# Inner automorphisms of Pauli strings in the unitary subgroup of matrices

## Statement

What is the set $$\mathcal{T}_n$$ of matrices $$T \in GL(2^n)$$ such that for all Pauli strings $$P \in \mathcal{P}_n=\{\otimes_{i=1}^n \sigma_{m_i}\mid m\in \{1,2,3\}^n, \sigma_{m_i} \text{ Pauli matrix}\}$$, $$T P T^{-1}$$ is a unitary matrix, i.e. $$T\mathcal{P}_nT^{-1} \subset U(2^n)$$?

## Possible direction

We know $$U(2^n) \subset \mathcal{T}_n$$, since all Pauli strings are unitary operators as well.

More generally, applying the QR decomposition on $$T$$, we can write $$T = Q R$$, for $$Q \in U(2^n)$$ and $$R$$ an upper triangular matrix. Hence, $$T \in \mathcal{T}_n \Leftrightarrow R P R^{-1} \in U(2^n)$$

### Case $$n=1$$

For $$n=1$$, everything is a 2 by 2 matrix, so we can write $$R = \pmatrix{a & c \\ 0 & b}$$, $$a,b \neq 0$$. The conjugation of $$\sigma_3 = \pmatrix{1 & 0 \\ 0 & -1}$$ by $$R$$ is, then, $$R \sigma_3 R^{-1} = \pmatrix{1 & -2 c / b \\ 0 & -1}$$. The unitarity condition $$R \sigma_3 R^{-1} \in U(2)$$ implies $$c = 0$$. Doing the same for $$\sigma_1$$ and $$\sigma_2$$ reveals that $$|a| = |b|$$, so $$\lambda R \in U(2)$$ for $$\lambda = \frac{1}{\sqrt{ab}} \in \mathbb{C}^*$$. Hence, $$\mathcal{T}_1 = \mathbb{C}^* \cdot U(2)$$.

### Conjecture

I conjecture that $$\mathcal{T}_n = \mathbb{C}^* \cdot U(2^n)$$. I checked that this is true for $$n=2$$ following the same line of argument of the $$n=1$$ case explained above.

Firstly, we can expand $$\mathcal{P}_n$$ to the subgroup $$\langle \mathcal{P}_n \rangle$$ generated by it, which also includes terms with the identity matrix $$1_{2\times2} = \sigma_i^2$$.
Then, $$T P T^{-1} \in U(2^n)$$ for $$P \in \langle \mathcal{P}_n \rangle$$ implies \begin{align} (TPT^{-1})^\dagger & = (T P T^{-1})^{-1} \\ (T^\dagger)^{-1} P T^{\dagger} & = T P T^{-1} \\ [P,T^\dagger T] & = 0 \end{align} where in the second line we used that $$P$$ is hermitian and unitary.
Since $$\mathrm{span}_\mathbb{C}(\langle \mathcal{P}_n \rangle) = M_\mathbb{C}(2^n)$$, the vector space of all matrices, then $$T^\dagger T \in Z(M_\mathbb{C}(2^n)) = \mathbb{C} \cdot 1_{2^n \times 2^n}$$.
Using the polar decomposition of $$T = U R$$, where $$U$$ is unitary and $$R$$ is positive semidefinite, the last conclusion implies $$R \in \mathbb{R}^+ \cdot 1_{2^n \times 2^n}$$ and thus $$T \in \mathbb{R}^+ \cdot U(2^n)$$.
EDIT: We don't even have to assume that $$T$$ is invertable. If for every $$P \in \mathbb{P}_n$$, $$T P = U_P T$$ for some $$U_P$$, then taking the hermitian conjugate gives $$P T^\dagger = T^\dagger U_P$$, so $$P T^\dagger T = T^\dagger U_P T = T^\dagger T P.$$