WO2020013665A1 - Quantum computer - Google Patents

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WO2020013665A1
WO2020013665A1 PCT/KR2019/008667 KR2019008667W WO2020013665A1 WO 2020013665 A1 WO2020013665 A1 WO 2020013665A1 KR 2019008667 W KR2019008667 W KR 2019008667W WO 2020013665 A1 WO2020013665 A1 WO 2020013665A1
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qubit
quantum computer
gate
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output
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박영준
박동우
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(주)더포스컨설팅
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Priority claimed from KR1020190084618A external-priority patent/KR20200007736A/en
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    • GPHYSICS
    • G06COMPUTING; CALCULATING OR COUNTING
    • G06NCOMPUTING ARRANGEMENTS BASED ON SPECIFIC COMPUTATIONAL MODELS
    • G06N10/00Quantum computing, i.e. information processing based on quantum-mechanical phenomena
    • BPERFORMING OPERATIONS; TRANSPORTING
    • B82NANOTECHNOLOGY
    • B82YSPECIFIC USES OR APPLICATIONS OF NANOSTRUCTURES; MEASUREMENT OR ANALYSIS OF NANOSTRUCTURES; MANUFACTURE OR TREATMENT OF NANOSTRUCTURES
    • B82Y10/00Nanotechnology for information processing, storage or transmission, e.g. quantum computing or single electron logic

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  • the technology relates to quantum computer implementations.
  • the present technology is to solve the difficulty of the quantum computer according to the prior art, and the present technology solves the difficulty of the quantum computer according to the prior art, which requires a large memory capacity, and uses a phase locked loop and / or a delay locked loop. Providing is one of the main goals.
  • This technology calculates and tracks the probabilities of any one of the states 1 and 0 of the qubits instead of storing the "quantum state vector" consisting of the information of the qubits that form the basis of the quantum computer. Save it. Also read the probabilities by qubit if necessary.
  • an output is provided in which a relationship between a probability of each qubit and an input of a quantum circuit is known in advance, so that an output of a quantum computer can be quickly known.
  • 1 is a diagram for explaining a qubit.
  • FIG. 3 is a schematic diagram of a Hadamard gate for explaining the present embodiment.
  • FIG. 4 is a schematic diagram of a CNOT gate for explaining the present embodiment.
  • FIG. 5 is a diagram showing an outline of Embodiment 3.
  • FIG. 5 is a diagram showing an outline of Embodiment 3.
  • each qubit instead of storing a "quantum state vector" consisting of qubit (quantum bit) information that forms the basis of a quantum computer, each qubit stores the probability of any one of states 1 and 0. do.
  • each qubit may further store connection relationship information for calculating a probability. It also reads probabilities for each qubit as output.
  • the qubit may be represented by a circle having a radius of 1 between a real axis and an imaginary axis, as illustrated in FIG. 1A, and may theoretically display infinite information.
  • the square of the abscissa component x and the square of the ordinate component y represent the probabilities, respectively, and the sum of x2 and y2 is one.
  • FIG. 1 (b) shows the qubit of the 0 state
  • FIG. 1 (c) shows the qubit of the 1 state.
  • the quantum computer calculates and traces a zero state probability or a one state probability to perform a desired operation. As described above, since the sum of the zero state probability and the one state probability of the qubit is 1, the probability of the other state can be known by tracking only one state probability.
  • the probability of each qubit component is calculated and stored for each phase.
  • the present technology tracks and stores only probabilities of the zero states of q # 1 and q # 2 in each phase. In this embodiment, since it is not necessary to store all the quantum states for output, information to be stored for each qubit can be reduced.
  • the number of memories required is 2N.
  • nN memory is sufficient.
  • the information for tracking one qubit probability may include connection information by a gate or the like in a previous phase.
  • Equation 1 the information of each qubit is shown in Equation 1 below.
  • the values of a, b in the states 1 and 0
  • qubit # 2 remains unchanged. Therefore, only the a and b values of each qubit may be stored.
  • one of the zero state probability and the one state probability of the qubit is tracked and stored for each phase.
  • Equation 2 is a schematic diagram of the Hadamard gate H for explaining the present embodiment, and Equations 2 and 3 below are equations for explaining the Hadamard gate H.
  • FIG. When a cubit having (cos ⁇ i and sin ⁇ i ) is provided as an input to the Hadamard gate, the output of the Hadamard gate is expressed by Equation 2 below.
  • the a i component may be expressed as in Equation 3 below.
  • the Hadamard gate may be represented by a 45 degree shift of the qubit zero component. For example, if a zero state input is input to the Hadamard gate H, the zero state probability value output is 1/2.
  • FIG. 3B is a schematic diagram of an X gate for explaining the present embodiment, and Equation 4 below is an expression representing an input / output relationship between the X gates.
  • the X gate is a gate that exchanges components of the input. Given the (a, b) input to the X gate, the zero state probability of the X gate output is b 2 .
  • FIG. 3C is a schematic diagram of an R gate for explaining the present embodiment, and Equation 5 below is an equation representing an input / output relationship between R gates.
  • the qubit may store the ⁇ value of the R gate.
  • the embodiment illustrated by FIG. 2 includes a CNOT gate.
  • the output of the CNOT gate is expressed by Equation 6 below.
  • the zero state probability of qubit # 1 is a 1 2
  • the zero state probability of qubit # 2 is a 1 2 a 2 2 + b 1 2 b 2 2 .
