[Physics] Does the entanglement depend on the basis

quantum mechanicsquantum-entanglementtensor-calculus

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?

Best Answer

The answer is no: whether or not the state can be written as a product state does not depend on the basis. And you are precisely correct: there is indeed a basis-independent invariant that characterizes the entanglement. It is called the "entanglement spectrum": the eigenvalue spectrum of the reduced density matrix produced by taking the partial trace over one part of the system. These are also called the "Schmidt weights" of the Schmidt decomposition, which is another name for the singular value decomposition of the matrix that characterizes the entangled state vector.

To reduce the entanglement spectrum down to a single number that quantifies the amount of entanglement, one calculates the "entanglement entropy" corresponding to the entanglement spectrum. There are several different types of entanglement entropies: the most common is the von Neumann entropy, but sometimes it's also useful to consider the Renyi entropy instead.

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.

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