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

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).

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.

• Thanks a lot for this great answer. I will look in further details next week and feedback to you! Cheers. Jul 16, 2022 at 14:04