Proof that you can't disentangle two parties if you only operate on one
Solution 1
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 $$\Psi\rangle=\sum_k \sqrt{p_k}(u_k\rangle\otimesv_k\rangle),$$ 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 $$(U\otimes V)\Psi\rangle = \sum_k \sqrt{p_k} ((Uu_k\rangle)\otimes (Vv_k\rangle)),$$ and remembering that $\{u_k\rangle\}_k$ is an orthonormal basis if and only if $\{Uu_k\rangle\}_k$ 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: nonunitary local operations can absolutely degrade the entanglement. The standard example of this being entanglementbreaking channels. An entanglementbreaking channel $\Phi$ is such that $(\Phi\otimes I)\rho$ is separable, for any (possibly entangled) $\rho$. See this answer for a proof of this.
Solution 2
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.
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FriendlyLagrangian
Updated on August 01, 2022Comments

FriendlyLagrangian over 1 year
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.

Jason Funderberker about 2 yearsPossible duplicate: Is quantum entanglement maintained after measurement?.

FriendlyLagrangian about 2 years@Jakob isn't that question about measurements?

Jason Funderberker about 2 yearsIn the first version of your question, you did not specify $U_B$, so a projection would be a valid choice. BTW the title of your question does not really match the question itself: can vs cant

FriendlyLagrangian about 2 years@Jakob Thanks for the feedback, you're right! I edited that as well

ComptonScattering about 2 yearsIsn't this question tautological? Entanglement is defined as a property that cannot increase with LOCC operations, and thus is monotone under any reversible LOCC operations.

glS about 2 years@ComptonScattering I'd say it's more common to define entanglement as the property of a state not being a convex mixture of product states, with its relations with LOCC operations derived from that

ComptonScattering about 2 years@glS I don't know if that is obviously true. Eg. "entanglement may be defined as the sort of correlations that may not be created by LOCC alone." (Plenio & Virmani 2006), or "Formally [an entanglement measure] is any nonnegative real function of a state which can not increase under LOCC, and is zero for separable states." (Quantiki)

glS about 2 years@ComptonScattering an entanglement measure is defined like that, sure, but that's not the same as the definition of entanglement. And I'd say you don't define entanglement via LOCC for the simple reason that it's not really trivial to define precisely what LOCC channels look like. You need to introduce quantum instruments and all that to be able to write the structure of LOCC operations. But even just the fact that you need to know the formalism of maps and channels to talk about LOCCs, while entanglement can be defined without it

ComptonScattering about 2 years@glS defining LOCC operations seems preferable to using nebulous phrases like "The entanglement of $\Psi\rangle$ is encoded in the Schmidt coefficients" without explanation.

glS about 2 years@ComptonScattering indeed, that's not how entanglement is defined. You simply define entanglement as impossibility of writing the state as convex mixture of product states. The relevance of Schmidt coefficients isn't even particularly obvious for nonpure states. Question is, do you know of a good way to define what LOCC operations are? And I mean formally, not just saying in words "operations that can be implemented with such and such", which is intuitive, but cannot be used to characterise such channels in practice


SolubleFish about 2 yearsThanks for the feedback ! I corrected this typo.

FriendlyLagrangian about 2 yearsI'm a bit confused because entanglement is basis dependent. So in principle if $\psi\rangle$ is entangled then $$\psi \rangle = \sum_n \phi_n\rangle \langle \phi_n  \psi\rangle$$ can be separable in the new basis. But can't we say that $$\mathbb{I} = \sum_n \phi_n \rangle \langle \phi_n \ ?$$Which in turn is $$\mathbb{I} = \mathbb{I}_A\otimes \mathbb{I}_B=U_A\otimes U_B$$

Jason Funderberker about 2 yearsWhy do you think entanglement is basis dependent? Cf. this PSE thread