Product Pauli Matrix is the basis

Theorem Any n qubtis state can be represented by combination of product of pauli matrix of 2 qubtis, that is \(\{I,X,Y,Z\}^{\otimes n}\). Further, any \(2^n\times 2^n\) hermitian matrix can be represented the same way.

We use mathmatical deduction to prove it.

Suppose \(n\) qubits cases is done, i.e., the n qubits has matrix basis, for example, the basis is \([1,0;0,0]\), \([0,1;0,0]\),\([0,i;0,0]...\) Now we want to show by directly product 4 pauli matrix \(I,X,Y,Z\), we can attain \(n+1\) qubits' basis. We use \(B_k^n\) stands for the \(k\)-th basis for \(n\)-qubits.

\(\begin{pmatrix} 1&0\\0&0 \end{pmatrix}\otimes B_k^n\) will attain some basis, \(\begin{pmatrix} 0&0\\0&1 \end{pmatrix}\otimes B_k^n\) will attain some basis, \(\begin{pmatrix} 0&1\\0&0 \end{pmatrix}\otimes B_k^n\) will attain some basis. Adding those basis together is the whole basis of \(n+1\) qubits system.

For example. We deduce from 1 qubit to 2 qubit.

\[\begin{pmatrix} 1&0\\0&0 \end{pmatrix}\otimes B_k^1=diag(1,0,0,0),diag(0,1,0,0),\begin{pmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix},\begin{pmatrix} 0&i&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix} \]

The same for \([0,0;0,1]\) and \([0,1;0,0]\).