Is quantum entanglement "observer dependent"?

Question

See also here. According to the answers of the mainstream entanglement should be objective. Once more this question asked in a more mathematical way...

Any vector $|{\psi}>$ of norm 1 in a Hilbert space is totally equivalent to any other vector of the same norm. They can be transformed into each other by a unitary transformation. As long as no operators are defined, there is no structure in the space and all vectors must be equivalent. Since entanglement measures do not depend on operators, all vectors must be equivalent concerning entanglement, too. Because there can be entanglement or not, every vector can be entangled or not, depending on the subjective view.

Example with 2 different position bases

Consider a solution of the hydrogen problem. $\vec{r}=\vec{r_2}-\vec{r_1}$ is the relative and $\vec{R}$ the center of mass position. In this base this example solution function is a product (not entangled) $$ \psi(\vec{r},\vec{R}) = N e^{ i \vec{k}\vec{R} } e^{ -\frac{|\vec{r}|}{a} } = N e^{ i \frac{m_1}{m_1+m_2} \vec{k}\vec{r_1} } e^{ i \frac{m_2}{m_1+m_2} \vec{k}\vec{r_2} } e^{-\frac{ \sqrt{ \vec{r_1}^2 - 2\vec{r_1}\vec{r_2} + \vec{r_2}^2 } }{a}} = \psi(\vec{r_1},\vec{r_2}) $$ while in the $r_1,r_2$ base it isn't a product (entangled).

Example with qubits

Consider the Bell state $$ |\psi> = \frac{1}{\sqrt{2}}(|00>+|11>) $$ The von Neumann entropy $$ S = -\mathrm{tr}_A [\mathrm{tr}_B( |\psi\rangle\langle\psi|) \ln \mathrm{tr}_B(|\psi\rangle\langle\psi|)] = -\mathrm{tr_B} [\mathrm{tr_A}( |\psi\rangle\langle\psi|) \ln \mathrm{tr_A}(|\psi\rangle\langle\psi|)]$$ is 1. Now transform from the $$\{|00>,|01>,|10>,|11>\} := \{|u_0>,|u_1>,|u_2>,|u_3>\}$$ base to the equivalent base $$ \{ \frac{1}{\sqrt{2}}(|u_0>+|u_3>),\frac{1}{\sqrt{2}}(|u_0>-|u_3>),|u_1>,|u_2> \} := \{|v_0>,|v_1>,|v_2>,|v_3>\} $$ so that $$|\psi> \equiv |v_0>$$ No define a split of the Hilbert space different from the first one by $$ |v_0> := |\tilde{0}\tilde{0}> \quad |v_1> := |\tilde{0}\tilde{1}> \quad |v_2> := |\tilde{1}\tilde{0}> \quad |v_3> := |\tilde{1}\tilde{1}> $$ and you see $|\psi>$ is no more entangled and has entropy 0.

Since there exist counterexamples, entanglement must depend on the subjective choice how to split the Hilbert space into sub spaces.

So: where is the error? In the arguments above or in the mainstream's belief?

Answer 1

The split of the space of states into subspaces is not arbitrary! Entanglement is "objective" in the following sense: Given a state $\lvert \psi\rangle\in H$ and a split $H = H_1\otimes H_2$ (or more generally a split into $n$ subspaces), everyone agrees on whether or not $\lvert \psi\rangle$ is entangled or not.

The split into the $H_i$ is not an observer-dependent choice, it is the nature of the physical system under consideration that dictates it: The $H_i$ are precisely the subspaces corresponding to physically meaningful subsystems of the system described by $H$, e.g. for a two-particle system the $H_i$ are the spaces of state of the single particles the pair consists of.