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The Pauli basis is \begin{align} I=\left[\begin{matrix} 1&0 \\ 0&1 \end{matrix} \right], \end{align} \begin{align} X=\left[\begin{matrix} 0&1 \\ 1&0 \end{matrix} \right], \end{align} \begin{align} Y=\left[\begin{matrix} 0&-i \\ i&0 \end{matrix} \right], \end{align} \begin{align} Z=\left[\begin{matrix} 1&0 \\ 0&-1 \end{matrix} \right]. \end{align} $I,X,Y,Z$ form a vector space. How to transform this pauli basis to another basis? How to perform this transformation? An example would be helpful.

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One way to do this consist in exploiting that the three Pauli matrices can be related to Cartesian axes of the Bloch sphere (SU(2) is the double cover of SO(3)).

That means that you can perform rotations of those axes and obtain a new basis. Define the Pauli operator $\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)=(X,Y,Z)$. You can construct any other operator in a different axis by selecting a unit vector in the Bloch sphere $\hat{n}=(\cos\varphi\sin\theta,\sin\varphi\sin\theta,\cos\theta)$ using polar coordinates ($\varphi$ azimuthal angle, $\theta$ polar angle).

Thus you can now replace $Z$ with a rotated operator $Z'=\vec{\sigma}\cdot \hat n$. The rest follows from the usual coordinate transformations between polar and Cartesian, you can get two other ortogonal operators as $$X'=X\cos\varphi\cos\theta+Y\sin\varphi \cos \theta-Z\sin\theta,$$ $$Y'=-X\sin\varphi+Y\cos\varphi.$$

For more general parametrisations you can use rotation matrices and Euler's angles.

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There is perhaps a better answer than this one.

Still, one approach is to note that $\mathbb{C}^{2\times 2}\cong \mathbb{C}^2\otimes \mathbb{C}^2 \cong \mathbb{C}^4$ and recall the vectorization isomorphism $vec:\mathbb{C}^2\otimes \mathbb{C}^2\to \mathbb{C}^4$; maps $vec(A)=\sum_i |v_i\rangle \otimes A|u_i\rangle$, for $|v_i\rangle$ and $|u_i\rangle$ orthonormal basis (o.n.b) for $\mathbb{C}^2$, note that each choice of o.n.b gives you an o.n.b for $\mathbb{C}^4$.For simplicity, consider the case where $|u_i\rangle=|v_i\rangle=|i\rangle$ for $i=1,2$.

Then the Pauli matrix basis is a basis for $\mathbb{C^4}$ under this isomorphism $vec(I)=(1,0,0,1)^T$, $vec(X)=(0,1,1,0)^T$, $vec(Y)=(0,i,-i,0)^T$, and $vec(Z)=(1,0,0,-1)^T$. From this, it is easy to construct a unitary that maps to or from this basis of vectors. For instance, if $e_1=(1,0,0,0)$, $e_2=(0,1,0,0)^T$, etc... the change of basis matrix from the standard basis to the Pauli basis is simply the unitary $$\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & i & 0 \\ 0 & 1 & -i & 0 \\ 1 & 0 & 0 & -1\\\end{pmatrix}.$$ Reversing the $vec$ construction would give you back the Pauli matrices. Note that this could be done for any basis for $\mathbb{C}^{2\times 2}$.

That being said, if all you want to do is express a $2\times 2$ complex matrix $A$, that is given to you in the basis $\{B_1,B_2,B_3,B_4\}$, in the basis $\{C_1,C_2,C_3,C_4\}$ then this can be done by recalling that any $A$ can be decomposed in that basis, the same way you would in any vector space, that is $$A= \sum_{i=1}^4 \frac{Tr(A^*C_i)}{\|C_i\|_F^2}C_i,$$ where $\|\cdot\|_F$ is the Frobenius norm.

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