IQ determination
We firstly discuss how to determine the qubit state from the readout data. A single-qubit with n levels (qudit) has quantum states
\[\vert\psi_n\rangle = \vert 0\rangle,\vert 1\rangle \cdots \vert N-1\rangle\]corresponding to the point in IQ space $x_n = (I_n,Q_n)$ when you readout it. If the noises involved, the readout result become a Gaussian distribution around the point, like a cloud. We should repeat the measurements to determine the center point of the IQ cloud.
After we save the readout data for the qubit, we can determine an unknown quantum state from its readout data $x = (I,Q)$. We regard the state as $\vert\psi_a\rangle$ if $\vert x-x_a\vert = \mathrm{Min}(\vert x-x_n \vert) $
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import numpy as np
def get_meas(data0, q, level):
"""return the state assignments of measurement
for example, level=2, return=0,1,1,0,...
"""
points = np.asarray(data0)
IQcenter = np.asarray(q['IQcenter'])
center_complex = (IQcenter[:level, 0] + 1j *
IQcenter[:level, 1]).reshape((level, 1))
tunnels = np.argmin(np.abs(points - center_complex), axis=0)
# element of tunnels, x, has 0<=x<level
return tunnels
多比特多能级
物理里，一般用张量积tensor写N比特的状态；在代码里，我们可以看成是一串长度为N的数字，若比特有M个能级，则数字的进制为M进制（通常qubit是2，如果叫qutrit就是3，更高的都叫qudit）。按照进制的转换方法，我们可以将N比特的状态表示成一个数字。因此我们创建数组binary_count，其中每一个元素都是一个数字，数字为M进制，长度为比特数。
然后利用循环，对每个态计数，最后求出重复实验所测量到每个态的概率，即布局数（population）。
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qNum = len(qubits)
counts_num = len(data[0])
binary_count = np.zeros((counts_num), dtype=int)
for i in np.arange(qNum):
binary_count += get_meas(
data[i], qubits[i], level) * (level**(qNum-1-i))
prob = np.bincount(binary_count, minlength=level**qNum)/counts_num
以下为实际代码示例
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import numpy as np
def tunneling(qubits, data, level=2):
""" get probability for 1,2,3...N qubits with level (2,3,4,....)
Args:
qubits (dict): qubit information in registry
data (list): list of IQ data (array of complex number) for N qubits
level (int): level of qubit
"""
qNum = len(qubits)
counts_num = len(data[0])
binary_count = np.zeros((counts_num), dtype=int)
def get_meas(data0, q, level):
"""return the state assignments of measurement
for example, level=2, return=0,1,1,0,...
"""
points = np.asarray(data0)
IQcenter = np.asarray(q['IQcenter'])
center_complex = (IQcenter[:level, 0] + 1j *
IQcenter[:level, 1]).reshape((level, 1))
tunnels = np.argmin(np.abs(points - center_complex), axis=0)
# element of tunnels, x, has 0<=x<level
return tunnels
for i in np.arange(qNum):
binary_count += get_meas(
data[i], qubits[i], level) * (level**(qNum-1-i))
prob = np.bincount(binary_count, minlength=level**qNum)/counts_num
return prob
2. 量子态层析
量子态层析(quantum state tomography)，一般简称tomo。对于N个量子比特，我们需要 $3^N$ 个操作 ${I,X_{\pi/2},Y_{\pi/2}}^{\otimes N}$ ，其中算符有时也写成 $I,X/2,Y/2$。而每个操作都同时对N个比特 $(q_1,q_2,q_3 \cdots)$ 作用，也就是包含N个微波脉冲（不同的DAC，或者同一个DAC对不同频率混频）。因此我们可以简洁的把它们写成两重循环。
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for tomo_idx in range(3**N):
for q_idx in range(N):
q_tomo = qlist[N-1-q_idx]
gate_idx = (tomo_idx//(3**(q_idx+1)))%(3**(q_idx)
q_tomo.add_gate(gates[gate_idx])
Math Tips
上面相当于把tomo_idx这个变量当成一个三进制数，其位数表示作用的比特，每一位上的数字表示需要进行的门操作。比如 “0212” 代表操作 $I \otimes Y/2 \otimes X/2 \otimes Y/2$。
Others:
- 因为我喜欢把比特从左往右排 $q_1,q_2,\cdots$，但是循环对应的进制是从右往左数，所以我们用
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q_tomo = qlist[N-1-q_idx]
然后还需注意的一点是：量子操作的顺序与物理公式书写顺序是相反的，比如 $XYZ$ 操作对应的量子操作的时序是 $Z,Y,X$。
一个量子门可以由二元组（two-tuple）表示，即 ($\theta$,$\phi$)，我们可以在程序中定义tomo所需的三个门
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thetas = [0,np.pi/2,np.pi/2]
phis = [0,0,np.pi/2]
## gates = [I,X/2,Y/2]
gates = [get_gate(thetas[i],phis[i]) for i in range(3)]
- 对于多能级的tomo，一般用的各有不同，需要根据所用的门来在循环里进行增改。