Does the entanglement depend on the basis?

Let's say, we have a composite system $A\otimes B$. We take the basis for $A$ as $|i\rangle,|j\rangle...,$ the basis for $B$ as $|\alpha\rangle,|\beta\rangle....$ Then an entangled state is a state which can not be expressed as a tensor direct product, e.g. a state like $$\frac{1}{\sqrt{2}}(|i\ \alpha\rangle+|j\ \beta\rangle).$$ My question is, can a state which can not be expressed as a tensor direct product in one basis be expressed as a tensor direct product in another basis?

If yes, then it means that the entanglement depends on the basis which I think is hard to accept. If no, then there should be an invariant under the basis transformation to character the entanglement. What's that?

• The most general tensor product looks like $\left(\lambda_i |i\rangle + \lambda_j |j\rangle + \lambda_k | k\rangle\right) \otimes \left( \mu_\alpha |\alpha\rangle + \mu_\beta |\beta\rangle +\mu_\gamma | \gamma \rangle \right)$. If you apply a change basis to $i,j,k$ and a different change of basis to $\alpha, \beta, \gamma$, it will still be of this form. Jul 10, 2016 at 12:47

For example, the state $|\uparrow \uparrow \rangle + |\uparrow \downarrow \rangle + |\downarrow \uparrow \rangle + |\downarrow \downarrow \rangle$ might look entangled upon quick inspection, but by performing the Schmidt decomposition it's easy to see that the state equals $(|\uparrow \rangle + | \downarrow \rangle) \otimes (|\uparrow \rangle + | \downarrow \rangle)$ and is therefore an unentangled product state.
A non-entangled state is a state of the form $|j\rangle \otimes |\alpha\rangle$ for some states $|j\rangle,|\alpha\rangle$ of the two subsystems; all other states in the composite Hilbert space are entangled. The previous statement clearly doesn't make any reference to any basis so it cannot depend on any choice of such a basis.