I would like to understand the meaning of applying permutation to a unitary matrix

Question

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

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

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.

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.

Answer 1

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)