  • the cubit # 2 of the zero component input is CNOT calculated by the qubit # 1 of the zero component input
  • Table 1 is a table showing a zero state probability stored in each qubit after each phase indicated in the embodiment of FIG. 2.
  • the present embodiment tracks and stores only one component (eg, 0 state) probability of qubit # 1 and qubit # 2 for each phase.
  • Table 2 illustrates the probability of a zero state component that each qubit should store when a zero state input is provided to qubit # 1 and a zero state input to qubit # 2 in the embodiment illustrated in FIG. 2.
  • phase 1 (p-1), qubit # 1 is not affected by the CNOT gate. Therefore, the component of qubit # 1 is The zero state probability is 1/2. However, since cubit # 2 performs a CNOT operation by qubit # 1, the zero state probability is to be.
  • phase 2 (p-2) the X gate swaps the zero state component and one state component of qubit # 1, so in phase 2 the component of qubit # 1 is The component of qubit # 2 does not change. Therefore, the zero state probability of qubit # 1 and the zero state probability of qubit # 2 in phase 2 (p-2) are both 1/2.
  • phase 3 (p-3), qubit # 1 is not affected by the CNOT gate, so the component of qubit # 1 is to be.
  • Qubit # 2 is a CNOT operation by qubit # 1, and the 0 state probability to be.
  • phase 4 qubit # 1 is rotated 45 degrees by the Hadamard (H) gate.
  • H Hadamard
  • the zero state probability of qubit # 1 is 1 or 1 because the component of qubit # 1 is (1,0).
  • the zero state probability of qubit # 2 is zero.
  • the quantum computer can obtain and output the zero state probability or the one state probability for each phase of the qubits even without using a quantum state vector.
  • This embodiment uses a phase locked loop (PLL) and a delay locked loop (DLL) as a method of calculating or storing the probability of each qubit. For example, the zero component of each qubit is cos ⁇ i (i: cubit number). This can reduce computation time rather than software operation.
  • PLL phase locked loop
  • DLL delay locked loop
  • Equation 4 is a schematic diagram of a CNOT (control NOT) gate for explaining the present embodiment, and Equation 5 shows a probability of the a ( ⁇ in the figure) component of the j bit controlled by the i-th qubit.
  • the probability of the a (Fig. 3 (b) ⁇ ) component of the j bit controlled by the i th qubit may be expressed as twice the ⁇ component, and may be implemented by an electronic circuit such as a phase locked loop or a delay locked loop. Can be.
  • Example 3 includes three qubits, and performs CNOT operation on phase 1 with cubit # 2 and CNOT operation on phase 2 with qubit # 1.
  • each qubit may further store a connection relationship by a CNOT gate for each phase.
  • the zero state probability of qubit # 1 is a 1 2 since it is not affected by the CNOT operation of phase 2.
  • the zero state probability of the qubit # 2 is calculated as a 1 2 a 2 2 + b 1 2 b 2 2 .
  • the zero state probability of the qubit # 3 is calculated as a 2 2 a 3 2 + b 2 2 b 3 2 .
  • a plurality of inputs I 1 (a 1 , b 1 ), I 2 (a 2 , b 2 ), ..., I N (a N , b N ) result in a plurality of probability values O 1 , O 2 , ..., O K
  • the quantum computer according to the present embodiment may display the output probability of each desired qubit as a function of the value of the input qubit, and the relation at this time is expressed by the following equation.
  • O represents the probability of having zero of each qubit.
  • the conventional quantum computer using the quantum spin and the like can acquire the output value and at the same time the output value is destroyed to ensure the safety of the information.
  • the quantum computer according to the present embodiment can provide an output and at the same time, erase the output value, the buffer, and the operation result stored in the memory to maintain information safety.
  • the quantum computer according to the present technology may be implemented in software or a semiconductor CMOS circuit.

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Abstract

The present technology calculates probabilities of the states of 0 and 1 of each qubit and minimum information for the probabilities only and stores same in each qubit, rather than storing "quantum state vectors" consisting of information on qubits which are the basis of a quantum computer. Also, the present technology reads the probabilities of each qubit if necessary. The present technology provides the advantages of implementing a quantum computer with a small memory capacity in comparison with conventional technologies, foreseeing the relation between probabilities of each qubit and an input of a quantum circuit from an output, and thus quickly knowing outputs of the quantum computer.

Description

양자 컴퓨터Quantum computer
본 기술은 양자 컴퓨터 구현과 관련된다.The technology relates to quantum computer implementations.
종래의 소프트웨어나 반도체 CMOS 회로로 양자 컴퓨터를 구현하는 경우, 상태 벡터를 기억해야 하므로, 과도한 메모리를 사용한다. 일 예로, 64 큐빗인 경우에, 총 기억해야 하는 상태의 수가 264이다. 즉, 큐빗의 수가 증가함에 따라 기억해야 하는 상태의 수가 기하급수적으로 증가함에 따라 필요한 메모리의 용량도 기하급수적으로 증가한다.When implementing a quantum computer with conventional software or a semiconductor CMOS circuit, it is necessary to store state vectors, thus using excessive memory. For example, in the case of 64 qubits, the total number of states to be stored is 264. That is, as the number of qubits increases, the amount of memory required increases exponentially as the number of states to be stored increases exponentially.
