# How to express $n$-qubit Hermitian operator with Pauli matrices

How can we prove that all $$n$$-qubit Hermitian matrices can be written in terms of Pauli matrices $$I$$, $$X$$, $$Y$$, and $$Z$$ as $$\sum_{W_k \in \{I, X, Y, Z\}} a_{W_1,\dots,W_n}W_{1}\otimes ... \otimes W_{n}$$ where $$a_{W_1,\dots,W_n} \in \mathbb{R}$$?

• See this question as well. Commented Jan 29, 2022 at 14:15
• And this Commented Jan 29, 2022 at 14:26

A simple approach is to overdo it a little and show that $$n$$-qubit Pauli operators form an orthogonal basis in the real vector space $$L_H(\mathbb{C}^{2^n})$$ of $$n$$-qubit Hermitian operators with appropriately defined inner product. To that end, we first show that $$\langle A,B\rangle_{HS}:=\mathrm{tr}(A^\dagger B)$$ is an inner product$$^1$$. Next, we calculate that for any two Pauli operators $$W=W_1\otimes\dots\otimes W_n$$ and $$W'=W'_1\otimes\dots\otimes W'_n$$ we have $$\langle W,W'\rangle_{HS}=0$$ if and only if $$W\ne W'$$. This implies$$^2$$ linear independence. Finally, there is a linear bijection between $$L_H(\mathbb{C}^k)$$ and $$\mathbb{R}^{k^2}$$, so $$\dim L_H(\mathbb{C}^k)=k^2$$. But there are $$4^n$$ Pauli operators in $$L_H(\mathbb{C}^{2^n})$$, so they form a basis.

$$^1$$ The function $$\langle .,.\rangle_{HS}$$ is known as the Hilbert-Schmidt inner product.
$$^2$$ Because the coefficient in front of $$W_k$$ in a linear combination $$A$$ is proportional to $$\langle W_k,A\rangle_{HS}$$.

• Thank you this is super helpful! Commented Jan 29, 2022 at 21:57