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

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 $$u_{i_1,i_2,i_3}$$.

• – glS
Oct 14 at 12:47
• I doubt there's much you can say about constraints on these coefficients; and even if there are some sort of relations I'm not sure what their value would be. In the end the Pauli tensor product is just a basis; there are many other basis that are sometimes more useful to work with. Take the "elementary" basis for example : these are matrices with 1 in one row/column and 0 otherwise. Then when you expand $U$ in this basis, the coefficients are just the entries of $U$ itself. So the unitarity condition on $U$ translates directly to a relation on the coefficients. Oct 14 at 18:27

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).