본 기술은 종래 기술에 의한 양자 컴퓨터의 난점을 해소하기 위한 것으로, 본 기술은 큰 메모리 용량이 필요한 종래 기술에 의한 양자 컴퓨터의 난점을 해소하고, 위상 고정 루프 및/또는 지연 고정 루프를 이용하는 양자 컴퓨터를 제공하는 것을 주된 목표 중 하나로 한다.The present technology is to solve the difficulty of the quantum computer according to the prior art, and the present technology solves the difficulty of the quantum computer according to the prior art, which requires a large memory capacity, and uses a phase locked loop and / or a delay locked loop. Providing is one of the main goals.
본 기술은 양자 컴퓨터의 기본을 이루는 큐빗(Qubit)의 정보로 이루어지는 “양자 상태 벡터”를 기억하는 대신 각각 큐빗이 가지는 1과 0의 상태 중 어느 한 상태 확률을 연산하여 추적하며, 각각의 큐빗에 저장한다. 또한, 필요한 경우 큐빗별로 확률을 읽는다. This technology calculates and tracks the probabilities of any one of the states 1 and 0 of the qubits instead of storing the "quantum state vector" consisting of the information of the qubits that form the basis of the quantum computer. Save it. Also read the probabilities by qubit if necessary.
본 기술에 의하면 종래 기술에 비하여 적은 메모리 용량으로 양자 컴퓨터를 구현할 수 있다. 본 기술의 일 태양에 의하면 출력으로 각 큐빗의 확률과 양자회로의 입력과의 관계를 미리 알아, 양자컴퓨터의 출력을 빠르게 알 수 있다는 장점이 제공된다. According to the present technology, it is possible to implement a quantum computer with a smaller memory capacity than the prior art. According to one aspect of the present technology, an output is provided in which a relationship between a probability of each qubit and an input of a quantum circuit is known in advance, so that an output of a quantum computer can be quickly known.
도 1은 큐빗을 설명하기 위한 도면이다. 1 is a diagram for explaining a qubit.
도 2는 2 큐빗을 슈퍼 덴스코딩(superdense coding)에 적용한 예이다. 2 shows an example of applying 2 qubits to superdense coding.
도 3은 본 실시예를 설명하기 위한 하다마드 게이트의 개요도이다.3 is a schematic diagram of a Hadamard gate for explaining the present embodiment.
도 4는 본 실시예를 설명하기 위한 CNOT 게이트의 개요도이다.4 is a schematic diagram of a CNOT gate for explaining the present embodiment.
도 5은 실시예 3의 개요를 도시한 도면이다. 5 is a diagram showing an outline of Embodiment 3. FIG.
실시예Example 1 One
본 실시예는 양자 컴퓨터의 기본을 이루는 큐빗(qubit, Quantum Bit)의 정보로 이루어지는 “양자 상태 벡터”를 기억하는 대신 각각 큐빗이 가지는 1과 0의 상태 중 어느 한 상태의 확률을 각 큐빗에 저장한다. 일 실시예로, 각 큐빗에는 확률을 계산하기 위한 연결 관계 정보가 더 저장될 수 있다. 또한 출력으로서 큐빗 별로 확률을 읽는다.In this embodiment, instead of storing a "quantum state vector" consisting of qubit (quantum bit) information that forms the basis of a quantum computer, each qubit stores the probability of any one of states 1 and 0. do. In one embodiment, each qubit may further store connection relationship information for calculating a probability. It also reads probabilities for each qubit as output.
일 예로, 큐빗은 도 1(a)로 예시된 것과 같이 실축(real axis)과 허축(imaginary axis) 사이에서 반지름이 1인 원으로 표시될 수 있으며, 이론적으로 무한대의 정보를 표시할 수 있다. 횡축 성분 x의 제곱 및 종축 성분 y의 제곱은 각각 확률을 나타내며, x2과 y2의 합은 1이다. 도 1(b)는 0 상태의 큐빗을 나타내며, 도 1(c)는 1 상태의 큐빗을 나타낸다. For example, the qubit may be represented by a circle having a radius of 1 between a real axis and an imaginary axis, as illustrated in FIG. 1A, and may theoretically display infinite information. The square of the abscissa component x and the square of the ordinate component y represent the probabilities, respectively, and the sum of x2 and y2 is one. FIG. 1 (b) shows the qubit of the 0 state, and FIG. 1 (c) shows the qubit of the 1 state.
본 실시예에 의한 양자 컴퓨터는 0 상태 확률 또는 1 상태 확률을 연산하고, 추적하여 목적하는 연산을 수행한다. 상기한 바와 같이 큐빗의 0 상태 확률과 1 상태 확률의 합은 1이므로, 어느 한 상태 확률만 추적하여도 나머지 한 상태의 확률은 알 수 있다. The quantum computer according to the present embodiment calculates and traces a zero state probability or a one state probability to perform a desired operation. As described above, since the sum of the zero state probability and the one state probability of the qubit is 1, the probability of the other state can be known by tracking only one state probability.
양자 연산을 위한 게이트를 지나는 경우를 한 개의 페이즈라고 하면 각 페이즈(phase)별로 각 큐빗 성분의 확률을 계산하고 저장한다. 도 2를 참조하면, 본 기술에서는 각 페이즈 마다 q#1, q#2의 0 상태의 확률만을 추적 및 저장한다. 본 실시예는 출력을 위해서 모든 양자 상태를 저장할 필요가 없으므로 각 qubit 마다 저장하여야 하는 정보를 감소시킬 수 있다.If one phase passes the gate for quantum operation, the probability of each qubit component is calculated and stored for each phase. Referring to FIG. 2, the present technology tracks and stores only probabilities of the zero states of q # 1 and q # 2 in each phase. In this embodiment, since it is not necessary to store all the quantum states for output, information to be stored for each qubit can be reduced.
