.1 Steane \(\overline{\mathtt{T}}\) gate
Based on Fig. 3 and previous explanations, the overall cost to implement an encoded \(\overline{\mathtt{T}}\) gate with Steane code is shown in Table 13. The breakdown of the overall cost is as follows.
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1.
Cost of preparing \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state
The average cost of preparing \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state is shown in Table 14. The cost consists of two parts: the cost of the TXT measurement and the cost of the error detection.
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(a)
Cost of the TXT measurement
The average cost of the TXT measurement is shown in Table 15. First, an encoded \(|\overline{\mathtt{0}}\rangle \) state should be prepared. After that, physical T gates are applied transversally for each physical qubit. At the same time, a \({\mathtt{CAT}}_7\) state should be prepared. \(\overline{\mathtt{CNOT}}\) should be applied between these two states. After that, another physical T \(^\dag \) gate should be applied. At the same time, the physical measurements should be done for the \({\mathtt{CAT}}_7\) state. If all measurements are zero, it succeeds. Otherwise, the entire process repeats.
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(b)
Cost of the error detection
The next step is a verification of whether or not there are any errors in the \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state. For the error detection, we simply need to check the stabilizer generators. For each stabilizer generator, we use the circuit shown in Fig. 12. Since each stabilizer generator should be measured three times, the process in the dotted box (colored red in the online version) should be repeated three times for each of the six stabilizer generators in the Steane code where each stabilizer generator checks four qubits. Therefore, the cost of the error detection is
$$\begin{aligned} Cost(Error\_Detection)=6\times AvgCost(S_{i,4}), \end{aligned}$$(1)when we assume \(Cost({\mathtt{CX}})=Cost({\mathtt{CZ}})\).
Thus, the total cost of the \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state preparation is the sum of the above cost. Since the process succeeds only when two error detections yield no error, the average cost to prepare the \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state should be divided by \({P\_Success}^2\). The overall cost is shown in Table 14.
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(a)
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2.
Cost of \(\overline{\mathtt{CNOT}}\).
Apply \(\overline{\mathtt{CNOT}}\) between the source and the \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state. Since \(\overline{\mathtt{CNOT}}\) is transversal, it is easy to implement.
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3.
Cost of \(\overline{\mathtt{Mz}}\)
Measure the source qubit with the Z basis.
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4.
Cost for additional operations
Apply \(\overline{\mathtt{S}}\) and \(\overline{\mathtt{X}}\), which are transversal, if the measurement is equal to one. This operation is conditional on the measurement output.
1.1.2 Bacon-Shor \(\overline{\mathtt{T}}\) gate
As shown in Fig. 11a, the cost of the \(\overline{\mathtt{T}}\) gate in the Bacon-Shor code is explained as follows. The cost breakdown is shown in Table 16 with the following summary.
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1.
Cost of fault-tolerantly preparing an encoded \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state
The cost of preparing a high-fidelity \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) state is shown in Table 17.
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2.
Cost for \(\overline{\mathtt{CNOT}}\)
This is transversal.
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3.
Cost for \(\overline{\mathtt{Mz}}\)
This is shown in Table 11.
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4.
Cost for additional \(\overline{\mathtt{SX}}\)
If the logical measurement output is equal to one, \(\overline{\mathtt{S}}\) and \(\overline{\mathtt{X}}\) should be applied. Therefore, it should equal the sum of the cost of \(\overline{\mathtt{S}}\) and \(\overline{\mathtt{X}}\). The cost of logical \(\overline{\mathtt{S}}\) is shown in Table 21.
1.2 Dual-code models
1.2.1 Steane \(\overline{\mathtt{T}}\) gate to Reed-Muller \(\overline{\mathtt{T}}\) gate using code teleporation
Table 23 shows the necessary cost. The details are as follows.
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1.
Cost of converting from the Steane state to the Reed-Muller state
This consists of several costs as follows. Table 24 shows the details.
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(a)
Cost of fault-tolerant preparation of the Steane \(|\overline{\mathtt{+}}\rangle \) state.
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(b)
Cost of fault-tolerant preparation of the Reed-Muller \(|\overline{\mathtt{+}}\rangle \) state.
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(c)
Cost of fault-tolerant preparation of \({\mathtt{CAT}}_{22}\) state.
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(d)
Cost of two \(\overline{\mathtt{CNOT}}\)s between the Steane state and a part of \({\mathtt{CAT}}_{22}\) state; and the Reed-Muller state and the other part of \({\mathtt{CAT}}_{22}\) state.
