What is the structure of coefficients of $2^N\times 2^N$ unitary matrices decomposed in terms of the Pauli basis?

Question

Any square $2^N\times 2^N$ matrix can be written as a sum of tensor products of pauli matrices. Eg a $8\times 8$ matrix can be written as $$U=\sum_{i_1,i_2,i_3}u_{i_1,i_2,i_3}\sigma_{i_1}\otimes\sigma_{i_2}\otimes\sigma_{i_3}.$$ If U is unitary, what does it imply about $u_{i_1,i_2,i_3}$ ?

We can write $U^\dagger U=1 \implies Tr[U^\dagger U\cdot(\sigma_{i_1}\otimes\sigma_{i_2}\otimes\sigma_{i_3})]=0$ for $\{i_1,i_2,i_3\}\neq \{0,0,0\}$ and $Tr(U^\dagger U)=2^N$. The second equality gives: \begin{align}Tr(U^\dagger U)&=Tr\sum_{i_1,i_2,i_3,j_1,j_2,j_3}u^*_{i_1,i_2,i_3}u_{j_1,j_2,j_3}\sigma_{i_1}\sigma_{j_1}\otimes\sigma_{i_2}\sigma_{j_2}\otimes\sigma_{i_3}\sigma_{j_3}\\&=\sum_{i_1,i_2,i_3,j_1,j_2,j_3}u^*_{i_1,i_2,i_3}u_{j_1,j_2,j_3}\delta_{i_1,j_1}\delta_{i_2,j_2}\delta_{i_3,j_3}\\ &=\sum_{i_1,i_2,i_3}|u_{i_1,i_2,i_3}|^2=8 \end{align} But I have hard time using the first equality to derive a useful constraint on $u_{i_1,i_2,i_3}$.

Answer 1

There is a mistake, we have $$\text{Tr}(\sigma_{i_1}\sigma_{j_1}\otimes\sigma_{i_2}\sigma_{j_2}\otimes\sigma_{i_3}\sigma_{j_3}) = \delta_{i_1,j_1}\delta_{i_2,j_2}\delta_{i_3,j_3}\text{Tr}(I) = 8\delta_{i_1,j_1}\delta_{i_2,j_2}\delta_{i_3,j_3}.$$

So that $\sum_{i_1,i_2,i_3}|u_{i_1,i_2,i_3}|^2=1$. The same is true for every $n$.

The other equalities that we can derive from $U^\dagger U=I$ are much more subtle.

I assume Pauli matrices are indexed by $0,1,2,3$, where $\sigma_0 = I$.

Note that for any $k,l$ we have $\sigma_k \sigma_l = i^{q}\sigma_m$ for some numbers $q,m$ that depend on $k,l$, so they are, in fact, functions. I'll denote the index function as $m = k\circ l$ and power function as $q = k\bullet l$.

Moreover, if we fix $m$ then for any $k$ there is exactly one $l$ such that $\sigma_k \sigma_l$ equals to $i^{q}\sigma_m$ for some $q$. It's easy to see that $l = k \circ m$ and $-q = k \bullet m$.

Thus, the expansion $$ I = U^\dagger U= \sum_{k_1,k_2,k_3,l_1,l_2,l_3}u^*_{k_1,k_2,k_3}u_{l_1,l_2,l_3}\sigma_{k_1}\sigma_{l_1}\otimes\sigma_{k_2}\sigma_{l_2}\otimes\sigma_{k_3}\sigma_{l_3} $$

equals to

$$ \sum_{k_1,k_2,k_3,l_1,l_2,l_3}u^*_{k_1,k_2,k_3}u_{l_1,l_2,l_3}\cdot i^{k_1\bullet l_1 + k_2\bullet l_2 + k_3\bullet l_3}\sigma_{k_1 \circ l_1}\otimes\sigma_{k_2 \circ l_2}\otimes\sigma_{k_3\circ l_3} $$

and we can group the summands in groups of size 8:

$$ \sum_{m_1,m_2,m_3} \bigg(\sum_{k_1,k_2,k_3} u^*_{k_1,k_2,k_3}u_{l_1,l_2,l_3}\cdot i^{k_1\bullet l_1 + k_2\bullet l_2 + k_3\bullet l_3} \bigg) \sigma_{m_1}\otimes\sigma_{m_2}\otimes\sigma_{m_3} = I, $$

where $l_j = k_j \circ m_j$ (also we have $k_j\bullet l_j = -k_j\bullet m_j ~\text{mod}~ 4 $).

Thus, the coefficient near $\sigma_{m_1}\otimes\sigma_{m_2}\otimes\sigma_{m_3}$ has to be 0 whenever $m_1+m_2+m_3 \neq 0$ (and 1 otherwise, but we've already got that).