Question

给定应用于单个量子位的操作的2x2酉矩阵表示，如何计算Bloch sphere上对应的旋转？

例如，Hadamard矩阵围绕X + Z轴旋转180度。如何从[[1,1],[1,-1]]*sqrt(0.5)转到(X+Z, 180 deg)？

Answer 1

单量子位操作基本上只是unit quaternions，但具有额外的相位因子。相似性是因为Pauli matrices，次sqrt(-1)，满足定义四元数的i^2=j^2=k^2=ijk=-1关系。

因此，转换方法的难点已经由任何＆＃34;四元数到轴角度＆＃34;码。只需拉出相位四元数分量，找出相位因子，然后应用四元数到角度轴方法。

import math
import cmath

def toBlochAngleAxis(matrix):
    """
    Breaksdown a matrix U into axis, angle, and phase_angle components satisfying
    U = exp(i phase_angle) (I cos(angle/2) - axis sigma i sin(angle/2))

    :param matrix: The 2x2 unitary matrix U
    :return: The breakdown (axis(x, y, z), angle, phase_angle)
    """
    [[a, b], [c, d]] = matrix

    # --- Part 1: convert to a quaternion ---

    # Phased components of quaternion.
    wp = (a + d) / 2.0
    xp = (b + c) / 2.0j
    yp = (b - c) / 2.0
    zp = (a - d) / 2.0j

    # Arbitrarily use largest value to determine the global phase factor.
    phase = max([wp, xp, yp, zp], key=abs)
    phase /= abs(phase)

    # Cancel global phase factor, recovering quaternion components.
    w = complex(wp / phase).real
    x = complex(xp / phase).real
    y = complex(yp / phase).real
    z = complex(zp / phase).real

    # --- Part 2: convert from quaternion to angle-axis ---

    # Floating point error may have pushed w outside of [-1, +1]. Fix that.
    w = min(max(w, -1), +1)

    # Recover angle.
    angle = -2*math.acos(w)

    # Normalize axis.
    n = math.sqrt(x*x + y*y + z*z);
    if n < 0.000001:
        # There's an axis singularity near angle=0.
        # Just default to no rotation around the Z axis in this case.
        angle = 0
        x = 0
        y = 0
        z = 1
        n = 1
    x /= n
    y /= n
    z /= n

    # --- Part 3: (optional) canonicalize ---

    # Prefer angle in [-pi, pi]
    if angle <= -math.pi:
        angle += 2*math.pi
        phase *= -1

    # Prefer axes that point positive-ward.
    if x + y + z < 0:
        x *= -1
        y *= -1
        z *= -1
        angle *= -1

    phase_angle = cmath.polar(phase)[1]

    return (x, y, z), angle, phase_angle

测试出来：

print(toBlochAngleAxis([[1, 0], [0, 1]])) # Identity
# ([0, 0, 1], 0, 0.0)

print(toBlochAngleAxis([[0, 1], [1, 0]])) # Pauli X, 180 deg around X
# ([1.0, -0.0, -0.0], 3.141592653589793, 1.5707963267948966)

print(toBlochAngleAxis([[0, -1j], [1j, 0]])) # Pauli Y, 180 deg around Y
# ([-0.0, 1.0, -0.0], 3.141592653589793, 1.5707963267948966)

print(toBlochAngleAxis([[1, 0], [0, -1]])) # Pauli Z, 180 deg around Z
# ([-0.0, -0.0, 1.0], 3.141592653589793, 1.5707963267948966)

s = math.sqrt(0.5)
print(toBlochAngleAxis([[s, s], [s, -s]])) # Hadamard, 180 deg around X+Z
# ([0.7071067811865476, -0.0, 0.7071067811865476], 3.141592653589793, 1.5707963267948966)

print(toBlochAngleAxis([[s, s*1j], [s*1j, s]])) # -90 deg X axis, no phase
# ((1.0, 0.0, 0.0), -1.5707963267948966, 0.0)

从量子位操作转换为轴角Bloch球体旋转

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