量子步行3D网格

Question

我试图将量子硬币步道应用于3D网格，并带有3个Hadamard硬币。但是，在3个步骤后，我似乎无法获得对称结果。根本不可能具有与这种硬币对称的概率分布吗？

谢谢您的

ps实施基于 http：http：http：//susan-stepney.blogsne.blogspot。 com/2014/02/Mathjax.html ，位置向量捕获了3D网格。

PPS是否尝试过Qiskit？由于某些原因，我无法使用硬编码矩阵来完全对称结果...

Answer 1

不确定我回答了您的问题，但是
从您提到的代码参考中，我仅将第30行更改为：ax = fig.add_subplot（111，投影='3D'）和第3行，至：from Mpl_toolkits.mplot3d import axes3d

from numpy import *
from matplotlib.pyplot import *
from mpl_toolkits.mplot3d import Axes3D
N = 100      # number of random steps
P = 2*N+1    # number of positions
coin0 = array([1, 0])  # |0>
coin1 = array([0, 1])  # |1>
C00 = outer(coin0, coin0)  # |0><0| 
C01 = outer(coin0, coin1)  # |0><1| 
C10 = outer(coin1, coin0)  # |1><0| 
C11 = outer(coin1, coin1)  # |1><1| 
C_hat = (C00 + C01 + C10 - C11)/sqrt(2.)
ShiftPlus = roll(eye(P), 1, axis=0)
ShiftMinus = roll(eye(P), -1, axis=0)
S_hat = kron(ShiftPlus, C00) + kron(ShiftMinus, C11)
U = S_hat.dot(kron(eye(P), C_hat))
posn0 = zeros(P)
posn0[N] = 1     # array indexing starts from 0, so index N is the central posn
psi0 = kron(posn0,(coin0+coin1*1j)/sqrt(2.))
psiN = linalg.matrix_power(U, N).dot(psi0)
prob = empty(P)
for k in range(P):
    posn = zeros(P)
    posn[k] = 1     
    M_hat_k = kron( outer(posn,posn), eye(2))
    proj = M_hat_k.dot(psiN)
    prob[k] = proj.dot(proj.conjugate()).real

fig = figure()
ax = fig.add_subplot(111, projection = '3d')

plot(arange(P), prob)
plot(arange(P), prob, 'o')
loc = range(0, P, P // 10) #Location of ticks
xticks(loc)
xlim(0, P)
ax.set_xticklabels(range(-N, N+1, P // 10))

show()