즉, 종래 기술에 따라 양자 상태 벡터를 저장할 때, 필요한 메모리의 수는 2N이다. 즉, 64 큐빗이면 264의 상태를 기억할 수 있는 메모리가 필요하다. 그러나, 본 기술에 의하면 nN의 메모리면 충분하다. 일 예로, N=64이고, n이 하나의 큐빗 확률을 추적하기 위한 정보라고 하면, 64n의 메모리이면 충분하다. 일 실시예로, 하나의 큐빗 확률을 추적하기 위한 정보는 이전 페이즈에서 게이트 등에 의한 연결 정보를 포함할 수 있다.That is, when storing quantum state vectors according to the prior art, the number of memories required is 2N. In other words, if 64 qubits, a memory capable of storing 264 states is required. However, according to the present technology, nN memory is sufficient. For example, if N = 64 and n is information for tracking one qubit probability, 64 n of memory is sufficient. In one embodiment, the information for tracking one qubit probability may include connection information by a gate or the like in a previous phase.
도 2는 2 개의 큐빗(q#1, q#2)을 슈퍼 덴스코딩(superdense coding)에 적용한 예이다. 도 2를 참조하면, a1, b1은 큐빗 #1(q#1)의 성분이고, a2, b2는 큐빗 #2(q#2)의 성분이다. 페이즈 0(p-0)를 지난 후, 각 큐빗의 정보는 아래의 수학식 1과 같다. 큐빗 #1은 하다마드 게이트(Hadamard gate)를 지나면서 a, b(1과 0의 상태의)의 값이 a’, b’으로 바뀐다. 퀀텀 인탱글먼트(quantum entanglement, CNOT)에 입력되기 이전이므로 큐빗 #2은 변화없다. 따라서, 큐빗 각각의 a, b 값만 저장하여도 무방하다. 2 illustrates an example in which two qubits q # 1 and q # 2 are applied to superdense coding. Referring to FIG. 2, a1 and b1 are components of qubit # 1 (q # 1), and a2 and b2 are components of qubit # 2 (q # 2). After the phase 0 (p-0), the information of each qubit is shown in Equation 1 below. In qubit # 1, the values of a, b (in the states 1 and 0) change to a 'and b' as they pass through the Hadamard gate. Since it is before entering quantum entanglement (CNOT), qubit # 2 remains unchanged. Therefore, only the a and b values of each qubit may be stored.
Figure PCTKR2019008667-appb-M000001
Figure PCTKR2019008667-appb-M000001
본 실시예에서는 상기한 바와 같이 각 페이즈 별로 큐빗의 0 상태 확률 또는 1 상태 확률 중 어느 하나를 추적하여 저장한다.In this embodiment, as described above, one of the zero state probability and the one state probability of the qubit is tracked and stored for each phase.
도 3(a)은 본 실시예를 설명하기 위한 하다마드 게이트(H)의 개요도이며, 아래의 수학식 2 및 수학식 3은 하다마드 게이트(H)를 설명하기 위한 수학식이다. 하다마드 게이트에 (cosθi, sinθi) 성분을 가지는 큐빗을 입력으로 제공하면, 하다마드 게이트의 출력은 아래의 수학식 2와 같다.3A is a schematic diagram of the Hadamard gate H for explaining the present embodiment, and Equations 2 and 3 below are equations for explaining the Hadamard gate H. FIG. When a cubit having (cosθ i and sinθ i ) is provided as an input to the Hadamard gate, the output of the Hadamard gate is expressed by Equation 2 below.
Figure PCTKR2019008667-appb-M000002
Figure PCTKR2019008667-appb-M000002
이 때, ai 성분은 아래의 수학식 3과 같이 표시될 수 있다.In this case, the a i component may be expressed as in Equation 3 below.
Figure PCTKR2019008667-appb-M000003
Figure PCTKR2019008667-appb-M000003
즉, 하다마드 게이트는 큐빗 0 성분의 45도 편이(shift)로 표시될 수 있다. 일 예로, 0 상태 입력이 하다마드 게이트(H)에 입력되면, 출력되는 0 상태 확률값은 1/2이다. That is, the Hadamard gate may be represented by a 45 degree shift of the qubit zero component. For example, if a zero state input is input to the Hadamard gate H, the zero state probability value output is 1/2.
도 3(b)는 본 실시예를 설명하기 위한 X 게이트의 개요도이며, 아래의 수학식 4는 X 게이트의 입출력 관계를 나타내는 식이다. FIG. 3B is a schematic diagram of an X gate for explaining the present embodiment, and Equation 4 below is an expression representing an input / output relationship between the X gates.
Figure PCTKR2019008667-appb-M000004
Figure PCTKR2019008667-appb-M000004
수학식 4를 참조하면, X 게이트는 입력의 성분을 상호 교환하는 게이트이다. X 게이트로 (a, b)입력이 주어진 경우에, X 게이트 출력의 0 상태 확률은 b2이다. Referring to Equation 4, the X gate is a gate that exchanges components of the input. Given the (a, b) input to the X gate, the zero state probability of the X gate output is b 2 .
도 3(c)는 본 실시예를 설명하기 위한 R 게이트의 개요도이며, 아래의 수학식 5는 R 게이트의 입출력 관계를 나타내는 식이다.FIG. 3C is a schematic diagram of an R gate for explaining the present embodiment, and Equation 5 below is an equation representing an input / output relationship between R gates.
