# Equivalence of two unitary transformation with respect to local operations

Suppose that $$U_1$$ and $$U_2$$ are two (entangling) operators that act on a quantum system consisting of several subsystems. Is there any criterion to tell if these two are equivalent up to applying operators acting only locally to the subsystem?

For example the 2-qubit Control-Z phase gate can be transform to the Control-NOT via applying (local) Hadamard gates to the second (target) qubit before and after. However this is trivial. How one can tell in more complicated cases?

Cross-posted on physics.SE

Equivalent $$U_1$$ and $$U_2$$ should have the same spectrum. If eigenvalues of $$U_1$$ are all different, then there is essentially a unique unitary $$V$$ (up to a set of phases) which gives the equivalence $$VU_1V^\dagger = U_2$$. If $$U_1,U_2$$ are written in the eigenbasis, then matrices $$V' = DV$$, where $$D$$ is any diagonal unitary, form the set of all unitaries that give the equivalence between $$U_1$$ and $$U_2$$. We then can check if $$V' = V_1 \otimes V_2$$ for some $$D$$, which should be easy.
On the other hand, I don't think there is an efficient algorithm in a general situation. Let $$S_1$$, $$S_2$$ be eigensubspaces of $$U_1$$, $$U_2$$ that correspond to some eigenvalue $$\lambda$$. Clearly, $$\dim S_1 = \dim S_2$$. If $$U_1$$ and $$U_2$$ are equivalent by some $$V_1 \otimes V_2$$, then the minimum (and maximum) Schmidt rank of a vector in $$S_1$$ and in $$S_2$$ should coincide. In general, finding Schmidt rank of a mixed state is a hard problem. The separability problem is simpler, but it's known that it's NP-hard. I don't know if it's proven that finding Schmidt rank of a projector $$P$$ is also NP-hard, but I haven't seen efficient solutions of this problem either.
I assumed that $$U_1$$ and $$U_2$$ are locally equivalent if $$(V_1 \otimes V_2)U_1(V_1 \otimes V_2)^\dagger = U_2$$ for some local unitaries $$V_1,V_2$$. More broadly, we can define local equivalence by $$(V_1 \otimes V_2)U_1(V_3 \otimes V_4) = U_2.$$ Under this definition the problem looks even harder, I think.
Even checking if a quantum circuit is equal to the identity circuit is already QMA-hard. Thus, the general problem on $$N$$ qubits, provided that the unitary is specified by a circuit, is a computationally hard problem.