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(e)
Cost of fault-tolerant measurements of \({\mathtt{CAT}}_{22}\) state.
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(f)
Additional cost of fault-tolerant \(\overline{\mathtt{X}}\) in the Reed-Muller part of \({\mathtt{CAT}}_{22}\) state.
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(g)
Cost of fault-tolerant \(\overline{\mathtt{CNOT}}\) between the Steane encoded source qubit and the Steane part of \({\mathtt{CAT}}_{22}\) state.
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(h)
Cost of fault-tolerant measurement \(\overline{\mathtt{Mx}}\) and \(\overline{\mathtt{Mz}}\).
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(i)
Additional cost of fault-tolerant \(\overline{\mathtt{X}}\) and \(\overline{\mathtt{Z}}\) for the Reed-Muller part of \({\mathtt{CAT}}_{22}\) state.
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(a)
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2.
Cost of applying Reed-Muller \(\overline{\mathtt{T}}\) gate
Since the Reed-Muller code allows the transversal \(\overline{\mathtt{T}}\) gate operation, it consists of the simple application of physical T gates on each physical qubit.
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3.
Cost of conversion from the Reed-Muller state to the Steane state
This is almost the same as the cost of conversion from the Steane state to the Reed-Muller state as shown in Table 25.
1.2.2 Steane \(\overline{\mathtt{T}}\) gate to Reed-Muller \(\overline{\mathtt{T}}\) gate using direct conversion
The cost of this approach is shown in Table 31 based on Fig. 4. The details of the cost follow.
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1.
Conversion from the Steane state to the Reed-Muller state
Table 29 shows the cost. Since this approach utilizes a maximally entangled state between the encoded Steane state and a physical qubit, another circuit should be used as shown at the bottom of Fig. 4.
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2.
Applying the Reed-Muller logical \(\overline{\mathtt{T}}\) gate
Since the Reed-Muller code allows transversal \(\overline{\mathtt{T}}\) gate implementation, it consists of 15 physical T gates on the physical qubits. Therefore, the number of gates is 15.
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3.
Conversion from the Reed-Muller state to the Bacon-Shor state
Table 30 shows the cost.
1.2.3 Bacon-Shor \(\overline{\mathtt{T}}\) gate to Reed-Muller \(\overline{\mathtt{T}}\) gate using code teleportation
Similar to code conversion between the Steane and Reed-Muller codes, this conversion also involves the cost of the three following steps. Table 32 summarizes the necessary cost.
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1.
Conversion from the Bacon-Shor state to the Reed-Muller state
Table 33 shows the cost.
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2.
Applying the Reed-Muller logical \(\overline{\mathtt{T}}\) gate
Since the Reed-Muller code allows transversal \(\overline{\mathtt{T}}\) gate implementation, it consists of 15 physical T gates on the physical qubits. Therefore, the number of gates is 15.
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3.
Conversion from the Reed-Muller state to the Bacon-Shor state
Table 34 shows the cost.
1.2.4 Bacon-Shor \(\overline{\mathtt{T}}\) gate with Reed-Muller \(T|\overline{+}\rangle \) state
Table 35 summarizes the necessary cost.
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1.
Prepare \(|\overline{+}\rangle \) state based on the Reed-Muller code
Table 9 shows the cost.
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2.
Apply 15 transversal \(\mathtt{T}\) gates on each qubit
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3.
Teleport this state to the Bacon-Shor encoded state
Table 34 shows the cost.
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4.
Apply the \(\overline{\mathtt{CNOT}}\) gate between the source and \(\mathtt{T}|\overline{\mathtt{+}}\rangle \) states.
Since \(\overline{\mathtt{CNOT}}\) is transversal, it is fault-tolerant.
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5.
Measure the source qubit in the Z basis.
To achieve fault-tolerance, we apply the method shown in “Fault-tolerant logical measurement and syndrome measurement.”
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6.
Apply \(\overline{\mathtt{S}}\) and \(\overline{\mathtt{X}}\) if the measurement outcome is equal to one.
These operations are conditional on the measurement output. For the \(\overline{\mathtt{S}}\) gate, we apply the above fault-tolerant \(\overline{\mathtt{S}}\) gate operation. In contrast, \(\overline{\mathtt{X}}\) is transversal. Therefore, this step is also fault-tolerant.
Appendix 4: Tables
See Tables 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37.
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Choi, BS. Dual-code quantum computation model. Quantum Inf Process 14, 2775–2818 (2015). https://doi.org/10.1007/s11128-015-1022-0
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DOI: https://doi.org/10.1007/s11128-015-1022-0