Figure PCTKR2019008667-appb-M000005
Figure PCTKR2019008667-appb-M000005
일 실시예로, 큐빗은 R 게이트의 θ값을 저장할 수 있다.In one embodiment, the qubit may store the θ value of the R gate.
도 2로 예시된 실시예는 CNOT 게이트를 포함한다. 도시된 CNOT 게이트에 의하여 (a1, b1)의 큐빗 #1로 (a2, b2)의 큐빗 #2가 CNOT 연산되면 CNOT 게이트의 출력은 아래의 수학식 6과 같다.The embodiment illustrated by FIG. 2 includes a CNOT gate. When the cubit # 2 of (a2, b2) is CNOT calculated with the cubit # 1 of (a1, b1) by the illustrated CNOT gate, the output of the CNOT gate is expressed by Equation 6 below.
Figure PCTKR2019008667-appb-M000006
Figure PCTKR2019008667-appb-M000006
따라서, 큐빗 #1의 0상태 확률은 a1 2이고, 큐빗 #2의 0상태 확률은 a1 2a2 2+b1 2b2 2이다. 일 예로, 0 성분 입력의 큐빗 #1에 의하여 0 성분 입력의 큐빗 #2 가 CNOT 연산되면, 출력되는 0 성분의 확률의 값은 12*12 +0 = 1이다.Accordingly, the zero state probability of qubit # 1 is a 1 2 , and the zero state probability of qubit # 2 is a 1 2 a 2 2 + b 1 2 b 2 2 . For example, if the cubit # 2 of the zero component input is CNOT calculated by the qubit # 1 of the zero component input, the probability value of the zero component to be output is 1 2 * 1 2 +0 = 1.
표 1은 도 2의 실시예에서 표시된 각 페이즈들을 지난 후, 각 큐빗에 저장되는 0 상태 확률을 도시한 표이다. Table 1 is a table showing a zero state probability stored in each qubit after each phase indicated in the embodiment of FIG. 2.
p-0p-0 p-1p-1 p-2p-2 p-3p-3 p-4p-4
q#1q # 1 a1`2 a 1 ` 2 a1`2 a 1 ` 2 b1`2 b 1 ` 2 b1`2 b 1 ` 2 (a1` + b1`)2 (a 1 ` + b 1 `) 2
q#2q # 2 a2`2 a 2 ` 2 a1`2a2`2 + b1`2b2`2 a 1 `2 a 2` 2 + b 1 `2 b 2` 2 a1`2a2`2 + b1`2b2`2 a 1 `2 a 2` 2 + b 1 `2 b 2` 2 b2 2 b 2 2 b2 2 b 2 2
위 표에서 알 수 있는 바와 같이 본 실시예는 각 페이즈 마다 큐빗 #1, 큐빗 #2의 한 성분(예, 0 상태) 확률만을 추적 및 저장한다. 이 방법에 의하여 각 큐빗마다 저장하여야 하는 정보를 최소화할 수 있으며, 이로부터 출력 확률을 얻을 수 있음을 알 수 있다.As can be seen from the table above, the present embodiment tracks and stores only one component (eg, 0 state) probability of qubit # 1 and qubit # 2 for each phase. By this method, it is possible to minimize the information to be stored for each qubit, and the output probability can be obtained from this.
아래의 표 2는 도 2로 예시된 실시예에서, 큐빗 #1에 0 상태 입력, 큐빗 #2에 0 상태 입력이 제공되었을 때, 각 큐빗이 저장해야 하는 0 상태 성분의 확률을 예시한 것이다.Table 2 below illustrates the probability of a zero state component that each qubit should store when a zero state input is provided to qubit # 1 and a zero state input to qubit # 2 in the embodiment illustrated in FIG. 2.
00 1One 22 33 44
q#1|0> q # 1 | 0> 1/21/2 1/21/2 1/21/2 1/21/2 1One |0>| 0>
q#2|0> q # 2 | 0> 1One 1/2Х1 +1/2Х0=1/21 / 2Х1 + 1 / 2Х0 = 1/2 1/21/2 00 00 |0>| 0>
표 2를 참조하면, 큐빗 #1, 큐빗 #2 모두에 0 상태 입력(|0>)을 제공한 경우에, 페이즈 0(p-0)에서, 하다마드 게이트(H)는 입력된 성분을 45도 편이하므로, 하다마드 게이트(H)가 출력하는 큐빗 #1의 성분은
Figure PCTKR2019008667-appb-I000001
이며, 0 상태 확률은 1/2이다. 페이즈 0에서 큐빗 #2의 성분은 입력과 동일하게 (1,0)이므로, 0 상태 확률은 1이다.
Referring to Table 2, in the case where 0 state input (| 0>) is provided to both qubit # 1 and qubit # 2, at phase 0 (p-0), the Hadamard gate (H) returns the input component. Also, since the component of qubit # 1 output by the Hadamard gate H is
Figure PCTKR2019008667-appb-I000001
The zero state probability is 1/2. At phase 0, the component of qubit # 2 is (1,0) equal to the input, so the zero state probability is one.
페이즈 1(p-1)에서, 큐빗 #1은 CNOT 게이트에 의한 영향을 받지 않는다. 따라서, 큐빗 #1의 성분은
Figure PCTKR2019008667-appb-I000002
이며, 0 상태 확률은 1/2이다. 그러나, 큐빗 #2는 큐빗 #1에 의하여 CNOT 연산이 수행되므로, 0상태 확률은
Figure PCTKR2019008667-appb-I000003
이다.
