I have a question about the choosing of basis that compose the Hamiltonian of multiple systems. For example if we have a system of N particles, that depend in our case of Pauli matrixes i will give the most simple case, $$H=A\sum_i^N \sigma_i^x+B\sum_i^N\sigma_i^z$$
Where the notation means $\sigma^z_3=1\otimes 1 \otimes \sigma^z \otimes 1 \otimes \cdots \otimes 1$
So if I want to find how will look the Hamiltonian in other basis, we need to do the transformation,
$$H'=PHP^{-1}$$
Where $P$ is the change of basis matrix. If we have the more simple system of 1 particle, so we choose the basis for the spin $|0\rangle,|1\rangle$, so if we wanted to do a change of basis from $z$ to the $x$ basis, one of the ways to do it is with a rotation matrix for changing the basis, $$H_x=\begin{pmatrix}\cos\theta/2&\sin\theta/2 \\ -\sin\theta/2&\cos\theta/2 \end{pmatrix}H_z\begin{pmatrix}\cos\theta/2&-\sin\theta/2 \\ \sin\theta/2&\cos\theta/2 \end{pmatrix}_{\theta=\pi/2}$$
We can decompose any system in 2D into Pauli matrixes and identities so for example we have the $z$ pauli matrix in the $x$ basis $\sigma_x^z$ $$\sigma_x^z=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1 \\ -1&1 \end{pmatrix}\begin{pmatrix}1&0 \\ 0&-1 \end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1 \\ 1&1 \end{pmatrix}=\begin{pmatrix}0&1 \\ 1&0 \end{pmatrix}=\sigma_z^x$$
With a 1 particle system for changing from $z$ to $x$ basis is easy that we have to change $\sigma_x \leftrightarrow \sigma_z$. My question is, if we have a N particle system, for changing the basis it's only necesary to change the Pauli matrixes on the definition, i.e. $$1\otimes1\otimes \sigma_z \leftrightarrow 1\otimes1\otimes \sigma_x$$
Or its necessary to construct a analog for the rotation matrix but in the corresponding dymension of the system. I don't know if it's even possible, but my question is this, if for multiple particle systems you can change the basis "for each spin separately" how I have shown, or this have some implications that don't allow you to do it.
