Proof that you can't disentangle two parties if you only operate on one

Question

Let $A$ and $B$ be two entangled systems. Can someone prove or sketch a proof of why you cant unentangle $A$ and $B$ by only acting on $A$ or $B$ alone? i.e. by only applying $\mathbb{I}_A\otimes U_B$, with $U_B$ unitary.

Answer 1

If there was some unitary operator factorized as $\mathbb I_A \otimes U_B$ that would send the entangled state $|\psi\rangle$ to a factorized state $|\phi_A\rangle\otimes |\phi_B\rangle$, ie : $$|\phi_A\rangle\otimes |\phi_B\rangle = (\mathbb I_A\otimes U_B)|\psi\rangle$$

Then, we would have : $$|\psi\rangle = (\mathbb I_A\otimes U_B^\dagger)(|\phi_A\rangle\otimes |\phi_B\rangle)= |\phi_A\rangle \otimes (U_B^\dagger|\phi_B\rangle)$$ contradicting the fact that $|\psi\rangle $ is entangled.

In short, factorized unitary operators preserve disentangled states, so they must preserve entangled states as well.

Edit : Entanglement is basis independent. A state is unentangled if it can be written as a product state in some basis, and entangled if it cannot be factorized in any basis.

Answer 2

For a "positive" proof (as in, a proof not by contradiction), write the Schmidt decomposition of a generic bipartite pure state $|\Psi\rangle\in\mathcal H_A\otimes\mathcal H_B$ as $$|\Psi\rangle=\sum_k \sqrt{p_k}(|u_k\rangle\otimes|v_k\rangle),$$ for some positive reals $p_k\ge0$ summing to the identity, and orthonormal bases $|u_k\rangle$ and $|v_k\rangle$. Note that you can always do it: this decomposition amounts to the SVD of $|\Psi\rangle$ when thought of as an operator $\mathcal H_B\to\mathcal H_A$.

The entanglement of $|\Psi\rangle$ is then encoded in the Schmidt coefficients $(\sqrt{p_k})_k$. This statement can be made precise e.g. via the theory of majorization and its relation to entanglement. What we care about here is simply that separable states (which are equal to product ones for pure states) are all and only those with Schmidt coefficients equal to $(1,0,0,...)$, up to permutation of the elements.

Now, observe that local unitary operations do not affect the Schmidt coefficients. This is trivially seen writing $$(U\otimes V)|\Psi\rangle = \sum_k \sqrt{p_k} ((U|u_k\rangle)\otimes (V|v_k\rangle)),$$ and remembering that $\{|u_k\rangle\}_k$ is an orthonormal basis if and only if $\{U|u_k\rangle\}_k$ is, for any unitary $U$.

So, not only local unitary operations cannot disentangle states: they cannot affect the entanglement in any way. It is worth noting that this is not the case for generic local operations: non-unitary local operations can absolutely degrade the entanglement. The standard example of this being entanglement-breaking channels. An entanglement-breaking channel $\Phi$ is such that $(\Phi\otimes I)\rho$ is separable, for any (possibly entangled) $\rho$. See this answer for a proof of this.