In phase 1 (p-1), qubit # 1 is not affected by the CNOT gate. Therefore, the component of qubit # 1 is
Figure PCTKR2019008667-appb-I000002
The zero state probability is 1/2. However, since cubit # 2 performs a CNOT operation by qubit # 1, the zero state probability is
Figure PCTKR2019008667-appb-I000003
to be.
페이즈 2(p-2)에서 X 게이트는 큐빗 #1의 0 상태 성분과 1 상태 성분을 교체하므로, 페이즈 2에서 큐빗 #1의 성분은
Figure PCTKR2019008667-appb-I000004
이며, 큐빗 #2의 성분은 변화하지 않는다. 따라서, 페이즈 2(p-2)에서의 큐빗 #1의 0 상태 확률 및 큐빗 #2의 0 상태 확률은 모두 1/2이다.
In phase 2 (p-2), the X gate swaps the zero state component and one state component of qubit # 1, so in phase 2 the component of qubit # 1 is
Figure PCTKR2019008667-appb-I000004
The component of qubit # 2 does not change. Therefore, the zero state probability of qubit # 1 and the zero state probability of qubit # 2 in phase 2 (p-2) are both 1/2.
페이즈 3(p-3)에서 큐빗 #1은 CNOT 게이트에 의한 영향을 받지 않아 큐빗 #1의 성분은
Figure PCTKR2019008667-appb-I000005
이다. 큐빗 #2은 큐빗 #1에 의하여 CNOT 연산이 이루어지며, 0 상태 확률은
Figure PCTKR2019008667-appb-I000006
이다.
In phase 3 (p-3), qubit # 1 is not affected by the CNOT gate, so the component of qubit # 1 is
Figure PCTKR2019008667-appb-I000005
to be. Qubit # 2 is a CNOT operation by qubit # 1, and the 0 state probability
Figure PCTKR2019008667-appb-I000006
to be.
페이즈 4(p-4)에서 큐빗 #1은 하다마드(H) 게이트에 의하여 45도 회전하며, 결과적으로, 큐빗 #1의 성분은 (1,0)이므로 큐빗 #1의 0 상태 확률은 1나, 큐빗 #2의 0 상태 확률은 0이다.In phase 4 (p-4), qubit # 1 is rotated 45 degrees by the Hadamard (H) gate. As a result, the zero state probability of qubit # 1 is 1 or 1 because the component of qubit # 1 is (1,0). , The zero state probability of qubit # 2 is zero.
표 1 및 표 2로 예시된 것과 같이 본 실시예에 의한 양자 컴퓨터는 양자 상태 벡터를 이용하지 않더라도 큐빗들의 각 페이즈별 0 상태 확률 또는 1 상태 확률을 추적 및 저장하여 출력을 얻을 수 있다. As illustrated in Tables 1 and 2, the quantum computer according to the present embodiment can obtain and output the zero state probability or the one state probability for each phase of the qubits even without using a quantum state vector.
실시예Example 2 2
본 실시예는 각 큐빗의 확률을 계산 혹은 저장하는 방법으로 위상 고정 루프(PLL), 지연 고정 루프(DLL)를 사용한다. 일 예로, 각 큐빗의 0성분을 cosθi (i: 큐빗 번호)로 표시한다. 이를 이용하여 소프트웨어로 연산하는 것 보다 연산 시간을 단축시킬 수 있다.This embodiment uses a phase locked loop (PLL) and a delay locked loop (DLL) as a method of calculating or storing the probability of each qubit. For example, the zero component of each qubit is cosθ i (i: cubit number). This can reduce computation time rather than software operation.
도 4는 본 실시예를 설명하기 위한 CNOT(control NOT) 게이트의 개요도이며, 수학식 5은 i번째 큐빗에 의해서 control NOT되는 j bit의 a(그림에서 α)성분의 확률을 표시한다. 4 is a schematic diagram of a CNOT (control NOT) gate for explaining the present embodiment, and Equation 5 shows a probability of the a (α in the figure) component of the j bit controlled by the i-th qubit.
Figure PCTKR2019008667-appb-M000007
Figure PCTKR2019008667-appb-M000007
따라서,i 번째 qubit에 의해서 control NOT되는 j bit의 a(도 3(b) α)성분의 확률은 θ성분의 2배와 같이 표현할 수 있으며, 위상 고정 루프 또는 지연 고정 루프 등의 전자 회로로 구현할 수 있다. Therefore, the probability of the a (Fig. 3 (b) α) component of the j bit controlled by the i th qubit may be expressed as twice the θ component, and may be implemented by an electronic circuit such as a phase locked loop or a delay locked loop. Can be.
실시예Example 3 3
도 5은 실시예 3의 개요를 도시한 도면이다. 도 5을 참조하면, 실시예 3은 세 개의 큐빗들을 포함하며, 페이즈 1에서 큐빗 #2으로 큐빗 #3을 CNOT 연산하며, 페이즈 2에서 큐빗 #1로 큐빗 #3을 CNOT 연산한다. 본 실시예에서, 각 큐빗에는 각 페이즈별로 CNOT 게이트에 의한 연결 관계를 더 저장할 수 있다. 5 is a diagram showing an outline of Embodiment 3. FIG. Referring to FIG. 5, Example 3 includes three qubits, and performs CNOT operation on phase 1 with cubit # 2 and CNOT operation on phase 2 with qubit # 1. In this embodiment, each qubit may further store a connection relationship by a CNOT gate for each phase.
