1
$\begingroup$

$$U = \frac{1}{2} \begin{pmatrix} -1 & -1 & 1 & 1 \\\\ 1 & -1 & 1 & -1 \\\\ 1 & -1 & -1 & 1 \\\\ 1 & -1 & 1 & 1 \end{pmatrix}$$

$$P = \frac{1}{2} \begin{pmatrix} 1 & -1 & -1 & 1 \\\\ 1 & -1 & 1 & -1 \\\\ -1 & -1 & 1 & 1 \\\\ 1 & -1 & 1 & 1 \end{pmatrix}$$

P matrix is a matrix obtained by swapping the 1st and 3rd columns of the U matrix.

Subsequently, I used Qiskit to apply U and P to a quantum circuit and observed the state vector.

%matploblib inline
import numpy as np
from qiskit import QuantumCircuit
from qiskit.extensions import *
from qiskit.quantum_info import Statevector

U = (1/*2) * np.array([
    [-1, -1, 1, 1],
    [1, -1, 1, -1],
    [1, -1, -1, 1],
    [1, -1, 1, 1]])
P = (1/*2) * np.array([
    [1, -1, -1, 1],
    [1, -1, 1, -1],
    [-1, -1, 1, 1],
    [1, -1, 1, 1]])

gate = UnitaryGate(U)

circuit = QuantumCircuit(2, 2)
circuit.append(gate, [0, 1])

ket = Statevector(circuit)
ket.draw('latex')

This is the resulting state vector obtained by applying U and P, while varying the initial state. enter image description here

As can be seen in the table above, the state vectors of U and P swap with each other when the initial vectors are |00> and |10>. However, I have realized that this is related to swapping the 1st and 3rd columns of U to create P, but I am not currently able to provide an accurate explanation for the reasons and interpretation of the results.

$\endgroup$

1 Answer 1

2
$\begingroup$

You have to review the definition of the matrix representation of an operator in a given basis: If the matrix $U$ is the representation of some operator $u$ in the basis $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$, it means that the first column of $U$ contains the coordinates of $u(|00\rangle)$ in that basis, the second column is $u(|01\rangle)$ and so on:

$\begin{array}{cc} \begin{array}{cccc}\phantom{2} \text{$ \tiny u(|00\rangle)$} & \text{$\tiny u(|01\rangle)$} & \text{$\tiny u(|10\rangle)$} & \text{$ \tiny u(|11\rangle)$} \\ \end{array} &\\ \frac{1}{2}\begin{bmatrix} \phantom{2}-1\phantom{2} & \phantom{}-1\phantom{}& \phantom{2}1\phantom{2}&\phantom{2}1\phantom{2}\\ 1 & -1 & 1& -1\\ 1 & -1& -1&1\\ 1 & -1 & 1& 1\\ \end{bmatrix}% % &\begin{array}{l} \left.\vphantom{\begin{bmatrix} 0\\ \end{bmatrix}}\right. \text{$\tiny |00\rangle$}\\ \left.\vphantom{\begin{bmatrix} 0\\ \end{bmatrix}}\right. \text{$\tiny |01\rangle$}\\ \left.\vphantom{\begin{bmatrix} 0\\ \end{bmatrix}}\right. \text{$\tiny |10\rangle$}\\ \left.\vphantom{\begin{bmatrix} 0\\ \end{bmatrix}}\right. \text{$\tiny |11\rangle$}\\ \end{array} \\ \end{array}$

So if you swap the 1st and 3rd columns, you swap $u(|00\rangle)$ and $u(|10\rangle)$, the others are left untouched.

(I used your operator $U$ in my answer but you should be aware it is not unitary)

$\endgroup$
3
  • $\begingroup$ Ah, so is it correct to understand that swapping the 1st and 3rd columns simply means exchanging the positions of the basis? And U is likely to be unitary because in the Qiskit code 'gate = UnitaryGate(U),' an error would occur if U is not unitary! Thank you for your response. $\endgroup$
    – junghyunHa
    Feb 15 at 2:33
  • $\begingroup$ You are not really changing the order of the basis of the 2-qubits state space: it is still $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\} $. You exchange the result of the operator $U$ applied to $|00\rangle$ (the 1st vector in the basis) and the result of the operator $U$ applied to $|10\rangle$ (the 3rd vector in the basis). I don't know what Qiskit should do but I confirm $U$ is not unitary; you can see that by taking the inner product of column 1 and 2 for example is is $-\frac{1}{2}$, it should be 0. $U \text{ Unitary} \Leftrightarrow UU^*=U^*U=I$ $\endgroup$ Feb 15 at 7:40
  • $\begingroup$ Ah, it turns out the matrix is not unitary. I've confirmed it. $\endgroup$
    – junghyunHa
    Feb 16 at 2:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.