How to check if a $n$-qubit unitary is the tensor product of single-qubit unitaries

Question

Let's assume I give you the expression of a unitary matrix acting on two qubits that is:

$$U=\sum_{i} A_i \otimes B_i$$

for some operators $A_i$ and $B_i$.

Is there a simple criterion allowing you to find out if it is actually simply $U=\widetilde{A} \otimes \widetilde{B}$ for some $\widetilde{A}$ and $\widetilde{B}$?

By simple I mean that I would like to avoid taking two generic single qubit unitary matrices and check if $U=\widetilde{A} \otimes \widetilde{B}$ admits solution (it would be relatively messy).

As a comment: to check if we have $U=Tr_2(U) \otimes Tr_1(U)$ is not sufficient in general as it can be that $Tr_1(U)$ or $Tr_2(U)$ vanish.

If such simple method exists, do there exist a more general criteria for $n$ qubits (i.e. you have an $n$-qubit unitary matrix and you want to see if it can be written as a tensor product of single qubit gates).

Answer 1

TL;DR: A two-qubit unitary operator $U$ is a product operator, i.e. $U=U_1\otimes U_2$ for some single-qubit unitaries $U_1$ and $U_2$ if and only if $U$ has Schmidt rank one

$$ U = U_1\otimes U_2 \iff \mathrm{Sch}(U) = 1.\tag1 $$

This generalizes to $n$-qubit unitaries via straightforward recursive applications of $(1)$ over $n-1$ partitionings of the subsystems.

Operator Schmidt decomposition

The space of linear operators on a vector space $\mathcal{H}$ is a vector space $L(\mathcal{H})$ equipped with an inner product $\langle A,B\rangle_{HS} = \mathrm{tr}(A^\dagger B)$ known as the Hilbert-Schmidt inner product. Therefore, Schmidt decomposition applies to linear operators. Explicitly, any operator $U$ can be written as

$$ U = \sum_{k=1}^r \lambda_k A_k\otimes B_k\tag2 $$

where $A_k$ are orthonormal operators on the first subsystem, $B_k$ are orthonormal operators on the second subsystem and $\lambda_k$ are positive real numbers. The integer $r=:\mathrm{Sch}(U)$ is called the Schmidt rank of $U$.

See e.g. $6.4.2$ in Nielsen's PhD thesis or this paper for more details on operator variant of Schmidt decomposition.

Unitary product is a product of unitaries

Now, suppose that $U=A\otimes B$ for some operators $A$ and $B$. We'll show that $U=U_1\otimes U_2$ for some unitary operators $U_1$ and $U_2$. We have

$$ I=U^\dagger U = A^\dagger A\otimes B^\dagger B,\tag2 $$

so $\mathrm{tr}(A^\dagger A)>0$ and $\mathrm{tr}(B^\dagger B)>0$. Moreover, $A^\dagger A=\alpha I$ for some positive real number $\alpha$ and $B^\dagger B=\beta I$ for some positive real number $\beta$. Consequently, $U=U_1\otimes U_2$ where $U_1:=\frac{A}{\sqrt{\alpha}}$ and $U_2:=\frac{B}{\sqrt{\beta}}$ are unitary.

Generalization to $n$-qubit unitaries

An $n$-qubit unitary operator $U$ is the tensor product of an $(n-1)$-qubit unitary $U_{1\dots(n-1)}$ and a single-qubit unitary $U_n$ if and only if its Schmidt rank with respect to the partitioning $\{1\dots(n-1)\}\cup\{n\}$ is one. Applying this check recursively to $U_{1\dots(n-1)}$, we can determine whether $U$ is the tensor product of $n$ single-qubit unitaries by computing $n-1$ Schmidt decompositions.