큐빗 #1의 0 상태 확률은 페이즈 2의 CNOT 연산으로부터 영항을 받지 않아 a1 2이다. 또한, 큐빗 #2은 큐빗 #1에 의하여 CNOT 연산이 수행되므로, 큐빗 #2의 0 상태 확률은 a1 2a2 2+b1 2b2 2로 연산된다. 또한, 큐빗 #3은 큐빗 #2에 의하여 CNOT 연산이 수행되므로, 큐빗 #3의 0 상태 확률은 a2 2a3 2+b2 2b3 2로 연산된다. The zero state probability of qubit # 1 is a 1 2 since it is not affected by the CNOT operation of phase 2. In addition, since the CNOT operation is performed by the qubit # 2 by the qubit # 2, the zero state probability of the qubit # 2 is calculated as a 1 2 a 2 2 + b 1 2 b 2 2 . In addition, since the CNOT operation is performed by the qubit # 3, the zero state probability of the qubit # 3 is calculated as a 2 2 a 3 2 + b 2 2 b 3 2 .
종래 기술에 의하면 3개 큐빗에 대한 23개의 상태 변화를 추적하여야 하나, 본 실시예에 의하면 각 큐빗의 0 상태 확률을 추적하여 간단하고 용이하게 출력을 구한다.According to the prior art, two or three state changes for three qubits should be tracked, but according to the present embodiment, the zero state probability of each qubit is traced to obtain an output simply and easily.
위에서 설명된 양자 컴퓨터에는 복수의 입력 I1(a1, b1), I2(a2, b2), ...,IN(aN, bN)들이 결과적으로 복수의 확률값들 O1, O2, ...,OK 출력한다. 본 실시예에 의한 양자 컴퓨터는 원하는 각 큐빗의 출력 확률을 입력 큐빗의 값의 함수로 표시할 수 있으며, 이 때의 관계식은 아래의 수학식과 같다. 아래 식에서 O는 각 큐빗의 0을 가지는 확률을 나타낸다.In the quantum computer described above, a plurality of inputs I 1 (a 1 , b 1 ), I 2 (a 2 , b 2 ), ..., I N (a N , b N ) result in a plurality of probability values O 1 , O 2 , ..., O K Output The quantum computer according to the present embodiment may display the output probability of each desired qubit as a function of the value of the input qubit, and the relation at this time is expressed by the following equation. In the following equation, O represents the probability of having zero of each qubit.
Figure PCTKR2019008667-appb-M000008
Figure PCTKR2019008667-appb-M000008
필요에 따라서 양자 컴퓨터에 새로운 입력을 제공하여 많은 메모리가 필요로 하는 양자 상태를 전부 계산하지 않고, 함수(F)에 새로운 변수를 입력하고, 연산을 수행하여 출력값으로 목적하는 각 큐빗의 확률을 얻을 수 있다. 따라서, 양자 컴퓨터의 동작을 더욱 신속하게 할 수 있다.Provide new inputs to the quantum computer as needed, rather than calculating all of the quantum states required by a lot of memory, input new variables into the function (F), and perform calculations to obtain the desired probability of each qubit as the output. Can be. Therefore, the operation of the quantum computer can be made faster.
나아가, 양자의 스핀 등을 이용하는 기존의 양자 컴퓨터는 출력값을 획득함과 동시에 출력값이 소멸하여 정보의 안전성이 담보될 수 있다. 본 실시예에 의한 양자 컴퓨터도 이와 마찬가지로 출력을 제공함과 동시에 해당 출력값 및 버퍼, 메모리에 저장된 연산 결과를 소거하여 정보의 안전성을 유지할 수 있다.Furthermore, the conventional quantum computer using the quantum spin and the like can acquire the output value and at the same time the output value is destroyed to ensure the safety of the information. Similarly, the quantum computer according to the present embodiment can provide an output and at the same time, erase the output value, the buffer, and the operation result stored in the memory to maintain information safety.
일 실시예로, 본 기술에 의한 양자 컴퓨터는 소프트웨어나 반도체 CMOS 회로로 구현하는 것도 가능하다.In one embodiment, the quantum computer according to the present technology may be implemented in software or a semiconductor CMOS circuit.
상기에 기재되어 있음.Listed above.

Claims (14)

  1. 양자 컴퓨터로, With quantum computer,
    상기 양자 컴퓨터는 The quantum computer
    큐빗(qubit)이 가지는 1과 0의 상태의 확률 중 어느 하나를 추적 및 저장하여 확률값을 연산하여 출력하는 양자 컴퓨터. A quantum computer that tracks and stores any one of the probabilities of 1 and 0 states of a qubit to calculate and output a probability value.
  2. 제1항에 있어서,The method of claim 1,
    상기 양자 컴퓨터는 복수의 페이즈 들을 포함하며, The quantum computer includes a plurality of phases,
    큐빗에는 각 페이즈 별로 상기 1과 0의 상태의 확률 중 어느 하나가 저장되는 양자 컴퓨터.A quantum computer in which a qubit stores one of the probabilities of states 1 and 0 for each phase.
  3. 제2항에 있어서,The method of claim 2,
    상기 큐빗에는 CNOT 게이트의 연결관계가 더 저장된 양자 컴퓨터.And a quantum computer in which the connection relationship of the CNOT gate is further stored in the qubit.
