# 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

## 3 Answers

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.

No, the entanglement (yes/no) doesn't depend on the basis of the two subsystems, only on the way how the two subsystems are separated from one another.

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.

Note that you can use the entanglement entropy to calculate the amount of entanglement in a bipartite pure state, but this is not a good measure for a general bipartite (mixed) state. In the general case there are several different entanglement measures currently used, which have certain desiderata: https://quantiki.org/wiki/axiomatic-approach. Invariance under local operations then immediately implies invariance under basis changes.