  4. 제1항에 있어서, The method of claim 1,
    상기 양자 컴퓨터는 하다마드 게이트를 포함하며, The quantum computer comprises a Hadamard gate,
    상기 하다마드 게이트의 입력으로 0 상태의 큐빗이 제공하면,If a qubit of 0 state is provided to the input of the Hadamard gate,
    상기 하다마드 게이트는 출력으로 1/2의 0 상태 확률을 제공하는 양자 컴퓨터.Wherein said Hadamard gate provides a one-half zero state probability as an output.
  5. 제1항에 있어서, The method of claim 1,
    상기 양자 컴퓨터는 CNOT(Controlled NOT) 게이트를 포함하며, The quantum computer includes a controlled NOT gate,
    (a1, b1) 성분의 제1 큐빗으로 (a2, b2) 성분의 제2 큐빗에 대하여 CNOT연산을 수행하면, When the CNOT operation is performed on the second qubit of the component (a2, b2) with the first qubit of the component (a1, b1),
    상기 CNOT게이트는 a1 2a2 2+b1 2b2 2를 0 상태 확률로 출력하는 양자 컴퓨터.And the CNOT gate outputs a 1 2 a 2 2 + b 1 2 b 2 2 with a 0 state probability.
  6. 제1항에 있어서, The method of claim 1,
    X 게이트를 포함하며, Includes an X gate,
    상기 X 게이트는 수학식
    Figure PCTKR2019008667-appb-I000007
    으로 정의되는 입출력 관계를 가지는 양자 컴퓨터. (X: X게이트의 특성 행렬,
    Figure PCTKR2019008667-appb-I000008
    : 입력)
    The X gate is the equation
    Figure PCTKR2019008667-appb-I000007
    A quantum computer having an input-output relationship defined by. (X: characteristic matrix of the X gate,
    Figure PCTKR2019008667-appb-I000008
    : input)
  7. 제6항에 있어서, The method of claim 6,
    상기 X 게이트에 (a, b)입력이 주어진 경우에, Given the (a, b) input to the X gate,
    X 게이트 출력의 0 상태 확률은 b2인 양자 컴퓨터.A quantum computer where the zero state probability of the X gate output is b 2 .
  8. 제1항에 있어서, The method of claim 1,
    R 게이트를 포함하며, Including an R gate,
    상기 R 게이트는 수학식
    Figure PCTKR2019008667-appb-I000009
    으로 정의되는 입출력 관계를 가지는 양자 컴퓨터. (R: R게이트의 특성 행렬,
    Figure PCTKR2019008667-appb-I000010
    : 입력)
    The R gate is
    Figure PCTKR2019008667-appb-I000009
    A quantum computer having an input-output relationship defined by. (R: characteristic matrix of R gate,
    Figure PCTKR2019008667-appb-I000010
    : input)
  9. 제8항에 있어서, The method of claim 8,
    상기 R 게이트에 (a, b)입력이 주어진 경우에, If the (a, b) input is given to the R gate,
    R 게이트 출력의 0 상태 확률은 a2인 양자 컴퓨터.A quantum computer where the zero state probability of the R gate output is a 2 .
  10. 제9항에 있어서, The method of claim 9,
    상기 큐빗은 상기 θ값도 저장하는 양자 컴퓨터.The qubit also stores the [theta] value.
  11. 제1항에 있어서, The method of claim 1,
    상기 양자 컴퓨터는 제1 입력에 대한 제1 출력을 연산하는 함수로 표시되며, The quantum computer is represented by a function that computes a first output for a first input,
    상기 양자 컴퓨터는 제2 입력에 대한 제2 출력 연산시 상기 제2 입력을 상기 함수에 제공하여 상기 제2 출력을 연산하는 양자 컴퓨터.The quantum computer is configured to compute the second output by providing the second input to the function during a second output operation on a second input.
  12. 제1항에 있어서, The method of claim 1,
    상기 양자 컴퓨터는 The quantum computer
    메모리를 포함하며, Memory,
    상기 확률값 출력 후, 상기 메모리에 저장된 정보를 삭제하는 양자 컴퓨터.A quantum computer for deleting information stored in the memory after outputting the probability value.
  13. 제1항에 있어서, The method of claim 1,
    상기 양자 컴퓨터는, The quantum computer,
    CNOT 게이트에 의한 연결 상태를 더 저장하는 양자 컴퓨터.A quantum computer that further stores the connection state by the CNOT gate.
  14. 제1항에 있어서, The method of claim 1,
    상기 양자 컴퓨터는 소프트웨어나 반도체 CMOS 회로로 구현된 양자 컴퓨터.The quantum computer is implemented by software or semiconductor CMOS circuit.
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US9275011B2 (en) * 2013-03-27 2016-03-01 Microsoft Technology Licensing, Llc Fast quantum and classical phase estimation
KR101656637B1 (en) * 2012-10-26 2016-09-09 노스롭 그루먼 시스템즈 코포레이션 Efficient toffoli state generation from low-fidelity single qubit magic states
EP3109803A1 (en) * 2015-06-22 2016-12-28 Leonardo S.p.A. Structure and method for processing quantum information
KR20180043157A (en) * 2016-10-18 2018-04-27 한국전자통신연구원 Quantum circuit and method for implementation a heterogeneously encoded logical bell state

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US20140280427A1 (en) * 2013-03-15 2014-09-18 Microsoft Corporation Method and system for decomposing single-qubit quantum circuits into a discrete basis
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EP3109803A1 (en) * 2015-06-22 2016-12-28 Leonardo S.p.A. Structure and method for processing quantum information
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